Integrand size = 21, antiderivative size = 161 \[ \int \csc ^3(c+d x) (a+b \tan (c+d x))^4 \, dx=-\frac {a^4 \text {arctanh}(\cos (c+d x))}{2 d}-\frac {6 a^2 b^2 \text {arctanh}(\cos (c+d x))}{d}+\frac {4 a^3 b \text {arctanh}(\sin (c+d x))}{d}+\frac {2 a b^3 \text {arctanh}(\sin (c+d x))}{d}-\frac {4 a^3 b \csc (c+d x)}{d}-\frac {a^4 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {6 a^2 b^2 \sec (c+d x)}{d}+\frac {b^4 \sec ^3(c+d x)}{3 d}+\frac {2 a b^3 \sec (c+d x) \tan (c+d x)}{d} \] Output:
-1/2*a^4*arctanh(cos(d*x+c))/d-6*a^2*b^2*arctanh(cos(d*x+c))/d+4*a^3*b*arc tanh(sin(d*x+c))/d+2*a*b^3*arctanh(sin(d*x+c))/d-4*a^3*b*csc(d*x+c)/d-1/2* a^4*cot(d*x+c)*csc(d*x+c)/d+6*a^2*b^2*sec(d*x+c)/d+1/3*b^4*sec(d*x+c)^3/d+ 2*a*b^3*sec(d*x+c)*tan(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(1128\) vs. \(2(161)=322\).
Time = 7.29 (sec) , antiderivative size = 1128, normalized size of antiderivative = 7.01 \[ \int \csc ^3(c+d x) (a+b \tan (c+d x))^4 \, dx =\text {Too large to display} \] Input:
Integrate[Csc[c + d*x]^3*(a + b*Tan[c + d*x])^4,x]
Output:
(b^2*(36*a^2 + b^2)*Cos[c + d*x]^4*(a + b*Tan[c + d*x])^4)/(6*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) - (2*a^3*b*Cos[c + d*x]^4*Cot[(c + d*x)/2]*(a + b*Tan[c + d*x])^4)/(d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) - (a^4*Cos[c + d*x]^4*Csc[(c + d*x)/2]^2*(a + b*Tan[c + d*x])^4)/(8*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + ((-a^4 - 12*a^2*b^2)*Cos[c + d*x]^4*Log[Cos[(c + d*x) /2]]*(a + b*Tan[c + d*x])^4)/(2*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) - ( 2*(2*a^3*b + 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) + ((a^4 + 12*a^2*b^2)*Cos[c + d*x]^4*Log[Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^4) /(2*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4) + (2*(2*a^3*b + 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) + (a^4*Cos[c + d*x]^4*Sec[(c + d*x)/ 2]^2*(a + b*Tan[c + d*x])^4)/(8*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*Co s[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] + ...
Time = 0.40 (sec) , antiderivative size = 161, 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.
| \(\displaystyle \int \csc ^3(c+d x) (a+b \tan (c+d x))^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^4}{\sin (c+d x)^3}dx\) |
| \(\Big \downarrow \) 4000 |
| \(\displaystyle \int \left (a^4 \csc ^3(c+d x)+4 a^3 b \csc ^2(c+d x) \sec (c+d x)+6 a^2 b^2 \csc (c+d x) \sec ^2(c+d x)+4 a b^3 \sec ^3(c+d x)+b^4 \tan (c+d x) \sec ^3(c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^4 \text {arctanh}(\cos (c+d x))}{2 d}-\frac {a^4 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {4 a^3 b \text {arctanh}(\sin (c+d x))}{d}-\frac {4 a^3 b \csc (c+d x)}{d}-\frac {6 a^2 b^2 \text {arctanh}(\cos (c+d x))}{d}+\frac {6 a^2 b^2 \sec (c+d x)}{d}+\frac {2 a b^3 \text {arctanh}(\sin (c+d x))}{d}+\frac {2 a b^3 \tan (c+d x) \sec (c+d x)}{d}+\frac {b^4 \sec ^3(c+d x)}{3 d}\) |
Input:
Int[Csc[c + d*x]^3*(a + b*Tan[c + d*x])^4,x]
Output:
-1/2*(a^4*ArcTanh[Cos[c + d*x]])/d - (6*a^2*b^2*ArcTanh[Cos[c + d*x]])/d + (4*a^3*b*ArcTanh[Sin[c + d*x]])/d + (2*a*b^3*ArcTanh[Sin[c + d*x]])/d - ( 4*a^3*b*Csc[c + d*x])/d - (a^4*Cot[c + d*x]*Csc[c + d*x])/(2*d) + (6*a^2*b ^2*Sec[c + d*x])/d + (b^4*Sec[c + d*x]^3)/(3*d) + (2*a*b^3*Sec[c + d*x]*Ta n[c + d*x])/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 = 9.93 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.98
| method | result | size |
| derivativedivides | \(\frac {\frac {b^{4}}{3 \cos \left (d x +c \right )^{3}}+4 a \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 b^{2} a^{2} \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+4 a^{3} b \left (-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{4} \left (-\frac {\cot \left (d x +c \right ) \csc \left (d x +c \right )}{2}+\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\) | \(157\) |
| default | \(\frac {\frac {b^{4}}{3 \cos \left (d x +c \right )^{3}}+4 a \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 b^{2} a^{2} \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+4 a^{3} b \left (-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{4} \left (-\frac {\cot \left (d x +c \right ) \csc \left (d x +c \right )}{2}+\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\) | \(157\) |
| risch | \(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (12 i a^{4} {\mathrm e}^{6 i \left (d x +c \right )}-16 i b^{4} {\mathrm e}^{4 i \left (d x +c \right )}+24 a^{3} b \,{\mathrm e}^{8 i \left (d x +c \right )}+12 a \,b^{3} {\mathrm e}^{8 i \left (d x +c \right )}+36 i a^{2} b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-72 i a^{2} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+48 a^{3} b \,{\mathrm e}^{6 i \left (d x +c \right )}-24 a \,b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+3 i a^{4}+12 i a^{4} {\mathrm e}^{2 i \left (d x +c \right )}+3 i a^{4} {\mathrm e}^{8 i \left (d x +c \right )}+36 i a^{2} b^{2}+8 i b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-48 a^{3} b \,{\mathrm e}^{2 i \left (d x +c \right )}+24 a \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+8 i b^{4} {\mathrm e}^{6 i \left (d x +c \right )}+18 i a^{4} {\mathrm e}^{4 i \left (d x +c \right )}-24 a^{3} b -12 a \,b^{3}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d}+\frac {6 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{2}}{d}-\frac {4 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {2 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {4 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {2 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d}-\frac {6 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{2}}{d}\) | \(510\) |
Input:
int(csc(d*x+c)^3*(a+b*tan(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
1/d*(1/3*b^4/cos(d*x+c)^3+4*a*b^3*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d* x+c)+tan(d*x+c)))+6*b^2*a^2*(1/cos(d*x+c)+ln(csc(d*x+c)-cot(d*x+c)))+4*a^3 *b*(-1/sin(d*x+c)+ln(sec(d*x+c)+tan(d*x+c)))+a^4*(-1/2*cot(d*x+c)*csc(d*x+ c)+1/2*ln(csc(d*x+c)-cot(d*x+c))))
Leaf count of result is larger than twice the leaf count of optimal. 346 vs. \(2 (155) = 310\).
Time = 0.16 (sec) , antiderivative size = 346, normalized size of antiderivative = 2.15 \[ \int \csc ^3(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {6 \, {\left (a^{4} + 12 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{4} - 4 \, b^{4} - 4 \, {\left (18 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left ({\left (a^{4} + 12 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{5} - {\left (a^{4} + 12 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left ({\left (a^{4} + 12 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{5} - {\left (a^{4} + 12 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 12 \, {\left ({\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{5} - {\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \, {\left ({\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{5} - {\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 24 \, {\left (a b^{3} \cos \left (d x + c\right ) - {\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{5} - d \cos \left (d x + c\right )^{3}\right )}} \] Input:
integrate(csc(d*x+c)^3*(a+b*tan(d*x+c))^4,x, algorithm="fricas")
Output:
1/12*(6*(a^4 + 12*a^2*b^2)*cos(d*x + c)^4 - 4*b^4 - 4*(18*a^2*b^2 - b^4)*c os(d*x + c)^2 - 3*((a^4 + 12*a^2*b^2)*cos(d*x + c)^5 - (a^4 + 12*a^2*b^2)* cos(d*x + c)^3)*log(1/2*cos(d*x + c) + 1/2) + 3*((a^4 + 12*a^2*b^2)*cos(d* x + c)^5 - (a^4 + 12*a^2*b^2)*cos(d*x + c)^3)*log(-1/2*cos(d*x + c) + 1/2) + 12*((2*a^3*b + a*b^3)*cos(d*x + c)^5 - (2*a^3*b + a*b^3)*cos(d*x + c)^3 )*log(sin(d*x + c) + 1) - 12*((2*a^3*b + a*b^3)*cos(d*x + c)^5 - (2*a^3*b + a*b^3)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 24*(a*b^3*cos(d*x + c) - (2*a^3*b + a*b^3)*cos(d*x + c)^3)*sin(d*x + c))/(d*cos(d*x + c)^5 - d*cos (d*x + c)^3)
\[ \int \csc ^3(c+d x) (a+b \tan (c+d x))^4 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{4} \csc ^{3}{\left (c + d x \right )}\, dx \] Input:
integrate(csc(d*x+c)**3*(a+b*tan(d*x+c))**4,x)
Output:
Integral((a + b*tan(c + d*x))**4*csc(c + d*x)**3, x)
Time = 0.04 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.17 \[ \int \csc ^3(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {3 \, a^{4} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 12 \, a b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, a^{2} b^{2} {\left (\frac {2}{\cos \left (d x + c\right )} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 24 \, a^{3} b {\left (\frac {2}{\sin \left (d x + c\right )} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac {4 \, b^{4}}{\cos \left (d x + c\right )^{3}}}{12 \, d} \] Input:
integrate(csc(d*x+c)^3*(a+b*tan(d*x+c))^4,x, algorithm="maxima")
Output:
1/12*(3*a^4*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)) - 12*a*b^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 36*a^2*b^2*(2/cos(d*x + c ) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)) - 24*a^3*b*(2/sin(d*x + c) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 4*b^4/cos(d*x + c)^ 3)/d
Time = 0.38 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.86 \[ \int \csc ^3(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 48 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 48 \, {\left (2 \, a^{3} b + a b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 48 \, {\left (2 \, a^{3} b + a b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 12 \, {\left (a^{4} + 12 \, a^{2} b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {3 \, {\left (6 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 16 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{4}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} + \frac {16 \, {\left (6 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 18 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 36 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 18 \, a^{2} b^{2} - b^{4}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{24 \, d} \] Input:
integrate(csc(d*x+c)^3*(a+b*tan(d*x+c))^4,x, algorithm="giac")
Output:
1/24*(3*a^4*tan(1/2*d*x + 1/2*c)^2 - 48*a^3*b*tan(1/2*d*x + 1/2*c) + 48*(2 *a^3*b + a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 48*(2*a^3*b + a*b^3)* log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 12*(a^4 + 12*a^2*b^2)*log(abs(tan(1/2 *d*x + 1/2*c))) - 3*(6*a^4*tan(1/2*d*x + 1/2*c)^2 + 72*a^2*b^2*tan(1/2*d*x + 1/2*c)^2 + 16*a^3*b*tan(1/2*d*x + 1/2*c) + a^4)/tan(1/2*d*x + 1/2*c)^2 + 16*(6*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 18*a^2*b^2*tan(1/2*d*x + 1/2*c)^4 - 3*b^4*tan(1/2*d*x + 1/2*c)^4 + 36*a^2*b^2*tan(1/2*d*x + 1/2*c)^2 - 6*a*b^ 3*tan(1/2*d*x + 1/2*c) - 18*a^2*b^2 - b^4)/(tan(1/2*d*x + 1/2*c)^2 - 1)^3) /d
Time = 1.20 (sec) , antiderivative size = 670, normalized size of antiderivative = 4.16 \[ \int \csc ^3(c+d x) (a+b \tan (c+d x))^4 \, dx =\text {Too large to display} \] Input:
int((a + b*tan(c + d*x))^4/sin(c + d*x)^3,x)
Output:
(a^4*tan(c/2 + (d*x)/2)^2)/(8*d) - (a^4/2 - tan(c/2 + (d*x)/2)^2*((3*a^4)/ 2 + (8*b^4)/3 + 48*a^2*b^2) - tan(c/2 + (d*x)/2)^6*(a^4/2 + 8*b^4 + 48*a^2 *b^2) + tan(c/2 + (d*x)/2)^4*((3*a^4)/2 + 96*a^2*b^2) + tan(c/2 + (d*x)/2) ^7*(16*a*b^3 - 8*a^3*b) - tan(c/2 + (d*x)/2)^3*(16*a*b^3 + 24*a^3*b) + 8*a ^3*b*tan(c/2 + (d*x)/2) + 24*a^3*b*tan(c/2 + (d*x)/2)^5)/(d*(4*tan(c/2 + ( d*x)/2)^2 - 12*tan(c/2 + (d*x)/2)^4 + 12*tan(c/2 + (d*x)/2)^6 - 4*tan(c/2 + (d*x)/2)^8)) + (log(tan(c/2 + (d*x)/2))*(a^4/2 + 6*a^2*b^2))/d - (2*a^3* b*tan(c/2 + (d*x)/2))/d - (a*b*atan((a*b*(2*a^2 + b^2)*(tan(c/2 + (d*x)/2) *(a^4 + 12*a^2*b^2) - 4*a*b^3 - 8*a^3*b + 12*a*b*tan(c/2 + (d*x)/2)*(2*a^2 + b^2))*2i - a*b*(2*a^2 + b^2)*(4*a*b^3 - tan(c/2 + (d*x)/2)*(a^4 + 12*a^ 2*b^2) + 8*a^3*b + 12*a*b*tan(c/2 + (d*x)/2)*(2*a^2 + b^2))*2i)/(2*tan(c/2 + (d*x)/2)*(16*a^2*b^6 + 64*a^4*b^4 + 64*a^6*b^2) + 8*a^7*b + 48*a^3*b^5 + 100*a^5*b^3 - 2*a*b*(2*a^2 + b^2)*(tan(c/2 + (d*x)/2)*(a^4 + 12*a^2*b^2) - 4*a*b^3 - 8*a^3*b + 12*a*b*tan(c/2 + (d*x)/2)*(2*a^2 + b^2)) - 2*a*b*(2 *a^2 + b^2)*(4*a*b^3 - tan(c/2 + (d*x)/2)*(a^4 + 12*a^2*b^2) + 8*a^3*b + 1 2*a*b*tan(c/2 + (d*x)/2)*(2*a^2 + b^2))))*(2*a^2 + b^2)*4i)/d
Time = 0.21 (sec) , antiderivative size = 666, normalized size of antiderivative = 4.14 \[ \int \csc ^3(c+d x) (a+b \tan (c+d x))^4 \, dx =\text {Too large to display} \] Input:
int(csc(d*x+c)^3*(a+b*tan(d*x+c))^4,x)
Output:
( - 96*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**3*b - 48* cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a*b**3 + 96*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3*b + 48*cos(c + d*x)*l og(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**3 + 96*cos(c + d*x)*log(tan( (c + d*x)/2) + 1)*sin(c + d*x)**4*a**3*b + 48*cos(c + d*x)*log(tan((c + d* x)/2) + 1)*sin(c + d*x)**4*a*b**3 - 96*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**3*b - 48*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))*sin(c + d*x)** 4*a**4 + 144*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**4*a**2*b**2 - 12*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**4 - 144*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**2*b**2 - 9*cos(c + d*x)*sin (c + d*x)**4*a**4 - 144*cos(c + d*x)*sin(c + d*x)**4*a**2*b**2 - 8*cos(c + d*x)*sin(c + d*x)**4*b**4 - 96*cos(c + d*x)*sin(c + d*x)**3*a**3*b - 48*c os(c + d*x)*sin(c + d*x)**3*a*b**3 + 9*cos(c + d*x)*sin(c + d*x)**2*a**4 + 144*cos(c + d*x)*sin(c + d*x)**2*a**2*b**2 + 8*cos(c + d*x)*sin(c + d*x)* *2*b**4 + 96*cos(c + d*x)*sin(c + d*x)*a**3*b + 12*sin(c + d*x)**4*a**4 + 144*sin(c + d*x)**4*a**2*b**2 - 24*sin(c + d*x)**2*a**4 - 144*sin(c + d*x) **2*a**2*b**2 - 8*sin(c + d*x)**2*b**4 + 12*a**4)/(24*cos(c + d*x)*sin(c + d*x)**2*d*(sin(c + d*x)**2 - 1))