Integrand size = 33, antiderivative size = 198 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {(2 A+13 C) \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {2 (11 A+76 C) \tan (c+d x)}{15 a^3 d}+\frac {(2 A+13 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(A+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(A+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(11 A+76 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \] Output:
1/2*(2*A+13*C)*arctanh(sin(d*x+c))/a^3/d-2/15*(11*A+76*C)*tan(d*x+c)/a^3/d +1/2*(2*A+13*C)*sec(d*x+c)*tan(d*x+c)/a^3/d-1/5*(A+C)*sec(d*x+c)^4*tan(d*x +c)/d/(a+a*sec(d*x+c))^3-1/15*(A+11*C)*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+a*se c(d*x+c))^2-1/15*(11*A+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. \(632\) vs. \(2(198)=396\).
Time = 3.21 (sec) , antiderivative size = 632, normalized size of antiderivative = 3.19 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:
Integrate[(Sec[c + d*x]^4*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^3,x ]
Output:
-1/240*(Cos[(c + d*x)/2]*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*(1920*(2*A + 13*C)*Cos[(c + d*x)/2]^5*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[C os[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c/2]*Sec[c]*Sec[c + d*x]^2*(-5* (98*A + 247*C)*Sin[(d*x)/2] + 5*(106*A + 761*C)*Sin[(3*d*x)/2] - 654*A*Sin [c - (d*x)/2] - 4329*C*Sin[c - (d*x)/2] + 654*A*Sin[c + (d*x)/2] + 1989*C* Sin[c + (d*x)/2] - 490*A*Sin[2*c + (d*x)/2] - 3575*C*Sin[2*c + (d*x)/2] - 350*A*Sin[c + (3*d*x)/2] - 475*C*Sin[c + (3*d*x)/2] + 530*A*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] - 2275*C *Sin[3*c + (3*d*x)/2] + 378*A*Sin[c + (5*d*x)/2] + 2673*C*Sin[c + (5*d*x)/ 2] - 150*A*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] + 1593*C*Sin[3*c + (5*d*x)/2] - 150*A*Sin[4*c + (5*d*x)/2] - 975*C*Sin[4*c + (5*d*x)/2] + 190*A*Sin[2*c + (7*d*x)/2] + 1325*C*Sin[2* c + (7*d*x)/2] - 30*A*Sin[3*c + (7*d*x)/2] + 255*C*Sin[3*c + (7*d*x)/2] + 190*A*Sin[4*c + (7*d*x)/2] + 875*C*Sin[4*c + (7*d*x)/2] - 30*A*Sin[5*c + ( 7*d*x)/2] - 195*C*Sin[5*c + (7*d*x)/2] + 44*A*Sin[3*c + (9*d*x)/2] + 304*C *Sin[3*c + (9*d*x)/2] + 90*C*Sin[4*c + (9*d*x)/2] + 44*A*Sin[5*c + (9*d*x) /2] + 214*C*Sin[5*c + (9*d*x)/2])))/(a^3*d*(A + 2*C + A*Cos[2*(c + d*x)])* (1 + Sec[c + d*x])^3)
Time = 1.40 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.03, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.485, Rules used = {3042, 4573, 25, 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.
| \(\displaystyle \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 4573 |
| \(\displaystyle -\frac {\int -\frac {\sec ^4(c+d x) (a (A-4 C)+a (2 A+7 C) \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\sec ^4(c+d x) (a (A-4 C)+a (2 A+7 C) \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (a (A-4 C)+a (2 A+7 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\frac {\int -\frac {\sec ^3(c+d x) \left (3 a^2 (A+11 C)-a^2 (8 A+43 C) \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {a (A+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\int \frac {\sec ^3(c+d x) \left (3 a^2 (A+11 C)-a^2 (8 A+43 C) \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {a (A+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (3 a^2 (A+11 C)-a^2 (8 A+43 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {a (A+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {-\frac {\frac {\int \sec ^2(c+d x) \left (2 a^3 (11 A+76 C)-15 a^3 (2 A+13 C) \sec (c+d x)\right )dx}{a^2}+\frac {a^2 (11 A+76 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {a (A+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (2 a^3 (11 A+76 C)-15 a^3 (2 A+13 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {a^2 (11 A+76 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {a (A+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {-\frac {\frac {2 a^3 (11 A+76 C) \int \sec ^2(c+d x)dx-15 a^3 (2 A+13 C) \int \sec ^3(c+d x)dx}{a^2}+\frac {a^2 (11 A+76 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {a (A+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {2 a^3 (11 A+76 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-15 a^3 (2 A+13 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {a^2 (11 A+76 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {a (A+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {-\frac {\frac {-\frac {2 a^3 (11 A+76 C) \int 1d(-\tan (c+d x))}{d}-15 a^3 (2 A+13 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {a^2 (11 A+76 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {a (A+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {-\frac {\frac {\frac {2 a^3 (11 A+76 C) \tan (c+d x)}{d}-15 a^3 (2 A+13 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {a^2 (11 A+76 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {a (A+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {-\frac {\frac {\frac {2 a^3 (11 A+76 C) \tan (c+d x)}{d}-15 a^3 (2 A+13 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}+\frac {a^2 (11 A+76 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {a (A+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {\frac {2 a^3 (11 A+76 C) \tan (c+d x)}{d}-15 a^3 (2 A+13 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}+\frac {a^2 (11 A+76 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {a (A+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {-\frac {\frac {a^2 (11 A+76 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}+\frac {\frac {2 a^3 (11 A+76 C) \tan (c+d x)}{d}-15 a^3 (2 A+13 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}}{3 a^2}-\frac {a (A+11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
Input:
Int[(Sec[c + d*x]^4*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^3,x]
Output:
-1/5*((A + C)*Sec[c + d*x]^4*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^3) + (- 1/3*(a*(A + 11*C)*Sec[c + d*x]^3*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^2) - ((a^2*(11*A + 76*C)*Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Sec[c + d*x]) ) + ((2*a^3*(11*A + 76*C)*Tan[c + d*x])/d - 15*a^3*(2*A + 13*C)*(ArcTanh[S in[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_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a) *(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*C sc[e + f*x])^n*Simp[b*C*n + A*b*(2*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, C, n}, x] && EqQ[ a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 0.49 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.88
| method | result | size |
| parallelrisch | \(\frac {-240 \left (\frac {13 C}{2}+A \right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+240 \left (\frac {13 C}{2}+A \right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-22 \left (\left (\frac {108 A}{11}+\frac {783 C}{11}\right ) \cos \left (2 d x +2 c \right )+\left (\frac {51 A}{11}+\frac {717 C}{22}\right ) \cos \left (3 d x +3 c \right )+\left (A +\frac {76 C}{11}\right ) \cos \left (4 d x +4 c \right )+\left (\frac {153 A}{11}+\frac {2331 C}{22}\right ) \cos \left (d x +c \right )+\frac {97 A}{11}+\frac {677 C}{11}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{240 d \,a^{3} \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(175\) |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\left (-26 C -4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {2 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {14 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (26 C +4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {14 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{4 d \,a^{3}}\) | \(194\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\left (-26 C -4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {2 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {14 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (26 C +4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {14 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{4 d \,a^{3}}\) | \(194\) |
| norman | \(\frac {-\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{20 a d}-\frac {\left (A +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{12 a d}-\frac {\left (7 A +59 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{12 a d}+\frac {\left (51 C +7 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (475 C +71 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 a d}-\frac {\left (721 C +101 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}-\frac {\left (1165 C +173 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{12 a d}+\frac {\left (6613 C +953 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5} a^{2}}-\frac {\left (2 A +13 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3} d}+\frac {\left (2 A +13 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3} d}\) | \(276\) |
| risch | \(-\frac {i \left (30 A \,{\mathrm e}^{8 i \left (d x +c \right )}+195 C \,{\mathrm e}^{8 i \left (d x +c \right )}+150 A \,{\mathrm e}^{7 i \left (d x +c \right )}+975 C \,{\mathrm e}^{7 i \left (d x +c \right )}+350 A \,{\mathrm e}^{6 i \left (d x +c \right )}+2275 C \,{\mathrm e}^{6 i \left (d x +c \right )}+490 A \,{\mathrm e}^{5 i \left (d x +c \right )}+3575 C \,{\mathrm e}^{5 i \left (d x +c \right )}+654 A \,{\mathrm e}^{4 i \left (d x +c \right )}+4329 C \,{\mathrm e}^{4 i \left (d x +c \right )}+530 A \,{\mathrm e}^{3 i \left (d x +c \right )}+3805 C \,{\mathrm e}^{3 i \left (d x +c \right )}+378 A \,{\mathrm e}^{2 i \left (d x +c \right )}+2673 C \,{\mathrm e}^{2 i \left (d x +c \right )}+190 A \,{\mathrm e}^{i \left (d x +c \right )}+1325 C \,{\mathrm e}^{i \left (d x +c \right )}+44 A +304 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{a^{3} d}-\frac {13 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a^{3} d}+\frac {13 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 a^{3} d}\) | \(323\) |
Input:
int(sec(d*x+c)^4*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x,method=_RETURNVER BOSE)
Output:
1/240*(-240*(13/2*C+A)*(1+cos(2*d*x+2*c))*ln(tan(1/2*d*x+1/2*c)-1)+240*(13 /2*C+A)*(1+cos(2*d*x+2*c))*ln(tan(1/2*d*x+1/2*c)+1)-22*((108/11*A+783/11*C )*cos(2*d*x+2*c)+(51/11*A+717/22*C)*cos(3*d*x+3*c)+(A+76/11*C)*cos(4*d*x+4 *c)+(153/11*A+2331/22*C)*cos(d*x+c)+97/11*A+677/11*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 = 0.10 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.46 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {15 \, {\left ({\left (2 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (2 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (2 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (2 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (2 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (2 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (11 \, A + 76 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (34 \, A + 239 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (64 \, A + 479 \, C\right )} \cos \left (d x + c\right )^{2} + 45 \, C \cos \left (d x + c\right ) - 15 \, C\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate(sec(d*x+c)^4*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm= "fricas")
Output:
1/60*(15*((2*A + 13*C)*cos(d*x + c)^5 + 3*(2*A + 13*C)*cos(d*x + c)^4 + 3* (2*A + 13*C)*cos(d*x + c)^3 + (2*A + 13*C)*cos(d*x + c)^2)*log(sin(d*x + c ) + 1) - 15*((2*A + 13*C)*cos(d*x + c)^5 + 3*(2*A + 13*C)*cos(d*x + c)^4 + 3*(2*A + 13*C)*cos(d*x + c)^3 + (2*A + 13*C)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(4*(11*A + 76*C)*cos(d*x + c)^4 + 3*(34*A + 239*C)*cos(d*x + c)^3 + (64*A + 479*C)*cos(d*x + c)^2 + 45*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)
\[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {A \sec ^{4}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{6}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate(sec(d*x+c)**4*(A+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(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
Time = 0.04 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.67 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=-\frac {C {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} + A {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, d} \] Input:
integrate(sec(d*x+c)^4*(A+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) + 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 = 0.32 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.05 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {30 \, {\left (2 \, A + 13 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {30 \, {\left (2 \, A + 13 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {60 \, {\left (7 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 465 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \] Input:
integrate(sec(d*x+c)^4*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm= "giac")
Output:
1/60*(30*(2*A + 13*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 30*(2*A + 1 3*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + 60*(7*C*tan(1/2*d*x + 1/2*c) ^3 - 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*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 + 40*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 105*A*a^12*ta n(1/2*d*x + 1/2*c) + 465*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d
Time = 12.76 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.98 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A+\frac {13\,C}{2}\right )}{a^3\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A+C\right )}{2\,a^3}+\frac {3\,\left (A+5\,C\right )}{4\,a^3}-\frac {2\,A-10\,C}{4\,a^3}\right )}{d}-\frac {5\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-7\,C\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{4\,a^3}+\frac {A+5\,C}{12\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A+C\right )}{20\,a^3\,d} \] Input:
int((A + 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 + (13*C)/2))/(a^3*d) - (tan(c/2 + (d*x)/2) *((3*(A + C))/(2*a^3) + (3*(A + 5*C))/(4*a^3) - (2*A - 10*C)/(4*a^3)))/d - (5*C*tan(c/2 + (d*x)/2) - 7*C*tan(c/2 + (d*x)/2)^3)/(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)^3*((A + C)/(4*a^3) + (A + 5*C)/(12*a^3)))/d - (tan(c/2 + (d*x)/2)^5*(A + C))/(20* a^3*d)
Time = 0.17 (sec) , antiderivative size = 441, normalized size of antiderivative = 2.23 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^4*(A+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 - 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 + 780*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*c - 60* log(tan((c + d*x)/2) - 1)*a - 390*log(tan((c + d*x)/2) - 1)*c + 60*log(tan ((c + d*x)/2) + 1)*tan((c + d*x)/2)**4*a + 390*log(tan((c + d*x)/2) + 1)*t an((c + d*x)/2)**4*c - 120*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*a - 780*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*c + 60*log(tan((c + d *x)/2) + 1)*a + 390*log(tan((c + d*x)/2) + 1)*c - 3*tan((c + d*x)/2)**9*a - 3*tan((c + d*x)/2)**9*c - 14*tan((c + d*x)/2)**7*a - 34*tan((c + d*x)/2) **7*c - 68*tan((c + d*x)/2)**5*a - 388*tan((c + d*x)/2)**5*c + 190*tan((c + d*x)/2)**3*a + 1310*tan((c + d*x)/2)**3*c - 105*tan((c + d*x)/2)*a - 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))