Integrand size = 21, antiderivative size = 178 \[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {\left (a^2+6 b^2\right ) \cot (c+d x)}{a^5 d}+\frac {3 b \cot ^2(c+d x)}{2 a^4 d}-\frac {\cot ^3(c+d x)}{3 a^3 d}-\frac {b \left (3 a^2+10 b^2\right ) \log (\tan (c+d x))}{a^6 d}+\frac {b \left (3 a^2+10 b^2\right ) \log (a+b \tan (c+d x))}{a^6 d}-\frac {b \left (a^2+b^2\right )}{2 a^4 d (a+b \tan (c+d x))^2}-\frac {2 b \left (a^2+2 b^2\right )}{a^5 d (a+b \tan (c+d x))} \] Output:
-(a^2+6*b^2)*cot(d*x+c)/a^5/d+3/2*b*cot(d*x+c)^2/a^4/d-1/3*cot(d*x+c)^3/a^ 3/d-b*(3*a^2+10*b^2)*ln(tan(d*x+c))/a^6/d+b*(3*a^2+10*b^2)*ln(a+b*tan(d*x+ c))/a^6/d-1/2*b*(a^2+b^2)/a^4/d/(a+b*tan(d*x+c))^2-2*b*(a^2+2*b^2)/a^5/d/( a+b*tan(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(456\) vs. \(2(178)=356\).
Time = 3.67 (sec) , antiderivative size = 456, normalized size of antiderivative = 2.56 \[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {b^3 \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))}{2 a^4 d (a+b \tan (c+d x))^3}-\frac {\csc ^3(c+d x) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3}{3 a^3 d (a+b \tan (c+d x))^3}-\frac {2 \left (a^2 \cos (c+d x)+9 b^2 \cos (c+d x)\right ) \csc (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3}{3 a^5 d (a+b \tan (c+d x))^3}+\frac {3 b \csc ^2(c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3}{2 a^4 d (a+b \tan (c+d x))^3}+\frac {\left (-3 a^2 b-10 b^3\right ) \log (\sin (c+d x)) \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3}{a^6 d (a+b \tan (c+d x))^3}+\frac {\left (3 a^2 b+10 b^3\right ) \log (a \cos (c+d x)+b \sin (c+d x)) \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3}{a^6 d (a+b \tan (c+d x))^3}+\frac {\sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \left (3 a^2 b^2 \sin (c+d x)+4 b^4 \sin (c+d x)\right )}{a^6 d (a+b \tan (c+d x))^3} \] Input:
Integrate[Csc[c + d*x]^4/(a + b*Tan[c + d*x])^3,x]
Output:
-1/2*(b^3*Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x]))/(a^4*d*(a + b* Tan[c + d*x])^3) - (Csc[c + d*x]^3*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[ c + d*x])^3)/(3*a^3*d*(a + b*Tan[c + d*x])^3) - (2*(a^2*Cos[c + d*x] + 9*b ^2*Cos[c + d*x])*Csc[c + d*x]*Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d *x])^3)/(3*a^5*d*(a + b*Tan[c + d*x])^3) + (3*b*Csc[c + d*x]^2*Sec[c + d*x ]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)/(2*a^4*d*(a + b*Tan[c + d*x])^3) + ((-3*a^2*b - 10*b^3)*Log[Sin[c + d*x]]*Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)/(a^6*d*(a + b*Tan[c + d*x])^3) + ((3*a^2*b + 10*b^3)*Lo g[a*Cos[c + d*x] + b*Sin[c + d*x]]*Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[ c + d*x])^3)/(a^6*d*(a + b*Tan[c + d*x])^3) + (Sec[c + d*x]^3*(a*Cos[c + d *x] + b*Sin[c + d*x])^2*(3*a^2*b^2*Sin[c + d*x] + 4*b^4*Sin[c + d*x]))/(a^ 6*d*(a + b*Tan[c + d*x])^3)
Time = 0.39 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3999, 522, 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.
| \(\displaystyle \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x)^4 (a+b \tan (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3999 |
| \(\displaystyle \frac {b \int \frac {\cot ^4(c+d x) \left (\tan ^2(c+d x) b^2+b^2\right )}{b^4 (a+b \tan (c+d x))^3}d(b \tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 522 |
| \(\displaystyle \frac {b \int \left (\frac {\cot ^4(c+d x)}{a^3 b^2}-\frac {3 \cot ^3(c+d x)}{a^4 b}+\frac {\left (a^2+6 b^2\right ) \cot ^2(c+d x)}{a^5 b^2}+\frac {\left (-3 a^2-10 b^2\right ) \cot (c+d x)}{a^6 b}+\frac {3 a^2+10 b^2}{a^6 (a+b \tan (c+d x))}+\frac {2 \left (a^2+2 b^2\right )}{a^5 (a+b \tan (c+d x))^2}+\frac {a^2+b^2}{a^4 (a+b \tan (c+d x))^3}\right )d(b \tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b \left (\frac {3 \cot ^2(c+d x)}{2 a^4}-\frac {\cot ^3(c+d x)}{3 a^3 b}-\frac {\left (3 a^2+10 b^2\right ) \log (b \tan (c+d x))}{a^6}+\frac {\left (3 a^2+10 b^2\right ) \log (a+b \tan (c+d x))}{a^6}-\frac {2 \left (a^2+2 b^2\right )}{a^5 (a+b \tan (c+d x))}-\frac {\left (a^2+6 b^2\right ) \cot (c+d x)}{a^5 b}-\frac {a^2+b^2}{2 a^4 (a+b \tan (c+d x))^2}\right )}{d}\) |
Input:
Int[Csc[c + d*x]^4/(a + b*Tan[c + d*x])^3,x]
Output:
(b*(-(((a^2 + 6*b^2)*Cot[c + d*x])/(a^5*b)) + (3*Cot[c + d*x]^2)/(2*a^4) - Cot[c + d*x]^3/(3*a^3*b) - ((3*a^2 + 10*b^2)*Log[b*Tan[c + d*x]])/a^6 + ( (3*a^2 + 10*b^2)*Log[a + b*Tan[c + d*x]])/a^6 - (a^2 + b^2)/(2*a^4*(a + b* Tan[c + d*x])^2) - (2*(a^2 + 2*b^2))/(a^5*(a + b*Tan[c + d*x]))))/d
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[b/f Subst[Int[x^m*((a + x)^n/(b^2 + x^2)^(m/2 + 1)), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/2]
Time = 3.98 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.89
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{3 a^{3} \tan \left (d x +c \right )^{3}}-\frac {a^{2}+6 b^{2}}{a^{5} \tan \left (d x +c \right )}+\frac {3 b}{2 a^{4} \tan \left (d x +c \right )^{2}}-\frac {b \left (3 a^{2}+10 b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{a^{6}}-\frac {b \left (a^{2}+b^{2}\right )}{2 a^{4} \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {b \left (3 a^{2}+10 b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{6}}-\frac {2 b \left (a^{2}+2 b^{2}\right )}{a^{5} \left (a +b \tan \left (d x +c \right )\right )}}{d}\) | \(158\) |
| default | \(\frac {-\frac {1}{3 a^{3} \tan \left (d x +c \right )^{3}}-\frac {a^{2}+6 b^{2}}{a^{5} \tan \left (d x +c \right )}+\frac {3 b}{2 a^{4} \tan \left (d x +c \right )^{2}}-\frac {b \left (3 a^{2}+10 b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{a^{6}}-\frac {b \left (a^{2}+b^{2}\right )}{2 a^{4} \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {b \left (3 a^{2}+10 b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{6}}-\frac {2 b \left (a^{2}+2 b^{2}\right )}{a^{5} \left (a +b \tan \left (d x +c \right )\right )}}{d}\) | \(158\) |
| risch | \(-\frac {2 i \left (60 i a \,b^{4} {\mathrm e}^{8 i \left (d x +c \right )}+18 i a^{3} b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-2 i a^{5}+30 b^{5}+29 a^{2} b^{3}+2 a^{4} b +6 i a^{5} {\mathrm e}^{6 i \left (d x +c \right )}-6 a^{2} b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-106 a^{2} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-30 i a \,b^{4}+9 a^{4} b \,{\mathrm e}^{6 i \left (d x +c \right )}-9 a^{4} b \,{\mathrm e}^{8 i \left (d x +c \right )}-21 a^{2} b^{3} {\mathrm e}^{8 i \left (d x +c \right )}+10 i a^{5} {\mathrm e}^{4 i \left (d x +c \right )}-150 i a \,b^{4} {\mathrm e}^{6 i \left (d x +c \right )}+27 i a^{3} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+180 b^{5} {\mathrm e}^{4 i \left (d x +c \right )}-120 b^{5} {\mathrm e}^{2 i \left (d x +c \right )}+2 i a^{5} {\mathrm e}^{2 i \left (d x +c \right )}+104 a^{2} b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-a^{4} b \,{\mathrm e}^{2 i \left (d x +c \right )}-29 i a^{3} b^{2}+15 a^{4} b \,{\mathrm e}^{4 i \left (d x +c \right )}-120 b^{5} {\mathrm e}^{6 i \left (d x +c \right )}+30 b^{5} {\mathrm e}^{8 i \left (d x +c \right )}+30 i a \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+90 i a \,b^{4} {\mathrm e}^{4 i \left (d x +c \right )}+29 i a^{3} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-45 i a^{3} b^{2} {\mathrm e}^{6 i \left (d x +c \right )}\right )}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right )^{2} \left (i a +b \right ) a^{5} d}+\frac {3 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{a^{4} d}+\frac {10 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{a^{6} d}-\frac {3 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{4} d}-\frac {10 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{6} d}\) | \(595\) |
Input:
int(csc(d*x+c)^4/(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/3/a^3/tan(d*x+c)^3-(a^2+6*b^2)/a^5/tan(d*x+c)+3/2/a^4*b/tan(d*x+c) ^2-b*(3*a^2+10*b^2)/a^6*ln(tan(d*x+c))-1/2*b*(a^2+b^2)/a^4/(a+b*tan(d*x+c) )^2+b*(3*a^2+10*b^2)/a^6*ln(a+b*tan(d*x+c))-2*b*(a^2+2*b^2)/a^5/(a+b*tan(d *x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 811 vs. \(2 (172) = 344\).
Time = 0.14 (sec) , antiderivative size = 811, normalized size of antiderivative = 4.56 \[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx =\text {Too large to display} \] Input:
integrate(csc(d*x+c)^4/(a+b*tan(d*x+c))^3,x, algorithm="fricas")
Output:
1/6*(2*(2*a^7 + 27*a^5*b^2 + a^3*b^4 - 30*a*b^6)*cos(d*x + c)^5 - 2*(3*a^7 + 43*a^5*b^2 - 8*a^3*b^4 - 60*a*b^6)*cos(d*x + c)^3 + 6*(5*a^5*b^2 - 3*a^ 3*b^4 - 10*a*b^6)*cos(d*x + c) + 3*(2*(3*a^5*b^2 + 13*a^3*b^4 + 10*a*b^6)* cos(d*x + c)^5 - 4*(3*a^5*b^2 + 13*a^3*b^4 + 10*a*b^6)*cos(d*x + c)^3 + 2* (3*a^5*b^2 + 13*a^3*b^4 + 10*a*b^6)*cos(d*x + c) + (3*a^4*b^3 + 13*a^2*b^5 + 10*b^7 - (3*a^6*b + 10*a^4*b^3 - 3*a^2*b^5 - 10*b^7)*cos(d*x + c)^4 + ( 3*a^6*b + 7*a^4*b^3 - 16*a^2*b^5 - 20*b^7)*cos(d*x + c)^2)*sin(d*x + c))*l og(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) - 3 *(2*(3*a^5*b^2 + 13*a^3*b^4 + 10*a*b^6)*cos(d*x + c)^5 - 4*(3*a^5*b^2 + 13 *a^3*b^4 + 10*a*b^6)*cos(d*x + c)^3 + 2*(3*a^5*b^2 + 13*a^3*b^4 + 10*a*b^6 )*cos(d*x + c) + (3*a^4*b^3 + 13*a^2*b^5 + 10*b^7 - (3*a^6*b + 10*a^4*b^3 - 3*a^2*b^5 - 10*b^7)*cos(d*x + c)^4 + (3*a^6*b + 7*a^4*b^3 - 16*a^2*b^5 - 20*b^7)*cos(d*x + c)^2)*sin(d*x + c))*log(-1/4*cos(d*x + c)^2 + 1/4) + (2 4*a^4*b^3 + 30*a^2*b^5 + 4*(2*a^6*b + 29*a^4*b^3 + 30*a^2*b^5)*cos(d*x + c )^4 - 3*(a^6*b + 45*a^4*b^3 + 50*a^2*b^5)*cos(d*x + c)^2)*sin(d*x + c))/(2 *(a^9*b + a^7*b^3)*d*cos(d*x + c)^5 - 4*(a^9*b + a^7*b^3)*d*cos(d*x + c)^3 + 2*(a^9*b + a^7*b^3)*d*cos(d*x + c) - ((a^10 - a^6*b^4)*d*cos(d*x + c)^4 - (a^10 - a^8*b^2 - 2*a^6*b^4)*d*cos(d*x + c)^2 - (a^8*b^2 + a^6*b^4)*d)* sin(d*x + c))
\[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\int \frac {\csc ^{4}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{3}}\, dx \] Input:
integrate(csc(d*x+c)**4/(a+b*tan(d*x+c))**3,x)
Output:
Integral(csc(c + d*x)**4/(a + b*tan(c + d*x))**3, x)
Time = 0.04 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.08 \[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {5 \, a^{3} b \tan \left (d x + c\right ) - 6 \, {\left (3 \, a^{2} b^{2} + 10 \, b^{4}\right )} \tan \left (d x + c\right )^{4} - 2 \, a^{4} - 9 \, {\left (3 \, a^{3} b + 10 \, a b^{3}\right )} \tan \left (d x + c\right )^{3} - 2 \, {\left (3 \, a^{4} + 10 \, a^{2} b^{2}\right )} \tan \left (d x + c\right )^{2}}{a^{5} b^{2} \tan \left (d x + c\right )^{5} + 2 \, a^{6} b \tan \left (d x + c\right )^{4} + a^{7} \tan \left (d x + c\right )^{3}} + \frac {6 \, {\left (3 \, a^{2} b + 10 \, b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6}} - \frac {6 \, {\left (3 \, a^{2} b + 10 \, b^{3}\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{6}}}{6 \, d} \] Input:
integrate(csc(d*x+c)^4/(a+b*tan(d*x+c))^3,x, algorithm="maxima")
Output:
1/6*((5*a^3*b*tan(d*x + c) - 6*(3*a^2*b^2 + 10*b^4)*tan(d*x + c)^4 - 2*a^4 - 9*(3*a^3*b + 10*a*b^3)*tan(d*x + c)^3 - 2*(3*a^4 + 10*a^2*b^2)*tan(d*x + c)^2)/(a^5*b^2*tan(d*x + c)^5 + 2*a^6*b*tan(d*x + c)^4 + a^7*tan(d*x + c )^3) + 6*(3*a^2*b + 10*b^3)*log(b*tan(d*x + c) + a)/a^6 - 6*(3*a^2*b + 10* b^3)*log(tan(d*x + c))/a^6)/d
Time = 0.22 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.04 \[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {{\left (3 \, a^{2} b + 10 \, b^{3}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{6} d} + \frac {{\left (3 \, a^{2} b^{2} + 10 \, b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b d} + \frac {5 \, a^{4} b \tan \left (d x + c\right ) - 2 \, a^{5} - 6 \, {\left (3 \, a^{3} b^{2} + 10 \, a b^{4}\right )} \tan \left (d x + c\right )^{4} - 9 \, {\left (3 \, a^{4} b + 10 \, a^{2} b^{3}\right )} \tan \left (d x + c\right )^{3} - 2 \, {\left (3 \, a^{5} + 10 \, a^{3} b^{2}\right )} \tan \left (d x + c\right )^{2}}{6 \, {\left (b \tan \left (d x + c\right ) + a\right )}^{2} a^{6} d \tan \left (d x + c\right )^{3}} \] Input:
integrate(csc(d*x+c)^4/(a+b*tan(d*x+c))^3,x, algorithm="giac")
Output:
-(3*a^2*b + 10*b^3)*log(abs(tan(d*x + c)))/(a^6*d) + (3*a^2*b^2 + 10*b^4)* log(abs(b*tan(d*x + c) + a))/(a^6*b*d) + 1/6*(5*a^4*b*tan(d*x + c) - 2*a^5 - 6*(3*a^3*b^2 + 10*a*b^4)*tan(d*x + c)^4 - 9*(3*a^4*b + 10*a^2*b^3)*tan( d*x + c)^3 - 2*(3*a^5 + 10*a^3*b^2)*tan(d*x + c)^2)/((b*tan(d*x + c) + a)^ 2*a^6*d*tan(d*x + c)^3)
Time = 1.44 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.12 \[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {2\,b\,\mathrm {atanh}\left (\frac {b\,\left (3\,a^2+10\,b^2\right )\,\left (a+2\,b\,\mathrm {tan}\left (c+d\,x\right )\right )}{a\,\left (3\,a^2\,b+10\,b^3\right )}\right )\,\left (3\,a^2+10\,b^2\right )}{a^6\,d}-\frac {\frac {1}{3\,a}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (3\,a^2+10\,b^2\right )}{3\,a^3}-\frac {5\,b\,\mathrm {tan}\left (c+d\,x\right )}{6\,a^2}+\frac {b^2\,{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (3\,a^2+10\,b^2\right )}{a^5}+\frac {3\,b\,{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (3\,a^2+10\,b^2\right )}{2\,a^4}}{d\,\left (a^2\,{\mathrm {tan}\left (c+d\,x\right )}^3+2\,a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^4+b^2\,{\mathrm {tan}\left (c+d\,x\right )}^5\right )} \] Input:
int(1/(sin(c + d*x)^4*(a + b*tan(c + d*x))^3),x)
Output:
(2*b*atanh((b*(3*a^2 + 10*b^2)*(a + 2*b*tan(c + d*x)))/(a*(3*a^2*b + 10*b^ 3)))*(3*a^2 + 10*b^2))/(a^6*d) - (1/(3*a) + (tan(c + d*x)^2*(3*a^2 + 10*b^ 2))/(3*a^3) - (5*b*tan(c + d*x))/(6*a^2) + (b^2*tan(c + d*x)^4*(3*a^2 + 10 *b^2))/a^5 + (3*b*tan(c + d*x)^3*(3*a^2 + 10*b^2))/(2*a^4))/(d*(a^2*tan(c + d*x)^3 + b^2*tan(c + d*x)^5 + 2*a*b*tan(c + d*x)^4))
Time = 0.19 (sec) , antiderivative size = 798, normalized size of antiderivative = 4.48 \[ \int \frac {\csc ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int(csc(d*x+c)^4/(a+b*tan(d*x+c))^3,x)
Output:
(288*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*si n(c + d*x)**4*a**3*b**3 + 960*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - 2*t an((c + d*x)/2)*b - a)*sin(c + d*x)**4*a*b**5 - 288*cos(c + d*x)*log(tan(( c + d*x)/2))*sin(c + d*x)**4*a**3*b**3 - 960*cos(c + d*x)*log(tan((c + d*x )/2))*sin(c + d*x)**4*a*b**5 + 6*cos(c + d*x)*sin(c + d*x)**4*a**5*b + 40* cos(c + d*x)*sin(c + d*x)**4*a**3*b**3 - 16*cos(c + d*x)*sin(c + d*x)**2*a **5*b - 160*cos(c + d*x)*sin(c + d*x)**2*a**3*b**3 - 16*cos(c + d*x)*a**5* b - 144*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*sin(c + d*x) **5*a**4*b**2 - 336*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)* sin(c + d*x)**5*a**2*b**4 + 480*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x )/2)*b - a)*sin(c + d*x)**5*b**6 + 144*log(tan((c + d*x)/2)**2*a - 2*tan(( c + d*x)/2)*b - a)*sin(c + d*x)**3*a**4*b**2 + 480*log(tan((c + d*x)/2)**2 *a - 2*tan((c + d*x)/2)*b - a)*sin(c + d*x)**3*a**2*b**4 + 144*log(tan((c + d*x)/2))*sin(c + d*x)**5*a**4*b**2 + 336*log(tan((c + d*x)/2))*sin(c + d *x)**5*a**2*b**4 - 480*log(tan((c + d*x)/2))*sin(c + d*x)**5*b**6 - 144*lo g(tan((c + d*x)/2))*sin(c + d*x)**3*a**4*b**2 - 480*log(tan((c + d*x)/2))* sin(c + d*x)**3*a**2*b**4 + 13*sin(c + d*x)**5*a**6 + 231*sin(c + d*x)**5* a**4*b**2 + 492*sin(c + d*x)**5*a**2*b**4 + 240*sin(c + d*x)**5*b**6 - 13* sin(c + d*x)**3*a**6 - 284*sin(c + d*x)**3*a**4*b**2 - 480*sin(c + d*x)**3 *a**2*b**4 + 40*sin(c + d*x)*a**4*b**2)/(48*sin(c + d*x)**3*a**6*b*d*(2...