Integrand size = 21, antiderivative size = 94 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {3 a \left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {3 a \sec ^2(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d} \] Output:
3/8*a*(a^2-b^2)*arctanh(sin(d*x+c))/d+3/8*a*sec(d*x+c)^2*(b+a*sin(d*x+c))* (a+b*sin(d*x+c))/d+1/4*sec(d*x+c)^3*(a+b*sin(d*x+c))^3*tan(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(318\) vs. \(2(94)=188\).
Time = 4.63 (sec) , antiderivative size = 318, normalized size of antiderivative = 3.38 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {-6 a \left (a^2-b^2\right )^3 (\log (1-\sin (c+d x))-\log (1+\sin (c+d x)))+a b \sec ^4(c+d x) \left (-8 a^5+8 a^3 b^2+\left (18 a^4 b-11 a^2 b^3+5 b^5\right ) \sin (3 (c+d x))\right )+a \left (8 a^6-22 a^4 b^2+29 a^2 b^4-3 b^6\right ) \sec ^3(c+d x) \tan (c+d x)+16 a^4 b \left (3 a^2-2 b^2\right ) \tan ^2(c+d x)+8 b^3 \left (4 a^4-5 a^2 b^2+b^4\right ) \tan ^4(c+d x)+4 a \sec (c+d x) \tan (c+d x) \left (3 \left (a^6-5 a^4 b^2\right )+4 b^2 \left (3 a^4-5 a^2 b^2+2 b^4\right ) \tan ^2(c+d x)\right )+16 a^2 b \sec ^2(c+d x) \left (-a^4+\left (2 a^4-5 a^2 b^2+3 b^4\right ) \tan ^2(c+d x)\right )}{32 \left (a^2-b^2\right )^2 d} \] Input:
Integrate[Sec[c + d*x]^5*(a + b*Sin[c + d*x])^3,x]
Output:
(-6*a*(a^2 - b^2)^3*(Log[1 - Sin[c + d*x]] - Log[1 + Sin[c + d*x]]) + a*b* Sec[c + d*x]^4*(-8*a^5 + 8*a^3*b^2 + (18*a^4*b - 11*a^2*b^3 + 5*b^5)*Sin[3 *(c + d*x)]) + a*(8*a^6 - 22*a^4*b^2 + 29*a^2*b^4 - 3*b^6)*Sec[c + d*x]^3* Tan[c + d*x] + 16*a^4*b*(3*a^2 - 2*b^2)*Tan[c + d*x]^2 + 8*b^3*(4*a^4 - 5* a^2*b^2 + b^4)*Tan[c + d*x]^4 + 4*a*Sec[c + d*x]*Tan[c + d*x]*(3*(a^6 - 5* a^4*b^2) + 4*b^2*(3*a^4 - 5*a^2*b^2 + 2*b^4)*Tan[c + d*x]^2) + 16*a^2*b*Se c[c + d*x]^2*(-a^4 + (2*a^4 - 5*a^2*b^2 + 3*b^4)*Tan[c + d*x]^2))/(32*(a^2 - b^2)^2*d)
Time = 0.38 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.49, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3147, 477, 2009}
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.
| \(\displaystyle \int \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (c+d x))^3}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle \frac {b^5 \int \frac {(a+b \sin (c+d x))^3}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 477 |
| \(\displaystyle \frac {\int \left (\frac {(a+b)^3 b^3}{8 (b-b \sin (c+d x))^3}+\frac {(a-b)^3 b^3}{8 (\sin (c+d x) b+b)^3}+\frac {3 a \left (a^2-b^2\right ) b^2}{8 \left (b^2-b^2 \sin ^2(c+d x)\right )}+\frac {3 (a-b) (a+b)^2 b^2}{16 (b-b \sin (c+d x))^2}+\frac {3 (a-b)^2 (a+b) b^2}{16 (\sin (c+d x) b+b)^2}\right )d(b \sin (c+d x))}{b d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {3}{8} a b \left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))+\frac {b^3 (a+b)^3}{16 (b-b \sin (c+d x))^2}-\frac {b^3 (a-b)^3}{16 (b \sin (c+d x)+b)^2}+\frac {3 b^2 (a-b) (a+b)^2}{16 (b-b \sin (c+d x))}-\frac {3 b^2 (a-b)^2 (a+b)}{16 (b \sin (c+d x)+b)}}{b d}\) |
Input:
Int[Sec[c + d*x]^5*(a + b*Sin[c + d*x])^3,x]
Output:
((3*a*b*(a^2 - b^2)*ArcTanh[Sin[c + d*x]])/8 + (b^3*(a + b)^3)/(16*(b - b* Sin[c + d*x])^2) + (3*(a - b)*b^2*(a + b)^2)/(16*(b - b*Sin[c + d*x])) - ( (a - b)^3*b^3)/(16*(b + b*Sin[c + d*x])^2) - (3*(a - b)^2*b^2*(a + b))/(16 *(b + b*Sin[c + d*x])))/(b*d)
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ a^p Int[ExpandIntegrand[(c + d*x)^n*(1 - Rt[-b/a, 2]*x)^p*(1 + Rt[-b/a, 2 ]*x)^p, x], x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[p, 0] && IntegerQ[n] & & NiceSqrtQ[-b/a] && !FractionalPowerFactorQ[Rt[-b/a, 2]]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 0.63 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.66
| method | result | size |
| derivativedivides | \(\frac {a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {3 a^{2} b}{4 \cos \left (d x +c \right )^{4}}+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {b^{3} \sin \left (d x +c \right )^{4}}{4 \cos \left (d x +c \right )^{4}}}{d}\) | \(156\) |
| default | \(\frac {a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {3 a^{2} b}{4 \cos \left (d x +c \right )^{4}}+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {b^{3} \sin \left (d x +c \right )^{4}}{4 \cos \left (d x +c \right )^{4}}}{d}\) | \(156\) |
| parallelrisch | \(\frac {-6 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -b \right ) a \left (a +b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -b \right ) a \left (a +b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+4 \left (-3 a^{2} b -b^{3}\right ) \cos \left (2 d x +2 c \right )+\left (-3 a^{2} b +b^{3}\right ) \cos \left (4 d x +4 c \right )+3 \left (a^{3}-a \,b^{2}\right ) \sin \left (3 d x +3 c \right )+\left (11 a^{3}+21 a \,b^{2}\right ) \sin \left (d x +c \right )+15 a^{2} b +3 b^{3}}{4 d \left (3+4 \cos \left (2 d x +2 c \right )+\cos \left (4 d x +4 c \right )\right )}\) | \(215\) |
| risch | \(-\frac {{\mathrm e}^{i \left (d x +c \right )} \left (3 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-3 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+11 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+21 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+8 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-11 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-21 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-48 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}-3 i a^{3}+3 i a \,b^{2}+8 b^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a \,b^{2}}{8 d}+\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a \,b^{2}}{8 d}\) | \(283\) |
| norman | \(\frac {\frac {6 a^{2} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {6 a^{2} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d}+\frac {\left (18 a^{2} b +4 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {\left (18 a^{2} b +4 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}+\frac {a \left (7 a^{2}+33 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {3 \left (8 a^{2} b +4 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}+\frac {3 \left (8 a^{2} b +4 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\frac {3 a \left (3 a^{2}+5 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 d}+\frac {3 a \left (3 a^{2}+5 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{2 d}+\frac {a \left (5 a^{2}+3 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a \left (5 a^{2}+3 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d}+\frac {3 a \left (9 a^{2}+31 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}+\frac {3 a \left (9 a^{2}+31 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-\frac {3 a \left (a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {3 a \left (a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(431\) |
Input:
int(sec(d*x+c)^5*(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(a^3*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c) +tan(d*x+c)))+3/4*a^2*b/cos(d*x+c)^4+3*a*b^2*(1/4*sin(d*x+c)^3/cos(d*x+c)^ 4+1/8*sin(d*x+c)^3/cos(d*x+c)^2+1/8*sin(d*x+c)-1/8*ln(sec(d*x+c)+tan(d*x+c )))+1/4*b^3*sin(d*x+c)^4/cos(d*x+c)^4)
Time = 0.10 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.47 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {3 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 8 \, b^{3} \cos \left (d x + c\right )^{2} + 12 \, a^{2} b + 4 \, b^{3} + 2 \, {\left (2 \, a^{3} + 6 \, a b^{2} + 3 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate(sec(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="fricas")
Output:
1/16*(3*(a^3 - a*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(a^3 - a*b^ 2)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) - 8*b^3*cos(d*x + c)^2 + 12*a^2*b + 4*b^3 + 2*(2*a^3 + 6*a*b^2 + 3*(a^3 - a*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^4)
Timed out. \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**5*(a+b*sin(d*x+c))**3,x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.45 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {3 \, {\left (a^{3} - a b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (a^{3} - a b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + \frac {2 \, {\left (4 \, b^{3} \sin \left (d x + c\right )^{2} - 3 \, {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )^{3} + 6 \, a^{2} b - 2 \, b^{3} + {\left (5 \, a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \] Input:
integrate(sec(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="maxima")
Output:
1/16*(3*(a^3 - a*b^2)*log(sin(d*x + c) + 1) - 3*(a^3 - a*b^2)*log(sin(d*x + c) - 1) + 2*(4*b^3*sin(d*x + c)^2 - 3*(a^3 - a*b^2)*sin(d*x + c)^3 + 6*a ^2*b - 2*b^3 + (5*a^3 + 3*a*b^2)*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d
Time = 0.13 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.52 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {3 \, {\left (a^{3} - a b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, d} - \frac {3 \, {\left (a^{3} - a b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, d} - \frac {3 \, a^{3} \sin \left (d x + c\right )^{3} - 3 \, a b^{2} \sin \left (d x + c\right )^{3} - 4 \, b^{3} \sin \left (d x + c\right )^{2} - 5 \, a^{3} \sin \left (d x + c\right ) - 3 \, a b^{2} \sin \left (d x + c\right ) - 6 \, a^{2} b + 2 \, b^{3}}{8 \, {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2} d} \] Input:
integrate(sec(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="giac")
Output:
3/16*(a^3 - a*b^2)*log(abs(sin(d*x + c) + 1))/d - 3/16*(a^3 - a*b^2)*log(a bs(sin(d*x + c) - 1))/d - 1/8*(3*a^3*sin(d*x + c)^3 - 3*a*b^2*sin(d*x + c) ^3 - 4*b^3*sin(d*x + c)^2 - 5*a^3*sin(d*x + c) - 3*a*b^2*sin(d*x + c) - 6* a^2*b + 2*b^3)/((sin(d*x + c)^2 - 1)^2*d)
Time = 15.38 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.21 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {{\sin \left (c+d\,x\right )}^3\,\left (\frac {3\,a\,b^2}{8}-\frac {3\,a^3}{8}\right )+\frac {3\,a^2\,b}{4}-\frac {b^3}{4}+\sin \left (c+d\,x\right )\,\left (\frac {5\,a^3}{8}+\frac {3\,a\,b^2}{8}\right )+\frac {b^3\,{\sin \left (c+d\,x\right )}^2}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2+1\right )}+\frac {3\,a\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (a^2-b^2\right )}{8\,d} \] Input:
int((a + b*sin(c + d*x))^3/cos(c + d*x)^5,x)
Output:
(sin(c + d*x)^3*((3*a*b^2)/8 - (3*a^3)/8) + (3*a^2*b)/4 - b^3/4 + sin(c + d*x)*((3*a*b^2)/8 + (5*a^3)/8) + (b^3*sin(c + d*x)^2)/2)/(d*(sin(c + d*x)^ 4 - 2*sin(c + d*x)^2 + 1)) + (3*a*atanh(sin(c + d*x))*(a^2 - b^2))/(8*d)
Time = 0.18 (sec) , antiderivative size = 393, normalized size of antiderivative = 4.18 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a^{3}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a \,b^{2}+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{3}-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a \,b^{2}-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{3}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a \,b^{2}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a^{3}-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a \,b^{2}-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{3}+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a \,b^{2}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{3}-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a \,b^{2}-6 \sin \left (d x +c \right )^{4} a^{2} b +2 \sin \left (d x +c \right )^{4} b^{3}-3 \sin \left (d x +c \right )^{3} a^{3}+3 \sin \left (d x +c \right )^{3} a \,b^{2}+12 \sin \left (d x +c \right )^{2} a^{2} b +5 \sin \left (d x +c \right ) a^{3}+3 \sin \left (d x +c \right ) a \,b^{2}}{8 d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(sec(d*x+c)^5*(a+b*sin(d*x+c))^3,x)
Output:
( - 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**3 + 3*log(tan((c + d*x) /2) - 1)*sin(c + d*x)**4*a*b**2 + 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x) **2*a**3 - 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**2 - 3*log(tan( (c + d*x)/2) - 1)*a**3 + 3*log(tan((c + d*x)/2) - 1)*a*b**2 + 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**3 - 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a*b**2 - 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**3 + 6*log (tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b**2 + 3*log(tan((c + d*x)/2) + 1 )*a**3 - 3*log(tan((c + d*x)/2) + 1)*a*b**2 - 6*sin(c + d*x)**4*a**2*b + 2 *sin(c + d*x)**4*b**3 - 3*sin(c + d*x)**3*a**3 + 3*sin(c + d*x)**3*a*b**2 + 12*sin(c + d*x)**2*a**2*b + 5*sin(c + d*x)*a**3 + 3*sin(c + d*x)*a*b**2) /(8*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))