3.218 \(\int \frac{\tanh ^4(x)}{(i+\sinh (x))^2} \, dx\)

Optimal. Leaf size=47 \[ \frac{2 \tanh ^7(x)}{7}-\frac{\tanh ^5(x)}{5}+\frac{2}{7} i \text{sech}^7(x)-\frac{4}{5} i \text{sech}^5(x)+\frac{2}{3} i \text{sech}^3(x) \]

[Out]

((2*I)/3)*Sech[x]^3 - ((4*I)/5)*Sech[x]^5 + ((2*I)/7)*Sech[x]^7 - Tanh[x]^5/5 + (2*Tanh[x]^7)/7

________________________________________________________________________________________

Rubi [A]  time = 0.122853, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {2711, 2607, 14, 2606, 270, 30} \[ \frac{2 \tanh ^7(x)}{7}-\frac{\tanh ^5(x)}{5}+\frac{2}{7} i \text{sech}^7(x)-\frac{4}{5} i \text{sech}^5(x)+\frac{2}{3} i \text{sech}^3(x) \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^4/(I + Sinh[x])^2,x]

[Out]

((2*I)/3)*Sech[x]^3 - ((4*I)/5)*Sech[x]^5 + ((2*I)/7)*Sech[x]^7 - Tanh[x]^5/5 + (2*Tanh[x]^7)/7

Rule 2711

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[a^(2*
m), Int[ExpandIntegrand[(g*Tan[e + f*x])^p/Sec[e + f*x]^m, (a*Sec[e + f*x] - b*Tan[e + f*x])^(-m), x], x], x]
/; FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\tanh ^4(x)}{(i+\sinh (x))^2} \, dx &=\int \left (-\text{sech}^4(x) \tanh ^4(x)-2 i \text{sech}^3(x) \tanh ^5(x)+\text{sech}^2(x) \tanh ^6(x)\right ) \, dx\\ &=-\left (2 i \int \text{sech}^3(x) \tanh ^5(x) \, dx\right )-\int \text{sech}^4(x) \tanh ^4(x) \, dx+\int \text{sech}^2(x) \tanh ^6(x) \, dx\\ &=i \operatorname{Subst}\left (\int x^6 \, dx,x,i \tanh (x)\right )+i \operatorname{Subst}\left (\int x^4 \left (1+x^2\right ) \, dx,x,i \tanh (x)\right )+2 i \operatorname{Subst}\left (\int x^2 \left (-1+x^2\right )^2 \, dx,x,\text{sech}(x)\right )\\ &=\frac{\tanh ^7(x)}{7}+i \operatorname{Subst}\left (\int \left (x^4+x^6\right ) \, dx,x,i \tanh (x)\right )+2 i \operatorname{Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\text{sech}(x)\right )\\ &=\frac{2}{3} i \text{sech}^3(x)-\frac{4}{5} i \text{sech}^5(x)+\frac{2}{7} i \text{sech}^7(x)-\frac{\tanh ^5(x)}{5}+\frac{2 \tanh ^7(x)}{7}\\ \end{align*}

Mathematica [B]  time = 0.146499, size = 112, normalized size = 2.38 \[ -\frac{1232 \sinh (x)+824 \sinh (2 x)-1896 \sinh (3 x)+412 \sinh (4 x)+72 \sinh (5 x)+1442 i \cosh (x)-1664 i \cosh (2 x)+309 i \cosh (3 x)+288 i \cosh (4 x)-103 i \cosh (5 x)-672 i}{13440 \left (\cosh \left (\frac{x}{2}\right )-i \sinh \left (\frac{x}{2}\right )\right )^7 \left (\cosh \left (\frac{x}{2}\right )+i \sinh \left (\frac{x}{2}\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^4/(I + Sinh[x])^2,x]

[Out]

-(-672*I + (1442*I)*Cosh[x] - (1664*I)*Cosh[2*x] + (309*I)*Cosh[3*x] + (288*I)*Cosh[4*x] - (103*I)*Cosh[5*x] +
1232*Sinh[x] + 824*Sinh[2*x] - 1896*Sinh[3*x] + 412*Sinh[4*x] + 72*Sinh[5*x])/(13440*(Cosh[x/2] - I*Sinh[x/2]
)^7*(Cosh[x/2] + I*Sinh[x/2])^3)

________________________________________________________________________________________

Maple [B]  time = 0.093, size = 116, normalized size = 2.5 \begin{align*}{-{\frac{i}{8}} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-2}}+{\frac{1}{12} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-3}}+{\frac{1}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-1}}+{2\,i \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-6}}-{i \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-4}}-{{\frac{i}{8}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-2}}+{\frac{4}{7} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-7}}-{\frac{12}{5} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-5}}-{\frac{1}{12} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-3}}-{\frac{1}{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)^4/(I+sinh(x))^2,x)

[Out]

-1/8*I/(tanh(1/2*x)-I)^2+1/12/(tanh(1/2*x)-I)^3+1/8/(tanh(1/2*x)-I)+2*I/(tanh(1/2*x)+I)^6-I/(tanh(1/2*x)+I)^4-
1/8*I/(tanh(1/2*x)+I)^2+4/7/(tanh(1/2*x)+I)^7-12/5/(tanh(1/2*x)+I)^5-1/12/(tanh(1/2*x)+I)^3-1/8/(tanh(1/2*x)+I
)

________________________________________________________________________________________

Maxima [B]  time = 1.11311, size = 774, normalized size = 16.47 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^4/(I+sinh(x))^2,x, algorithm="maxima")

[Out]

72*I*e^(-x)/(420*I*e^(-x) - 315*e^(-2*x) + 840*I*e^(-3*x) - 1470*e^(-4*x) - 1470*e^(-6*x) - 840*I*e^(-7*x) - 3
15*e^(-8*x) - 420*I*e^(-9*x) + 105*e^(-10*x) + 105) - 264*e^(-2*x)/(420*I*e^(-x) - 315*e^(-2*x) + 840*I*e^(-3*
x) - 1470*e^(-4*x) - 1470*e^(-6*x) - 840*I*e^(-7*x) - 315*e^(-8*x) - 420*I*e^(-9*x) + 105*e^(-10*x) + 105) - 1
36*I*e^(-3*x)/(420*I*e^(-x) - 315*e^(-2*x) + 840*I*e^(-3*x) - 1470*e^(-4*x) - 1470*e^(-6*x) - 840*I*e^(-7*x) -
315*e^(-8*x) - 420*I*e^(-9*x) + 105*e^(-10*x) + 105) + 28*e^(-4*x)/(420*I*e^(-x) - 315*e^(-2*x) + 840*I*e^(-3
*x) - 1470*e^(-4*x) - 1470*e^(-6*x) - 840*I*e^(-7*x) - 315*e^(-8*x) - 420*I*e^(-9*x) + 105*e^(-10*x) + 105) -
168*I*e^(-5*x)/(420*I*e^(-x) - 315*e^(-2*x) + 840*I*e^(-3*x) - 1470*e^(-4*x) - 1470*e^(-6*x) - 840*I*e^(-7*x)
- 315*e^(-8*x) - 420*I*e^(-9*x) + 105*e^(-10*x) + 105) - 280*e^(-6*x)/(420*I*e^(-x) - 315*e^(-2*x) + 840*I*e^(
-3*x) - 1470*e^(-4*x) - 1470*e^(-6*x) - 840*I*e^(-7*x) - 315*e^(-8*x) - 420*I*e^(-9*x) + 105*e^(-10*x) + 105)
- 280*I*e^(-7*x)/(420*I*e^(-x) - 315*e^(-2*x) + 840*I*e^(-3*x) - 1470*e^(-4*x) - 1470*e^(-6*x) - 840*I*e^(-7*x
) - 315*e^(-8*x) - 420*I*e^(-9*x) + 105*e^(-10*x) + 105) + 210*e^(-8*x)/(420*I*e^(-x) - 315*e^(-2*x) + 840*I*e
^(-3*x) - 1470*e^(-4*x) - 1470*e^(-6*x) - 840*I*e^(-7*x) - 315*e^(-8*x) - 420*I*e^(-9*x) + 105*e^(-10*x) + 105
) + 18/(420*I*e^(-x) - 315*e^(-2*x) + 840*I*e^(-3*x) - 1470*e^(-4*x) - 1470*e^(-6*x) - 840*I*e^(-7*x) - 315*e^
(-8*x) - 420*I*e^(-9*x) + 105*e^(-10*x) + 105)

________________________________________________________________________________________

Fricas [B]  time = 2.00328, size = 351, normalized size = 7.47 \begin{align*} -\frac{210 \, e^{\left (8 \, x\right )} + 280 i \, e^{\left (7 \, x\right )} - 280 \, e^{\left (6 \, x\right )} + 168 i \, e^{\left (5 \, x\right )} + 28 \, e^{\left (4 \, x\right )} + 136 i \, e^{\left (3 \, x\right )} - 264 \, e^{\left (2 \, x\right )} - 72 i \, e^{x} + 18}{105 \, e^{\left (10 \, x\right )} + 420 i \, e^{\left (9 \, x\right )} - 315 \, e^{\left (8 \, x\right )} + 840 i \, e^{\left (7 \, x\right )} - 1470 \, e^{\left (6 \, x\right )} - 1470 \, e^{\left (4 \, x\right )} - 840 i \, e^{\left (3 \, x\right )} - 315 \, e^{\left (2 \, x\right )} - 420 i \, e^{x} + 105} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^4/(I+sinh(x))^2,x, algorithm="fricas")

[Out]

-(210*e^(8*x) + 280*I*e^(7*x) - 280*e^(6*x) + 168*I*e^(5*x) + 28*e^(4*x) + 136*I*e^(3*x) - 264*e^(2*x) - 72*I*
e^x + 18)/(105*e^(10*x) + 420*I*e^(9*x) - 315*e^(8*x) + 840*I*e^(7*x) - 1470*e^(6*x) - 1470*e^(4*x) - 840*I*e^
(3*x) - 315*e^(2*x) - 420*I*e^x + 105)

________________________________________________________________________________________

Sympy [B]  time = 2.53294, size = 139, normalized size = 2.96 \begin{align*} \frac{- 2 e^{8 x} - \frac{8 i e^{7 x}}{3} + \frac{8 e^{6 x}}{3} - \frac{8 i e^{5 x}}{5} - \frac{4 e^{4 x}}{15} - \frac{136 i e^{3 x}}{105} + \frac{88 e^{2 x}}{35} + \frac{24 i e^{x}}{35} - \frac{6}{35}}{e^{10 x} + 4 i e^{9 x} - 3 e^{8 x} + 8 i e^{7 x} - 14 e^{6 x} - 14 e^{4 x} - 8 i e^{3 x} - 3 e^{2 x} - 4 i e^{x} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)**4/(I+sinh(x))**2,x)

[Out]

(-2*exp(8*x) - 8*I*exp(7*x)/3 + 8*exp(6*x)/3 - 8*I*exp(5*x)/5 - 4*exp(4*x)/15 - 136*I*exp(3*x)/105 + 88*exp(2*
x)/35 + 24*I*exp(x)/35 - 6/35)/(exp(10*x) + 4*I*exp(9*x) - 3*exp(8*x) + 8*I*exp(7*x) - 14*exp(6*x) - 14*exp(4*
x) - 8*I*exp(3*x) - 3*exp(2*x) - 4*I*exp(x) + 1)

________________________________________________________________________________________

Giac [B]  time = 1.16407, size = 88, normalized size = 1.87 \begin{align*} -\frac{-6 i \, e^{\left (2 \, x\right )} - 9 \, e^{x} + 5 i}{24 \,{\left (e^{x} - i\right )}^{3}} - \frac{210 i \, e^{\left (6 \, x\right )} - 105 \, e^{\left (5 \, x\right )} + 175 i \, e^{\left (4 \, x\right )} - 910 \, e^{\left (3 \, x\right )} - 756 i \, e^{\left (2 \, x\right )} + 427 \, e^{x} + 31 i}{840 \,{\left (e^{x} + i\right )}^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^4/(I+sinh(x))^2,x, algorithm="giac")

[Out]

-1/24*(-6*I*e^(2*x) - 9*e^x + 5*I)/(e^x - I)^3 - 1/840*(210*I*e^(6*x) - 105*e^(5*x) + 175*I*e^(4*x) - 910*e^(3
*x) - 756*I*e^(2*x) + 427*e^x + 31*I)/(e^x + I)^7