Integrand size = 24, antiderivative size = 486 \[ \int (c+d x)^3 \csc ^2(a+b x) \sec ^3(a+b x) \, dx=-\frac {6 i d^2 (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^3}-\frac {3 i (c+d x)^3 \arctan \left (e^{i (a+b x)}\right )}{b}-\frac {6 d (c+d x)^2 \text {arctanh}\left (e^{i (a+b x)}\right )}{b^2}-\frac {3 (c+d x)^3 \csc (a+b x)}{2 b}+\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^3}+\frac {3 i d^3 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^4}+\frac {9 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{2 b^2}-\frac {3 i d^3 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^4}-\frac {9 i d (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{2 b^2}-\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^3}-\frac {6 d^3 \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^4}-\frac {9 d^2 (c+d x) \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^3}+\frac {9 d^2 (c+d x) \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^3}+\frac {6 d^3 \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )}{b^4}-\frac {9 i d^3 \operatorname {PolyLog}\left (4,-i e^{i (a+b x)}\right )}{b^4}+\frac {9 i d^3 \operatorname {PolyLog}\left (4,i e^{i (a+b x)}\right )}{b^4}-\frac {3 d (c+d x)^2 \sec (a+b x)}{2 b^2}+\frac {(c+d x)^3 \csc (a+b x) \sec ^2(a+b x)}{2 b} \]
3*I*d^3*polylog(2,-I*exp(I*(b*x+a)))/b^4-3*I*(d*x+c)^3*arctan(exp(I*(b*x+a )))/b-6*d*(d*x+c)^2*arctanh(exp(I*(b*x+a)))/b^2-3/2*(d*x+c)^3*csc(b*x+a)/b -9*I*d^3*polylog(4,-I*exp(I*(b*x+a)))/b^4+9*I*d^3*polylog(4,I*exp(I*(b*x+a )))/b^4-6*I*d^2*(d*x+c)*arctan(exp(I*(b*x+a)))/b^3-3*I*d^3*polylog(2,I*exp (I*(b*x+a)))/b^4-9/2*I*d*(d*x+c)^2*polylog(2,I*exp(I*(b*x+a)))/b^2+9/2*I*d *(d*x+c)^2*polylog(2,-I*exp(I*(b*x+a)))/b^2-6*d^3*polylog(3,-exp(I*(b*x+a) ))/b^4-9*d^2*(d*x+c)*polylog(3,-I*exp(I*(b*x+a)))/b^3+9*d^2*(d*x+c)*polylo g(3,I*exp(I*(b*x+a)))/b^3+6*d^3*polylog(3,exp(I*(b*x+a)))/b^4+6*I*d^2*(d*x +c)*polylog(2,-exp(I*(b*x+a)))/b^3-6*I*d^2*(d*x+c)*polylog(2,exp(I*(b*x+a) ))/b^3-3/2*d*(d*x+c)^2*sec(b*x+a)/b^2+1/2*(d*x+c)^3*csc(b*x+a)*sec(b*x+a)^ 2/b
Time = 7.51 (sec) , antiderivative size = 819, normalized size of antiderivative = 1.69 \[ \int (c+d x)^3 \csc ^2(a+b x) \sec ^3(a+b x) \, dx=\frac {3 d \left ((c+d x)^2 \log \left (1-e^{i (a+b x)}\right )-(c+d x)^2 \log \left (1+e^{i (a+b x)}\right )+\frac {2 i d \left (b (c+d x) \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )+i d \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )\right )}{b^2}+\frac {2 d \left (-i b (c+d x) \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )+d \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )\right )}{b^2}\right )}{b^2}-\frac {3 \left (2 i b^3 c^3 \arctan \left (e^{i (a+b x)}\right )+4 i b c d^2 \arctan \left (e^{i (a+b x)}\right )-3 b^3 c^2 d x \log \left (1-i e^{i (a+b x)}\right )-2 b d^3 x \log \left (1-i e^{i (a+b x)}\right )-3 b^3 c d^2 x^2 \log \left (1-i e^{i (a+b x)}\right )-b^3 d^3 x^3 \log \left (1-i e^{i (a+b x)}\right )+3 b^3 c^2 d x \log \left (1+i e^{i (a+b x)}\right )+2 b d^3 x \log \left (1+i e^{i (a+b x)}\right )+3 b^3 c d^2 x^2 \log \left (1+i e^{i (a+b x)}\right )+b^3 d^3 x^3 \log \left (1+i e^{i (a+b x)}\right )-i d \left (2 d^2+3 b^2 (c+d x)^2\right ) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )+i d \left (2 d^2+3 b^2 (c+d x)^2\right ) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )+6 b c d^2 \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )+6 b d^3 x \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )-6 b c d^2 \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )-6 b d^3 x \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )+6 i d^3 \operatorname {PolyLog}\left (4,-i e^{i (a+b x)}\right )-6 i d^3 \operatorname {PolyLog}\left (4,i e^{i (a+b x)}\right )\right )}{2 b^4}-\frac {\csc (a+b x) \sec ^2(a+b x) \left (b c^3+3 b c^2 d x+3 b c d^2 x^2+b d^3 x^3+3 b c^3 \cos (2 a+2 b x)+9 b c^2 d x \cos (2 a+2 b x)+9 b c d^2 x^2 \cos (2 a+2 b x)+3 b d^3 x^3 \cos (2 a+2 b x)+3 c^2 d \sin (2 a+2 b x)+6 c d^2 x \sin (2 a+2 b x)+3 d^3 x^2 \sin (2 a+2 b x)\right )}{4 b^2} \]
(3*d*((c + d*x)^2*Log[1 - E^(I*(a + b*x))] - (c + d*x)^2*Log[1 + E^(I*(a + b*x))] + ((2*I)*d*(b*(c + d*x)*PolyLog[2, -E^(I*(a + b*x))] + I*d*PolyLog [3, -E^(I*(a + b*x))]))/b^2 + (2*d*((-I)*b*(c + d*x)*PolyLog[2, E^(I*(a + b*x))] + d*PolyLog[3, E^(I*(a + b*x))]))/b^2))/b^2 - (3*((2*I)*b^3*c^3*Arc Tan[E^(I*(a + b*x))] + (4*I)*b*c*d^2*ArcTan[E^(I*(a + b*x))] - 3*b^3*c^2*d *x*Log[1 - I*E^(I*(a + b*x))] - 2*b*d^3*x*Log[1 - I*E^(I*(a + b*x))] - 3*b ^3*c*d^2*x^2*Log[1 - I*E^(I*(a + b*x))] - b^3*d^3*x^3*Log[1 - I*E^(I*(a + b*x))] + 3*b^3*c^2*d*x*Log[1 + I*E^(I*(a + b*x))] + 2*b*d^3*x*Log[1 + I*E^ (I*(a + b*x))] + 3*b^3*c*d^2*x^2*Log[1 + I*E^(I*(a + b*x))] + b^3*d^3*x^3* Log[1 + I*E^(I*(a + b*x))] - I*d*(2*d^2 + 3*b^2*(c + d*x)^2)*PolyLog[2, (- I)*E^(I*(a + b*x))] + I*d*(2*d^2 + 3*b^2*(c + d*x)^2)*PolyLog[2, I*E^(I*(a + b*x))] + 6*b*c*d^2*PolyLog[3, (-I)*E^(I*(a + b*x))] + 6*b*d^3*x*PolyLog [3, (-I)*E^(I*(a + b*x))] - 6*b*c*d^2*PolyLog[3, I*E^(I*(a + b*x))] - 6*b* d^3*x*PolyLog[3, I*E^(I*(a + b*x))] + (6*I)*d^3*PolyLog[4, (-I)*E^(I*(a + b*x))] - (6*I)*d^3*PolyLog[4, I*E^(I*(a + b*x))]))/(2*b^4) - (Csc[a + b*x] *Sec[a + b*x]^2*(b*c^3 + 3*b*c^2*d*x + 3*b*c*d^2*x^2 + b*d^3*x^3 + 3*b*c^3 *Cos[2*a + 2*b*x] + 9*b*c^2*d*x*Cos[2*a + 2*b*x] + 9*b*c*d^2*x^2*Cos[2*a + 2*b*x] + 3*b*d^3*x^3*Cos[2*a + 2*b*x] + 3*c^2*d*Sin[2*a + 2*b*x] + 6*c*d^ 2*x*Sin[2*a + 2*b*x] + 3*d^3*x^2*Sin[2*a + 2*b*x]))/(4*b^2)
Time = 1.53 (sec) , antiderivative size = 513, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4920, 27, 7292, 27, 7293, 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.
\(\displaystyle \int (c+d x)^3 \csc ^2(a+b x) \sec ^3(a+b x) \, dx\) |
\(\Big \downarrow \) 4920 |
\(\displaystyle -3 d \int \frac {1}{2} (c+d x)^2 \left (\frac {\csc (a+b x) \sec ^2(a+b x)}{b}+\frac {3 \text {arctanh}(\sin (a+b x))}{b}-\frac {3 \csc (a+b x)}{b}\right )dx+\frac {3 (c+d x)^3 \text {arctanh}(\sin (a+b x))}{2 b}-\frac {3 (c+d x)^3 \csc (a+b x)}{2 b}+\frac {(c+d x)^3 \csc (a+b x) \sec ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {3}{2} d \int (c+d x)^2 \left (\frac {\csc (a+b x) \sec ^2(a+b x)}{b}+\frac {3 \text {arctanh}(\sin (a+b x))}{b}-\frac {3 \csc (a+b x)}{b}\right )dx+\frac {3 (c+d x)^3 \text {arctanh}(\sin (a+b x))}{2 b}-\frac {3 (c+d x)^3 \csc (a+b x)}{2 b}+\frac {(c+d x)^3 \csc (a+b x) \sec ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -\frac {3}{2} d \int \frac {(c+d x)^2 \left (\csc (a+b x) \sec ^2(a+b x)+3 \text {arctanh}(\sin (a+b x))-3 \csc (a+b x)\right )}{b}dx+\frac {3 (c+d x)^3 \text {arctanh}(\sin (a+b x))}{2 b}-\frac {3 (c+d x)^3 \csc (a+b x)}{2 b}+\frac {(c+d x)^3 \csc (a+b x) \sec ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {3 d \int (c+d x)^2 \left (\csc (a+b x) \sec ^2(a+b x)+3 \text {arctanh}(\sin (a+b x))-3 \csc (a+b x)\right )dx}{2 b}+\frac {3 (c+d x)^3 \text {arctanh}(\sin (a+b x))}{2 b}-\frac {3 (c+d x)^3 \csc (a+b x)}{2 b}+\frac {(c+d x)^3 \csc (a+b x) \sec ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {3 d \int \left (\csc (a+b x) \sec ^2(a+b x) (c+d x)^2+3 (\text {arctanh}(\sin (a+b x))-\csc (a+b x)) (c+d x)^2\right )dx}{2 b}+\frac {3 (c+d x)^3 \text {arctanh}(\sin (a+b x))}{2 b}-\frac {3 (c+d x)^3 \csc (a+b x)}{2 b}+\frac {(c+d x)^3 \csc (a+b x) \sec ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3 d \left (\frac {4 i d (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^2}+\frac {2 i (c+d x)^3 \arctan \left (e^{i (a+b x)}\right )}{d}+\frac {4 (c+d x)^2 \text {arctanh}\left (e^{i (a+b x)}\right )}{b}+\frac {(c+d x)^3 \text {arctanh}(\sin (a+b x))}{d}-\frac {2 i d^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}+\frac {2 i d^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}+\frac {4 d^2 \operatorname {PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^3}-\frac {4 d^2 \operatorname {PolyLog}\left (3,e^{i (a+b x)}\right )}{b^3}+\frac {6 i d^2 \operatorname {PolyLog}\left (4,-i e^{i (a+b x)}\right )}{b^3}-\frac {6 i d^2 \operatorname {PolyLog}\left (4,i e^{i (a+b x)}\right )}{b^3}-\frac {4 i d (c+d x) \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}+\frac {4 i d (c+d x) \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}+\frac {6 d (c+d x) \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^2}-\frac {6 d (c+d x) \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^2}-\frac {3 i (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b}+\frac {3 i (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b}+\frac {(c+d x)^2 \sec (a+b x)}{b}\right )}{2 b}+\frac {3 (c+d x)^3 \text {arctanh}(\sin (a+b x))}{2 b}-\frac {3 (c+d x)^3 \csc (a+b x)}{2 b}+\frac {(c+d x)^3 \csc (a+b x) \sec ^2(a+b x)}{2 b}\) |
(3*(c + d*x)^3*ArcTanh[Sin[a + b*x]])/(2*b) - (3*(c + d*x)^3*Csc[a + b*x]) /(2*b) + ((c + d*x)^3*Csc[a + b*x]*Sec[a + b*x]^2)/(2*b) - (3*d*(((4*I)*d* (c + d*x)*ArcTan[E^(I*(a + b*x))])/b^2 + ((2*I)*(c + d*x)^3*ArcTan[E^(I*(a + b*x))])/d + (4*(c + d*x)^2*ArcTanh[E^(I*(a + b*x))])/b + ((c + d*x)^3*A rcTanh[Sin[a + b*x]])/d - ((4*I)*d*(c + d*x)*PolyLog[2, -E^(I*(a + b*x))]) /b^2 - ((2*I)*d^2*PolyLog[2, (-I)*E^(I*(a + b*x))])/b^3 - ((3*I)*(c + d*x) ^2*PolyLog[2, (-I)*E^(I*(a + b*x))])/b + ((2*I)*d^2*PolyLog[2, I*E^(I*(a + b*x))])/b^3 + ((3*I)*(c + d*x)^2*PolyLog[2, I*E^(I*(a + b*x))])/b + ((4*I )*d*(c + d*x)*PolyLog[2, E^(I*(a + b*x))])/b^2 + (4*d^2*PolyLog[3, -E^(I*( a + b*x))])/b^3 + (6*d*(c + d*x)*PolyLog[3, (-I)*E^(I*(a + b*x))])/b^2 - ( 6*d*(c + d*x)*PolyLog[3, I*E^(I*(a + b*x))])/b^2 - (4*d^2*PolyLog[3, E^(I* (a + b*x))])/b^3 + ((6*I)*d^2*PolyLog[4, (-I)*E^(I*(a + b*x))])/b^3 - ((6* I)*d^2*PolyLog[4, I*E^(I*(a + b*x))])/b^3 + ((c + d*x)^2*Sec[a + b*x])/b)) /(2*b)
3.4.17.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b _.)*(x_)]^(p_.), x_Symbol] :> Module[{u = IntHide[Csc[a + b*x]^n*Sec[a + b* x]^p, x]}, Simp[(c + d*x)^m u, x] - Simp[d*m Int[(c + d*x)^(m - 1)*u, x ], x]] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p] && GtQ[m, 0] && NeQ[n, p]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1628 vs. \(2 (428 ) = 856\).
Time = 1.36 (sec) , antiderivative size = 1629, normalized size of antiderivative = 3.35
-I/b^2/(exp(2*I*(b*x+a))+1)^2/(exp(2*I*(b*x+a))-1)*(3*d^3*x^3*b*exp(5*I*(b *x+a))+9*c*d^2*x^2*b*exp(5*I*(b*x+a))+9*c^2*d*x*b*exp(5*I*(b*x+a))+2*d^3*x ^3*b*exp(3*I*(b*x+a))+3*c^3*b*exp(5*I*(b*x+a))+6*c*d^2*x^2*b*exp(3*I*(b*x+ a))-3*I*d^3*x^2*exp(5*I*(b*x+a))+6*c^2*d*x*b*exp(3*I*(b*x+a))+3*d^3*x^3*b* exp(I*(b*x+a))+3*I*c^2*d*exp(I*(b*x+a))+2*c^3*b*exp(3*I*(b*x+a))+9*c*d^2*x ^2*b*exp(I*(b*x+a))+3*I*d^3*x^2*exp(I*(b*x+a))+9*c^2*d*x*b*exp(I*(b*x+a))+ 3*c^3*b*exp(I*(b*x+a))-6*I*c*d^2*x*exp(5*I*(b*x+a))+6*I*c*d^2*x*exp(I*(b*x +a))-3*I*c^2*d*exp(5*I*(b*x+a)))-9*I/b^3*a^2*c*d^2*arctan(exp(I*(b*x+a)))- 3*I*d^3*polylog(2,I*exp(I*(b*x+a)))/b^4-9*I*d^3*polylog(4,-I*exp(I*(b*x+a) ))/b^4+9/b^3*c*d^2*polylog(3,I*exp(I*(b*x+a)))+3/b^3*d^3*ln(1-I*exp(I*(b*x +a)))*x+3/b^4*d^3*ln(1-I*exp(I*(b*x+a)))*a+3/b^2*d^3*ln(1-exp(I*(b*x+a)))* x^2+3/b^4*d^3*ln(1-exp(I*(b*x+a)))*a^2-3/2/b*d^3*ln(1+I*exp(I*(b*x+a)))*x^ 3-3/b^2*d^3*ln(exp(I*(b*x+a))+1)*x^2+3/2/b*d^3*ln(1-I*exp(I*(b*x+a)))*x^3+ 9/b^3*d^3*polylog(3,I*exp(I*(b*x+a)))*x+3/b^4*d^3*a^2*ln(exp(I*(b*x+a))-1) -3/b^2*c^2*d*ln(exp(I*(b*x+a))+1)+3/b^2*c^2*d*ln(exp(I*(b*x+a))-1)+3*I*d^3 *polylog(2,-I*exp(I*(b*x+a)))/b^4+9*I*d^3*polylog(4,I*exp(I*(b*x+a)))/b^4- 9*I/b^2*c*d^2*polylog(2,I*exp(I*(b*x+a)))*x+9*I/b^2*c*d^2*polylog(2,-I*exp (I*(b*x+a)))*x+9*I/b^2*a*c^2*d*arctan(exp(I*(b*x+a)))-3/b^3*d^3*ln(1+I*exp (I*(b*x+a)))*x-3/b^4*d^3*ln(1+I*exp(I*(b*x+a)))*a-3/2/b^4*a^3*d^3*ln(1+I*e xp(I*(b*x+a)))-9/b^3*c*d^2*polylog(3,-I*exp(I*(b*x+a)))+3/2/b^4*a^3*d^3...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2226 vs. \(2 (404) = 808\).
Time = 0.43 (sec) , antiderivative size = 2226, normalized size of antiderivative = 4.58 \[ \int (c+d x)^3 \csc ^2(a+b x) \sec ^3(a+b x) \, dx=\text {Too large to display} \]
1/4*(2*b^3*d^3*x^3 + 6*b^3*c*d^2*x^2 + 6*b^3*c^2*d*x + 18*I*d^3*cos(b*x + a)^2*polylog(4, I*cos(b*x + a) + sin(b*x + a))*sin(b*x + a) + 18*I*d^3*cos (b*x + a)^2*polylog(4, I*cos(b*x + a) - sin(b*x + a))*sin(b*x + a) - 18*I* d^3*cos(b*x + a)^2*polylog(4, -I*cos(b*x + a) + sin(b*x + a))*sin(b*x + a) - 18*I*d^3*cos(b*x + a)^2*polylog(4, -I*cos(b*x + a) - sin(b*x + a))*sin( b*x + a) + 12*d^3*cos(b*x + a)^2*polylog(3, cos(b*x + a) + I*sin(b*x + a)) *sin(b*x + a) + 12*d^3*cos(b*x + a)^2*polylog(3, cos(b*x + a) - I*sin(b*x + a))*sin(b*x + a) - 12*d^3*cos(b*x + a)^2*polylog(3, -cos(b*x + a) + I*si n(b*x + a))*sin(b*x + a) - 12*d^3*cos(b*x + a)^2*polylog(3, -cos(b*x + a) - I*sin(b*x + a))*sin(b*x + a) + 2*b^3*c^3 - 12*(I*b*d^3*x + I*b*c*d^2)*co s(b*x + a)^2*dilog(cos(b*x + a) + I*sin(b*x + a))*sin(b*x + a) - 12*(-I*b* d^3*x - I*b*c*d^2)*cos(b*x + a)^2*dilog(cos(b*x + a) - I*sin(b*x + a))*sin (b*x + a) - 3*(3*I*b^2*d^3*x^2 + 6*I*b^2*c*d^2*x + 3*I*b^2*c^2*d + 2*I*d^3 )*cos(b*x + a)^2*dilog(I*cos(b*x + a) + sin(b*x + a))*sin(b*x + a) - 3*(3* I*b^2*d^3*x^2 + 6*I*b^2*c*d^2*x + 3*I*b^2*c^2*d + 2*I*d^3)*cos(b*x + a)^2* dilog(I*cos(b*x + a) - sin(b*x + a))*sin(b*x + a) - 3*(-3*I*b^2*d^3*x^2 - 6*I*b^2*c*d^2*x - 3*I*b^2*c^2*d - 2*I*d^3)*cos(b*x + a)^2*dilog(-I*cos(b*x + a) + sin(b*x + a))*sin(b*x + a) - 3*(-3*I*b^2*d^3*x^2 - 6*I*b^2*c*d^2*x - 3*I*b^2*c^2*d - 2*I*d^3)*cos(b*x + a)^2*dilog(-I*cos(b*x + a) - sin(b*x + a))*sin(b*x + a) - 12*(I*b*d^3*x + I*b*c*d^2)*cos(b*x + a)^2*dilog(-...
Timed out. \[ \int (c+d x)^3 \csc ^2(a+b x) \sec ^3(a+b x) \, dx=\text {Timed out} \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 8046 vs. \(2 (404) = 808\).
Time = 3.83 (sec) , antiderivative size = 8046, normalized size of antiderivative = 16.56 \[ \int (c+d x)^3 \csc ^2(a+b x) \sec ^3(a+b x) \, dx=\text {Too large to display} \]
-1/4*(c^3*(2*(3*sin(b*x + a)^2 - 2)/(sin(b*x + a)^3 - sin(b*x + a)) - 3*lo g(sin(b*x + a) + 1) + 3*log(sin(b*x + a) - 1)) - 3*a*c^2*d*(2*(3*sin(b*x + a)^2 - 2)/(sin(b*x + a)^3 - sin(b*x + a)) - 3*log(sin(b*x + a) + 1) + 3*l og(sin(b*x + a) - 1))/b + 3*a^2*c*d^2*(2*(3*sin(b*x + a)^2 - 2)/(sin(b*x + a)^3 - sin(b*x + a)) - 3*log(sin(b*x + a) + 1) + 3*log(sin(b*x + a) - 1)) /b^2 - a^3*d^3*(2*(3*sin(b*x + a)^2 - 2)/(sin(b*x + a)^3 - sin(b*x + a)) - 3*log(sin(b*x + a) + 1) + 3*log(sin(b*x + a) - 1))/b^3 - 4*(6*((b*x + a)^ 3*d^3 + 2*b*c*d^2 - 2*a*d^3 + 3*(b*c*d^2 - a*d^3)*(b*x + a)^2 + (3*b^2*c^2 *d - 6*a*b*c*d^2 + (3*a^2 + 2)*d^3)*(b*x + a) - ((b*x + a)^3*d^3 + 2*b*c*d ^2 - 2*a*d^3 + 3*(b*c*d^2 - a*d^3)*(b*x + a)^2 + (3*b^2*c^2*d - 6*a*b*c*d^ 2 + (3*a^2 + 2)*d^3)*(b*x + a))*cos(6*b*x + 6*a) - ((b*x + a)^3*d^3 + 2*b* c*d^2 - 2*a*d^3 + 3*(b*c*d^2 - a*d^3)*(b*x + a)^2 + (3*b^2*c^2*d - 6*a*b*c *d^2 + (3*a^2 + 2)*d^3)*(b*x + a))*cos(4*b*x + 4*a) + ((b*x + a)^3*d^3 + 2 *b*c*d^2 - 2*a*d^3 + 3*(b*c*d^2 - a*d^3)*(b*x + a)^2 + (3*b^2*c^2*d - 6*a* b*c*d^2 + (3*a^2 + 2)*d^3)*(b*x + a))*cos(2*b*x + 2*a) - (I*(b*x + a)^3*d^ 3 + 2*I*b*c*d^2 - 2*I*a*d^3 + 3*(I*b*c*d^2 - I*a*d^3)*(b*x + a)^2 + (3*I*b ^2*c^2*d - 6*I*a*b*c*d^2 + (3*I*a^2 + 2*I)*d^3)*(b*x + a))*sin(6*b*x + 6*a ) - (I*(b*x + a)^3*d^3 + 2*I*b*c*d^2 - 2*I*a*d^3 + 3*(I*b*c*d^2 - I*a*d^3) *(b*x + a)^2 + (3*I*b^2*c^2*d - 6*I*a*b*c*d^2 + (3*I*a^2 + 2*I)*d^3)*(b*x + a))*sin(4*b*x + 4*a) - (-I*(b*x + a)^3*d^3 - 2*I*b*c*d^2 + 2*I*a*d^3 ...
\[ \int (c+d x)^3 \csc ^2(a+b x) \sec ^3(a+b x) \, dx=\int { {\left (d x + c\right )}^{3} \csc \left (b x + a\right )^{2} \sec \left (b x + a\right )^{3} \,d x } \]
Timed out. \[ \int (c+d x)^3 \csc ^2(a+b x) \sec ^3(a+b x) \, dx=\text {Hanged} \]