Integrand size = 41, antiderivative size = 216 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {(2 A-6 B+13 C) \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {2 (11 A-36 B+76 C) \tan (c+d x)}{15 a^3 d}+\frac {(2 A-6 B+13 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(A-6 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(11 A-36 B+76 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \] Output:
1/2*(2*A-6*B+13*C)*arctanh(sin(d*x+c))/a^3/d-2/15*(11*A-36*B+76*C)*tan(d*x +c)/a^3/d+1/2*(2*A-6*B+13*C)*sec(d*x+c)*tan(d*x+c)/a^3/d-1/5*(A-B+C)*sec(d *x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^3-1/15*(A-6*B+11*C)*sec(d*x+c)^3*tan (d*x+c)/a/d/(a+a*sec(d*x+c))^2-1/15*(11*A-36*B+76*C)*sec(d*x+c)^2*tan(d*x+ c)/d/(a^3+a^3*sec(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(1081\) vs. \(2(216)=432\).
Time = 7.63 (sec) , antiderivative size = 1081, normalized size of antiderivative = 5.00 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:
Integrate[(Sec[c + d*x]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a* Sec[c + d*x])^3,x]
Output:
(-8*(2*A - 6*B + 13*C)*Cos[c/2 + (d*x)/2]^6*Log[Cos[c/2 + (d*x)/2] - Sin[c /2 + (d*x)/2]]*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^3) + (8*(2*A - 6*B + 13*C)*Cos[c/2 + (d*x)/2]^6*Log[Cos[c/2 + (d*x)/2] + Sin[c/ 2 + (d*x)/2]]*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^3) + ( Cos[c/2 + (d*x)/2]*Sec[c/2]*Sec[c]*Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C* Sec[c + d*x]^2)*(490*A*Sin[(d*x)/2] - 870*B*Sin[(d*x)/2] + 1235*C*Sin[(d*x )/2] - 530*A*Sin[(3*d*x)/2] + 1830*B*Sin[(3*d*x)/2] - 3805*C*Sin[(3*d*x)/2 ] + 654*A*Sin[c - (d*x)/2] - 2094*B*Sin[c - (d*x)/2] + 4329*C*Sin[c - (d*x )/2] - 654*A*Sin[c + (d*x)/2] + 1314*B*Sin[c + (d*x)/2] - 1989*C*Sin[c + ( d*x)/2] + 490*A*Sin[2*c + (d*x)/2] - 1650*B*Sin[2*c + (d*x)/2] + 3575*C*Si n[2*c + (d*x)/2] + 350*A*Sin[c + (3*d*x)/2] - 450*B*Sin[c + (3*d*x)/2] + 4 75*C*Sin[c + (3*d*x)/2] - 530*A*Sin[2*c + (3*d*x)/2] + 1230*B*Sin[2*c + (3 *d*x)/2] - 2005*C*Sin[2*c + (3*d*x)/2] + 350*A*Sin[3*c + (3*d*x)/2] - 1050 *B*Sin[3*c + (3*d*x)/2] + 2275*C*Sin[3*c + (3*d*x)/2] - 378*A*Sin[c + (5*d *x)/2] + 1278*B*Sin[c + (5*d*x)/2] - 2673*C*Sin[c + (5*d*x)/2] + 150*A*Sin [2*c + (5*d*x)/2] - 90*B*Sin[2*c + (5*d*x)/2] - 105*C*Sin[2*c + (5*d*x)/2] - 378*A*Sin[3*c + (5*d*x)/2] + 918*B*Sin[3*c + (5*d*x)/2] - 1593*C*Sin[3* c + (5*d*x)/2] + 150*A*Sin[4*c + (5*d*x)/2] - 450*B*Sin[4*c + (5*d*x)/2...
Time = 1.48 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.01, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.366, Rules used = {3042, 4572, 3042, 4507, 25, 3042, 4507, 3042, 4274, 3042, 4254, 24, 4255, 3042, 4257}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 4572 |
| \(\displaystyle \frac {\int \frac {\sec ^4(c+d x) (a (A+4 B-4 C)+a (2 A-2 B+7 C) \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (a (A+4 B-4 C)+a (2 A-2 B+7 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\frac {\int -\frac {\sec ^3(c+d x) \left (3 a^2 (A-6 B+11 C)-a^2 (8 A-18 B+43 C) \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {a (A-6 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\int \frac {\sec ^3(c+d x) \left (3 a^2 (A-6 B+11 C)-a^2 (8 A-18 B+43 C) \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {a (A-6 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (3 a^2 (A-6 B+11 C)-a^2 (8 A-18 B+43 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {a (A-6 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {-\frac {\frac {\int \sec ^2(c+d x) \left (2 a^3 (11 A-36 B+76 C)-15 a^3 (2 A-6 B+13 C) \sec (c+d x)\right )dx}{a^2}+\frac {a^2 (11 A-36 B+76 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {a (A-6 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (2 a^3 (11 A-36 B+76 C)-15 a^3 (2 A-6 B+13 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {a^2 (11 A-36 B+76 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {a (A-6 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {-\frac {\frac {2 a^3 (11 A-36 B+76 C) \int \sec ^2(c+d x)dx-15 a^3 (2 A-6 B+13 C) \int \sec ^3(c+d x)dx}{a^2}+\frac {a^2 (11 A-36 B+76 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {a (A-6 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {2 a^3 (11 A-36 B+76 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-15 a^3 (2 A-6 B+13 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {a^2 (11 A-36 B+76 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {a (A-6 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {-\frac {\frac {-\frac {2 a^3 (11 A-36 B+76 C) \int 1d(-\tan (c+d x))}{d}-15 a^3 (2 A-6 B+13 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {a^2 (11 A-36 B+76 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {a (A-6 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {-\frac {\frac {\frac {2 a^3 (11 A-36 B+76 C) \tan (c+d x)}{d}-15 a^3 (2 A-6 B+13 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {a^2 (11 A-36 B+76 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {a (A-6 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {-\frac {\frac {\frac {2 a^3 (11 A-36 B+76 C) \tan (c+d x)}{d}-15 a^3 (2 A-6 B+13 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}+\frac {a^2 (11 A-36 B+76 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {a (A-6 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {\frac {2 a^3 (11 A-36 B+76 C) \tan (c+d x)}{d}-15 a^3 (2 A-6 B+13 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}+\frac {a^2 (11 A-36 B+76 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {a (A-6 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {-\frac {\frac {a^2 (11 A-36 B+76 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}+\frac {\frac {2 a^3 (11 A-36 B+76 C) \tan (c+d x)}{d}-15 a^3 (2 A-6 B+13 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}}{3 a^2}-\frac {a (A-6 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
Input:
Int[(Sec[c + d*x]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^3,x]
Output:
-1/5*((A - B + C)*Sec[c + d*x]^4*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^3) + (-1/3*(a*(A - 6*B + 11*C)*Sec[c + d*x]^3*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^2) - ((a^2*(11*A - 36*B + 76*C)*Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])) + ((2*a^3*(11*A - 36*B + 76*C)*Tan[c + d*x])/d - 15*a^3 *(2*A - 6*B + 13*C)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d *x])/(2*d)))/a^2)/(3*a^2))/(5*a^2)
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(a*f*( 2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)* (d*Csc[e + f*x])^(n - 1)*Simp[A*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && G tQ[n, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[(-(a*A - b*B + a*C))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m - n)))*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 0.66 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.92
| method | result | size |
| parallelrisch | \(\frac {-40 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A -3 B +\frac {13 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+40 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A -3 B +\frac {13 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-17 \left (\frac {9 \left (4 A -14 B +29 C \right ) \cos \left (2 d x +2 c \right )}{17}+\left (A -\frac {57 B}{17}+\frac {239 C}{34}\right ) \cos \left (3 d x +3 c \right )+\frac {\left (\frac {11 A}{3}-12 B +\frac {76 C}{3}\right ) \cos \left (4 d x +4 c \right )}{17}+\left (3 A -\frac {191 B}{17}+\frac {777 C}{34}\right ) \cos \left (d x +c \right )+\frac {97 A}{51}-\frac {114 B}{17}+\frac {677 C}{51}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{40 d \,a^{3} \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(198\) |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B -\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {-14 C +4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-26 C +12 B -4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {2 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-14 C +4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (26 C -12 B +4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}{4 d \,a^{3}}\) | \(252\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B -\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {-14 C +4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-26 C +12 B -4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {2 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-14 C +4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (26 C -12 B +4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}{4 d \,a^{3}}\) | \(252\) |
| norman | \(\frac {-\frac {\left (A -3 B +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{12 a d}-\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{20 a d}-\frac {\left (7 A -27 B +59 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{12 a d}+\frac {\left (7 A -25 B +51 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (71 A -225 B +475 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 a d}-\frac {\left (101 A -345 B +721 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}-\frac {\left (173 A -549 B +1165 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{12 a d}+\frac {\left (953 A -3123 B +6613 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5} a^{2}}-\frac {\left (2 A -6 B +13 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3} d}+\frac {\left (2 A -6 B +13 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3} d}\) | \(306\) |
| risch | \(-\frac {i \left (30 A \,{\mathrm e}^{8 i \left (d x +c \right )}+2275 C \,{\mathrm e}^{6 i \left (d x +c \right )}+304 C +44 A -144 B -90 B \,{\mathrm e}^{8 i \left (d x +c \right )}-1050 B \,{\mathrm e}^{6 i \left (d x +c \right )}+350 A \,{\mathrm e}^{6 i \left (d x +c \right )}+654 A \,{\mathrm e}^{4 i \left (d x +c \right )}+4329 C \,{\mathrm e}^{4 i \left (d x +c \right )}+378 A \,{\mathrm e}^{2 i \left (d x +c \right )}+2673 C \,{\mathrm e}^{2 i \left (d x +c \right )}+150 A \,{\mathrm e}^{7 i \left (d x +c \right )}+490 A \,{\mathrm e}^{5 i \left (d x +c \right )}+530 A \,{\mathrm e}^{3 i \left (d x +c \right )}+190 A \,{\mathrm e}^{i \left (d x +c \right )}+975 C \,{\mathrm e}^{7 i \left (d x +c \right )}+3575 C \,{\mathrm e}^{5 i \left (d x +c \right )}+3805 C \,{\mathrm e}^{3 i \left (d x +c \right )}+1325 C \,{\mathrm e}^{i \left (d x +c \right )}-630 B \,{\mathrm e}^{i \left (d x +c \right )}+195 C \,{\mathrm e}^{8 i \left (d x +c \right )}-1278 B \,{\mathrm e}^{2 i \left (d x +c \right )}-1650 B \,{\mathrm e}^{5 i \left (d x +c \right )}-2094 B \,{\mathrm e}^{4 i \left (d x +c \right )}-1830 B \,{\mathrm e}^{3 i \left (d x +c \right )}-450 B \,{\mathrm e}^{7 i \left (d x +c \right )}\right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a^{3} d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{3} d}+\frac {13 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 a^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{a^{3} d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a^{3} d}-\frac {13 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 a^{3} d}\) | \(466\) |
Input:
int(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x,meth od=_RETURNVERBOSE)
Output:
1/40*(-40*(1+cos(2*d*x+2*c))*(A-3*B+13/2*C)*ln(tan(1/2*d*x+1/2*c)-1)+40*(1 +cos(2*d*x+2*c))*(A-3*B+13/2*C)*ln(tan(1/2*d*x+1/2*c)+1)-17*(9/17*(4*A-14* B+29*C)*cos(2*d*x+2*c)+(A-57/17*B+239/34*C)*cos(3*d*x+3*c)+1/17*(11/3*A-12 *B+76/3*C)*cos(4*d*x+4*c)+(3*A-191/17*B+777/34*C)*cos(d*x+c)+97/51*A-114/1 7*B+677/51*C)*sec(1/2*d*x+1/2*c)^4*tan(1/2*d*x+1/2*c))/d/a^3/(1+cos(2*d*x+ 2*c))
Time = 0.09 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.52 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {15 \, {\left ({\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (11 \, A - 36 \, B + 76 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (34 \, A - 114 \, B + 239 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (64 \, A - 234 \, B + 479 \, C\right )} \cos \left (d x + c\right )^{2} - 15 \, {\left (2 \, B - 3 \, C\right )} \cos \left (d x + c\right ) - 15 \, C\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3, x, algorithm="fricas")
Output:
1/60*(15*((2*A - 6*B + 13*C)*cos(d*x + c)^5 + 3*(2*A - 6*B + 13*C)*cos(d*x + c)^4 + 3*(2*A - 6*B + 13*C)*cos(d*x + c)^3 + (2*A - 6*B + 13*C)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 15*((2*A - 6*B + 13*C)*cos(d*x + c)^5 + 3 *(2*A - 6*B + 13*C)*cos(d*x + c)^4 + 3*(2*A - 6*B + 13*C)*cos(d*x + c)^3 + (2*A - 6*B + 13*C)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(4*(11*A - 36*B + 76*C)*cos(d*x + c)^4 + 3*(34*A - 114*B + 239*C)*cos(d*x + c)^3 + (6 4*A - 234*B + 479*C)*cos(d*x + c)^2 - 15*(2*B - 3*C)*cos(d*x + c) - 15*C)* sin(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos (d*x + c)^3 + a^3*d*cos(d*x + c)^2)
\[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {A \sec ^{4}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{5}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{6}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate(sec(d*x+c)**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))* *3,x)
Output:
(Integral(A*sec(c + d*x)**4/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)**5/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**6/(sec(c + d *x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x))/a**3
Leaf count of result is larger than twice the leaf count of optimal. 493 vs. \(2 (204) = 408\).
Time = 0.05 (sec) , antiderivative size = 493, normalized size of antiderivative = 2.28 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:
integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3, x, algorithm="maxima")
Output:
-1/60*(C*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) - 7*sin(d*x + c)^3/(cos(d* x + c) + 1)^3)/(a^3 - 2*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^3*sin( d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1) + 40*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 390*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 390*log(sin (d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) - 3*B*(40*sin(d*x + c)/((a^3 - a^3* sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x + c )/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 60*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) + A*((105*sin(d* x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin (d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 60*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3))/d
Time = 0.31 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.33 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {30 \, {\left (2 \, A - 6 \, B + 13 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {30 \, {\left (2 \, A - 6 \, B + 13 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {60 \, {\left (2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 255 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 465 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \] Input:
integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3, x, algorithm="giac")
Output:
1/60*(30*(2*A - 6*B + 13*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 30*(2 *A - 6*B + 13*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 - 60*(2*B*tan(1/2* d*x + 1/2*c)^3 - 7*C*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c) + 5 *C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3) - (3*A*a^12* tan(1/2*d*x + 1/2*c)^5 - 3*B*a^12*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/ 2*d*x + 1/2*c)^5 + 20*A*a^12*tan(1/2*d*x + 1/2*c)^3 - 30*B*a^12*tan(1/2*d* x + 1/2*c)^3 + 40*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 105*A*a^12*tan(1/2*d*x + 1/2*c) - 255*B*a^12*tan(1/2*d*x + 1/2*c) + 465*C*a^12*tan(1/2*d*x + 1/2*c ))/a^15)/d
Time = 13.04 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.05 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A-3\,B+\frac {13\,C}{2}\right )}{a^3\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-3\,B+5\,C\right )}{4\,a^3}-\frac {2\,A+2\,B-10\,C}{4\,a^3}+\frac {3\,\left (A-B+C\right )}{2\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,B-7\,C\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,B-5\,C\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B+C\right )}{20\,a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-3\,B+5\,C}{12\,a^3}+\frac {A-B+C}{4\,a^3}\right )}{d} \] Input:
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a/cos(c + d*x))^3),x)
Output:
(2*atanh(tan(c/2 + (d*x)/2))*(A - 3*B + (13*C)/2))/(a^3*d) - (tan(c/2 + (d *x)/2)*((3*(A - 3*B + 5*C))/(4*a^3) - (2*A + 2*B - 10*C)/(4*a^3) + (3*(A - B + C))/(2*a^3)))/d - (tan(c/2 + (d*x)/2)^3*(2*B - 7*C) - tan(c/2 + (d*x) /2)*(2*B - 5*C))/(d*(a^3*tan(c/2 + (d*x)/2)^4 - 2*a^3*tan(c/2 + (d*x)/2)^2 + a^3)) - (tan(c/2 + (d*x)/2)^5*(A - B + C))/(20*a^3*d) - (tan(c/2 + (d*x )/2)^3*((A - 3*B + 5*C)/(12*a^3) + (A - B + C)/(4*a^3)))/d
Time = 0.20 (sec) , antiderivative size = 643, normalized size of antiderivative = 2.98 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x)
Output:
( - 60*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**4*a + 180*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**4*b - 390*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**4*c + 120*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*a - 360 *log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*b + 780*log(tan((c + d*x)/2 ) - 1)*tan((c + d*x)/2)**2*c - 60*log(tan((c + d*x)/2) - 1)*a + 180*log(ta n((c + d*x)/2) - 1)*b - 390*log(tan((c + d*x)/2) - 1)*c + 60*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**4*a - 180*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**4*b + 390*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**4*c - 120 *log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*a + 360*log(tan((c + d*x)/2 ) + 1)*tan((c + d*x)/2)**2*b - 780*log(tan((c + d*x)/2) + 1)*tan((c + d*x) /2)**2*c + 60*log(tan((c + d*x)/2) + 1)*a - 180*log(tan((c + d*x)/2) + 1)* b + 390*log(tan((c + d*x)/2) + 1)*c - 3*tan((c + d*x)/2)**9*a + 3*tan((c + d*x)/2)**9*b - 3*tan((c + d*x)/2)**9*c - 14*tan((c + d*x)/2)**7*a + 24*ta n((c + d*x)/2)**7*b - 34*tan((c + d*x)/2)**7*c - 68*tan((c + d*x)/2)**5*a + 198*tan((c + d*x)/2)**5*b - 388*tan((c + d*x)/2)**5*c + 190*tan((c + d*x )/2)**3*a - 600*tan((c + d*x)/2)**3*b + 1310*tan((c + d*x)/2)**3*c - 105*t an((c + d*x)/2)*a + 375*tan((c + d*x)/2)*b - 765*tan((c + d*x)/2)*c)/(60*a **3*d*(tan((c + d*x)/2)**4 - 2*tan((c + d*x)/2)**2 + 1))