3.290 \(\int \sinh ^2(\frac{a}{c+d x}) \, dx\)

Optimal. Leaf size=39 \[ \frac{(c+d x) \sinh ^2\left (\frac{a}{c+d x}\right )}{d}-\frac{a \text{Shi}\left (\frac{2 a}{c+d x}\right )}{d} \]

[Out]

((c + d*x)*Sinh[a/(c + d*x)]^2)/d - (a*SinhIntegral[(2*a)/(c + d*x)])/d

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Rubi [A]  time = 0.0636995, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {5310, 5302, 3313, 12, 3298} \[ \frac{(c+d x) \sinh ^2\left (\frac{a}{c+d x}\right )}{d}-\frac{a \text{Shi}\left (\frac{2 a}{c+d x}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a/(c + d*x)]^2,x]

[Out]

((c + d*x)*Sinh[a/(c + d*x)]^2)/d - (a*SinhIntegral[(2*a)/(c + d*x)])/d

Rule 5310

Int[((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(u_)^(n_)])^(p_.), x_Symbol] :> Dist[1/Coefficient[u, x, 1], Subst[Int[(
a + b*Sinh[c + d*x^n])^p, x], x, u], x] /; FreeQ[{a, b, c, d, n}, x] && IntegerQ[p] && LinearQ[u, x] && NeQ[u,
x]

Rule 5302

Int[((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> -Subst[Int[(a + b*Sinh[c + d/x^n])^p/x^2
, x], x, 1/x] /; FreeQ[{a, b, c, d}, x] && ILtQ[n, 0] && IntegerQ[p]

Rule 3313

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x]^
n)/(d*(m + 1)), x] - Dist[(f*n)/(d*(m + 1)), Int[ExpandTrigReduce[(c + d*x)^(m + 1), Cos[e + f*x]*Sin[e + f*x]
^(n - 1), x], x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && GeQ[m, -2] && LtQ[m, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rubi steps

\begin{align*} \int \sinh ^2\left (\frac{a}{c+d x}\right ) \, dx &=\frac{\operatorname{Subst}\left (\int \sinh ^2\left (\frac{a}{x}\right ) \, dx,x,c+d x\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{\sinh ^2(a x)}{x^2} \, dx,x,\frac{1}{c+d x}\right )}{d}\\ &=\frac{(c+d x) \sinh ^2\left (\frac{a}{c+d x}\right )}{d}+\frac{(2 i a) \operatorname{Subst}\left (\int \frac{i \sinh (2 a x)}{2 x} \, dx,x,\frac{1}{c+d x}\right )}{d}\\ &=\frac{(c+d x) \sinh ^2\left (\frac{a}{c+d x}\right )}{d}-\frac{a \operatorname{Subst}\left (\int \frac{\sinh (2 a x)}{x} \, dx,x,\frac{1}{c+d x}\right )}{d}\\ &=\frac{(c+d x) \sinh ^2\left (\frac{a}{c+d x}\right )}{d}-\frac{a \text{Shi}\left (\frac{2 a}{c+d x}\right )}{d}\\ \end{align*}

Mathematica [A]  time = 0.0403094, size = 37, normalized size = 0.95 \[ \frac{(c+d x) \sinh ^2\left (\frac{a}{c+d x}\right )-a \text{Shi}\left (\frac{2 a}{c+d x}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a/(c + d*x)]^2,x]

[Out]

((c + d*x)*Sinh[a/(c + d*x)]^2 - a*SinhIntegral[(2*a)/(c + d*x)])/d

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Maple [A]  time = 0.016, size = 50, normalized size = 1.3 \begin{align*} -{\frac{a}{d} \left ({\frac{dx+c}{2\,a}}-{\frac{dx+c}{2\,a}\cosh \left ( 2\,{\frac{a}{dx+c}} \right ) }+{\it Shi} \left ( 2\,{\frac{a}{dx+c}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a/(d*x+c))^2,x)

[Out]

-1/d*a*(1/2/a*(d*x+c)-1/2/a*(d*x+c)*cosh(2*a/(d*x+c))+Shi(2*a/(d*x+c)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, a d \int \frac{x e^{\left (\frac{2 \, a}{d x + c}\right )}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\,{d x} - \frac{1}{2} \, a d \int \frac{x e^{\left (-\frac{2 \, a}{d x + c}\right )}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\,{d x} + \frac{1}{4} \, x e^{\left (\frac{2 \, a}{d x + c}\right )} + \frac{1}{4} \, x e^{\left (-\frac{2 \, a}{d x + c}\right )} - \frac{1}{2} \, x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a/(d*x+c))^2,x, algorithm="maxima")

[Out]

1/2*a*d*integrate(x*e^(2*a/(d*x + c))/(d^2*x^2 + 2*c*d*x + c^2), x) - 1/2*a*d*integrate(x*e^(-2*a/(d*x + c))/(
d^2*x^2 + 2*c*d*x + c^2), x) + 1/4*x*e^(2*a/(d*x + c)) + 1/4*x*e^(-2*a/(d*x + c)) - 1/2*x

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Fricas [A]  time = 2.04376, size = 165, normalized size = 4.23 \begin{align*} \frac{{\left (d x + c\right )} \cosh \left (\frac{a}{d x + c}\right )^{2} +{\left (d x + c\right )} \sinh \left (\frac{a}{d x + c}\right )^{2} - d x - a{\rm Ei}\left (\frac{2 \, a}{d x + c}\right ) + a{\rm Ei}\left (-\frac{2 \, a}{d x + c}\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a/(d*x+c))^2,x, algorithm="fricas")

[Out]

1/2*((d*x + c)*cosh(a/(d*x + c))^2 + (d*x + c)*sinh(a/(d*x + c))^2 - d*x - a*Ei(2*a/(d*x + c)) + a*Ei(-2*a/(d*
x + c)))/d

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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(a/(d*x+c))**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sinh \left (\frac{a}{d x + c}\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a/(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(sinh(a/(d*x + c))^2, x)