Integrand size = 21, antiderivative size = 211 \[ \int (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx=a^3 x+\frac {15}{2} a b^2 x-\frac {3 a^2 b \cos (c+d x)}{d}-\frac {3 b^3 \cos (c+d x)}{d}+\frac {b^3 \cos ^3(c+d x)}{3 d}-\frac {6 a^2 b \sec (c+d x)}{d}-\frac {3 b^3 \sec (c+d x)}{d}+\frac {a^2 b \sec ^3(c+d x)}{d}+\frac {b^3 \sec ^3(c+d x)}{3 d}-\frac {3 a b^2 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {a^3 \tan (c+d x)}{d}-\frac {6 a b^2 \tan (c+d x)}{d}+\frac {a^3 \tan ^3(c+d x)}{3 d}+\frac {a b^2 \tan ^3(c+d x)}{d} \] Output:
a^3*x+15/2*a*b^2*x-3*a^2*b*cos(d*x+c)/d-3*b^3*cos(d*x+c)/d+1/3*b^3*cos(d*x +c)^3/d-6*a^2*b*sec(d*x+c)/d-3*b^3*sec(d*x+c)/d+a^2*b*sec(d*x+c)^3/d+1/3*b ^3*sec(d*x+c)^3/d-3/2*a*b^2*cos(d*x+c)*sin(d*x+c)/d-a^3*tan(d*x+c)/d-6*a*b ^2*tan(d*x+c)/d+1/3*a^3*tan(d*x+c)^3/d+a*b^2*tan(d*x+c)^3/d
Time = 1.05 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.07 \[ \int (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx=\frac {\sec ^3(c+d x) \left (-300 a^2 b-210 b^3+36 a \left (2 a^2+15 b^2\right ) (c+d x) \cos (c+d x)-3 \left (144 a^2 b+91 b^3\right ) \cos (2 (c+d x))+24 a^3 c \cos (3 (c+d x))+180 a b^2 c \cos (3 (c+d x))+24 a^3 d x \cos (3 (c+d x))+180 a b^2 d x \cos (3 (c+d x))-36 a^2 b \cos (4 (c+d x))-30 b^3 \cos (4 (c+d x))+b^3 \cos (6 (c+d x))-90 a b^2 \sin (c+d x)-32 a^3 \sin (3 (c+d x))-195 a b^2 \sin (3 (c+d x))-9 a b^2 \sin (5 (c+d x))\right )}{96 d} \] Input:
Integrate[(a + b*Sin[c + d*x])^3*Tan[c + d*x]^4,x]
Output:
(Sec[c + d*x]^3*(-300*a^2*b - 210*b^3 + 36*a*(2*a^2 + 15*b^2)*(c + d*x)*Co s[c + d*x] - 3*(144*a^2*b + 91*b^3)*Cos[2*(c + d*x)] + 24*a^3*c*Cos[3*(c + d*x)] + 180*a*b^2*c*Cos[3*(c + d*x)] + 24*a^3*d*x*Cos[3*(c + d*x)] + 180* a*b^2*d*x*Cos[3*(c + d*x)] - 36*a^2*b*Cos[4*(c + d*x)] - 30*b^3*Cos[4*(c + d*x)] + b^3*Cos[6*(c + d*x)] - 90*a*b^2*Sin[c + d*x] - 32*a^3*Sin[3*(c + d*x)] - 195*a*b^2*Sin[3*(c + d*x)] - 9*a*b^2*Sin[5*(c + d*x)]))/(96*d)
Time = 0.51 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 3201, 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.
| \(\displaystyle \int \tan ^4(c+d x) (a+b \sin (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (c+d x)^4 (a+b \sin (c+d x))^3dx\) |
| \(\Big \downarrow \) 3201 |
| \(\displaystyle \int \left (a^3 \tan ^4(c+d x)+3 a^2 b \sin (c+d x) \tan ^4(c+d x)+3 a b^2 \sin ^2(c+d x) \tan ^4(c+d x)+b^3 \sin ^3(c+d x) \tan ^4(c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^3 \tan ^3(c+d x)}{3 d}-\frac {a^3 \tan (c+d x)}{d}+a^3 x-\frac {3 a^2 b \cos (c+d x)}{d}+\frac {a^2 b \sec ^3(c+d x)}{d}-\frac {6 a^2 b \sec (c+d x)}{d}+\frac {5 a b^2 \tan ^3(c+d x)}{2 d}-\frac {15 a b^2 \tan (c+d x)}{2 d}-\frac {3 a b^2 \sin ^2(c+d x) \tan ^3(c+d x)}{2 d}+\frac {15}{2} a b^2 x+\frac {b^3 \cos ^3(c+d x)}{3 d}-\frac {3 b^3 \cos (c+d x)}{d}+\frac {b^3 \sec ^3(c+d x)}{3 d}-\frac {3 b^3 \sec (c+d x)}{d}\) |
Input:
Int[(a + b*Sin[c + d*x])^3*Tan[c + d*x]^4,x]
Output:
a^3*x + (15*a*b^2*x)/2 - (3*a^2*b*Cos[c + d*x])/d - (3*b^3*Cos[c + d*x])/d + (b^3*Cos[c + d*x]^3)/(3*d) - (6*a^2*b*Sec[c + d*x])/d - (3*b^3*Sec[c + d*x])/d + (a^2*b*Sec[c + d*x]^3)/d + (b^3*Sec[c + d*x]^3)/(3*d) - (a^3*Tan [c + d*x])/d - (15*a*b^2*Tan[c + d*x])/(2*d) + (a^3*Tan[c + d*x]^3)/(3*d) + (5*a*b^2*Tan[c + d*x]^3)/(2*d) - (3*a*b^2*Sin[c + d*x]^2*Tan[c + d*x]^3) /(2*d)
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((g_.)*tan[(e_.) + (f_.)*( x_)])^(p_.), x_Symbol] :> Int[ExpandIntegrand[(g*Tan[e + f*x])^p, (a + b*Si n[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]
Time = 8.57 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.27
| method | result | size |
| derivativedivides | \(\frac {a^{3} \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+d x +c \right )+3 a^{2} b \left (\frac {\sin \left (d x +c \right )^{6}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )+b^{3} \left (\frac {\sin \left (d x +c \right )^{8}}{3 \cos \left (d x +c \right )^{3}}-\frac {5 \sin \left (d x +c \right )^{8}}{3 \cos \left (d x +c \right )}-\frac {5 \left (\frac {16}{5}+\sin \left (d x +c \right )^{6}+\frac {6 \sin \left (d x +c \right )^{4}}{5}+\frac {8 \sin \left (d x +c \right )^{2}}{5}\right ) \cos \left (d x +c \right )}{3}\right )}{d}\) | \(268\) |
| default | \(\frac {a^{3} \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+d x +c \right )+3 a^{2} b \left (\frac {\sin \left (d x +c \right )^{6}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )+b^{3} \left (\frac {\sin \left (d x +c \right )^{8}}{3 \cos \left (d x +c \right )^{3}}-\frac {5 \sin \left (d x +c \right )^{8}}{3 \cos \left (d x +c \right )}-\frac {5 \left (\frac {16}{5}+\sin \left (d x +c \right )^{6}+\frac {6 \sin \left (d x +c \right )^{4}}{5}+\frac {8 \sin \left (d x +c \right )^{2}}{5}\right ) \cos \left (d x +c \right )}{3}\right )}{d}\) | \(268\) |
| parts | \(\frac {a^{3} \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {b^{3} \left (\frac {\sin \left (d x +c \right )^{8}}{3 \cos \left (d x +c \right )^{3}}-\frac {5 \sin \left (d x +c \right )^{8}}{3 \cos \left (d x +c \right )}-\frac {5 \left (\frac {16}{5}+\sin \left (d x +c \right )^{6}+\frac {6 \sin \left (d x +c \right )^{4}}{5}+\frac {8 \sin \left (d x +c \right )^{2}}{5}\right ) \cos \left (d x +c \right )}{3}\right )}{d}+\frac {3 a^{2} b \left (\frac {\sin \left (d x +c \right )^{6}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )}{d}+\frac {3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )}{d}\) | \(279\) |
| risch | \(a^{3} x +\frac {15 a \,b^{2} x}{2}+\frac {b^{3} {\mathrm e}^{3 i \left (d x +c \right )}}{24 d}+\frac {3 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {3 b \,{\mathrm e}^{i \left (d x +c \right )} a^{2}}{2 d}-\frac {11 b^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {3 b \,{\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 d}-\frac {11 b^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {3 i a \,b^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {b^{3} {\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}-\frac {2 \left (6 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+27 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+18 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}+9 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+6 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+36 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+24 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+14 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+4 i a^{3}+21 i a \,b^{2}+18 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+9 b^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}\) | \(337\) |
Input:
int((a+b*sin(d*x+c))^3*tan(d*x+c)^4,x,method=_RETURNVERBOSE)
Output:
1/d*(a^3*(1/3*tan(d*x+c)^3-tan(d*x+c)+d*x+c)+3*a^2*b*(1/3*sin(d*x+c)^6/cos (d*x+c)^3-sin(d*x+c)^6/cos(d*x+c)-(8/3+sin(d*x+c)^4+4/3*sin(d*x+c)^2)*cos( d*x+c))+3*a*b^2*(1/3*sin(d*x+c)^7/cos(d*x+c)^3-4/3*sin(d*x+c)^7/cos(d*x+c) -4/3*(sin(d*x+c)^5+5/4*sin(d*x+c)^3+15/8*sin(d*x+c))*cos(d*x+c)+5/2*d*x+5/ 2*c)+b^3*(1/3*sin(d*x+c)^8/cos(d*x+c)^3-5/3*sin(d*x+c)^8/cos(d*x+c)-5/3*(1 6/5+sin(d*x+c)^6+6/5*sin(d*x+c)^4+8/5*sin(d*x+c)^2)*cos(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.74 \[ \int (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx=\frac {2 \, b^{3} \cos \left (d x + c\right )^{6} + 3 \, {\left (2 \, a^{3} + 15 \, a b^{2}\right )} d x \cos \left (d x + c\right )^{3} - 18 \, {\left (a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{4} + 6 \, a^{2} b + 2 \, b^{3} - 18 \, {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} - {\left (9 \, a b^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{3} - 6 \, a b^{2} + 2 \, {\left (4 \, a^{3} + 21 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate((a+b*sin(d*x+c))^3*tan(d*x+c)^4,x, algorithm="fricas")
Output:
1/6*(2*b^3*cos(d*x + c)^6 + 3*(2*a^3 + 15*a*b^2)*d*x*cos(d*x + c)^3 - 18*( a^2*b + b^3)*cos(d*x + c)^4 + 6*a^2*b + 2*b^3 - 18*(2*a^2*b + b^3)*cos(d*x + c)^2 - (9*a*b^2*cos(d*x + c)^4 - 2*a^3 - 6*a*b^2 + 2*(4*a^3 + 21*a*b^2) *cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^3)
\[ \int (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{3} \tan ^{4}{\left (c + d x \right )}\, dx \] Input:
integrate((a+b*sin(d*x+c))**3*tan(d*x+c)**4,x)
Output:
Integral((a + b*sin(c + d*x))**3*tan(c + d*x)**4, x)
Time = 0.12 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.79 \[ \int (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx=\frac {2 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{3} + 3 \, {\left (2 \, \tan \left (d x + c\right )^{3} + 15 \, d x + 15 \, c - \frac {3 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 12 \, \tan \left (d x + c\right )\right )} a b^{2} + 2 \, {\left (\cos \left (d x + c\right )^{3} - \frac {9 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} - 9 \, \cos \left (d x + c\right )\right )} b^{3} - 6 \, a^{2} b {\left (\frac {6 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} + 3 \, \cos \left (d x + c\right )\right )}}{6 \, d} \] Input:
integrate((a+b*sin(d*x+c))^3*tan(d*x+c)^4,x, algorithm="maxima")
Output:
1/6*(2*(tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a^3 + 3*(2*tan(d*x + c)^3 + 15*d*x + 15*c - 3*tan(d*x + c)/(tan(d*x + c)^2 + 1) - 12*tan(d*x + c))*a*b^2 + 2*(cos(d*x + c)^3 - (9*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 - 9*cos(d*x + c))*b^3 - 6*a^2*b*((6*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 + 3*c os(d*x + c)))/d
Timed out. \[ \int (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*sin(d*x+c))^3*tan(d*x+c)^4,x, algorithm="giac")
Output:
Timed out
Time = 19.28 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.41 \[ \int (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx=\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2+15\,b^2\right )}{2\,a^3+15\,a\,b^2}\right )\,\left (2\,a^2+15\,b^2\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {2\,a^3}{3}+5\,a\,b^2\right )-16\,a^2\,b-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^3+15\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {2\,a^3}{3}+5\,a\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (2\,a^3+15\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (12\,a^3+42\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (12\,a^3+42\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (48\,a^2\,b+32\,b^3\right )-\frac {32\,b^3}{3}+32\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-1\right )} \] Input:
int(tan(c + d*x)^4*(a + b*sin(c + d*x))^3,x)
Output:
(a*atan((a*tan(c/2 + (d*x)/2)*(2*a^2 + 15*b^2))/(15*a*b^2 + 2*a^3))*(2*a^2 + 15*b^2))/d - (tan(c/2 + (d*x)/2)^3*(5*a*b^2 + (2*a^3)/3) - 16*a^2*b - t an(c/2 + (d*x)/2)*(15*a*b^2 + 2*a^3) + tan(c/2 + (d*x)/2)^9*(5*a*b^2 + (2* a^3)/3) - tan(c/2 + (d*x)/2)^11*(15*a*b^2 + 2*a^3) + tan(c/2 + (d*x)/2)^5* (42*a*b^2 + 12*a^3) + tan(c/2 + (d*x)/2)^7*(42*a*b^2 + 12*a^3) + tan(c/2 + (d*x)/2)^4*(48*a^2*b + 32*b^3) - (32*b^3)/3 + 32*a^2*b*tan(c/2 + (d*x)/2) ^6)/(d*(3*tan(c/2 + (d*x)/2)^4 - 3*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/ 2)^12 - 1))
Time = 0.16 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.89 \[ \int (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx=\frac {2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} \tan \left (d x +c \right )^{3} a^{3}-6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} \tan \left (d x +c \right ) a^{3}+6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{3} d x +48 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{2} b +45 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{2} c +45 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{2} d x +32 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{3}-2 \cos \left (d x +c \right ) \tan \left (d x +c \right )^{3} a^{3}+6 \cos \left (d x +c \right ) \tan \left (d x +c \right ) a^{3}-6 \cos \left (d x +c \right ) a^{3} d x -48 \cos \left (d x +c \right ) a^{2} b -45 \cos \left (d x +c \right ) a \,b^{2} c -45 \cos \left (d x +c \right ) a \,b^{2} d x -32 \cos \left (d x +c \right ) b^{3}+2 \sin \left (d x +c \right )^{6} b^{3}+9 \sin \left (d x +c \right )^{5} a \,b^{2}+18 \sin \left (d x +c \right )^{4} a^{2} b +12 \sin \left (d x +c \right )^{4} b^{3}-60 \sin \left (d x +c \right )^{3} a \,b^{2}-72 \sin \left (d x +c \right )^{2} a^{2} b -48 \sin \left (d x +c \right )^{2} b^{3}+45 \sin \left (d x +c \right ) a \,b^{2}+48 a^{2} b +32 b^{3}}{6 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int((a+b*sin(d*x+c))^3*tan(d*x+c)^4,x)
Output:
(2*cos(c + d*x)*sin(c + d*x)**2*tan(c + d*x)**3*a**3 - 6*cos(c + d*x)*sin( c + d*x)**2*tan(c + d*x)*a**3 + 6*cos(c + d*x)*sin(c + d*x)**2*a**3*d*x + 48*cos(c + d*x)*sin(c + d*x)**2*a**2*b + 45*cos(c + d*x)*sin(c + d*x)**2*a *b**2*c + 45*cos(c + d*x)*sin(c + d*x)**2*a*b**2*d*x + 32*cos(c + d*x)*sin (c + d*x)**2*b**3 - 2*cos(c + d*x)*tan(c + d*x)**3*a**3 + 6*cos(c + d*x)*t an(c + d*x)*a**3 - 6*cos(c + d*x)*a**3*d*x - 48*cos(c + d*x)*a**2*b - 45*c os(c + d*x)*a*b**2*c - 45*cos(c + d*x)*a*b**2*d*x - 32*cos(c + d*x)*b**3 + 2*sin(c + d*x)**6*b**3 + 9*sin(c + d*x)**5*a*b**2 + 18*sin(c + d*x)**4*a* *2*b + 12*sin(c + d*x)**4*b**3 - 60*sin(c + d*x)**3*a*b**2 - 72*sin(c + d* x)**2*a**2*b - 48*sin(c + d*x)**2*b**3 + 45*sin(c + d*x)*a*b**2 + 48*a**2* b + 32*b**3)/(6*cos(c + d*x)*d*(sin(c + d*x)**2 - 1))