Integrand size = 21, antiderivative size = 229 \[ \int \csc ^5(c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {3 a^3 \text {arctanh}(\cos (c+d x))}{8 d}-\frac {9 a b^2 \text {arctanh}(\cos (c+d x))}{2 d}+\frac {3 a^2 b \text {arctanh}(\sin (c+d x))}{d}+\frac {3 b^3 \text {arctanh}(\sin (c+d x))}{2 d}-\frac {3 a^2 b \csc (c+d x)}{d}-\frac {3 b^3 \csc (c+d x)}{2 d}-\frac {3 a^3 \cot (c+d x) \csc (c+d x)}{8 d}-\frac {a^2 b \csc ^3(c+d x)}{d}-\frac {a^3 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac {9 a b^2 \sec (c+d x)}{2 d}-\frac {3 a b^2 \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac {b^3 \csc (c+d x) \sec ^2(c+d x)}{2 d} \] Output:
-3/8*a^3*arctanh(cos(d*x+c))/d-9/2*a*b^2*arctanh(cos(d*x+c))/d+3*a^2*b*arc tanh(sin(d*x+c))/d+3/2*b^3*arctanh(sin(d*x+c))/d-3*a^2*b*csc(d*x+c)/d-3/2* b^3*csc(d*x+c)/d-3/8*a^3*cot(d*x+c)*csc(d*x+c)/d-a^2*b*csc(d*x+c)^3/d-1/4* a^3*cot(d*x+c)*csc(d*x+c)^3/d+9/2*a*b^2*sec(d*x+c)/d-3/2*a*b^2*csc(d*x+c)^ 2*sec(d*x+c)/d+1/2*b^3*csc(d*x+c)*sec(d*x+c)^2/d
Leaf count is larger than twice the leaf count of optimal. \(1229\) vs. \(2(229)=458\).
Time = 7.01 (sec) , antiderivative size = 1229, normalized size of antiderivative = 5.37 \[ \int \csc ^5(c+d x) (a+b \tan (c+d x))^3 \, dx =\text {Too large to display} \] Input:
Integrate[Csc[c + d*x]^5*(a + b*Tan[c + d*x])^3,x]
Output:
(3*a*b^2*Cos[c + d*x]^3*(a + b*Tan[c + d*x])^3)/(d*(a*Cos[c + d*x] + b*Sin [c + d*x])^3) + ((-7*a^2*b*Cos[(c + d*x)/2] - 2*b^3*Cos[(c + d*x)/2])*Cos[ c + d*x]^3*Csc[(c + d*x)/2]*(a + b*Tan[c + d*x])^3)/(4*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) - (3*(a^3 + 4*a*b^2)*Cos[c + d*x]^3*Csc[(c + d*x)/2]^2 *(a + b*Tan[c + d*x])^3)/(32*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) - (a^2 *b*Cos[c + d*x]^3*Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2*(a + b*Tan[c + d*x]) ^3)/(8*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) - (a^3*Cos[c + d*x]^3*Csc[(c + d*x)/2]^4*(a + b*Tan[c + d*x])^3)/(64*d*(a*Cos[c + d*x] + b*Sin[c + d*x ])^3) - (3*(a^3 + 12*a*b^2)*Cos[c + d*x]^3*Log[Cos[(c + d*x)/2]]*(a + b*Ta n[c + d*x])^3)/(8*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) - (3*(2*a^2*b + b ^3)*Cos[c + d*x]^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^3)/(2*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + (3*(a^3 + 12*a*b^2)* Cos[c + d*x]^3*Log[Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^3)/(8*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + (3*(2*a^2*b + b^3)*Cos[c + d*x]^3*Log[Cos[( c + d*x)/2] + Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^3)/(2*d*(a*Cos[c + d* x] + b*Sin[c + d*x])^3) + (3*(a^3 + 4*a*b^2)*Cos[c + d*x]^3*Sec[(c + d*x)/ 2]^2*(a + b*Tan[c + d*x])^3)/(32*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + (a^3*Cos[c + d*x]^3*Sec[(c + d*x)/2]^4*(a + b*Tan[c + d*x])^3)/(64*d*(a*Co s[c + d*x] + b*Sin[c + d*x])^3) + (b^3*Cos[c + d*x]^3*(a + b*Tan[c + d*x]) ^3)/(4*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2*(a*Cos[c + d*x] + b*Si...
Time = 0.47 (sec) , antiderivative size = 229, 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.
| \(\displaystyle \int \csc ^5(c+d x) (a+b \tan (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^3}{\sin (c+d x)^5}dx\) |
| \(\Big \downarrow \) 4000 |
| \(\displaystyle \int \left (a^3 \csc ^5(c+d x)+3 a^2 b \csc ^4(c+d x) \sec (c+d x)+3 a b^2 \csc ^3(c+d x) \sec ^2(c+d x)+b^3 \csc ^2(c+d x) \sec ^3(c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 a^3 \text {arctanh}(\cos (c+d x))}{8 d}-\frac {a^3 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {3 a^3 \cot (c+d x) \csc (c+d x)}{8 d}+\frac {3 a^2 b \text {arctanh}(\sin (c+d x))}{d}-\frac {a^2 b \csc ^3(c+d x)}{d}-\frac {3 a^2 b \csc (c+d x)}{d}-\frac {9 a b^2 \text {arctanh}(\cos (c+d x))}{2 d}+\frac {9 a b^2 \sec (c+d x)}{2 d}-\frac {3 a b^2 \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac {3 b^3 \text {arctanh}(\sin (c+d x))}{2 d}-\frac {3 b^3 \csc (c+d x)}{2 d}+\frac {b^3 \csc (c+d x) \sec ^2(c+d x)}{2 d}\) |
Input:
Int[Csc[c + d*x]^5*(a + b*Tan[c + d*x])^3,x]
Output:
(-3*a^3*ArcTanh[Cos[c + d*x]])/(8*d) - (9*a*b^2*ArcTanh[Cos[c + d*x]])/(2* d) + (3*a^2*b*ArcTanh[Sin[c + d*x]])/d + (3*b^3*ArcTanh[Sin[c + d*x]])/(2* d) - (3*a^2*b*Csc[c + d*x])/d - (3*b^3*Csc[c + d*x])/(2*d) - (3*a^3*Cot[c + d*x]*Csc[c + d*x])/(8*d) - (a^2*b*Csc[c + d*x]^3)/d - (a^3*Cot[c + d*x]* Csc[c + d*x]^3)/(4*d) + (9*a*b^2*Sec[c + d*x])/(2*d) - (3*a*b^2*Csc[c + d* x]^2*Sec[c + d*x])/(2*d) + (b^3*Csc[c + d*x]*Sec[c + d*x]^2)/(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 = 16.48 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.86
| method | result | size |
| derivativedivides | \(\frac {b^{3} \left (\frac {1}{2 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a \,b^{2} \left (-\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {3}{2 \cos \left (d x +c \right )}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+3 a^{2} b \left (-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{3} \left (\left (-\frac {\csc \left (d x +c \right )^{3}}{4}-\frac {3 \csc \left (d x +c \right )}{8}\right ) \cot \left (d x +c \right )+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )}{d}\) | \(198\) |
| default | \(\frac {b^{3} \left (\frac {1}{2 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a \,b^{2} \left (-\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {3}{2 \cos \left (d x +c \right )}+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+3 a^{2} b \left (-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{3} \left (\left (-\frac {\csc \left (d x +c \right )^{3}}{4}-\frac {3 \csc \left (d x +c \right )}{8}\right ) \cot \left (d x +c \right )+\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )}{d}\) | \(198\) |
| risch | \(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (-5 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-60 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+24 a^{2} b \,{\mathrm e}^{10 i \left (d x +c \right )}+12 b^{3} {\mathrm e}^{10 i \left (d x +c \right )}-30 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+3 i a^{3}-56 a^{2} b \,{\mathrm e}^{8 i \left (d x +c \right )}-28 b^{3} {\mathrm e}^{8 i \left (d x +c \right )}-5 i a^{3} {\mathrm e}^{8 i \left (d x +c \right )}+3 i a^{3} {\mathrm e}^{10 i \left (d x +c \right )}-80 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+24 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+36 i a \,b^{2}+36 i a \,b^{2} {\mathrm e}^{10 i \left (d x +c \right )}+80 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}-24 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-30 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-60 i a \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+56 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+28 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+24 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+24 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-24 a^{2} b -12 b^{3}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2}}{d}+\frac {3 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d}+\frac {9 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{2}}{2 d}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d}-\frac {9 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{2}}{2 d}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{d}-\frac {3 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\) | \(580\) |
Input:
int(csc(d*x+c)^5*(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(b^3*(1/2/sin(d*x+c)/cos(d*x+c)^2-3/2/sin(d*x+c)+3/2*ln(sec(d*x+c)+tan (d*x+c)))+3*a*b^2*(-1/2/sin(d*x+c)^2/cos(d*x+c)+3/2/cos(d*x+c)+3/2*ln(csc( d*x+c)-cot(d*x+c)))+3*a^2*b*(-1/3/sin(d*x+c)^3-1/sin(d*x+c)+ln(sec(d*x+c)+ tan(d*x+c)))+a^3*((-1/4*csc(d*x+c)^3-3/8*csc(d*x+c))*cot(d*x+c)+3/8*ln(csc (d*x+c)-cot(d*x+c))))
Leaf count of result is larger than twice the leaf count of optimal. 427 vs. \(2 (211) = 422\).
Time = 0.18 (sec) , antiderivative size = 427, normalized size of antiderivative = 1.86 \[ \int \csc ^5(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {6 \, {\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} + 48 \, a b^{2} \cos \left (d x + c\right ) - 10 \, {\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left ({\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{6} - 2 \, {\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + {\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left ({\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{6} - 2 \, {\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + {\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 12 \, {\left ({\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{6} - 2 \, {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{4} + {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \, {\left ({\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{6} - 2 \, {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{4} + {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 8 \, {\left (3 \, {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{4} + b^{3} - 4 \, {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (d \cos \left (d x + c\right )^{6} - 2 \, d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate(csc(d*x+c)^5*(a+b*tan(d*x+c))^3,x, algorithm="fricas")
Output:
1/16*(6*(a^3 + 12*a*b^2)*cos(d*x + c)^5 + 48*a*b^2*cos(d*x + c) - 10*(a^3 + 12*a*b^2)*cos(d*x + c)^3 - 3*((a^3 + 12*a*b^2)*cos(d*x + c)^6 - 2*(a^3 + 12*a*b^2)*cos(d*x + c)^4 + (a^3 + 12*a*b^2)*cos(d*x + c)^2)*log(1/2*cos(d *x + c) + 1/2) + 3*((a^3 + 12*a*b^2)*cos(d*x + c)^6 - 2*(a^3 + 12*a*b^2)*c os(d*x + c)^4 + (a^3 + 12*a*b^2)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1 /2) + 12*((2*a^2*b + b^3)*cos(d*x + c)^6 - 2*(2*a^2*b + b^3)*cos(d*x + c)^ 4 + (2*a^2*b + b^3)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 12*((2*a^2*b + b^3)*cos(d*x + c)^6 - 2*(2*a^2*b + b^3)*cos(d*x + c)^4 + (2*a^2*b + b^3)* cos(d*x + c)^2)*log(-sin(d*x + c) + 1) + 8*(3*(2*a^2*b + b^3)*cos(d*x + c) ^4 + b^3 - 4*(2*a^2*b + b^3)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c) ^6 - 2*d*cos(d*x + c)^4 + d*cos(d*x + c)^2)
\[ \int \csc ^5(c+d x) (a+b \tan (c+d x))^3 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \csc ^{5}{\left (c + d x \right )}\, dx \] Input:
integrate(csc(d*x+c)**5*(a+b*tan(d*x+c))**3,x)
Output:
Integral((a + b*tan(c + d*x))**3*csc(c + d*x)**5, x)
Time = 0.05 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.09 \[ \int \csc ^5(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {a^{3} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 12 \, a b^{2} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 2\right )}}{\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 4 \, b^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{2} - 2\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 8 \, a^{2} b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{2} + 1\right )}}{\sin \left (d x + c\right )^{3}} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{16 \, d} \] Input:
integrate(csc(d*x+c)^5*(a+b*tan(d*x+c))^3,x, algorithm="maxima")
Output:
1/16*(a^3*(2*(3*cos(d*x + c)^3 - 5*cos(d*x + c))/(cos(d*x + c)^4 - 2*cos(d *x + c)^2 + 1) - 3*log(cos(d*x + c) + 1) + 3*log(cos(d*x + c) - 1)) + 12*a *b^2*(2*(3*cos(d*x + c)^2 - 2)/(cos(d*x + c)^3 - cos(d*x + c)) - 3*log(cos (d*x + c) + 1) + 3*log(cos(d*x + c) - 1)) - 4*b^3*(2*(3*sin(d*x + c)^2 - 2 )/(sin(d*x + c)^3 - sin(d*x + c)) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d* x + c) - 1)) - 8*a^2*b*(2*(3*sin(d*x + c)^2 + 1)/sin(d*x + c)^3 - 3*log(si n(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)))/d
Time = 0.31 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.63 \[ \int \csc ^5(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 8 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 120 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 32 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 96 \, {\left (2 \, a^{2} b + b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 96 \, {\left (2 \, a^{2} b + b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 24 \, {\left (a^{3} + 12 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {64 \, {\left (b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} - \frac {50 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 600 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 120 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 32 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{64 \, d} \] Input:
integrate(csc(d*x+c)^5*(a+b*tan(d*x+c))^3,x, algorithm="giac")
Output:
1/64*(a^3*tan(1/2*d*x + 1/2*c)^4 - 8*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 8*a^3* tan(1/2*d*x + 1/2*c)^2 + 24*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 120*a^2*b*tan(1 /2*d*x + 1/2*c) - 32*b^3*tan(1/2*d*x + 1/2*c) + 96*(2*a^2*b + b^3)*log(abs (tan(1/2*d*x + 1/2*c) + 1)) - 96*(2*a^2*b + b^3)*log(abs(tan(1/2*d*x + 1/2 *c) - 1)) + 24*(a^3 + 12*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c))) + 64*(b^3*t an(1/2*d*x + 1/2*c)^3 - 6*a*b^2*tan(1/2*d*x + 1/2*c)^2 + b^3*tan(1/2*d*x + 1/2*c) + 6*a*b^2)/(tan(1/2*d*x + 1/2*c)^2 - 1)^2 - (50*a^3*tan(1/2*d*x + 1/2*c)^4 + 600*a*b^2*tan(1/2*d*x + 1/2*c)^4 + 120*a^2*b*tan(1/2*d*x + 1/2* c)^3 + 32*b^3*tan(1/2*d*x + 1/2*c)^3 + 8*a^3*tan(1/2*d*x + 1/2*c)^2 + 24*a *b^2*tan(1/2*d*x + 1/2*c)^2 + 8*a^2*b*tan(1/2*d*x + 1/2*c) + a^3)/tan(1/2* d*x + 1/2*c)^4)/d
Time = 1.11 (sec) , antiderivative size = 698, normalized size of antiderivative = 3.05 \[ \int \csc ^5(c+d x) (a+b \tan (c+d x))^3 \, dx =\text {Too large to display} \] Input:
int((a + b*tan(c + d*x))^3/sin(c + d*x)^5,x)
Output:
(a^3*tan(c/2 + (d*x)/2)^4)/(64*d) + (tan(c/2 + (d*x)/2)^2*((3*a*b^2)/8 + a ^3/8))/d - (atan(((3*a^2*b + (3*b^3)/2)*(tan(c/2 + (d*x)/2)*(9*a*b^2 + (3* a^3)/4) + 6*tan(c/2 + (d*x)/2)*(3*a^2*b + (3*b^3)/2) - 6*a^2*b - 3*b^3)*1i - (3*a^2*b + (3*b^3)/2)*(6*tan(c/2 + (d*x)/2)*(3*a^2*b + (3*b^3)/2) - tan (c/2 + (d*x)/2)*(9*a*b^2 + (3*a^3)/4) + 6*a^2*b + 3*b^3)*1i)/(2*tan(c/2 + (d*x)/2)*(9*b^6 + 36*a^2*b^4 + 36*a^4*b^2) + 27*a*b^5 + (9*a^5*b)/2 - (3*a ^2*b + (3*b^3)/2)*(tan(c/2 + (d*x)/2)*(9*a*b^2 + (3*a^3)/4) + 6*tan(c/2 + (d*x)/2)*(3*a^2*b + (3*b^3)/2) - 6*a^2*b - 3*b^3) - (3*a^2*b + (3*b^3)/2)* (6*tan(c/2 + (d*x)/2)*(3*a^2*b + (3*b^3)/2) - tan(c/2 + (d*x)/2)*(9*a*b^2 + (3*a^3)/4) + 6*a^2*b + 3*b^3) + (225*a^3*b^3)/4))*(a^2*b*6i + b^3*3i))/d - (tan(c/2 + (d*x)/2)^2*(6*a*b^2 + (3*a^3)/2) + tan(c/2 + (d*x)/2)^6*(102 *a*b^2 + 2*a^3) - tan(c/2 + (d*x)/2)^4*(108*a*b^2 + (15*a^3)/4) + tan(c/2 + (d*x)/2)^3*(26*a^2*b + 8*b^3) + tan(c/2 + (d*x)/2)^7*(30*a^2*b - 8*b^3) - tan(c/2 + (d*x)/2)^5*(58*a^2*b + 32*b^3) + a^3/4 + 2*a^2*b*tan(c/2 + (d* x)/2))/(d*(16*tan(c/2 + (d*x)/2)^4 - 32*tan(c/2 + (d*x)/2)^6 + 16*tan(c/2 + (d*x)/2)^8)) - (tan(c/2 + (d*x)/2)*((15*a^2*b)/8 + b^3/2))/d + (3*a*log( tan(c/2 + (d*x)/2))*(a^2 + 12*b^2))/(8*d) - (a^2*b*tan(c/2 + (d*x)/2)^3)/( 8*d)
Time = 0.21 (sec) , antiderivative size = 533, normalized size of antiderivative = 2.33 \[ \int \csc ^5(c+d x) (a+b \tan (c+d x))^3 \, dx =\text {Too large to display} \] Input:
int(csc(d*x+c)^5*(a+b*tan(d*x+c))^3,x)
Output:
( - 24*cos(c + d*x)*sin(c + d*x)**4*a**3 - 288*cos(c + d*x)*sin(c + d*x)** 4*a*b**2 + 8*cos(c + d*x)*sin(c + d*x)**2*a**3 + 96*cos(c + d*x)*sin(c + d *x)**2*a*b**2 + 16*cos(c + d*x)*a**3 - 192*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6*a**2*b - 96*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6*b**3 + 19 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2*b + 96*log(tan((c + d*x)/ 2) - 1)*sin(c + d*x)**4*b**3 + 192*log(tan((c + d*x)/2) + 1)*sin(c + d*x)* *6*a**2*b + 96*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**6*b**3 - 192*log(ta n((c + d*x)/2) + 1)*sin(c + d*x)**4*a**2*b - 96*log(tan((c + d*x)/2) + 1)* sin(c + d*x)**4*b**3 + 24*log(tan((c + d*x)/2))*sin(c + d*x)**6*a**3 + 288 *log(tan((c + d*x)/2))*sin(c + d*x)**6*a*b**2 - 24*log(tan((c + d*x)/2))*s in(c + d*x)**4*a**3 - 288*log(tan((c + d*x)/2))*sin(c + d*x)**4*a*b**2 - 1 5*sin(c + d*x)**6*a**3 - 240*sin(c + d*x)**6*a*b**2 - 192*sin(c + d*x)**5* a**2*b - 96*sin(c + d*x)**5*b**3 + 15*sin(c + d*x)**4*a**3 + 240*sin(c + d *x)**4*a*b**2 + 128*sin(c + d*x)**3*a**2*b + 64*sin(c + d*x)**3*b**3 + 64* sin(c + d*x)*a**2*b)/(64*sin(c + d*x)**4*d*(sin(c + d*x)**2 - 1))