Integrand size = 20, antiderivative size = 556 \[ \int (c+d x)^3 (a+b \coth (e+f x))^3 \, dx=-\frac {3 b^3 d (c+d x)^2}{2 f^2}-\frac {3 a b^2 (c+d x)^3}{f}+\frac {b^3 (c+d x)^3}{2 f}+\frac {a^3 (c+d x)^4}{4 d}-\frac {3 a^2 b (c+d x)^4}{4 d}+\frac {3 a b^2 (c+d x)^4}{4 d}-\frac {b^3 (c+d x)^4}{4 d}-\frac {3 b^3 d (c+d x)^2 \coth (e+f x)}{2 f^2}-\frac {3 a b^2 (c+d x)^3 \coth (e+f x)}{f}-\frac {b^3 (c+d x)^3 \coth ^2(e+f x)}{2 f}+\frac {3 b^3 d^2 (c+d x) \log \left (1-e^{2 (e+f x)}\right )}{f^3}+\frac {9 a b^2 d (c+d x)^2 \log \left (1-e^{2 (e+f x)}\right )}{f^2}+\frac {3 a^2 b (c+d x)^3 \log \left (1-e^{2 (e+f x)}\right )}{f}+\frac {b^3 (c+d x)^3 \log \left (1-e^{2 (e+f x)}\right )}{f}+\frac {3 b^3 d^3 \operatorname {PolyLog}\left (2,e^{2 (e+f x)}\right )}{2 f^4}+\frac {9 a b^2 d^2 (c+d x) \operatorname {PolyLog}\left (2,e^{2 (e+f x)}\right )}{f^3}+\frac {9 a^2 b d (c+d x)^2 \operatorname {PolyLog}\left (2,e^{2 (e+f x)}\right )}{2 f^2}+\frac {3 b^3 d (c+d x)^2 \operatorname {PolyLog}\left (2,e^{2 (e+f x)}\right )}{2 f^2}-\frac {9 a b^2 d^3 \operatorname {PolyLog}\left (3,e^{2 (e+f x)}\right )}{2 f^4}-\frac {9 a^2 b d^2 (c+d x) \operatorname {PolyLog}\left (3,e^{2 (e+f x)}\right )}{2 f^3}-\frac {3 b^3 d^2 (c+d x) \operatorname {PolyLog}\left (3,e^{2 (e+f x)}\right )}{2 f^3}+\frac {9 a^2 b d^3 \operatorname {PolyLog}\left (4,e^{2 (e+f x)}\right )}{4 f^4}+\frac {3 b^3 d^3 \operatorname {PolyLog}\left (4,e^{2 (e+f x)}\right )}{4 f^4} \]
-3/2*b^3*d*(d*x+c)^2/f^2-3*a*b^2*(d*x+c)^3/f+1/2*b^3*(d*x+c)^3/f+1/4*a^3*( d*x+c)^4/d-3/4*a^2*b*(d*x+c)^4/d+3/4*a*b^2*(d*x+c)^4/d-1/4*b^3*(d*x+c)^4/d -3/2*b^3*d*(d*x+c)^2*coth(f*x+e)/f^2-3*a*b^2*(d*x+c)^3*coth(f*x+e)/f-1/2*b ^3*(d*x+c)^3*coth(f*x+e)^2/f+3*b^3*d^2*(d*x+c)*ln(1-exp(2*f*x+2*e))/f^3+9* a*b^2*d*(d*x+c)^2*ln(1-exp(2*f*x+2*e))/f^2+3*a^2*b*(d*x+c)^3*ln(1-exp(2*f* x+2*e))/f+b^3*(d*x+c)^3*ln(1-exp(2*f*x+2*e))/f+3/2*b^3*d^3*polylog(2,exp(2 *f*x+2*e))/f^4+9*a*b^2*d^2*(d*x+c)*polylog(2,exp(2*f*x+2*e))/f^3+9/2*a^2*b *d*(d*x+c)^2*polylog(2,exp(2*f*x+2*e))/f^2+3/2*b^3*d*(d*x+c)^2*polylog(2,e xp(2*f*x+2*e))/f^2-9/2*a*b^2*d^3*polylog(3,exp(2*f*x+2*e))/f^4-9/2*a^2*b*d ^2*(d*x+c)*polylog(3,exp(2*f*x+2*e))/f^3-3/2*b^3*d^2*(d*x+c)*polylog(3,exp (2*f*x+2*e))/f^3+9/4*a^2*b*d^3*polylog(4,exp(2*f*x+2*e))/f^4+3/4*b^3*d^3*p olylog(4,exp(2*f*x+2*e))/f^4
Leaf count is larger than twice the leaf count of optimal. \(2043\) vs. \(2(556)=1112\).
Time = 8.91 (sec) , antiderivative size = 2043, normalized size of antiderivative = 3.67 \[ \int (c+d x)^3 (a+b \coth (e+f x))^3 \, dx=\text {Result too large to show} \]
((-(b^3*c^3) - 3*b^3*c^2*d*x - 3*b^3*c*d^2*x^2 - b^3*d^3*x^3)*Csch[e + f*x ]^2)/(2*f) - (b*E^(2*e)*(24*b^2*c*d^2*x + 72*a*b*c^2*d*f*x + 24*a^2*c^3*f^ 2*x + 8*b^2*c^3*f^2*x + 12*b^2*d^3*x^2 + 72*a*b*c*d^2*f*x^2 + 36*a^2*c^2*d *f^2*x^2 + 12*b^2*c^2*d*f^2*x^2 + 24*a*b*d^3*f*x^3 + 24*a^2*c*d^2*f^2*x^3 + 8*b^2*c*d^2*f^2*x^3 + 6*a^2*d^3*f^2*x^4 + 2*b^2*d^3*f^2*x^4 - 36*a*b*c^2 *d*Log[1 - E^(2*(e + f*x))] + (36*a*b*c^2*d*Log[1 - E^(2*(e + f*x))])/E^(2 *e) - (12*b^2*c*d^2*Log[1 - E^(2*(e + f*x))])/f + (12*b^2*c*d^2*Log[1 - E^ (2*(e + f*x))])/(E^(2*e)*f) - 12*a^2*c^3*f*Log[1 - E^(2*(e + f*x))] - 4*b^ 2*c^3*f*Log[1 - E^(2*(e + f*x))] + (12*a^2*c^3*f*Log[1 - E^(2*(e + f*x))]) /E^(2*e) + (4*b^2*c^3*f*Log[1 - E^(2*(e + f*x))])/E^(2*e) - 72*a*b*c*d^2*x *Log[1 - E^(2*(e + f*x))] + (72*a*b*c*d^2*x*Log[1 - E^(2*(e + f*x))])/E^(2 *e) - (12*b^2*d^3*x*Log[1 - E^(2*(e + f*x))])/f + (12*b^2*d^3*x*Log[1 - E^ (2*(e + f*x))])/(E^(2*e)*f) - 36*a^2*c^2*d*f*x*Log[1 - E^(2*(e + f*x))] - 12*b^2*c^2*d*f*x*Log[1 - E^(2*(e + f*x))] + (36*a^2*c^2*d*f*x*Log[1 - E^(2 *(e + f*x))])/E^(2*e) + (12*b^2*c^2*d*f*x*Log[1 - E^(2*(e + f*x))])/E^(2*e ) - 36*a*b*d^3*x^2*Log[1 - E^(2*(e + f*x))] + (36*a*b*d^3*x^2*Log[1 - E^(2 *(e + f*x))])/E^(2*e) - 36*a^2*c*d^2*f*x^2*Log[1 - E^(2*(e + f*x))] - 12*b ^2*c*d^2*f*x^2*Log[1 - E^(2*(e + f*x))] + (36*a^2*c*d^2*f*x^2*Log[1 - E^(2 *(e + f*x))])/E^(2*e) + (12*b^2*c*d^2*f*x^2*Log[1 - E^(2*(e + f*x))])/E^(2 *e) - 12*a^2*d^3*f*x^3*Log[1 - E^(2*(e + f*x))] - 4*b^2*d^3*f*x^3*Log[1...
Time = 1.43 (sec) , antiderivative size = 556, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3042, 4205, 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.
\(\displaystyle \int (c+d x)^3 (a+b \coth (e+f x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (c+d x)^3 \left (a-i b \tan \left (i e+i f x+\frac {\pi }{2}\right )\right )^3dx\) |
\(\Big \downarrow \) 4205 |
\(\displaystyle \int \left (a^3 (c+d x)^3+3 a^2 b (c+d x)^3 \coth (e+f x)+3 a b^2 (c+d x)^3 \coth ^2(e+f x)+b^3 (c+d x)^3 \coth ^3(e+f x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 (c+d x)^4}{4 d}-\frac {9 a^2 b d^2 (c+d x) \operatorname {PolyLog}\left (3,e^{2 (e+f x)}\right )}{2 f^3}+\frac {9 a^2 b d (c+d x)^2 \operatorname {PolyLog}\left (2,e^{2 (e+f x)}\right )}{2 f^2}+\frac {3 a^2 b (c+d x)^3 \log \left (1-e^{2 (e+f x)}\right )}{f}-\frac {3 a^2 b (c+d x)^4}{4 d}+\frac {9 a^2 b d^3 \operatorname {PolyLog}\left (4,e^{2 (e+f x)}\right )}{4 f^4}+\frac {9 a b^2 d^2 (c+d x) \operatorname {PolyLog}\left (2,e^{2 (e+f x)}\right )}{f^3}+\frac {9 a b^2 d (c+d x)^2 \log \left (1-e^{2 (e+f x)}\right )}{f^2}-\frac {3 a b^2 (c+d x)^3 \coth (e+f x)}{f}-\frac {3 a b^2 (c+d x)^3}{f}+\frac {3 a b^2 (c+d x)^4}{4 d}-\frac {9 a b^2 d^3 \operatorname {PolyLog}\left (3,e^{2 (e+f x)}\right )}{2 f^4}-\frac {3 b^3 d^2 (c+d x) \operatorname {PolyLog}\left (3,e^{2 (e+f x)}\right )}{2 f^3}+\frac {3 b^3 d^2 (c+d x) \log \left (1-e^{2 (e+f x)}\right )}{f^3}+\frac {3 b^3 d (c+d x)^2 \operatorname {PolyLog}\left (2,e^{2 (e+f x)}\right )}{2 f^2}-\frac {3 b^3 d (c+d x)^2 \coth (e+f x)}{2 f^2}+\frac {b^3 (c+d x)^3 \log \left (1-e^{2 (e+f x)}\right )}{f}-\frac {b^3 (c+d x)^3 \coth ^2(e+f x)}{2 f}-\frac {3 b^3 d (c+d x)^2}{2 f^2}+\frac {b^3 (c+d x)^3}{2 f}-\frac {b^3 (c+d x)^4}{4 d}+\frac {3 b^3 d^3 \operatorname {PolyLog}\left (2,e^{2 (e+f x)}\right )}{2 f^4}+\frac {3 b^3 d^3 \operatorname {PolyLog}\left (4,e^{2 (e+f x)}\right )}{4 f^4}\) |
(-3*b^3*d*(c + d*x)^2)/(2*f^2) - (3*a*b^2*(c + d*x)^3)/f + (b^3*(c + d*x)^ 3)/(2*f) + (a^3*(c + d*x)^4)/(4*d) - (3*a^2*b*(c + d*x)^4)/(4*d) + (3*a*b^ 2*(c + d*x)^4)/(4*d) - (b^3*(c + d*x)^4)/(4*d) - (3*b^3*d*(c + d*x)^2*Coth [e + f*x])/(2*f^2) - (3*a*b^2*(c + d*x)^3*Coth[e + f*x])/f - (b^3*(c + d*x )^3*Coth[e + f*x]^2)/(2*f) + (3*b^3*d^2*(c + d*x)*Log[1 - E^(2*(e + f*x))] )/f^3 + (9*a*b^2*d*(c + d*x)^2*Log[1 - E^(2*(e + f*x))])/f^2 + (3*a^2*b*(c + d*x)^3*Log[1 - E^(2*(e + f*x))])/f + (b^3*(c + d*x)^3*Log[1 - E^(2*(e + f*x))])/f + (3*b^3*d^3*PolyLog[2, E^(2*(e + f*x))])/(2*f^4) + (9*a*b^2*d^ 2*(c + d*x)*PolyLog[2, E^(2*(e + f*x))])/f^3 + (9*a^2*b*d*(c + d*x)^2*Poly Log[2, E^(2*(e + f*x))])/(2*f^2) + (3*b^3*d*(c + d*x)^2*PolyLog[2, E^(2*(e + f*x))])/(2*f^2) - (9*a*b^2*d^3*PolyLog[3, E^(2*(e + f*x))])/(2*f^4) - ( 9*a^2*b*d^2*(c + d*x)*PolyLog[3, E^(2*(e + f*x))])/(2*f^3) - (3*b^3*d^2*(c + d*x)*PolyLog[3, E^(2*(e + f*x))])/(2*f^3) + (9*a^2*b*d^3*PolyLog[4, E^( 2*(e + f*x))])/(4*f^4) + (3*b^3*d^3*PolyLog[4, E^(2*(e + f*x))])/(4*f^4)
3.1.47.3.1 Defintions of rubi rules used
Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.) , x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(2822\) vs. \(2(524)=1048\).
Time = 0.78 (sec) , antiderivative size = 2823, normalized size of antiderivative = 5.08
-3/f^4*b^3*e^2*d^3-3/f^2*b^3*d^3*x^2-3/2/f^4*b^3*e^4*d^3+3/f^4*b^3*d^3*pol ylog(2,exp(f*x+e))+3/f^4*b^3*d^3*polylog(2,-exp(f*x+e))+6/f^4*b^3*d^3*poly log(4,exp(f*x+e))+6/f^4*b^3*d^3*polylog(4,-exp(f*x+e))-18/f*b^2*a*c*d^2*x^ 2+12/f^3*b*e^3*d^2*c*a^2+6/f^2*b^3*d^2*c*polylog(2,exp(f*x+e))*x+3/f*b^3*d ^2*c*ln(1+exp(f*x+e))*x^2+6/f^2*b^3*d^2*c*polylog(2,-exp(f*x+e))*x+3/f*b*a ^2*d^3*ln(1-exp(f*x+e))*x^3+9/f^2*b*a^2*d^3*polylog(2,exp(f*x+e))*x^2-18/f ^3*b*a^2*d^3*polylog(3,exp(f*x+e))*x+3/f*b*a^2*d^3*ln(1+exp(f*x+e))*x^3+9/ f^2*b*a^2*d^3*polylog(2,-exp(f*x+e))*x^2-18/f^3*b*a^2*d^3*polylog(3,-exp(f *x+e))*x+9/f^4*b^2*e^2*a*d^3*ln(exp(f*x+e)-1)-6/f^3*b^3*d^3*e*x+12/f^4*b^2 *e^3*a*d^3-6/f*b^2*a*d^3*x^3-2/f^3*b^3*e^3*d^3*x+3*a^2*b*c^3*x+3*a*b^2*c^3 *x+1/4*d^3*a^3*x^4-1/4*d^3*b^3*x^4+1/4/d*c^4*a^3+1/4/d*c^4*b^3-18/f^4*b^2* e^2*a*d^3*ln(exp(f*x+e))-3/f^3*b^3*e^2*d^2*c*ln(1-exp(f*x+e))-9/f^4*b^2*e^ 2*a*d^3*ln(1-exp(f*x+e))-3/f^4*b*e^3*a^2*d^3*ln(exp(f*x+e)-1)+6/f^4*b*e^3* a^2*d^3*ln(exp(f*x+e))+18/f^3*b^2*a*c*d^2*polylog(2,exp(f*x+e))+18/f^3*b^2 *a*c*d^2*polylog(2,-exp(f*x+e))+9/f^2*b^2*a*d^3*ln(1-exp(f*x+e))*x^2+18/f^ 3*b^2*a*d^3*polylog(2,exp(f*x+e))*x+9/f^2*b^2*a*d^3*ln(1+exp(f*x+e))*x^2+1 8/f^3*b^2*a*d^3*polylog(2,-exp(f*x+e))*x+3/f*b^3*c^2*d*ln(1-exp(f*x+e))*x+ 3/f^2*b^3*c^2*d*ln(1-exp(f*x+e))*e+3/f*b^3*c^2*d*ln(1+exp(f*x+e))*x+3/f^4* b*e^3*a^2*d^3*ln(1-exp(f*x+e))+9/f^2*b*d*c^2*a^2*polylog(2,exp(f*x+e))+9/f ^2*b*d*c^2*a^2*polylog(2,-exp(f*x+e))+3/f^3*b^3*e^2*d^2*c*ln(exp(f*x+e)...
Leaf count of result is larger than twice the leaf count of optimal. 11137 vs. \(2 (520) = 1040\).
Time = 0.39 (sec) , antiderivative size = 11137, normalized size of antiderivative = 20.03 \[ \int (c+d x)^3 (a+b \coth (e+f x))^3 \, dx=\text {Too large to display} \]
\[ \int (c+d x)^3 (a+b \coth (e+f x))^3 \, dx=\int \left (a + b \coth {\left (e + f x \right )}\right )^{3} \left (c + d x\right )^{3}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 1531 vs. \(2 (520) = 1040\).
Time = 0.30 (sec) , antiderivative size = 1531, normalized size of antiderivative = 2.75 \[ \int (c+d x)^3 (a+b \coth (e+f x))^3 \, dx=\text {Too large to display} \]
1/4*a^3*d^3*x^4 + a^3*c*d^2*x^3 + 3/2*a^3*c^2*d*x^2 + a^3*c^3*x + 3*a^2*b* c^3*log(sinh(f*x + e))/f + 1/4*(24*a*b^2*c^3*f + 12*b^3*c^2*d + (3*a^2*b*d ^3*f^2 + 3*a*b^2*d^3*f^2 + b^3*d^3*f^2)*x^4 + 4*(3*a^2*b*c*d^2*f^2 + b^3*c *d^2*f^2 + 3*(c*d^2*f^2 + 2*d^3*f)*a*b^2)*x^3 + 6*(3*a^2*b*c^2*d*f^2 + 3*( c^2*d*f^2 + 4*c*d^2*f)*a*b^2 + (c^2*d*f^2 + 2*d^3)*b^3)*x^2 + 4*(3*(c^3*f^ 2 + 6*c^2*d*f)*a*b^2 + (c^3*f^2 + 6*c*d^2)*b^3)*x + ((3*a^2*b*d^3*f^2*e^(4 *e) + 3*a*b^2*d^3*f^2*e^(4*e) + b^3*d^3*f^2*e^(4*e))*x^4 + 4*(3*a^2*b*c*d^ 2*f^2*e^(4*e) + 3*a*b^2*c*d^2*f^2*e^(4*e) + b^3*c*d^2*f^2*e^(4*e))*x^3 + 6 *(3*a^2*b*c^2*d*f^2*e^(4*e) + 3*a*b^2*c^2*d*f^2*e^(4*e) + b^3*c^2*d*f^2*e^ (4*e))*x^2 + 4*(3*a*b^2*c^3*f^2*e^(4*e) + b^3*c^3*f^2*e^(4*e))*x)*e^(4*f*x ) - 2*(12*a*b^2*c^3*f*e^(2*e) + (3*a^2*b*d^3*f^2*e^(2*e) + 3*a*b^2*d^3*f^2 *e^(2*e) + b^3*d^3*f^2*e^(2*e))*x^4 + 2*(2*c^3*f*e^(2*e) + 3*c^2*d*e^(2*e) )*b^3 + 4*(3*a^2*b*c*d^2*f^2*e^(2*e) + 3*(c*d^2*f^2*e^(2*e) + d^3*f*e^(2*e ))*a*b^2 + (c*d^2*f^2*e^(2*e) + d^3*f*e^(2*e))*b^3)*x^3 + 6*(3*a^2*b*c^2*d *f^2*e^(2*e) + 3*(c^2*d*f^2*e^(2*e) + 2*c*d^2*f*e^(2*e))*a*b^2 + (c^2*d*f^ 2*e^(2*e) + 2*c*d^2*f*e^(2*e) + d^3*e^(2*e))*b^3)*x^2 + 4*(3*(c^3*f^2*e^(2 *e) + 3*c^2*d*f*e^(2*e))*a*b^2 + (c^3*f^2*e^(2*e) + 3*c^2*d*f*e^(2*e) + 3* c*d^2*e^(2*e))*b^3)*x)*e^(2*f*x))/(f^2*e^(4*f*x + 4*e) - 2*f^2*e^(2*f*x + 2*e) + f^2) - 2*(9*a*b^2*c^2*d*f + (c^3*f^2 + 3*c*d^2)*b^3)*x/f^2 + (9*a*b ^2*c^2*d*f + (c^3*f^2 + 3*c*d^2)*b^3)*log(e^(f*x + e) + 1)/f^3 + (9*a*b...
\[ \int (c+d x)^3 (a+b \coth (e+f x))^3 \, dx=\int { {\left (d x + c\right )}^{3} {\left (b \coth \left (f x + e\right ) + a\right )}^{3} \,d x } \]
Timed out. \[ \int (c+d x)^3 (a+b \coth (e+f x))^3 \, dx=\int {\left (a+b\,\mathrm {coth}\left (e+f\,x\right )\right )}^3\,{\left (c+d\,x\right )}^3 \,d x \]