\(\int \frac {\tanh ^3(x)}{i+\text {csch}(x)} \, dx\) [104]

Optimal result

Integrand size = 13, antiderivative size = 77 \[ \int \frac {\tanh ^3(x)}{i+\text {csch}(x)} \, dx=-\frac {11}{16} i \log (i-\sinh (x))-\frac {5}{16} i \log (i+\sinh (x))-\frac {i}{8 (1-i \sinh (x))}+\frac {i}{8 (1+i \sinh (x))^2}-\frac {3 i}{4 (1+i \sinh (x))} \]

[Out]

-11/16*I*ln(I-sinh(x))-5/16*I*ln(I+sinh(x))-1/8*I/(1-I*sinh(x))+1/8*I/(1+I*sinh(x))^2-3/4*I/(1+I*sinh(x))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3964, 90} \[ \int \frac {\tanh ^3(x)}{i+\text {csch}(x)} \, dx=-\frac {i}{8 (1-i \sinh (x))}-\frac {3 i}{4 (1+i \sinh (x))}+\frac {i}{8 (1+i \sinh (x))^2}-\frac {11}{16} i \log (-\sinh (x)+i)-\frac {5}{16} i \log (\sinh (x)+i) \]

[In]

Int[Tanh[x]^3/(I + Csch[x]),x]

[Out]

((-11*I)/16)*Log[I - Sinh[x]] - ((5*I)/16)*Log[I + Sinh[x]] - (I/8)/(1 - I*Sinh[x]) + (I/8)/(1 + I*Sinh[x])^2
- ((3*I)/4)/(1 + I*Sinh[x])

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 3964

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[1/(a^(m - n
- 1)*b^n*d), Subst[Int[(a - b*x)^((m - 1)/2)*((a + b*x)^((m - 1)/2 + n)/x^(m + n)), x], x, Sin[c + d*x]], x] /
; FreeQ[{a, b, c, d}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {x^4}{(i-i x)^2 (i+i x)^3} \, dx,x,i \sinh (x)\right ) \\ & = \text {Subst}\left (\int \left (-\frac {i}{8 (-1+x)^2}-\frac {5 i}{16 (-1+x)}-\frac {i}{4 (1+x)^3}+\frac {3 i}{4 (1+x)^2}-\frac {11 i}{16 (1+x)}\right ) \, dx,x,i \sinh (x)\right ) \\ & = -\frac {11}{16} i \log (i-\sinh (x))-\frac {5}{16} i \log (i+\sinh (x))-\frac {i}{8 (1-i \sinh (x))}+\frac {i}{8 (1+i \sinh (x))^2}-\frac {3 i}{4 (1+i \sinh (x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.79 \[ \int \frac {\tanh ^3(x)}{i+\text {csch}(x)} \, dx=\frac {1}{16} \left (-11 i \log (i-\sinh (x))-5 i \log (i+\sinh (x))-\frac {2 \left (6+3 i \sinh (x)+5 \sinh ^2(x)\right )}{(-i+\sinh (x))^2 (i+\sinh (x))}\right ) \]

[In]

Integrate[Tanh[x]^3/(I + Csch[x]),x]

[Out]

((-11*I)*Log[I - Sinh[x]] - (5*I)*Log[I + Sinh[x]] - (2*(6 + (3*I)*Sinh[x] + 5*Sinh[x]^2))/((-I + Sinh[x])^2*(
I + Sinh[x])))/16

Maple [A] (verified)

Time = 1.67 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.92

method	result	size
risch	\(i x -\frac {5 \,{\mathrm e}^{x}-6 i {\mathrm e}^{2 x}+14 \,{\mathrm e}^{3 x}+6 i {\mathrm e}^{4 x}+5 \,{\mathrm e}^{5 x}}{4 \left ({\mathrm e}^{x}+i\right )^{2} \left ({\mathrm e}^{x}-i\right )^{4}}-\frac {5 i \ln \left ({\mathrm e}^{x}+i\right )}{8}-\frac {11 i \ln \left ({\mathrm e}^{x}-i\right )}{8}\)	\(71\)
default	\(i \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )-\frac {5 i \ln \left (\tanh \left (\frac {x}{2}\right )+i\right )}{8}+\frac {i}{4 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{2}}-\frac {1}{4 \left (\tanh \left (\frac {x}{2}\right )+i\right )}+i \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )-\frac {11 i \ln \left (\tanh \left (\frac {x}{2}\right )-i\right )}{8}+\frac {i}{2 \left (\tanh \left (\frac {x}{2}\right )-i\right )^{4}}+\frac {i}{2 \left (\tanh \left (\frac {x}{2}\right )-i\right )^{2}}+\frac {1}{\left (\tanh \left (\frac {x}{2}\right )-i\right )^{3}}+\frac {1}{\tanh \left (\frac {x}{2}\right )-i}\)	\(109\)

[In]

int(tanh(x)^3/(I+csch(x)),x,method=_RETURNVERBOSE)

[Out]

I*x-1/4*(5*exp(x)-6*I*exp(x)^2+14*exp(x)^3+6*I*exp(x)^4+5*exp(x)^5)/(exp(x)+I)^2/(exp(x)-I)^4-5/8*I*ln(exp(x)+
I)-11/8*I*ln(exp(x)-I)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 186 vs. \(2 (47) = 94\).

Time = 0.29 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.42 \[ \int \frac {\tanh ^3(x)}{i+\text {csch}(x)} \, dx=\frac {8 i \, x e^{\left (6 \, x\right )} + 2 \, {\left (8 \, x - 5\right )} e^{\left (5 \, x\right )} - 4 \, {\left (-2 i \, x + 3 i\right )} e^{\left (4 \, x\right )} + 4 \, {\left (8 \, x - 7\right )} e^{\left (3 \, x\right )} - 4 \, {\left (2 i \, x - 3 i\right )} e^{\left (2 \, x\right )} + 2 \, {\left (8 \, x - 5\right )} e^{x} - 5 \, {\left (i \, e^{\left (6 \, x\right )} + 2 \, e^{\left (5 \, x\right )} + i \, e^{\left (4 \, x\right )} + 4 \, e^{\left (3 \, x\right )} - i \, e^{\left (2 \, x\right )} + 2 \, e^{x} - i\right )} \log \left (e^{x} + i\right ) - 11 \, {\left (i \, e^{\left (6 \, x\right )} + 2 \, e^{\left (5 \, x\right )} + i \, e^{\left (4 \, x\right )} + 4 \, e^{\left (3 \, x\right )} - i \, e^{\left (2 \, x\right )} + 2 \, e^{x} - i\right )} \log \left (e^{x} - i\right ) - 8 i \, x}{8 \, {\left (e^{\left (6 \, x\right )} - 2 i \, e^{\left (5 \, x\right )} + e^{\left (4 \, x\right )} - 4 i \, e^{\left (3 \, x\right )} - e^{\left (2 \, x\right )} - 2 i \, e^{x} - 1\right )}} \]

[In]

integrate(tanh(x)^3/(I+csch(x)),x, algorithm="fricas")

[Out]

1/8*(8*I*x*e^(6*x) + 2*(8*x - 5)*e^(5*x) - 4*(-2*I*x + 3*I)*e^(4*x) + 4*(8*x - 7)*e^(3*x) - 4*(2*I*x - 3*I)*e^
(2*x) + 2*(8*x - 5)*e^x - 5*(I*e^(6*x) + 2*e^(5*x) + I*e^(4*x) + 4*e^(3*x) - I*e^(2*x) + 2*e^x - I)*log(e^x +
I) - 11*(I*e^(6*x) + 2*e^(5*x) + I*e^(4*x) + 4*e^(3*x) - I*e^(2*x) + 2*e^x - I)*log(e^x - I) - 8*I*x)/(e^(6*x)
- 2*I*e^(5*x) + e^(4*x) - 4*I*e^(3*x) - e^(2*x) - 2*I*e^x - 1)

Sympy [F]

\[ \int \frac {\tanh ^3(x)}{i+\text {csch}(x)} \, dx=\int \frac {\tanh ^{3}{\left (x \right )}}{\operatorname {csch}{\left (x \right )} + i}\, dx \]

[In]

integrate(tanh(x)**3/(I+csch(x)),x)

[Out]

Integral(tanh(x)**3/(csch(x) + I), x)

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 96 vs. \(2 (47) = 94\).

Time = 0.20 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.25 \[ \int \frac {\tanh ^3(x)}{i+\text {csch}(x)} \, dx=-i \, x + \frac {5 \, e^{\left (-x\right )} + 6 i \, e^{\left (-2 \, x\right )} + 14 \, e^{\left (-3 \, x\right )} - 6 i \, e^{\left (-4 \, x\right )} + 5 \, e^{\left (-5 \, x\right )}}{8 i \, e^{\left (-x\right )} - 4 \, e^{\left (-2 \, x\right )} + 16 i \, e^{\left (-3 \, x\right )} + 4 \, e^{\left (-4 \, x\right )} + 8 i \, e^{\left (-5 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} - 4} - \frac {5}{8} i \, \log \left (e^{\left (-x\right )} - i\right ) - \frac {11}{8} i \, \log \left (i \, e^{\left (-x\right )} - 1\right ) \]

[In]

integrate(tanh(x)^3/(I+csch(x)),x, algorithm="maxima")

[Out]

-I*x + (5*e^(-x) + 6*I*e^(-2*x) + 14*e^(-3*x) - 6*I*e^(-4*x) + 5*e^(-5*x))/(8*I*e^(-x) - 4*e^(-2*x) + 16*I*e^(
-3*x) + 4*e^(-4*x) + 8*I*e^(-5*x) + 4*e^(-6*x) - 4) - 5/8*I*log(e^(-x) - I) - 11/8*I*log(I*e^(-x) - 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.22 \[ \int \frac {\tanh ^3(x)}{i+\text {csch}(x)} \, dx=\frac {5 \, e^{\left (-x\right )} - 5 \, e^{x} - 6 i}{16 \, {\left (-i \, e^{\left (-x\right )} + i \, e^{x} - 2\right )}} + \frac {33 i \, {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} - 84 \, e^{\left (-x\right )} + 84 \, e^{x} - 52 i}{32 \, {\left (e^{\left (-x\right )} - e^{x} + 2 i\right )}^{2}} - \frac {5}{16} i \, \log \left (-e^{\left (-x\right )} + e^{x} + 2 i\right ) - \frac {11}{16} i \, \log \left (-e^{\left (-x\right )} + e^{x} - 2 i\right ) \]

[In]

integrate(tanh(x)^3/(I+csch(x)),x, algorithm="giac")

[Out]

1/16*(5*e^(-x) - 5*e^x - 6*I)/(-I*e^(-x) + I*e^x - 2) + 1/32*(33*I*(e^(-x) - e^x)^2 - 84*e^(-x) + 84*e^x - 52*
I)/(e^(-x) - e^x + 2*I)^2 - 5/16*I*log(-e^(-x) + e^x + 2*I) - 11/16*I*log(-e^(-x) + e^x - 2*I)

Mupad [B] (verification not implemented)

Time = 0.62 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.82 \[ \int \frac {\tanh ^3(x)}{i+\text {csch}(x)} \, dx=x\,1{}\mathrm {i}-\ln \left (\left (\frac {3\,{\mathrm {e}}^x}{4}-\frac {3}{4}{}\mathrm {i}\right )\,\left (\frac {3\,{\mathrm {e}}^x}{4}+\frac {3}{4}{}\mathrm {i}\right )\right )\,1{}\mathrm {i}+\frac {3\,\mathrm {atan}\left ({\mathrm {e}}^x\right )}{4}-\frac {1}{{\mathrm {e}}^{2\,x}\,3{}\mathrm {i}-{\mathrm {e}}^{3\,x}+3\,{\mathrm {e}}^x-\mathrm {i}}-\frac {1{}\mathrm {i}}{4\,\left ({\mathrm {e}}^{2\,x}-1+{\mathrm {e}}^x\,2{}\mathrm {i}\right )}+\frac {1{}\mathrm {i}}{2\,\left ({\mathrm {e}}^{4\,x}-6\,{\mathrm {e}}^{2\,x}+1-{\mathrm {e}}^{3\,x}\,4{}\mathrm {i}+{\mathrm {e}}^x\,4{}\mathrm {i}\right )}+\frac {2{}\mathrm {i}}{1-{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x\,2{}\mathrm {i}}-\frac {3}{2\,\left ({\mathrm {e}}^x-\mathrm {i}\right )}+\frac {1}{4\,\left ({\mathrm {e}}^x+1{}\mathrm {i}\right )} \]

[In]

int(tanh(x)^3/(1/sinh(x) + 1i),x)

[Out]

x*1i - log(((3*exp(x))/4 - 3i/4)*((3*exp(x))/4 + 3i/4))*1i + (3*atan(exp(x)))/4 - 1/(exp(2*x)*3i - exp(3*x) +
3*exp(x) - 1i) - 1i/(4*(exp(2*x) + exp(x)*2i - 1)) + 1i/(2*(exp(4*x) - exp(3*x)*4i - 6*exp(2*x) + exp(x)*4i +
1)) + 2i/(exp(x)*2i - exp(2*x) + 1) - 3/(2*(exp(x) - 1i)) + 1/(4*(exp(x) + 1i))