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]

-a*Shi(2*a/(d*x+c))/d+(d*x+c)*sinh(a/(d*x+c))^2/d

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Rubi [A]  time = 0.06, antiderivative size = 39, normalized size of antiderivative = 1.00, 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.04, 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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fricas [A]  time = 1.00, size = 73, normalized size = 1.87 \[ \frac {{\left (d x + c\right )} \cosh \left (\frac {a}{d x + c}\right )^{2} + {\left (d x + c\right )} \sinh \left (\frac {a}{d x + c}\right )^{2} - d x - a {\rm Ei}\left (\frac {2 \, a}{d x + c}\right ) + a {\rm Ei}\left (-\frac {2 \, a}{d x + c}\right )}{2 \, d} \]

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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giac [B]  time = 3.08, size = 97, normalized size = 2.49 \[ -\frac {{\left (\frac {2 \, a^{3} {\rm Ei}\left (\frac {2 \, a}{d x + c}\right )}{d x + c} - \frac {2 \, a^{3} {\rm Ei}\left (-\frac {2 \, a}{d x + c}\right )}{d x + c} - a^{2} e^{\left (\frac {2 \, a}{d x + c}\right )} - a^{2} e^{\left (-\frac {2 \, a}{d x + c}\right )} + 2 \, a^{2}\right )} {\left (d x + c\right )}}{4 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*a^3*Ei(2*a/(d*x + c))/(d*x + c) - 2*a^3*Ei(-2*a/(d*x + c))/(d*x + c) - a^2*e^(2*a/(d*x + c)) - a^2*e^(
-2*a/(d*x + c)) + 2*a^2)*(d*x + c)/(a^2*d)

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maple [A]  time = 0.04, size = 50, normalized size = 1.28 \[ -\frac {a \left (\frac {d x +c}{2 a}-\frac {\left (d x +c \right ) \cosh \left (\frac {2 a}{d x +c}\right )}{2 a}+\Shi \left (\frac {2 a}{d x +c}\right )\right )}{d} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, a d \int \frac {x e^{\left (\frac {2 \, a}{d x + c}\right )}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\,{d x} - \frac {1}{2} \, a d \int \frac {x e^{\left (-\frac {2 \, a}{d x + c}\right )}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\,{d x} + \frac {1}{4} \, x e^{\left (\frac {2 \, a}{d x + c}\right )} + \frac {1}{4} \, x e^{\left (-\frac {2 \, a}{d x + c}\right )} - \frac {1}{2} \, x \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int {\mathrm {sinh}\left (\frac {a}{c+d\,x}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a/(c + d*x))^2,x)

[Out]

int(sinh(a/(c + d*x))^2, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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