3.291 \(\int \sinh ^3(\frac {a}{c+d x}) \, dx\)

Optimal. Leaf size=59 \[ \frac {3 a \text {Chi}\left (\frac {a}{c+d x}\right )}{4 d}-\frac {3 a \text {Chi}\left (\frac {3 a}{c+d x}\right )}{4 d}+\frac {(c+d x) \sinh ^3\left (\frac {a}{c+d x}\right )}{d} \]

[Out]

3/4*a*Chi(a/(d*x+c))/d-3/4*a*Chi(3*a/(d*x+c))/d+(d*x+c)*sinh(a/(d*x+c))^3/d

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Rubi [A]  time = 0.09, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5310, 5302, 3313, 3301} \[ \frac {3 a \text {Chi}\left (\frac {a}{c+d x}\right )}{4 d}-\frac {3 a \text {Chi}\left (\frac {3 a}{c+d x}\right )}{4 d}+\frac {(c+d x) \sinh ^3\left (\frac {a}{c+d x}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a/(c + d*x)]^3,x]

[Out]

(3*a*CoshIntegral[a/(c + d*x)])/(4*d) - (3*a*CoshIntegral[(3*a)/(c + d*x)])/(4*d) + ((c + d*x)*Sinh[a/(c + d*x
)]^3)/d

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - 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 ^3\left (\frac {a}{c+d x}\right ) \, dx &=\frac {\operatorname {Subst}\left (\int \sinh ^3\left (\frac {a}{x}\right ) \, dx,x,c+d x\right )}{d}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {\sinh ^3(a x)}{x^2} \, dx,x,\frac {1}{c+d x}\right )}{d}\\ &=\frac {(c+d x) \sinh ^3\left (\frac {a}{c+d x}\right )}{d}+\frac {(3 a) \operatorname {Subst}\left (\int \left (\frac {\cosh (a x)}{4 x}-\frac {\cosh (3 a x)}{4 x}\right ) \, dx,x,\frac {1}{c+d x}\right )}{d}\\ &=\frac {(c+d x) \sinh ^3\left (\frac {a}{c+d x}\right )}{d}+\frac {(3 a) \operatorname {Subst}\left (\int \frac {\cosh (a x)}{x} \, dx,x,\frac {1}{c+d x}\right )}{4 d}-\frac {(3 a) \operatorname {Subst}\left (\int \frac {\cosh (3 a x)}{x} \, dx,x,\frac {1}{c+d x}\right )}{4 d}\\ &=\frac {3 a \text {Chi}\left (\frac {a}{c+d x}\right )}{4 d}-\frac {3 a \text {Chi}\left (\frac {3 a}{c+d x}\right )}{4 d}+\frac {(c+d x) \sinh ^3\left (\frac {a}{c+d x}\right )}{d}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 54, normalized size = 0.92 \[ \frac {3 a \text {Chi}\left (\frac {a}{c+d x}\right )-3 a \text {Chi}\left (\frac {3 a}{c+d x}\right )+4 (c+d x) \sinh ^3\left (\frac {a}{c+d x}\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a/(c + d*x)]^3,x]

[Out]

(3*a*CoshIntegral[a/(c + d*x)] - 3*a*CoshIntegral[(3*a)/(c + d*x)] + 4*(c + d*x)*Sinh[a/(c + d*x)]^3)/(4*d)

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fricas [B]  time = 1.14, size = 118, normalized size = 2.00 \[ \frac {2 \, {\left (d x + c\right )} \sinh \left (\frac {a}{d x + c}\right )^{3} - 3 \, a {\rm Ei}\left (\frac {3 \, a}{d x + c}\right ) + 3 \, a {\rm Ei}\left (\frac {a}{d x + c}\right ) + 3 \, a {\rm Ei}\left (-\frac {a}{d x + c}\right ) - 3 \, a {\rm Ei}\left (-\frac {3 \, a}{d x + c}\right ) + 6 \, {\left ({\left (d x + c\right )} \cosh \left (\frac {a}{d x + c}\right )^{2} - d x - c\right )} \sinh \left (\frac {a}{d x + c}\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*(d*x + c)*sinh(a/(d*x + c))^3 - 3*a*Ei(3*a/(d*x + c)) + 3*a*Ei(a/(d*x + c)) + 3*a*Ei(-a/(d*x + c)) - 3*
a*Ei(-3*a/(d*x + c)) + 6*((d*x + c)*cosh(a/(d*x + c))^2 - d*x - c)*sinh(a/(d*x + c)))/d

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giac [B]  time = 3.90, size = 167, normalized size = 2.83 \[ -\frac {{\left (\frac {3 \, a^{3} {\rm Ei}\left (\frac {3 \, a}{d x + c}\right )}{d x + c} - \frac {3 \, a^{3} {\rm Ei}\left (\frac {a}{d x + c}\right )}{d x + c} - \frac {3 \, a^{3} {\rm Ei}\left (-\frac {a}{d x + c}\right )}{d x + c} + \frac {3 \, a^{3} {\rm Ei}\left (-\frac {3 \, a}{d x + c}\right )}{d x + c} - a^{2} e^{\left (\frac {3 \, a}{d x + c}\right )} + 3 \, a^{2} e^{\left (\frac {a}{d x + c}\right )} - 3 \, a^{2} e^{\left (-\frac {a}{d x + c}\right )} + a^{2} e^{\left (-\frac {3 \, a}{d x + c}\right )}\right )} {\left (d x + c\right )}}{8 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(3*a^3*Ei(3*a/(d*x + c))/(d*x + c) - 3*a^3*Ei(a/(d*x + c))/(d*x + c) - 3*a^3*Ei(-a/(d*x + c))/(d*x + c) +
3*a^3*Ei(-3*a/(d*x + c))/(d*x + c) - a^2*e^(3*a/(d*x + c)) + 3*a^2*e^(a/(d*x + c)) - 3*a^2*e^(-a/(d*x + c)) +
a^2*e^(-3*a/(d*x + c)))*(d*x + c)/(a^2*d)

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maple [A]  time = 0.03, size = 74, normalized size = 1.25 \[ -\frac {a \left (\frac {3 \left (d x +c \right ) \sinh \left (\frac {a}{d x +c}\right )}{4 a}-\frac {3 \Chi \left (\frac {a}{d x +c}\right )}{4}-\frac {\left (d x +c \right ) \sinh \left (\frac {3 a}{d x +c}\right )}{4 a}+\frac {3 \Chi \left (\frac {3 a}{d x +c}\right )}{4}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a/(d*x+c))^3,x)

[Out]

-1/d*a*(3/4/a*(d*x+c)*sinh(a/(d*x+c))-3/4*Chi(a/(d*x+c))-1/4/a*(d*x+c)*sinh(3*a/(d*x+c))+3/4*Chi(3*a/(d*x+c)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {3}{8} \, a d \int \frac {x e^{\left (\frac {3 \, a}{d x + c}\right )}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\,{d x} - \frac {3}{8} \, a d \int \frac {x e^{\left (\frac {a}{d x + c}\right )}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\,{d x} - \frac {3}{8} \, a d \int \frac {x e^{\left (-\frac {a}{d x + c}\right )}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\,{d x} + \frac {3}{8} \, a d \int \frac {x e^{\left (-\frac {3 \, a}{d x + c}\right )}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\,{d x} + \frac {1}{8} \, x e^{\left (\frac {3 \, a}{d x + c}\right )} - \frac {3}{8} \, x e^{\left (\frac {a}{d x + c}\right )} + \frac {3}{8} \, x e^{\left (-\frac {a}{d x + c}\right )} - \frac {1}{8} \, x e^{\left (-\frac {3 \, a}{d x + c}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*a*d*integrate(x*e^(3*a/(d*x + c))/(d^2*x^2 + 2*c*d*x + c^2), x) - 3/8*a*d*integrate(x*e^(a/(d*x + c))/(d^2
*x^2 + 2*c*d*x + c^2), x) - 3/8*a*d*integrate(x*e^(-a/(d*x + c))/(d^2*x^2 + 2*c*d*x + c^2), x) + 3/8*a*d*integ
rate(x*e^(-3*a/(d*x + c))/(d^2*x^2 + 2*c*d*x + c^2), x) + 1/8*x*e^(3*a/(d*x + c)) - 3/8*x*e^(a/(d*x + c)) + 3/
8*x*e^(-a/(d*x + c)) - 1/8*x*e^(-3*a/(d*x + c))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\mathrm {sinh}\left (\frac {a}{c+d\,x}\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a/(c + d*x))^3,x)

[Out]

int(sinh(a/(c + d*x))^3, 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))**3,x)

[Out]

Timed out

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