Integrand size = 41, antiderivative size = 169 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {(2 A-4 B+7 C) \text {arctanh}(\sin (c+d x))}{2 a^2 d}-\frac {2 (2 A-5 B+8 C) \tan (c+d x)}{3 a^2 d}+\frac {(2 A-4 B+7 C) \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac {(2 A-5 B+8 C) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2} \] Output:
1/2*(2*A-4*B+7*C)*arctanh(sin(d*x+c))/a^2/d-2/3*(2*A-5*B+8*C)*tan(d*x+c)/a ^2/d+1/2*(2*A-4*B+7*C)*sec(d*x+c)*tan(d*x+c)/a^2/d-1/3*(2*A-5*B+8*C)*sec(d *x+c)^2*tan(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*(A-B+C)*sec(d*x+c)^3*tan(d*x+c )/d/(a+a*sec(d*x+c))^2
Leaf count is larger than twice the leaf count of optimal. \(901\) vs. \(2(169)=338\).
Time = 7.51 (sec) , antiderivative size = 901, normalized size of antiderivative = 5.33 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:
Integrate[(Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a* Sec[c + d*x])^2,x]
Output:
(-4*(2*A - 4*B + 7*C)*Cos[c/2 + (d*x)/2]^4*Log[Cos[c/2 + (d*x)/2] - Sin[c/ 2 + (d*x)/2]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*(A + 2*C + 2*B*C os[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) + (4*(2*A - 4*B + 7*C)*Cos[c/2 + (d*x)/2]^4*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*( 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])^2) + (Cos[c/2 + (d*x)/2]*Sec[c/2]* Sec[c]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(20*A*Sin[(d *x)/2] - 14*B*Sin[(d*x)/2] + 14*C*Sin[(d*x)/2] - 22*A*Sin[(3*d*x)/2] + 64* B*Sin[(3*d*x)/2] - 97*C*Sin[(3*d*x)/2] + 36*A*Sin[c - (d*x)/2] - 84*B*Sin[ c - (d*x)/2] + 126*C*Sin[c - (d*x)/2] - 36*A*Sin[c + (d*x)/2] + 42*B*Sin[c + (d*x)/2] - 42*C*Sin[c + (d*x)/2] + 20*A*Sin[2*c + (d*x)/2] - 56*B*Sin[2 *c + (d*x)/2] + 98*C*Sin[2*c + (d*x)/2] + 18*A*Sin[c + (3*d*x)/2] - 6*B*Si n[c + (3*d*x)/2] + 3*C*Sin[c + (3*d*x)/2] - 22*A*Sin[2*c + (3*d*x)/2] + 34 *B*Sin[2*c + (3*d*x)/2] - 37*C*Sin[2*c + (3*d*x)/2] + 18*A*Sin[3*c + (3*d* x)/2] - 36*B*Sin[3*c + (3*d*x)/2] + 63*C*Sin[3*c + (3*d*x)/2] - 18*A*Sin[c + (5*d*x)/2] + 48*B*Sin[c + (5*d*x)/2] - 75*C*Sin[c + (5*d*x)/2] + 6*A*Si n[2*c + (5*d*x)/2] + 6*B*Sin[2*c + (5*d*x)/2] - 15*C*Sin[2*c + (5*d*x)/2] - 18*A*Sin[3*c + (5*d*x)/2] + 30*B*Sin[3*c + (5*d*x)/2] - 39*C*Sin[3*c + ( 5*d*x)/2] + 6*A*Sin[4*c + (5*d*x)/2] - 12*B*Sin[4*c + (5*d*x)/2] + 21*C*Si n[4*c + (5*d*x)/2] - 8*A*Sin[2*c + (7*d*x)/2] + 20*B*Sin[2*c + (7*d*x)/...
Time = 1.05 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.98, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.317, Rules used = {3042, 4572, 3042, 4507, 25, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 4572 |
| \(\displaystyle \frac {\int \frac {\sec ^3(c+d x) (3 a (B-C)+a (2 A-2 B+5 C) \sec (c+d x))}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (3 a (B-C)+a (2 A-2 B+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\frac {\int -\sec ^2(c+d x) \left (2 a^2 (2 A-5 B+8 C)-3 a^2 (2 A-4 B+7 C) \sec (c+d x)\right )dx}{a^2}-\frac {(2 A-5 B+8 C) \tan (c+d x) \sec ^2(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\int \sec ^2(c+d x) \left (2 a^2 (2 A-5 B+8 C)-3 a^2 (2 A-4 B+7 C) \sec (c+d x)\right )dx}{a^2}-\frac {(2 A-5 B+8 C) \tan (c+d x) \sec ^2(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (2 a^2 (2 A-5 B+8 C)-3 a^2 (2 A-4 B+7 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}-\frac {(2 A-5 B+8 C) \tan (c+d x) \sec ^2(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {-\frac {2 a^2 (2 A-5 B+8 C) \int \sec ^2(c+d x)dx-3 a^2 (2 A-4 B+7 C) \int \sec ^3(c+d x)dx}{a^2}-\frac {(2 A-5 B+8 C) \tan (c+d x) \sec ^2(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {2 a^2 (2 A-5 B+8 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-3 a^2 (2 A-4 B+7 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}-\frac {(2 A-5 B+8 C) \tan (c+d x) \sec ^2(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {-\frac {-\frac {2 a^2 (2 A-5 B+8 C) \int 1d(-\tan (c+d x))}{d}-3 a^2 (2 A-4 B+7 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}-\frac {(2 A-5 B+8 C) \tan (c+d x) \sec ^2(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {-\frac {\frac {2 a^2 (2 A-5 B+8 C) \tan (c+d x)}{d}-3 a^2 (2 A-4 B+7 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}-\frac {(2 A-5 B+8 C) \tan (c+d x) \sec ^2(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {-\frac {\frac {2 a^2 (2 A-5 B+8 C) \tan (c+d x)}{d}-3 a^2 (2 A-4 B+7 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}-\frac {(2 A-5 B+8 C) \tan (c+d x) \sec ^2(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {2 a^2 (2 A-5 B+8 C) \tan (c+d x)}{d}-3 a^2 (2 A-4 B+7 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}-\frac {(2 A-5 B+8 C) \tan (c+d x) \sec ^2(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {-\frac {\frac {2 a^2 (2 A-5 B+8 C) \tan (c+d x)}{d}-3 a^2 (2 A-4 B+7 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}-\frac {(2 A-5 B+8 C) \tan (c+d x) \sec ^2(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
Input:
Int[(Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^2,x]
Output:
-1/3*((A - B + C)*Sec[c + d*x]^3*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^2) + (-(((2*A - 5*B + 8*C)*Sec[c + d*x]^2*Tan[c + d*x])/(d*(1 + Sec[c + d*x]) )) - ((2*a^2*(2*A - 5*B + 8*C)*Tan[c + d*x])/d - 3*a^2*(2*A - 4*B + 7*C)*( ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/a^2)/(3* 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 = 0.56 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.04
| method | result | size |
| parallelrisch | \(\frac {-3 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A -2 B +\frac {7 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A -2 B +\frac {7 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (\left (\frac {5 A}{2}-7 B +\frac {43 C}{4}\right ) \cos \left (2 d x +2 c \right )+\left (A -\frac {5 B}{2}+4 C \right ) \cos \left (3 d x +3 c \right )+3 \left (A -\frac {7 B}{2}+5 C \right ) \cos \left (d x +c \right )+\frac {5 A}{2}-7 B +\frac {37 C}{4}\right )}{3 d \,a^{2} \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(175\) |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {2 B -5 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-7 C +4 B -2 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 B -5 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (2 A -4 B +7 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}{2 d \,a^{2}}\) | \(209\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {2 B -5 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-7 C +4 B -2 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 B -5 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (2 A -4 B +7 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}{2 d \,a^{2}}\) | \(209\) |
| norman | \(\frac {\frac {\left (5 A -11 B +18 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}-\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{6 a d}-\frac {\left (3 A -9 B +13 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {\left (5 A -11 B +17 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{6 a d}-\frac {\left (25 A -61 B +100 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 a d}+\frac {\left (35 A -95 B +149 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4} a}-\frac {\left (2 A -4 B +7 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{2} d}+\frac {\left (2 A -4 B +7 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{2} d}\) | \(249\) |
| risch | \(-\frac {i \left (6 A \,{\mathrm e}^{6 i \left (d x +c \right )}-12 B \,{\mathrm e}^{6 i \left (d x +c \right )}+21 C \,{\mathrm e}^{6 i \left (d x +c \right )}+18 A \,{\mathrm e}^{5 i \left (d x +c \right )}-36 B \,{\mathrm e}^{5 i \left (d x +c \right )}+63 C \,{\mathrm e}^{5 i \left (d x +c \right )}+20 A \,{\mathrm e}^{4 i \left (d x +c \right )}-56 B \,{\mathrm e}^{4 i \left (d x +c \right )}+98 C \,{\mathrm e}^{4 i \left (d x +c \right )}+36 A \,{\mathrm e}^{3 i \left (d x +c \right )}-84 B \,{\mathrm e}^{3 i \left (d x +c \right )}+126 C \,{\mathrm e}^{3 i \left (d x +c \right )}+22 A \,{\mathrm e}^{2 i \left (d x +c \right )}-64 B \,{\mathrm e}^{2 i \left (d x +c \right )}+97 C \,{\mathrm e}^{2 i \left (d x +c \right )}+18 A \,{\mathrm e}^{i \left (d x +c \right )}-48 B \,{\mathrm e}^{i \left (d x +c \right )}+75 C \,{\mathrm e}^{i \left (d x +c \right )}+8 A -20 B +32 C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a^{2} d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{2} d}+\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 a^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{a^{2} d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a^{2} d}-\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 a^{2} d}\) | \(394\) |
Input:
int(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x,meth od=_RETURNVERBOSE)
Output:
1/3*(-3*(1+cos(2*d*x+2*c))*(A-2*B+7/2*C)*ln(tan(1/2*d*x+1/2*c)-1)+3*(1+cos (2*d*x+2*c))*(A-2*B+7/2*C)*ln(tan(1/2*d*x+1/2*c)+1)-tan(1/2*d*x+1/2*c)*sec (1/2*d*x+1/2*c)^2*((5/2*A-7*B+43/4*C)*cos(2*d*x+2*c)+(A-5/2*B+4*C)*cos(3*d *x+3*c)+3*(A-7/2*B+5*C)*cos(d*x+c)+5/2*A-7*B+37/4*C))/d/a^2/(1+cos(2*d*x+2 *c))
Time = 0.09 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.49 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {3 \, {\left ({\left (2 \, A - 4 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (2 \, A - 4 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A - 4 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (2 \, A - 4 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (2 \, A - 4 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A - 4 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (2 \, A - 5 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (10 \, A - 28 \, B + 43 \, C\right )} \cos \left (d x + c\right )^{2} - 6 \, {\left (B - C\right )} \cos \left (d x + c\right ) - 3 \, C\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2, x, algorithm="fricas")
Output:
1/12*(3*((2*A - 4*B + 7*C)*cos(d*x + c)^4 + 2*(2*A - 4*B + 7*C)*cos(d*x + c)^3 + (2*A - 4*B + 7*C)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 3*((2*A - 4*B + 7*C)*cos(d*x + c)^4 + 2*(2*A - 4*B + 7*C)*cos(d*x + c)^3 + (2*A - 4 *B + 7*C)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(4*(2*A - 5*B + 8*C)* cos(d*x + c)^3 + (10*A - 28*B + 43*C)*cos(d*x + c)^2 - 6*(B - C)*cos(d*x + c) - 3*C)*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)
\[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {A \sec ^{3}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{4}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{5}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(sec(d*x+c)**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))* *2,x)
Output:
(Integral(A*sec(c + d*x)**3/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + I ntegral(B*sec(c + d*x)**4/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Int egral(C*sec(c + d*x)**5/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x))/a**2
Leaf count of result is larger than twice the leaf count of optimal. 431 vs. \(2 (159) = 318\).
Time = 0.05 (sec) , antiderivative size = 431, normalized size of antiderivative = 2.55 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=-\frac {C {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} - \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} - B {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + A {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )}}{6 \, d} \] Input:
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2, x, algorithm="maxima")
Output:
-1/6*(C*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 - 2*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d* x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) + sin (d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 21*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 21*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2) - B*((15* sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 12*log(sin(d*x + c)/( cos(d*x + c) + 1) - 1)/a^2 + 12*sin(d*x + c)/((a^2 - a^2*sin(d*x + c)^2/(c os(d*x + c) + 1)^2)*(cos(d*x + c) + 1))) + A*((9*sin(d*x + c)/(cos(d*x + c ) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 6*log(sin(d*x + c)/(co s(d*x + c) + 1) + 1)/a^2 + 6*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2) )/d
Time = 0.30 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.39 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {3 \, {\left (2 \, A - 4 \, B + 7 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {3 \, {\left (2 \, A - 4 \, B + 7 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac {6 \, {\left (2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2}} - \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \] Input:
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2, x, algorithm="giac")
Output:
1/6*(3*(2*A - 4*B + 7*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 3*(2*A - 4*B + 7*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 - 6*(2*B*tan(1/2*d*x + 1/2*c)^3 - 5*C*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c) + 3*C*tan (1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^2) - (A*a^4*tan(1/2*d *x + 1/2*c)^3 - B*a^4*tan(1/2*d*x + 1/2*c)^3 + C*a^4*tan(1/2*d*x + 1/2*c)^ 3 + 9*A*a^4*tan(1/2*d*x + 1/2*c) - 15*B*a^4*tan(1/2*d*x + 1/2*c) + 21*C*a^ 4*tan(1/2*d*x + 1/2*c))/a^6)/d
Time = 12.69 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.01 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A-2\,B+\frac {7\,C}{2}\right )}{a^2\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-B+C\right )}{2\,a^2}-\frac {2\,B-4\,C}{2\,a^2}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B+C\right )}{6\,a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,B-5\,C\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,B-3\,C\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )} \] Input:
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a/cos(c + d*x))^2),x)
Output:
(2*atanh(tan(c/2 + (d*x)/2))*(A - 2*B + (7*C)/2))/(a^2*d) - (tan(c/2 + (d* x)/2)*((3*(A - B + C))/(2*a^2) - (2*B - 4*C)/(2*a^2)))/d - (tan(c/2 + (d*x )/2)^3*(A - B + C))/(6*a^2*d) - (tan(c/2 + (d*x)/2)^3*(2*B - 5*C) - tan(c/ 2 + (d*x)/2)*(2*B - 3*C))/(d*(a^2*tan(c/2 + (d*x)/2)^4 - 2*a^2*tan(c/2 + ( d*x)/2)^2 + a^2))
Time = 0.16 (sec) , antiderivative size = 600, normalized size of antiderivative = 3.55 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x)
Output:
( - 6*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**4*a + 12*log(tan((c + d* x)/2) - 1)*tan((c + d*x)/2)**4*b - 21*log(tan((c + d*x)/2) - 1)*tan((c + d *x)/2)**4*c + 12*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*a - 24*log( tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*b + 42*log(tan((c + d*x)/2) - 1) *tan((c + d*x)/2)**2*c - 6*log(tan((c + d*x)/2) - 1)*a + 12*log(tan((c + d *x)/2) - 1)*b - 21*log(tan((c + d*x)/2) - 1)*c + 6*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**4*a - 12*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)** 4*b + 21*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**4*c - 12*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*a + 24*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*b - 42*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*c + 6*lo g(tan((c + d*x)/2) + 1)*a - 12*log(tan((c + d*x)/2) + 1)*b + 21*log(tan((c + d*x)/2) + 1)*c - tan((c + d*x)/2)**7*a + tan((c + d*x)/2)**7*b - tan((c + d*x)/2)**7*c - 7*tan((c + d*x)/2)**5*a + 13*tan((c + d*x)/2)**5*b - 19* tan((c + d*x)/2)**5*c + 17*tan((c + d*x)/2)**3*a - 41*tan((c + d*x)/2)**3* b + 71*tan((c + d*x)/2)**3*c - 9*tan((c + d*x)/2)*a + 27*tan((c + d*x)/2)* b - 39*tan((c + d*x)/2)*c)/(6*a**2*d*(tan((c + d*x)/2)**4 - 2*tan((c + d*x )/2)**2 + 1))