### 3.702 $$\int \frac{\sinh (x)}{(a \cosh (x)+b \sinh (x))^3} \, dx$$

Optimal. Leaf size=19 $\frac{\tanh ^2(x)}{2 a (a+b \tanh (x))^2}$

[Out]

Tanh[x]^2/(2*a*(a + b*Tanh[x])^2)

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Rubi [A]  time = 0.0321619, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.143, Rules used = {3087, 37} $\frac{\tanh ^2(x)}{2 a (a+b \tanh (x))^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sinh[x]/(a*Cosh[x] + b*Sinh[x])^3,x]

[Out]

Tanh[x]^2/(2*a*(a + b*Tanh[x])^2)

Rule 3087

Int[sin[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> Dist[1/d, Subst[Int[(x^m*(a + b*x)^n)/(1 + x^2)^((m + n + 2)/2), x], x, Tan[c + d*x]], x] /; FreeQ[{a,
b, c, d}, x] && IntegerQ[n] && IntegerQ[(m + n)/2] && NeQ[n, -1] &&  !(GtQ[n, 0] && GtQ[m, 1])

Rule 37

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

Rubi steps

\begin{align*} \int \frac{\sinh (x)}{(a \cosh (x)+b \sinh (x))^3} \, dx &=-\operatorname{Subst}\left (\int \frac{x}{(a-i b x)^3} \, dx,x,i \tanh (x)\right )\\ &=\frac{\tanh ^2(x)}{2 a (a+b \tanh (x))^2}\\ \end{align*}

Mathematica [B]  time = 0.114617, size = 54, normalized size = 2.84 $-\frac{a^2+a b \sinh (2 x)+b^2 \cosh (2 x)-b^2}{2 a (a-b) (a+b) (a \cosh (x)+b \sinh (x))^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sinh[x]/(a*Cosh[x] + b*Sinh[x])^3,x]

[Out]

-(a^2 - b^2 + b^2*Cosh[2*x] + a*b*Sinh[2*x])/(2*a*(a - b)*(a + b)*(a*Cosh[x] + b*Sinh[x])^2)

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Maple [A]  time = 0.068, size = 31, normalized size = 1.6 \begin{align*} 2\,{\frac{ \left ( \tanh \left ( x/2 \right ) \right ) ^{2}}{a \left ( a+2\,\tanh \left ( x/2 \right ) b+a \left ( \tanh \left ( x/2 \right ) \right ) ^{2} \right ) ^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)/(a*cosh(x)+b*sinh(x))^3,x)

[Out]

2/a*tanh(1/2*x)^2/(a+2*tanh(1/2*x)*b+a*tanh(1/2*x)^2)^2

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Maxima [B]  time = 1.26658, size = 225, normalized size = 11.84 \begin{align*} -\frac{2 \,{\left (a - b\right )} e^{\left (-2 \, x\right )}}{a^{4} - 2 \, a^{2} b^{2} + b^{4} + 2 \,{\left (a^{4} - 2 \, a^{3} b + 2 \, a b^{3} - b^{4}\right )} e^{\left (-2 \, x\right )} +{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} e^{\left (-4 \, x\right )}} - \frac{2 \, b}{a^{4} - 2 \, a^{2} b^{2} + b^{4} + 2 \,{\left (a^{4} - 2 \, a^{3} b + 2 \, a b^{3} - b^{4}\right )} e^{\left (-2 \, x\right )} +{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} e^{\left (-4 \, x\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)/(a*cosh(x)+b*sinh(x))^3,x, algorithm="maxima")

[Out]

-2*(a - b)*e^(-2*x)/(a^4 - 2*a^2*b^2 + b^4 + 2*(a^4 - 2*a^3*b + 2*a*b^3 - b^4)*e^(-2*x) + (a^4 - 4*a^3*b + 6*a
^2*b^2 - 4*a*b^3 + b^4)*e^(-4*x)) - 2*b/(a^4 - 2*a^2*b^2 + b^4 + 2*(a^4 - 2*a^3*b + 2*a*b^3 - b^4)*e^(-2*x) +
(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*e^(-4*x))

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Fricas [B]  time = 1.78601, size = 518, normalized size = 27.26 \begin{align*} -\frac{2 \,{\left (a \cosh \left (x\right ) +{\left (a + 2 \, b\right )} \sinh \left (x\right )\right )}}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \cosh \left (x\right )^{3} + 3 \,{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{2} +{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \sinh \left (x\right )^{3} +{\left (3 \, a^{4} + 4 \, a^{3} b - 2 \, a^{2} b^{2} - 4 \, a b^{3} - b^{4}\right )} \cosh \left (x\right ) +{\left (a^{4} + 4 \, a^{3} b + 2 \, a^{2} b^{2} - 4 \, a b^{3} - 3 \, b^{4} + 3 \,{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \cosh \left (x\right )^{2}\right )} \sinh \left (x\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)/(a*cosh(x)+b*sinh(x))^3,x, algorithm="fricas")

[Out]

-2*(a*cosh(x) + (a + 2*b)*sinh(x))/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*cosh(x)^3 + 3*(a^4 + 4*a^3*b +
6*a^2*b^2 + 4*a*b^3 + b^4)*cosh(x)*sinh(x)^2 + (a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*sinh(x)^3 + (3*a^4
+ 4*a^3*b - 2*a^2*b^2 - 4*a*b^3 - b^4)*cosh(x) + (a^4 + 4*a^3*b + 2*a^2*b^2 - 4*a*b^3 - 3*b^4 + 3*(a^4 + 4*a^
3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*cosh(x)^2)*sinh(x))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)/(a*cosh(x)+b*sinh(x))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.16817, size = 68, normalized size = 3.58 \begin{align*} -\frac{2 \,{\left (a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} - b\right )}}{{\left (a^{2} + 2 \, a b + b^{2}\right )}{\left (a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)/(a*cosh(x)+b*sinh(x))^3,x, algorithm="giac")

[Out]

-2*(a*e^(2*x) + b*e^(2*x) - b)/((a^2 + 2*a*b + b^2)*(a*e^(2*x) + b*e^(2*x) + a - b)^2)