3.102 \(\int \frac{\tanh ^5(x)}{i+\text{csch}(x)} \, dx\)

Optimal. Leaf size=109 \[ -\frac{i}{4 (1-i \sinh (x))}-\frac{15 i}{16 (1+i \sinh (x))}+\frac{i}{32 (1-i \sinh (x))^2}+\frac{9 i}{32 (1+i \sinh (x))^2}-\frac{i}{24 (1+i \sinh (x))^3}-\frac{21}{32} i \log (-\sinh (x)+i)-\frac{11}{32} i \log (\sinh (x)+i) \]

[Out]

((-21*I)/32)*Log[I - Sinh[x]] - ((11*I)/32)*Log[I + Sinh[x]] + (I/32)/(1 - I*Sinh[x])^2 - (I/4)/(1 - I*Sinh[x]
) - (I/24)/(1 + I*Sinh[x])^3 + ((9*I)/32)/(1 + I*Sinh[x])^2 - ((15*I)/16)/(1 + I*Sinh[x])

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Rubi [A]  time = 0.0860622, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3879, 88} \[ -\frac{i}{4 (1-i \sinh (x))}-\frac{15 i}{16 (1+i \sinh (x))}+\frac{i}{32 (1-i \sinh (x))^2}+\frac{9 i}{32 (1+i \sinh (x))^2}-\frac{i}{24 (1+i \sinh (x))^3}-\frac{21}{32} i \log (-\sinh (x)+i)-\frac{11}{32} i \log (\sinh (x)+i) \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^5/(I + Csch[x]),x]

[Out]

((-21*I)/32)*Log[I - Sinh[x]] - ((11*I)/32)*Log[I + Sinh[x]] + (I/32)/(1 - I*Sinh[x])^2 - (I/4)/(1 - I*Sinh[x]
) - (I/24)/(1 + I*Sinh[x])^3 + ((9*I)/32)/(1 + I*Sinh[x])^2 - ((15*I)/16)/(1 + I*Sinh[x])

Rule 3879

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]

Rule 88

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]))

Rubi steps

\begin{align*} \int \frac{\tanh ^5(x)}{i+\text{csch}(x)} \, dx &=\operatorname{Subst}\left (\int \frac{x^6}{(i-i x)^3 (i+i x)^4} \, dx,x,i \sinh (x)\right )\\ &=\operatorname{Subst}\left (\int \left (-\frac{i}{16 (-1+x)^3}-\frac{i}{4 (-1+x)^2}-\frac{11 i}{32 (-1+x)}+\frac{i}{8 (1+x)^4}-\frac{9 i}{16 (1+x)^3}+\frac{15 i}{16 (1+x)^2}-\frac{21 i}{32 (1+x)}\right ) \, dx,x,i \sinh (x)\right )\\ &=-\frac{21}{32} i \log (i-\sinh (x))-\frac{11}{32} i \log (i+\sinh (x))+\frac{i}{32 (1-i \sinh (x))^2}-\frac{i}{4 (1-i \sinh (x))}-\frac{i}{24 (1+i \sinh (x))^3}+\frac{9 i}{32 (1+i \sinh (x))^2}-\frac{15 i}{16 (1+i \sinh (x))}\\ \end{align*}

Mathematica [A]  time = 0.187191, size = 75, normalized size = 0.69 \[ \frac{1}{96} \left (-\frac{2 \left (33 \sinh ^4(x)+39 i \sinh ^3(x)+79 \sinh ^2(x)+29 i \sinh (x)+44\right )}{(\sinh (x)-i)^3 (\sinh (x)+i)^2}-63 i \log (-\sinh (x)+i)-33 i \log (\sinh (x)+i)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^5/(I + Csch[x]),x]

[Out]

((-63*I)*Log[I - Sinh[x]] - (33*I)*Log[I + Sinh[x]] - (2*(44 + (29*I)*Sinh[x] + 79*Sinh[x]^2 + (39*I)*Sinh[x]^
3 + 33*Sinh[x]^4))/((-I + Sinh[x])^3*(I + Sinh[x])^2))/96

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Maple [A]  time = 0.062, size = 155, normalized size = 1.4 \begin{align*} i\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) -{\frac{21\,i}{16}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) +{{\frac{3\,i}{8}} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-2}}+{{\frac{i}{3}} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-6}}-{{\frac{3\,i}{8}} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-4}}+ \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-5}+{\frac{11}{12} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-3}}+ \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-1}+i\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) -{\frac{11\,i}{16}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) +{{\frac{i}{8}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-4}}+{{\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 ) ^{-3}}-{\frac{3}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^5/(I+csch(x)),x)

[Out]

I*ln(tanh(1/2*x)+1)-21/16*I*ln(tanh(1/2*x)-I)+3/8*I/(tanh(1/2*x)-I)^2+1/3*I/(tanh(1/2*x)-I)^6-3/8*I/(tanh(1/2*
x)-I)^4+1/(tanh(1/2*x)-I)^5+11/12/(tanh(1/2*x)-I)^3+1/(tanh(1/2*x)-I)+I*ln(tanh(1/2*x)-1)-11/16*I*ln(tanh(1/2*
x)+I)+1/8*I/(tanh(1/2*x)+I)^4+1/4*I/(tanh(1/2*x)+I)^2-1/4/(tanh(1/2*x)+I)^3-3/8/(tanh(1/2*x)+I)

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Maxima [B]  time = 1.02221, size = 194, normalized size = 1.78 \begin{align*} -i \, x + \frac{33 \, e^{\left (-x\right )} + 78 i \, e^{\left (-2 \, x\right )} + 184 \, e^{\left (-3 \, x\right )} - 2 i \, e^{\left (-4 \, x\right )} + 270 \, e^{\left (-5 \, x\right )} + 2 i \, e^{\left (-6 \, x\right )} + 184 \, e^{\left (-7 \, x\right )} - 78 i \, e^{\left (-8 \, x\right )} + 33 \, e^{\left (-9 \, x\right )}}{48 i \, e^{\left (-x\right )} - 72 \, e^{\left (-2 \, x\right )} + 192 i \, e^{\left (-3 \, x\right )} - 48 \, e^{\left (-4 \, x\right )} + 288 i \, e^{\left (-5 \, x\right )} + 48 \, e^{\left (-6 \, x\right )} + 192 i \, e^{\left (-7 \, x\right )} + 72 \, e^{\left (-8 \, x\right )} + 48 i \, e^{\left (-9 \, x\right )} + 24 \, e^{\left (-10 \, x\right )} - 24} - \frac{11}{16} i \, \log \left (e^{\left (-x\right )} - i\right ) - \frac{21}{16} i \, \log \left (i \, e^{\left (-x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^5/(I+csch(x)),x, algorithm="maxima")

[Out]

-I*x + (33*e^(-x) + 78*I*e^(-2*x) + 184*e^(-3*x) - 2*I*e^(-4*x) + 270*e^(-5*x) + 2*I*e^(-6*x) + 184*e^(-7*x) -
78*I*e^(-8*x) + 33*e^(-9*x))/(48*I*e^(-x) - 72*e^(-2*x) + 192*I*e^(-3*x) - 48*e^(-4*x) + 288*I*e^(-5*x) + 48*
e^(-6*x) + 192*I*e^(-7*x) + 72*e^(-8*x) + 48*I*e^(-9*x) + 24*e^(-10*x) - 24) - 11/16*I*log(e^(-x) - I) - 21/16
*I*log(I*e^(-x) - 1)

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Fricas [B]  time = 1.73703, size = 991, normalized size = 9.09 \begin{align*} \frac{48 i \, x e^{\left (10 \, x\right )} + 6 \,{\left (16 \, x - 11\right )} e^{\left (9 \, x\right )} +{\left (144 i \, x - 156 i\right )} e^{\left (8 \, x\right )} + 16 \,{\left (24 \, x - 23\right )} e^{\left (7 \, x\right )} +{\left (96 i \, x + 4 i\right )} e^{\left (6 \, x\right )} + 36 \,{\left (16 \, x - 15\right )} e^{\left (5 \, x\right )} +{\left (-96 i \, x - 4 i\right )} e^{\left (4 \, x\right )} + 16 \,{\left (24 \, x - 23\right )} e^{\left (3 \, x\right )} +{\left (-144 i \, x + 156 i\right )} e^{\left (2 \, x\right )} + 6 \,{\left (16 \, x - 11\right )} e^{x} +{\left (-33 i \, e^{\left (10 \, x\right )} - 66 \, e^{\left (9 \, x\right )} - 99 i \, e^{\left (8 \, x\right )} - 264 \, e^{\left (7 \, x\right )} - 66 i \, e^{\left (6 \, x\right )} - 396 \, e^{\left (5 \, x\right )} + 66 i \, e^{\left (4 \, x\right )} - 264 \, e^{\left (3 \, x\right )} + 99 i \, e^{\left (2 \, x\right )} - 66 \, e^{x} + 33 i\right )} \log \left (e^{x} + i\right ) +{\left (-63 i \, e^{\left (10 \, x\right )} - 126 \, e^{\left (9 \, x\right )} - 189 i \, e^{\left (8 \, x\right )} - 504 \, e^{\left (7 \, x\right )} - 126 i \, e^{\left (6 \, x\right )} - 756 \, e^{\left (5 \, x\right )} + 126 i \, e^{\left (4 \, x\right )} - 504 \, e^{\left (3 \, x\right )} + 189 i \, e^{\left (2 \, x\right )} - 126 \, e^{x} + 63 i\right )} \log \left (e^{x} - i\right ) - 48 i \, x}{48 \, e^{\left (10 \, x\right )} - 96 i \, e^{\left (9 \, x\right )} + 144 \, e^{\left (8 \, x\right )} - 384 i \, e^{\left (7 \, x\right )} + 96 \, e^{\left (6 \, x\right )} - 576 i \, e^{\left (5 \, x\right )} - 96 \, e^{\left (4 \, x\right )} - 384 i \, e^{\left (3 \, x\right )} - 144 \, e^{\left (2 \, x\right )} - 96 i \, e^{x} - 48} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^5/(I+csch(x)),x, algorithm="fricas")

[Out]

(48*I*x*e^(10*x) + 6*(16*x - 11)*e^(9*x) + (144*I*x - 156*I)*e^(8*x) + 16*(24*x - 23)*e^(7*x) + (96*I*x + 4*I)
*e^(6*x) + 36*(16*x - 15)*e^(5*x) + (-96*I*x - 4*I)*e^(4*x) + 16*(24*x - 23)*e^(3*x) + (-144*I*x + 156*I)*e^(2
*x) + 6*(16*x - 11)*e^x + (-33*I*e^(10*x) - 66*e^(9*x) - 99*I*e^(8*x) - 264*e^(7*x) - 66*I*e^(6*x) - 396*e^(5*
x) + 66*I*e^(4*x) - 264*e^(3*x) + 99*I*e^(2*x) - 66*e^x + 33*I)*log(e^x + I) + (-63*I*e^(10*x) - 126*e^(9*x) -
189*I*e^(8*x) - 504*e^(7*x) - 126*I*e^(6*x) - 756*e^(5*x) + 126*I*e^(4*x) - 504*e^(3*x) + 189*I*e^(2*x) - 126
*e^x + 63*I)*log(e^x - I) - 48*I*x)/(48*e^(10*x) - 96*I*e^(9*x) + 144*e^(8*x) - 384*I*e^(7*x) + 96*e^(6*x) - 5
76*I*e^(5*x) - 96*e^(4*x) - 384*I*e^(3*x) - 144*e^(2*x) - 96*I*e^x - 48)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{5}{\left (x \right )}}{\operatorname{csch}{\left (x \right )} + i}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)**5/(I+csch(x)),x)

[Out]

Integral(tanh(x)**5/(csch(x) + I), x)

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Giac [A]  time = 1.16329, size = 162, normalized size = 1.49 \begin{align*} -\frac{33 i \,{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 100 \, e^{\left (-x\right )} - 100 \, e^{x} - 76 i}{64 \,{\left (-i \, e^{\left (-x\right )} + i \, e^{x} - 2\right )}^{2}} - \frac{-231 i \,{\left (e^{\left (-x\right )} - e^{x}\right )}^{3} + 1026 \,{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 1548 i \, e^{\left (-x\right )} - 1548 i \, e^{x} - 776}{192 \,{\left (e^{\left (-x\right )} - e^{x} + 2 i\right )}^{3}} - \frac{11}{32} i \, \log \left (-e^{\left (-x\right )} + e^{x} + 2 i\right ) - \frac{21}{32} i \, \log \left (-e^{\left (-x\right )} + e^{x} - 2 i\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^5/(I+csch(x)),x, algorithm="giac")

[Out]

-1/64*(33*I*(e^(-x) - e^x)^2 + 100*e^(-x) - 100*e^x - 76*I)/(-I*e^(-x) + I*e^x - 2)^2 - 1/192*(-231*I*(e^(-x)
- e^x)^3 + 1026*(e^(-x) - e^x)^2 + 1548*I*e^(-x) - 1548*I*e^x - 776)/(e^(-x) - e^x + 2*I)^3 - 11/32*I*log(-e^(
-x) + e^x + 2*I) - 21/32*I*log(-e^(-x) + e^x - 2*I)