Integrand size = 21, antiderivative size = 219 \[ \int \frac {\csc ^6(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {\left (a^2+b^2\right ) \left (a^2+5 b^2\right ) \cot (c+d x)}{a^6 d}+\frac {2 b \left (a^2+b^2\right ) \cot ^2(c+d x)}{a^5 d}-\frac {\left (2 a^2+3 b^2\right ) \cot ^3(c+d x)}{3 a^4 d}+\frac {b \cot ^4(c+d x)}{2 a^3 d}-\frac {\cot ^5(c+d x)}{5 a^2 d}-\frac {2 b \left (a^2+b^2\right ) \left (a^2+3 b^2\right ) \log (\tan (c+d x))}{a^7 d}+\frac {2 b \left (a^2+b^2\right ) \left (a^2+3 b^2\right ) \log (a+b \tan (c+d x))}{a^7 d}-\frac {b \left (a^2+b^2\right )^2}{a^6 d (a+b \tan (c+d x))} \] Output:
-(a^2+b^2)*(a^2+5*b^2)*cot(d*x+c)/a^6/d+2*b*(a^2+b^2)*cot(d*x+c)^2/a^5/d-1 /3*(2*a^2+3*b^2)*cot(d*x+c)^3/a^4/d+1/2*b*cot(d*x+c)^4/a^3/d-1/5*cot(d*x+c )^5/a^2/d-2*b*(a^2+b^2)*(a^2+3*b^2)*ln(tan(d*x+c))/a^7/d+2*b*(a^2+b^2)*(a^ 2+3*b^2)*ln(a+b*tan(d*x+c))/a^7/d-b*(a^2+b^2)^2/a^6/d/(a+b*tan(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(589\) vs. \(2(219)=438\).
Time = 7.03 (sec) , antiderivative size = 589, normalized size of antiderivative = 2.69 \[ \int \frac {\csc ^6(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {\csc ^5(c+d x) \sec (c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{5 a^2 d (a+b \tan (c+d x))^2}+\frac {\left (-8 a^4 \cos (c+d x)-75 a^2 b^2 \cos (c+d x)-75 b^4 \cos (c+d x)\right ) \csc (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{15 a^6 d (a+b \tan (c+d x))^2}+\frac {b \left (a^2+2 b^2\right ) \csc ^2(c+d x) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{a^5 d (a+b \tan (c+d x))^2}+\frac {\left (-4 a^2 \cos (c+d x)-15 b^2 \cos (c+d x)\right ) \csc ^3(c+d x) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{15 a^4 d (a+b \tan (c+d x))^2}+\frac {b \csc ^4(c+d x) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{2 a^3 d (a+b \tan (c+d x))^2}-\frac {2 \left (a^4 b+4 a^2 b^3+3 b^5\right ) \log (\sin (c+d x)) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{a^7 d (a+b \tan (c+d x))^2}+\frac {2 \left (a^4 b+4 a^2 b^3+3 b^5\right ) \log (a \cos (c+d x)+b \sin (c+d x)) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{a^7 d (a+b \tan (c+d x))^2}+\frac {\sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \left (a^4 b^2 \sin (c+d x)+2 a^2 b^4 \sin (c+d x)+b^6 \sin (c+d x)\right )}{a^7 d (a+b \tan (c+d x))^2} \] Input:
Integrate[Csc[c + d*x]^6/(a + b*Tan[c + d*x])^2,x]
Output:
-1/5*(Csc[c + d*x]^5*Sec[c + d*x]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(a^ 2*d*(a + b*Tan[c + d*x])^2) + ((-8*a^4*Cos[c + d*x] - 75*a^2*b^2*Cos[c + d *x] - 75*b^4*Cos[c + d*x])*Csc[c + d*x]*Sec[c + d*x]^2*(a*Cos[c + d*x] + b *Sin[c + d*x])^2)/(15*a^6*d*(a + b*Tan[c + d*x])^2) + (b*(a^2 + 2*b^2)*Csc [c + d*x]^2*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(a^5*d*(a + b*Tan[c + d*x])^2) + ((-4*a^2*Cos[c + d*x] - 15*b^2*Cos[c + d*x])*Csc[c + d*x]^3*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(15*a^4*d*(a + b*Tan[c + d*x])^2) + (b*Csc[c + d*x]^4*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(2*a^3*d*(a + b*Tan[c + d*x])^2) - (2*(a^4*b + 4*a^2*b^ 3 + 3*b^5)*Log[Sin[c + d*x]]*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d* x])^2)/(a^7*d*(a + b*Tan[c + d*x])^2) + (2*(a^4*b + 4*a^2*b^3 + 3*b^5)*Log [a*Cos[c + d*x] + b*Sin[c + d*x]]*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(a^7*d*(a + b*Tan[c + d*x])^2) + (Sec[c + d*x]^2*(a*Cos[c + d* x] + b*Sin[c + d*x])*(a^4*b^2*Sin[c + d*x] + 2*a^2*b^4*Sin[c + d*x] + b^6* Sin[c + d*x]))/(a^7*d*(a + b*Tan[c + d*x])^2)
Time = 0.44 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.94, 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.
| \(\displaystyle \int \frac {\csc ^6(c+d x)}{(a+b \tan (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x)^6 (a+b \tan (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3999 |
| \(\displaystyle \frac {b \int \frac {\cot ^6(c+d x) \left (\tan ^2(c+d x) b^2+b^2\right )^2}{b^6 (a+b \tan (c+d x))^2}d(b \tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 522 |
| \(\displaystyle \frac {b \int \left (\frac {\cot ^6(c+d x)}{a^2 b^2}-\frac {2 \cot ^5(c+d x)}{a^3 b}+\frac {\left (3 b^4+2 a^2 b^2\right ) \cot ^4(c+d x)}{a^4 b^4}-\frac {4 \left (a^2+b^2\right ) \cot ^3(c+d x)}{a^5 b}+\frac {\left (a^4+6 b^2 a^2+5 b^4\right ) \cot ^2(c+d x)}{a^6 b^2}-\frac {2 \left (a^4+4 b^2 a^2+3 b^4\right ) \cot (c+d x)}{a^7 b}+\frac {2 \left (a^4+4 b^2 a^2+3 b^4\right )}{a^7 (a+b \tan (c+d x))}+\frac {\left (a^2+b^2\right )^2}{a^6 (a+b \tan (c+d x))^2}\right )d(b \tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b \left (\frac {\cot ^4(c+d x)}{2 a^3}-\frac {\cot ^5(c+d x)}{5 a^2 b}-\frac {2 \left (a^2+b^2\right ) \left (a^2+3 b^2\right ) \log (b \tan (c+d x))}{a^7}+\frac {2 \left (a^2+b^2\right ) \left (a^2+3 b^2\right ) \log (a+b \tan (c+d x))}{a^7}-\frac {\left (a^2+b^2\right )^2}{a^6 (a+b \tan (c+d x))}-\frac {\left (a^2+b^2\right ) \left (a^2+5 b^2\right ) \cot (c+d x)}{a^6 b}+\frac {2 \left (a^2+b^2\right ) \cot ^2(c+d x)}{a^5}-\frac {\left (2 a^2+3 b^2\right ) \cot ^3(c+d x)}{3 a^4 b}\right )}{d}\) |
Input:
Int[Csc[c + d*x]^6/(a + b*Tan[c + d*x])^2,x]
Output:
(b*(-(((a^2 + b^2)*(a^2 + 5*b^2)*Cot[c + d*x])/(a^6*b)) + (2*(a^2 + b^2)*C ot[c + d*x]^2)/a^5 - ((2*a^2 + 3*b^2)*Cot[c + d*x]^3)/(3*a^4*b) + Cot[c + d*x]^4/(2*a^3) - Cot[c + d*x]^5/(5*a^2*b) - (2*(a^2 + b^2)*(a^2 + 3*b^2)*L og[b*Tan[c + d*x]])/a^7 + (2*(a^2 + b^2)*(a^2 + 3*b^2)*Log[a + b*Tan[c + d *x]])/a^7 - (a^2 + b^2)^2/(a^6*(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 = 4.70 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.94
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{5 a^{2} \tan \left (d x +c \right )^{5}}-\frac {2 a^{2}+3 b^{2}}{3 a^{4} \tan \left (d x +c \right )^{3}}-\frac {a^{4}+6 b^{2} a^{2}+5 b^{4}}{a^{6} \tan \left (d x +c \right )}+\frac {b}{2 a^{3} \tan \left (d x +c \right )^{4}}+\frac {2 b \left (a^{2}+b^{2}\right )}{a^{5} \tan \left (d x +c \right )^{2}}-\frac {2 b \left (a^{4}+4 b^{2} a^{2}+3 b^{4}\right ) \ln \left (\tan \left (d x +c \right )\right )}{a^{7}}-\frac {\left (a^{4}+2 b^{2} a^{2}+b^{4}\right ) b}{a^{6} \left (a +b \tan \left (d x +c \right )\right )}+\frac {2 b \left (a^{4}+4 b^{2} a^{2}+3 b^{4}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{7}}}{d}\) | \(205\) |
| default | \(\frac {-\frac {1}{5 a^{2} \tan \left (d x +c \right )^{5}}-\frac {2 a^{2}+3 b^{2}}{3 a^{4} \tan \left (d x +c \right )^{3}}-\frac {a^{4}+6 b^{2} a^{2}+5 b^{4}}{a^{6} \tan \left (d x +c \right )}+\frac {b}{2 a^{3} \tan \left (d x +c \right )^{4}}+\frac {2 b \left (a^{2}+b^{2}\right )}{a^{5} \tan \left (d x +c \right )^{2}}-\frac {2 b \left (a^{4}+4 b^{2} a^{2}+3 b^{4}\right ) \ln \left (\tan \left (d x +c \right )\right )}{a^{7}}-\frac {\left (a^{4}+2 b^{2} a^{2}+b^{4}\right ) b}{a^{6} \left (a +b \tan \left (d x +c \right )\right )}+\frac {2 b \left (a^{4}+4 b^{2} a^{2}+3 b^{4}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{7}}}{d}\) | \(205\) |
| risch | \(-\frac {4 i \left (4 i a^{5}-45 b^{5}-45 a^{2} b^{3}-4 a^{4} b +40 i a^{5} {\mathrm e}^{6 i \left (d x +c \right )}+450 a^{2} b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+210 a^{2} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+15 a^{4} b \,{\mathrm e}^{10 i \left (d x +c \right )}-16 i a^{5} {\mathrm e}^{2 i \left (d x +c \right )}+60 a^{2} b^{3} {\mathrm e}^{10 i \left (d x +c \right )}+45 i a \,b^{4}+45 i a^{3} b^{2}+40 a^{4} b \,{\mathrm e}^{6 i \left (d x +c \right )}-60 a^{4} b \,{\mathrm e}^{8 i \left (d x +c \right )}-255 a^{2} b^{3} {\mathrm e}^{8 i \left (d x +c \right )}+9 a^{4} b \,{\mathrm e}^{2 i \left (d x +c \right )}+20 i a^{5} {\mathrm e}^{4 i \left (d x +c \right )}-420 a^{2} b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-450 b^{5} {\mathrm e}^{4 i \left (d x +c \right )}+225 b^{5} {\mathrm e}^{2 i \left (d x +c \right )}+450 b^{5} {\mathrm e}^{6 i \left (d x +c \right )}-225 b^{5} {\mathrm e}^{8 i \left (d x +c \right )}+45 b^{5} {\mathrm e}^{10 i \left (d x +c \right )}-180 i a \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-180 i a \,b^{4} {\mathrm e}^{6 i \left (d x +c \right )}+240 i a^{3} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+270 i a \,b^{4} {\mathrm e}^{4 i \left (d x +c \right )}-180 i a^{3} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+15 i a^{3} b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+45 i a \,b^{4} {\mathrm e}^{8 i \left (d x +c \right )}-120 i a^{3} b^{2} {\mathrm e}^{6 i \left (d x +c \right )}\right )}{15 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5} \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 ) a^{6} d}-\frac {2 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{3} d}-\frac {8 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{5} d}-\frac {6 b^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{7} d}+\frac {2 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{a^{3} d}+\frac {8 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{a^{5} d}+\frac {6 b^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{a^{7} d}\) | \(681\) |
Input:
int(csc(d*x+c)^6/(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/5/a^2/tan(d*x+c)^5-1/3*(2*a^2+3*b^2)/a^4/tan(d*x+c)^3-(a^4+6*a^2*b ^2+5*b^4)/a^6/tan(d*x+c)+1/2/a^3*b/tan(d*x+c)^4+2*b*(a^2+b^2)/a^5/tan(d*x+ c)^2-2*b*(a^4+4*a^2*b^2+3*b^4)/a^7*ln(tan(d*x+c))-(a^4+2*a^2*b^2+b^4)*b/a^ 6/(a+b*tan(d*x+c))+2*b*(a^4+4*a^2*b^2+3*b^4)/a^7*ln(a+b*tan(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 787 vs. \(2 (213) = 426\).
Time = 0.13 (sec) , antiderivative size = 787, normalized size of antiderivative = 3.59 \[ \int \frac {\csc ^6(c+d x)}{(a+b \tan (c+d x))^2} \, dx =\text {Too large to display} \] Input:
integrate(csc(d*x+c)^6/(a+b*tan(d*x+c))^2,x, algorithm="fricas")
Output:
1/30*(4*(4*a^6 + 45*a^4*b^2 + 45*a^2*b^4)*cos(d*x + c)^6 - 75*a^4*b^2 - 90 *a^2*b^4 - 10*(4*a^6 + 45*a^4*b^2 + 45*a^2*b^4)*cos(d*x + c)^4 + 15*(2*a^6 + 23*a^4*b^2 + 24*a^2*b^4)*cos(d*x + c)^2 + 30*((a^4*b^2 + 4*a^2*b^4 + 3* b^6)*cos(d*x + c)^6 - a^4*b^2 - 4*a^2*b^4 - 3*b^6 - 3*(a^4*b^2 + 4*a^2*b^4 + 3*b^6)*cos(d*x + c)^4 + 3*(a^4*b^2 + 4*a^2*b^4 + 3*b^6)*cos(d*x + c)^2 - ((a^5*b + 4*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^5 - 2*(a^5*b + 4*a^3*b^3 + 3 *a*b^5)*cos(d*x + c)^3 + (a^5*b + 4*a^3*b^3 + 3*a*b^5)*cos(d*x + c))*sin(d *x + c))*log(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) - 30*((a^4*b^2 + 4*a^2*b^4 + 3*b^6)*cos(d*x + c)^6 - a^4*b^2 - 4*a^ 2*b^4 - 3*b^6 - 3*(a^4*b^2 + 4*a^2*b^4 + 3*b^6)*cos(d*x + c)^4 + 3*(a^4*b^ 2 + 4*a^2*b^4 + 3*b^6)*cos(d*x + c)^2 - ((a^5*b + 4*a^3*b^3 + 3*a*b^5)*cos (d*x + c)^5 - 2*(a^5*b + 4*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^3 + (a^5*b + 4* a^3*b^3 + 3*a*b^5)*cos(d*x + c))*sin(d*x + c))*log(-1/4*cos(d*x + c)^2 + 1 /4) + (4*(4*a^5*b + 45*a^3*b^3 + 45*a*b^5)*cos(d*x + c)^5 - 10*(a^5*b + 33 *a^3*b^3 + 36*a*b^5)*cos(d*x + c)^3 - 15*(a^5*b - 10*a^3*b^3 - 12*a*b^5)*c os(d*x + c))*sin(d*x + c))/(a^7*b*d*cos(d*x + c)^6 - 3*a^7*b*d*cos(d*x + c )^4 + 3*a^7*b*d*cos(d*x + c)^2 - a^7*b*d - (a^8*d*cos(d*x + c)^5 - 2*a^8*d *cos(d*x + c)^3 + a^8*d*cos(d*x + c))*sin(d*x + c))
\[ \int \frac {\csc ^6(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\int \frac {\csc ^{6}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{2}}\, dx \] Input:
integrate(csc(d*x+c)**6/(a+b*tan(d*x+c))**2,x)
Output:
Integral(csc(c + d*x)**6/(a + b*tan(c + d*x))**2, x)
Time = 0.04 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.03 \[ \int \frac {\csc ^6(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {\frac {9 \, a^{4} b \tan \left (d x + c\right ) - 60 \, {\left (a^{4} b + 4 \, a^{2} b^{3} + 3 \, b^{5}\right )} \tan \left (d x + c\right )^{5} - 6 \, a^{5} - 30 \, {\left (a^{5} + 4 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \tan \left (d x + c\right )^{4} + 10 \, {\left (4 \, a^{4} b + 3 \, a^{2} b^{3}\right )} \tan \left (d x + c\right )^{3} - 5 \, {\left (4 \, a^{5} + 3 \, a^{3} b^{2}\right )} \tan \left (d x + c\right )^{2}}{a^{6} b \tan \left (d x + c\right )^{6} + a^{7} \tan \left (d x + c\right )^{5}} + \frac {60 \, {\left (a^{4} b + 4 \, a^{2} b^{3} + 3 \, b^{5}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{7}} - \frac {60 \, {\left (a^{4} b + 4 \, a^{2} b^{3} + 3 \, b^{5}\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{7}}}{30 \, d} \] Input:
integrate(csc(d*x+c)^6/(a+b*tan(d*x+c))^2,x, algorithm="maxima")
Output:
1/30*((9*a^4*b*tan(d*x + c) - 60*(a^4*b + 4*a^2*b^3 + 3*b^5)*tan(d*x + c)^ 5 - 6*a^5 - 30*(a^5 + 4*a^3*b^2 + 3*a*b^4)*tan(d*x + c)^4 + 10*(4*a^4*b + 3*a^2*b^3)*tan(d*x + c)^3 - 5*(4*a^5 + 3*a^3*b^2)*tan(d*x + c)^2)/(a^6*b*t an(d*x + c)^6 + a^7*tan(d*x + c)^5) + 60*(a^4*b + 4*a^2*b^3 + 3*b^5)*log(b *tan(d*x + c) + a)/a^7 - 60*(a^4*b + 4*a^2*b^3 + 3*b^5)*log(tan(d*x + c))/ a^7)/d
Time = 0.19 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.07 \[ \int \frac {\csc ^6(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {2 \, {\left (a^{4} b + 4 \, a^{2} b^{3} + 3 \, b^{5}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{7} d} + \frac {2 \, {\left (a^{4} b^{2} + 4 \, a^{2} b^{4} + 3 \, b^{6}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{7} b d} + \frac {9 \, a^{5} b \tan \left (d x + c\right ) - 6 \, a^{6} - 60 \, {\left (a^{5} b + 4 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \tan \left (d x + c\right )^{5} - 30 \, {\left (a^{6} + 4 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \tan \left (d x + c\right )^{4} + 10 \, {\left (4 \, a^{5} b + 3 \, a^{3} b^{3}\right )} \tan \left (d x + c\right )^{3} - 5 \, {\left (4 \, a^{6} + 3 \, a^{4} b^{2}\right )} \tan \left (d x + c\right )^{2}}{30 \, {\left (b \tan \left (d x + c\right ) + a\right )} a^{7} d \tan \left (d x + c\right )^{5}} \] Input:
integrate(csc(d*x+c)^6/(a+b*tan(d*x+c))^2,x, algorithm="giac")
Output:
-2*(a^4*b + 4*a^2*b^3 + 3*b^5)*log(abs(tan(d*x + c)))/(a^7*d) + 2*(a^4*b^2 + 4*a^2*b^4 + 3*b^6)*log(abs(b*tan(d*x + c) + a))/(a^7*b*d) + 1/30*(9*a^5 *b*tan(d*x + c) - 6*a^6 - 60*(a^5*b + 4*a^3*b^3 + 3*a*b^5)*tan(d*x + c)^5 - 30*(a^6 + 4*a^4*b^2 + 3*a^2*b^4)*tan(d*x + c)^4 + 10*(4*a^5*b + 3*a^3*b^ 3)*tan(d*x + c)^3 - 5*(4*a^6 + 3*a^4*b^2)*tan(d*x + c)^2)/((b*tan(d*x + c) + a)*a^7*d*tan(d*x + c)^5)
Time = 2.17 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.08 \[ \int \frac {\csc ^6(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {4\,b\,\mathrm {atanh}\left (\frac {2\,b\,\left (a^2+3\,b^2\right )\,\left (a^2+b^2\right )\,\left (a+2\,b\,\mathrm {tan}\left (c+d\,x\right )\right )}{a\,\left (2\,a^4\,b+8\,a^2\,b^3+6\,b^5\right )}\right )\,\left (a^2+3\,b^2\right )\,\left (a^2+b^2\right )}{a^7\,d}-\frac {\frac {1}{5\,a}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (a^4+4\,a^2\,b^2+3\,b^4\right )}{a^5}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (4\,a^2+3\,b^2\right )}{6\,a^3}-\frac {3\,b\,\mathrm {tan}\left (c+d\,x\right )}{10\,a^2}+\frac {2\,b\,{\mathrm {tan}\left (c+d\,x\right )}^5\,\left (a^4+4\,a^2\,b^2+3\,b^4\right )}{a^6}-\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (4\,a^2+3\,b^2\right )}{3\,a^4}}{d\,\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^6+a\,{\mathrm {tan}\left (c+d\,x\right )}^5\right )} \] Input:
int(1/(sin(c + d*x)^6*(a + b*tan(c + d*x))^2),x)
Output:
(4*b*atanh((2*b*(a^2 + 3*b^2)*(a^2 + b^2)*(a + 2*b*tan(c + d*x)))/(a*(2*a^ 4*b + 6*b^5 + 8*a^2*b^3)))*(a^2 + 3*b^2)*(a^2 + b^2))/(a^7*d) - (1/(5*a) + (tan(c + d*x)^4*(a^4 + 3*b^4 + 4*a^2*b^2))/a^5 + (tan(c + d*x)^2*(4*a^2 + 3*b^2))/(6*a^3) - (3*b*tan(c + d*x))/(10*a^2) + (2*b*tan(c + d*x)^5*(a^4 + 3*b^4 + 4*a^2*b^2))/a^6 - (b*tan(c + d*x)^3*(4*a^2 + 3*b^2))/(3*a^4))/(d *(a*tan(c + d*x)^5 + b*tan(c + d*x)^6))
Time = 0.17 (sec) , antiderivative size = 772, normalized size of antiderivative = 3.53 \[ \int \frac {\csc ^6(c+d x)}{(a+b \tan (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int(csc(d*x+c)^6/(a+b*tan(d*x+c))^2,x)
Output:
(480*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)**5*a**5*b**2 + 1920*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - 2* tan((c + d*x)/2)*b - a)*sin(c + d*x)**5*a**3*b**4 + 1440*cos(c + d*x)*log( tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*sin(c + d*x)**5*a*b**6 - 480*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**5*a**5*b**2 - 1920*c os(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**5*a**3*b**4 - 1440*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**5*a*b**6 - 75*cos(c + d*x)*sin(c + d*x)**5*a**7 - 1163*cos(c + d*x)*sin(c + d*x)**5*a**5*b**2 - 2520*cos(c + d*x)*sin(c + d*x)**5*a**3*b**4 - 1440*cos(c + d*x)*sin(c + d*x)**5*a*b* *6 + 176*cos(c + d*x)*sin(c + d*x)**3*a**5*b**2 + 240*cos(c + d*x)*sin(c + d*x)**3*a**3*b**4 + 72*cos(c + d*x)*sin(c + d*x)*a**5*b**2 + 480*log(tan( (c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*sin(c + d*x)**6*a**4*b**3 + 1920*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*sin(c + d*x)**6 *a**2*b**5 + 1440*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*si n(c + d*x)**6*b**7 - 480*log(tan((c + d*x)/2))*sin(c + d*x)**6*a**4*b**3 - 1920*log(tan((c + d*x)/2))*sin(c + d*x)**6*a**2*b**5 - 1440*log(tan((c + d*x)/2))*sin(c + d*x)**6*b**7 + 53*sin(c + d*x)**6*a**6*b + 405*sin(c + d* x)**6*a**4*b**3 + 360*sin(c + d*x)**6*a**2*b**5 - 64*sin(c + d*x)**4*a**6* b - 720*sin(c + d*x)**4*a**4*b**3 - 720*sin(c + d*x)**4*a**2*b**5 - 16*sin (c + d*x)**2*a**6*b - 120*sin(c + d*x)**2*a**4*b**3 - 48*a**6*b)/(240*s...