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]

-21/32*I*ln(I-sinh(x))-11/32*I*ln(I+sinh(x))+1/32*I/(1-I*sinh(x))^2-1/4*I/(1-I*sinh(x))-1/24*I/(1+I*sinh(x))^3
+9/32*I/(1+I*sinh(x))^2-15/16*I/(1+I*sinh(x))

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Rubi [A]  time = 0.09, antiderivative size = 109, normalized size of antiderivative = 1.00, 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 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]))

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]

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*}

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Mathematica [A]  time = 0.20, 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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fricas [B]  time = 1.37, size = 299, normalized size = 2.74 \[ \frac {48 i \, x e^{\left (10 \, x\right )} + 6 \, {\left (16 \, x - 11\right )} e^{\left (9 \, x\right )} + {\left (144 i \, x - 156 i\right )} e^{\left (8 \, x\right )} + 16 \, {\left (24 \, x - 23\right )} e^{\left (7 \, x\right )} + {\left (96 i \, x + 4 i\right )} e^{\left (6 \, x\right )} + 36 \, {\left (16 \, x - 15\right )} e^{\left (5 \, x\right )} + {\left (-96 i \, x - 4 i\right )} e^{\left (4 \, x\right )} + 16 \, {\left (24 \, x - 23\right )} e^{\left (3 \, x\right )} + {\left (-144 i \, x + 156 i\right )} e^{\left (2 \, x\right )} + 6 \, {\left (16 \, x - 11\right )} e^{x} + {\left (-33 i \, e^{\left (10 \, x\right )} - 66 \, e^{\left (9 \, x\right )} - 99 i \, e^{\left (8 \, x\right )} - 264 \, e^{\left (7 \, x\right )} - 66 i \, e^{\left (6 \, x\right )} - 396 \, e^{\left (5 \, x\right )} + 66 i \, e^{\left (4 \, x\right )} - 264 \, e^{\left (3 \, x\right )} + 99 i \, e^{\left (2 \, x\right )} - 66 \, e^{x} + 33 i\right )} \log \left (e^{x} + i\right ) + {\left (-63 i \, e^{\left (10 \, x\right )} - 126 \, e^{\left (9 \, x\right )} - 189 i \, e^{\left (8 \, x\right )} - 504 \, e^{\left (7 \, x\right )} - 126 i \, e^{\left (6 \, x\right )} - 756 \, e^{\left (5 \, x\right )} + 126 i \, e^{\left (4 \, x\right )} - 504 \, e^{\left (3 \, x\right )} + 189 i \, e^{\left (2 \, x\right )} - 126 \, e^{x} + 63 i\right )} \log \left (e^{x} - i\right ) - 48 i \, x}{48 \, e^{\left (10 \, x\right )} - 96 i \, e^{\left (9 \, x\right )} + 144 \, e^{\left (8 \, x\right )} - 384 i \, e^{\left (7 \, x\right )} + 96 \, e^{\left (6 \, x\right )} - 576 i \, e^{\left (5 \, x\right )} - 96 \, e^{\left (4 \, x\right )} - 384 i \, e^{\left (3 \, x\right )} - 144 \, e^{\left (2 \, x\right )} - 96 i \, e^{x} - 48} \]

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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giac [A]  time = 0.15, size = 120, normalized size = 1.10 \[ -\frac {33 i \, {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 100 \, e^{\left (-x\right )} - 100 \, e^{x} - 76 i}{64 \, {\left (-i \, e^{\left (-x\right )} + i \, e^{x} - 2\right )}^{2}} - \frac {-231 i \, {\left (e^{\left (-x\right )} - e^{x}\right )}^{3} + 1026 \, {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 1548 i \, e^{\left (-x\right )} - 1548 i \, e^{x} - 776}{192 \, {\left (e^{\left (-x\right )} - e^{x} + 2 i\right )}^{3}} - \frac {11}{32} i \, \log \left (-e^{\left (-x\right )} + e^{x} + 2 i\right ) - \frac {21}{32} i \, \log \left (-e^{\left (-x\right )} + e^{x} - 2 i\right ) \]

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)

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maple [A]  time = 0.25, size = 155, normalized size = 1.42 \[ i \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )+i \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )-\frac {11 i \ln \left (\tanh \left (\frac {x}{2}\right )+i\right )}{16}+\frac {i}{8 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{4}}+\frac {i}{4 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{2}}-\frac {1}{4 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{3}}-\frac {3}{8 \left (\tanh \left (\frac {x}{2}\right )+i\right )}-\frac {21 i \ln \left (\tanh \left (\frac {x}{2}\right )-i\right )}{16}+\frac {3 i}{8 \left (\tanh \left (\frac {x}{2}\right )-i\right )^{2}}+\frac {i}{3 \left (\tanh \left (\frac {x}{2}\right )-i\right )^{6}}-\frac {3 i}{8 \left (\tanh \left (\frac {x}{2}\right )-i\right )^{4}}+\frac {1}{\left (\tanh \left (\frac {x}{2}\right )-i\right )^{5}}+\frac {11}{12 \left (\tanh \left (\frac {x}{2}\right )-i\right )^{3}}+\frac {1}{\tanh \left (\frac {x}{2}\right )-i} \]

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)+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)-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)

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maxima [B]  time = 0.33, size = 144, normalized size = 1.32 \[ -i \, x + \frac {33 \, e^{\left (-x\right )} + 78 i \, e^{\left (-2 \, x\right )} + 184 \, e^{\left (-3 \, x\right )} - 2 i \, e^{\left (-4 \, x\right )} + 270 \, e^{\left (-5 \, x\right )} + 2 i \, e^{\left (-6 \, x\right )} + 184 \, e^{\left (-7 \, x\right )} - 78 i \, e^{\left (-8 \, x\right )} + 33 \, e^{\left (-9 \, x\right )}}{48 i \, e^{\left (-x\right )} - 72 \, e^{\left (-2 \, x\right )} + 192 i \, e^{\left (-3 \, x\right )} - 48 \, e^{\left (-4 \, x\right )} + 288 i \, e^{\left (-5 \, x\right )} + 48 \, e^{\left (-6 \, x\right )} + 192 i \, e^{\left (-7 \, x\right )} + 72 \, e^{\left (-8 \, x\right )} + 48 i \, e^{\left (-9 \, x\right )} + 24 \, e^{\left (-10 \, x\right )} - 24} - \frac {11}{16} i \, \log \left (e^{\left (-x\right )} - i\right ) - \frac {21}{16} i \, \log \left (i \, e^{\left (-x\right )} - 1\right ) \]

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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mupad [B]  time = 4.11, size = 274, normalized size = 2.51 \[ x\,1{}\mathrm {i}-\ln \left (\left (\frac {5\,{\mathrm {e}}^x}{8}-\frac {5}{8}{}\mathrm {i}\right )\,\left (\frac {5\,{\mathrm {e}}^x}{8}+\frac {5}{8}{}\mathrm {i}\right )\right )\,1{}\mathrm {i}+\frac {5\,\mathrm {atan}\left ({\mathrm {e}}^x\right )}{8}+\frac {1{}\mathrm {i}}{3\,\left (15\,{\mathrm {e}}^{4\,x}-15\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^{6\,x}+1-{\mathrm {e}}^{3\,x}\,20{}\mathrm {i}+{\mathrm {e}}^{5\,x}\,6{}\mathrm {i}+{\mathrm {e}}^x\,6{}\mathrm {i}\right )}-\frac {1}{{\mathrm {e}}^{2\,x}\,10{}\mathrm {i}-10\,{\mathrm {e}}^{3\,x}-{\mathrm {e}}^{4\,x}\,5{}\mathrm {i}+{\mathrm {e}}^{5\,x}+5\,{\mathrm {e}}^x-\mathrm {i}}-\frac {31}{12\,\left ({\mathrm {e}}^{2\,x}\,3{}\mathrm {i}-{\mathrm {e}}^{3\,x}+3\,{\mathrm {e}}^x-\mathrm {i}\right )}-\frac {5{}\mathrm {i}}{8\,\left ({\mathrm {e}}^{2\,x}-1+{\mathrm {e}}^x\,2{}\mathrm {i}\right )}+\frac {17{}\mathrm {i}}{8\,\left ({\mathrm {e}}^{4\,x}-6\,{\mathrm {e}}^{2\,x}+1-{\mathrm {e}}^{3\,x}\,4{}\mathrm {i}+{\mathrm {e}}^x\,4{}\mathrm {i}\right )}+\frac {1{}\mathrm {i}}{8\,\left ({\mathrm {e}}^{4\,x}-6\,{\mathrm {e}}^{2\,x}+1+{\mathrm {e}}^{3\,x}\,4{}\mathrm {i}-{\mathrm {e}}^x\,4{}\mathrm {i}\right )}+\frac {3{}\mathrm {i}}{1-{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x\,2{}\mathrm {i}}-\frac {15}{8\,\left ({\mathrm {e}}^x-\mathrm {i}\right )}+\frac {1}{2\,\left ({\mathrm {e}}^x+1{}\mathrm {i}\right )}-\frac {1}{4\,\left ({\mathrm {e}}^{2\,x}\,3{}\mathrm {i}+{\mathrm {e}}^{3\,x}-3\,{\mathrm {e}}^x-\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^5/(1/sinh(x) + 1i),x)

[Out]

x*1i - log(((5*exp(x))/8 - 5i/8)*((5*exp(x))/8 + 5i/8))*1i + (5*atan(exp(x)))/8 + 1i/(3*(15*exp(4*x) - exp(3*x
)*20i - 15*exp(2*x) + exp(5*x)*6i - exp(6*x) + exp(x)*6i + 1)) - 1/(exp(2*x)*10i - 10*exp(3*x) - exp(4*x)*5i +
exp(5*x) + 5*exp(x) - 1i) - 31/(12*(exp(2*x)*3i - exp(3*x) + 3*exp(x) - 1i)) - 5i/(8*(exp(2*x) + exp(x)*2i -
1)) + 17i/(8*(exp(4*x) - exp(3*x)*4i - 6*exp(2*x) + exp(x)*4i + 1)) + 1i/(8*(exp(3*x)*4i - 6*exp(2*x) + exp(4*
x) - exp(x)*4i + 1)) + 3i/(exp(x)*2i - exp(2*x) + 1) - 15/(8*(exp(x) - 1i)) + 1/(2*(exp(x) + 1i)) - 1/(4*(exp(
2*x)*3i + exp(3*x) - 3*exp(x) - 1i))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh ^{5}{\relax (x )}}{\operatorname {csch}{\relax (x )} + i}\, dx \]

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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