Integrand size = 21, antiderivative size = 275 \[ \int \sin ^3(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {4 a^3 b \text {arctanh}(\sin (c+d x))}{d}-\frac {10 a b^3 \text {arctanh}(\sin (c+d x))}{d}-\frac {a^4 \cos (c+d x)}{d}+\frac {12 a^2 b^2 \cos (c+d x)}{d}-\frac {3 b^4 \cos (c+d x)}{d}+\frac {a^4 \cos ^3(c+d x)}{3 d}-\frac {2 a^2 b^2 \cos ^3(c+d x)}{d}+\frac {b^4 \cos ^3(c+d x)}{3 d}+\frac {6 a^2 b^2 \sec (c+d x)}{d}-\frac {3 b^4 \sec (c+d x)}{d}+\frac {b^4 \sec ^3(c+d x)}{3 d}-\frac {4 a^3 b \sin (c+d x)}{d}+\frac {10 a b^3 \sin (c+d x)}{d}-\frac {4 a^3 b \sin ^3(c+d x)}{3 d}+\frac {10 a b^3 \sin ^3(c+d x)}{3 d}+\frac {2 a b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{d} \] Output:
4*a^3*b*arctanh(sin(d*x+c))/d-10*a*b^3*arctanh(sin(d*x+c))/d-a^4*cos(d*x+c )/d+12*a^2*b^2*cos(d*x+c)/d-3*b^4*cos(d*x+c)/d+1/3*a^4*cos(d*x+c)^3/d-2*a^ 2*b^2*cos(d*x+c)^3/d+1/3*b^4*cos(d*x+c)^3/d+6*a^2*b^2*sec(d*x+c)/d-3*b^4*s ec(d*x+c)/d+1/3*b^4*sec(d*x+c)^3/d-4*a^3*b*sin(d*x+c)/d+10*a*b^3*sin(d*x+c )/d-4/3*a^3*b*sin(d*x+c)^3/d+10/3*a*b^3*sin(d*x+c)^3/d+2*a*b^3*sin(d*x+c)^ 3*tan(d*x+c)^2/d
Leaf count is larger than twice the leaf count of optimal. \(1017\) vs. \(2(275)=550\).
Time = 7.17 (sec) , antiderivative size = 1017, normalized size of antiderivative = 3.70 \[ \int \sin ^3(c+d x) (a+b \tan (c+d x))^4 \, dx =\text {Too large to display} \] Input:
Integrate[Sin[c + d*x]^3*(a + b*Tan[c + d*x])^4,x]
Output:
-1/6*(b^2*(-36*a^2 + 17*b^2)*Cos[c + d*x]^4*(a + b*Tan[c + d*x])^4)/(d*(a* Cos[c + d*x] + b*Sin[c + d*x])^4) - ((3*a^4 - 42*a^2*b^2 + 11*b^4)*Cos[c + d*x]^5*(a + b*Tan[c + d*x])^4)/(4*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + ((a^4 - 6*a^2*b^2 + b^4)*Cos[c + d*x]^4*Cos[3*(c + d*x)]*(a + b*Tan[c + d*x])^4)/(12*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) - (2*(2*a^3*b - 5*a*b^ 3)*Cos[c + d*x]^4*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^4)/(d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (2*(2*a^3*b - 5*a*b^3)* Cos[c + d*x]^4*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x ])^4)/(d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + ((12*a*b^3 + b^4)*Cos[c + d*x]^4*(a + b*Tan[c + d*x])^4)/(12*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) ^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (b^4*Cos[c + d*x]^4*Sin[(c + d*x )/2]*(a + b*Tan[c + d*x])^4)/(6*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3* (a*Cos[c + d*x] + b*Sin[c + d*x])^4) - (b^4*Cos[c + d*x]^4*Sin[(c + d*x)/2 ]*(a + b*Tan[c + d*x])^4)/(6*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(a* Cos[c + d*x] + b*Sin[c + d*x])^4) + ((-12*a*b^3 + b^4)*Cos[c + d*x]^4*(a + b*Tan[c + d*x])^4)/(12*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (Cos[c + d*x]^4*(36*a^2*b^2*Sin[(c + d*x)/2 ] - 17*b^4*Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^4)/(6*d*(Cos[(c + d*x)/2 ] - Sin[(c + d*x)/2])*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (Cos[c + d*x] ^4*(-36*a^2*b^2*Sin[(c + d*x)/2] + 17*b^4*Sin[(c + d*x)/2])*(a + b*Tan[...
Time = 0.55 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 4000, 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.
| \(\displaystyle \int \sin ^3(c+d x) (a+b \tan (c+d x))^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (c+d x)^3 (a+b \tan (c+d x))^4dx\) |
| \(\Big \downarrow \) 4000 |
| \(\displaystyle \int \left (a^4 \sin ^3(c+d x)+4 a^3 b \sin ^3(c+d x) \tan (c+d x)+6 a^2 b^2 \sin ^3(c+d x) \tan ^2(c+d x)+4 a b^3 \sin ^3(c+d x) \tan ^3(c+d x)+b^4 \sin ^3(c+d x) \tan ^4(c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^4 \cos ^3(c+d x)}{3 d}-\frac {a^4 \cos (c+d x)}{d}+\frac {4 a^3 b \text {arctanh}(\sin (c+d x))}{d}-\frac {4 a^3 b \sin ^3(c+d x)}{3 d}-\frac {4 a^3 b \sin (c+d x)}{d}-\frac {2 a^2 b^2 \cos ^3(c+d x)}{d}+\frac {12 a^2 b^2 \cos (c+d x)}{d}+\frac {6 a^2 b^2 \sec (c+d x)}{d}-\frac {10 a b^3 \text {arctanh}(\sin (c+d x))}{d}+\frac {10 a b^3 \sin ^3(c+d x)}{3 d}+\frac {10 a b^3 \sin (c+d x)}{d}+\frac {2 a b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{d}+\frac {b^4 \cos ^3(c+d x)}{3 d}-\frac {3 b^4 \cos (c+d x)}{d}+\frac {b^4 \sec ^3(c+d x)}{3 d}-\frac {3 b^4 \sec (c+d x)}{d}\) |
Input:
Int[Sin[c + d*x]^3*(a + b*Tan[c + d*x])^4,x]
Output:
(4*a^3*b*ArcTanh[Sin[c + d*x]])/d - (10*a*b^3*ArcTanh[Sin[c + d*x]])/d - ( a^4*Cos[c + d*x])/d + (12*a^2*b^2*Cos[c + d*x])/d - (3*b^4*Cos[c + d*x])/d + (a^4*Cos[c + d*x]^3)/(3*d) - (2*a^2*b^2*Cos[c + d*x]^3)/d + (b^4*Cos[c + d*x]^3)/(3*d) + (6*a^2*b^2*Sec[c + d*x])/d - (3*b^4*Sec[c + d*x])/d + (b ^4*Sec[c + d*x]^3)/(3*d) - (4*a^3*b*Sin[c + d*x])/d + (10*a*b^3*Sin[c + d* x])/d - (4*a^3*b*Sin[c + d*x]^3)/(3*d) + (10*a*b^3*Sin[c + d*x]^3)/(3*d) + (2*a*b^3*Sin[c + d*x]^3*Tan[c + d*x]^2)/d
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n _.), x_Symbol] :> Int[Expand[Sin[e + f*x]^m*(a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IGtQ[n, 0]
Time = 8.03 (sec) , antiderivative size = 267, normalized size of antiderivative = 0.97
| method | result | size |
| derivativedivides | \(\frac {-\frac {a^{4} \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}+4 a^{3} b \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+6 b^{2} a^{2} \left (\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )+4 a \,b^{3} \left (\frac {\sin \left (d x +c \right )^{7}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{5}}{2}+\frac {5 \sin \left (d x +c \right )^{3}}{6}+\frac {5 \sin \left (d x +c \right )}{2}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+b^{4} \left (\frac {\sin \left (d x +c \right )^{8}}{3 \cos \left (d x +c \right )^{3}}-\frac {5 \sin \left (d x +c \right )^{8}}{3 \cos \left (d x +c \right )}-\frac {5 \left (\frac {16}{5}+\sin \left (d x +c \right )^{6}+\frac {6 \sin \left (d x +c \right )^{4}}{5}+\frac {8 \sin \left (d x +c \right )^{2}}{5}\right ) \cos \left (d x +c \right )}{3}\right )}{d}\) | \(267\) |
| default | \(\frac {-\frac {a^{4} \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}+4 a^{3} b \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+6 b^{2} a^{2} \left (\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )+4 a \,b^{3} \left (\frac {\sin \left (d x +c \right )^{7}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{5}}{2}+\frac {5 \sin \left (d x +c \right )^{3}}{6}+\frac {5 \sin \left (d x +c \right )}{2}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+b^{4} \left (\frac {\sin \left (d x +c \right )^{8}}{3 \cos \left (d x +c \right )^{3}}-\frac {5 \sin \left (d x +c \right )^{8}}{3 \cos \left (d x +c \right )}-\frac {5 \left (\frac {16}{5}+\sin \left (d x +c \right )^{6}+\frac {6 \sin \left (d x +c \right )^{4}}{5}+\frac {8 \sin \left (d x +c \right )^{2}}{5}\right ) \cos \left (d x +c \right )}{3}\right )}{d}\) | \(267\) |
| risch | \(-\frac {3 \,{\mathrm e}^{i \left (d x +c \right )} a^{4}}{8 d}-\frac {11 \,{\mathrm e}^{i \left (d x +c \right )} b^{4}}{8 d}+\frac {{\mathrm e}^{3 i \left (d x +c \right )} a^{4}}{24 d}+\frac {{\mathrm e}^{3 i \left (d x +c \right )} b^{4}}{24 d}+\frac {{\mathrm e}^{-3 i \left (d x +c \right )} a^{4}}{24 d}+\frac {{\mathrm e}^{-3 i \left (d x +c \right )} b^{4}}{24 d}-\frac {3 \,{\mathrm e}^{-i \left (d x +c \right )} a^{4}}{8 d}-\frac {11 \,{\mathrm e}^{-i \left (d x +c \right )} b^{4}}{8 d}+\frac {5 i {\mathrm e}^{i \left (d x +c \right )} a^{3} b}{2 d}-\frac {9 i {\mathrm e}^{i \left (d x +c \right )} a \,b^{3}}{2 d}-\frac {5 i {\mathrm e}^{-i \left (d x +c \right )} a^{3} b}{2 d}+\frac {9 i {\mathrm e}^{-i \left (d x +c \right )} a \,b^{3}}{2 d}+\frac {i {\mathrm e}^{-3 i \left (d x +c \right )} a^{3} b}{6 d}-\frac {i {\mathrm e}^{-3 i \left (d x +c \right )} a \,b^{3}}{6 d}-\frac {i {\mathrm e}^{3 i \left (d x +c \right )} a^{3} b}{6 d}+\frac {i {\mathrm e}^{3 i \left (d x +c \right )} a \,b^{3}}{6 d}+\frac {21 \,{\mathrm e}^{i \left (d x +c \right )} b^{2} a^{2}}{4 d}-\frac {4 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {10 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {4 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {10 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {21 \,{\mathrm e}^{-i \left (d x +c \right )} b^{2} a^{2}}{4 d}-\frac {{\mathrm e}^{-3 i \left (d x +c \right )} b^{2} a^{2}}{4 d}-\frac {{\mathrm e}^{3 i \left (d x +c \right )} b^{2} a^{2}}{4 d}-\frac {2 b^{2} {\mathrm e}^{i \left (d x +c \right )} \left (-18 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+9 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+6 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}-36 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+14 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-18 a^{2}+9 b^{2}-6 i a b \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}\) | \(616\) |
Input:
int(sin(d*x+c)^3*(a+b*tan(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/3*a^4*(2+sin(d*x+c)^2)*cos(d*x+c)+4*a^3*b*(-1/3*sin(d*x+c)^3-sin(d *x+c)+ln(sec(d*x+c)+tan(d*x+c)))+6*b^2*a^2*(sin(d*x+c)^6/cos(d*x+c)+(8/3+s in(d*x+c)^4+4/3*sin(d*x+c)^2)*cos(d*x+c))+4*a*b^3*(1/2*sin(d*x+c)^7/cos(d* x+c)^2+1/2*sin(d*x+c)^5+5/6*sin(d*x+c)^3+5/2*sin(d*x+c)-5/2*ln(sec(d*x+c)+ tan(d*x+c)))+b^4*(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*(16/5+sin(d*x+c)^6+6/5*sin(d*x+c)^4+8/5*sin(d*x+c)^2)*cos(d*x+c)))
Time = 0.10 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.81 \[ \int \sin ^3(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{6} - 3 \, {\left (a^{4} - 12 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (2 \, a^{3} b - 5 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, a^{3} b - 5 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + b^{4} + 9 \, {\left (2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{5} + 3 \, a b^{3} \cos \left (d x + c\right ) - 2 \, {\left (4 \, a^{3} b - 7 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate(sin(d*x+c)^3*(a+b*tan(d*x+c))^4,x, algorithm="fricas")
Output:
1/3*((a^4 - 6*a^2*b^2 + b^4)*cos(d*x + c)^6 - 3*(a^4 - 12*a^2*b^2 + 3*b^4) *cos(d*x + c)^4 + 3*(2*a^3*b - 5*a*b^3)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(2*a^3*b - 5*a*b^3)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + b^4 + 9 *(2*a^2*b^2 - b^4)*cos(d*x + c)^2 + 2*(2*(a^3*b - a*b^3)*cos(d*x + c)^5 + 3*a*b^3*cos(d*x + c) - 2*(4*a^3*b - 7*a*b^3)*cos(d*x + c)^3)*sin(d*x + c)) /(d*cos(d*x + c)^3)
\[ \int \sin ^3(c+d x) (a+b \tan (c+d x))^4 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{4} \sin ^{3}{\left (c + d x \right )}\, dx \] Input:
integrate(sin(d*x+c)**3*(a+b*tan(d*x+c))**4,x)
Output:
Integral((a + b*tan(c + d*x))**4*sin(c + d*x)**3, x)
Time = 0.04 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.79 \[ \int \sin ^3(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{4} - 2 \, {\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a^{3} b - 6 \, {\left (\cos \left (d x + c\right )^{3} - \frac {3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} a^{2} b^{2} + {\left (4 \, \sin \left (d x + c\right )^{3} - \frac {6 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 24 \, \sin \left (d x + c\right )\right )} a b^{3} + {\left (\cos \left (d x + c\right )^{3} - \frac {9 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} - 9 \, \cos \left (d x + c\right )\right )} b^{4}}{3 \, d} \] Input:
integrate(sin(d*x+c)^3*(a+b*tan(d*x+c))^4,x, algorithm="maxima")
Output:
1/3*((cos(d*x + c)^3 - 3*cos(d*x + c))*a^4 - 2*(2*sin(d*x + c)^3 - 3*log(s in(d*x + c) + 1) + 3*log(sin(d*x + c) - 1) + 6*sin(d*x + c))*a^3*b - 6*(co s(d*x + c)^3 - 3/cos(d*x + c) - 6*cos(d*x + c))*a^2*b^2 + (4*sin(d*x + c)^ 3 - 6*sin(d*x + c)/(sin(d*x + c)^2 - 1) - 15*log(sin(d*x + c) + 1) + 15*lo g(sin(d*x + c) - 1) + 24*sin(d*x + c))*a*b^3 + (cos(d*x + c)^3 - (9*cos(d* x + c)^2 - 1)/cos(d*x + c)^3 - 9*cos(d*x + c))*b^4)/d
Timed out. \[ \int \sin ^3(c+d x) (a+b \tan (c+d x))^4 \, dx=\text {Timed out} \] Input:
integrate(sin(d*x+c)^3*(a+b*tan(d*x+c))^4,x, algorithm="giac")
Output:
Timed out
Time = 4.53 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.16 \[ \int \sin ^3(c+d x) (a+b \tan (c+d x))^4 \, dx=-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (8\,a^4-96\,a^2\,b^2+32\,b^4\right )+4\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (20\,a\,b^3-8\,a^3\,b\right )-\frac {4\,a^4}{3}-\frac {32\,b^4}{3}+32\,a^2\,b^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {32\,a^4}{3}-64\,a^2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {20\,a\,b^3}{3}-\frac {8\,a^3\,b}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (20\,a\,b^3-8\,a^3\,b\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {20\,a\,b^3}{3}-\frac {8\,a^3\,b}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (56\,a\,b^3-48\,a^3\,b\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (56\,a\,b^3-48\,a^3\,b\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-1\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (20\,a\,b^3-8\,a^3\,b\right )}{d} \] Input:
int(sin(c + d*x)^3*(a + b*tan(c + d*x))^4,x)
Output:
- (tan(c/2 + (d*x)/2)^4*(8*a^4 + 32*b^4 - 96*a^2*b^2) + 4*a^4*tan(c/2 + (d *x)/2)^8 + tan(c/2 + (d*x)/2)*(20*a*b^3 - 8*a^3*b) - (4*a^4)/3 - (32*b^4)/ 3 + 32*a^2*b^2 - tan(c/2 + (d*x)/2)^6*((32*a^4)/3 - 64*a^2*b^2) + tan(c/2 + (d*x)/2)^3*((20*a*b^3)/3 - (8*a^3*b)/3) - tan(c/2 + (d*x)/2)^11*(20*a*b^ 3 - 8*a^3*b) - tan(c/2 + (d*x)/2)^9*((20*a*b^3)/3 - (8*a^3*b)/3) - tan(c/2 + (d*x)/2)^5*(56*a*b^3 - 48*a^3*b) + tan(c/2 + (d*x)/2)^7*(56*a*b^3 - 48* a^3*b))/(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)) - (atanh(tan(c/2 + (d*x)/2))*(20*a*b^3 - 8*a^3*b))/d
Time = 0.22 (sec) , antiderivative size = 591, normalized size of antiderivative = 2.15 \[ \int \sin ^3(c+d x) (a+b \tan (c+d x))^4 \, dx =\text {Too large to display} \] Input:
int(sin(d*x+c)^3*(a+b*tan(d*x+c))^4,x)
Output:
( - 12*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3*b + 30* cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**3 + 12*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**3*b - 30*cos(c + d*x)*log(tan((c + d*x) /2) - 1)*a*b**3 + 12*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)** 2*a**3*b - 30*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b** 3 - 12*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**3*b + 30*cos(c + d*x)*log (tan((c + d*x)/2) + 1)*a*b**3 - 4*cos(c + d*x)*sin(c + d*x)**5*a**3*b + 4* cos(c + d*x)*sin(c + d*x)**5*a*b**3 - 8*cos(c + d*x)*sin(c + d*x)**3*a**3* b + 20*cos(c + d*x)*sin(c + d*x)**3*a*b**3 + 2*cos(c + d*x)*sin(c + d*x)** 2*a**4 - 48*cos(c + d*x)*sin(c + d*x)**2*a**2*b**2 + 16*cos(c + d*x)*sin(c + d*x)**2*b**4 + 12*cos(c + d*x)*sin(c + d*x)*a**3*b - 30*cos(c + d*x)*si n(c + d*x)*a*b**3 - 2*cos(c + d*x)*a**4 + 48*cos(c + d*x)*a**2*b**2 - 16*c os(c + d*x)*b**4 + sin(c + d*x)**6*a**4 - 6*sin(c + d*x)**6*a**2*b**2 + si n(c + d*x)**6*b**4 - 18*sin(c + d*x)**4*a**2*b**2 + 6*sin(c + d*x)**4*b**4 - 3*sin(c + d*x)**2*a**4 + 72*sin(c + d*x)**2*a**2*b**2 - 24*sin(c + d*x) **2*b**4 + 2*a**4 - 48*a**2*b**2 + 16*b**4)/(3*cos(c + d*x)*d*(sin(c + d*x )**2 - 1))