3.3.93 \(\int \sinh ^2(\frac {b x}{c+d x}) \, dx\) [293]

Optimal. Leaf size=80 \[ \frac {b c \text {Chi}\left (\frac {2 b c}{d (c+d x)}\right ) \sinh \left (\frac {2 b}{d}\right )}{d^2}+\frac {(c+d x) \sinh ^2\left (\frac {b x}{c+d x}\right )}{d}-\frac {b c \cosh \left (\frac {2 b}{d}\right ) \text {Shi}\left (\frac {2 b c}{d (c+d x)}\right )}{d^2} \]

[Out]

-b*c*cosh(2*b/d)*Shi(2*b*c/d/(d*x+c))/d^2+b*c*Chi(2*b*c/d/(d*x+c))*sinh(2*b/d)/d^2+(d*x+c)*sinh(b*x/(d*x+c))^2
/d

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Rubi [A]
time = 0.11, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {5726, 3394, 12, 3384, 3379, 3382} \begin {gather*} \frac {b c \sinh \left (\frac {2 b}{d}\right ) \text {Chi}\left (\frac {2 b c}{d (c+d x)}\right )}{d^2}-\frac {b c \cosh \left (\frac {2 b}{d}\right ) \text {Shi}\left (\frac {2 b c}{d (c+d x)}\right )}{d^2}+\frac {(c+d x) \sinh ^2\left (\frac {b x}{c+d x}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sinh[(b*x)/(c + d*x)]^2,x]

[Out]

(b*c*CoshIntegral[(2*b*c)/(d*(c + d*x))]*Sinh[(2*b)/d])/d^2 + ((c + d*x)*Sinh[(b*x)/(c + d*x)]^2)/d - (b*c*Cos
h[(2*b)/d]*SinhIntegral[(2*b*c)/(d*(c + d*x))])/d^2

Rule 12

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

Rule 3379

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 3382

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 3384

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3394

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 5726

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

Rubi steps

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

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Mathematica [A]
time = 0.29, size = 85, normalized size = 1.06 \begin {gather*} \frac {d \left (-d x+(c+d x) \cosh \left (\frac {2 b x}{c+d x}\right )\right )+2 b c \text {Chi}\left (\frac {2 b c}{d (c+d x)}\right ) \sinh \left (\frac {2 b}{d}\right )-2 b c \cosh \left (\frac {2 b}{d}\right ) \text {Shi}\left (\frac {2 b c}{d (c+d x)}\right )}{2 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sinh[(b*x)/(c + d*x)]^2,x]

[Out]

(d*(-(d*x) + (c + d*x)*Cosh[(2*b*x)/(c + d*x)]) + 2*b*c*CoshIntegral[(2*b*c)/(d*(c + d*x))]*Sinh[(2*b)/d] - 2*
b*c*Cosh[(2*b)/d]*SinhIntegral[(2*b*c)/(d*(c + d*x))])/(2*d^2)

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Maple [A]
time = 18.35, size = 120, normalized size = 1.50

method	result	size
risch	\(-\frac {x}{2}+\frac {{\mathrm e}^{-\frac {2 b x}{d x +c}} \left (d x +c \right )}{4 d}+\frac {b c \,{\mathrm e}^{-\frac {2 b}{d}} \expIntegral \left (1, -\frac {2 b c}{d \left (d x +c \right )}\right )}{2 d^{2}}+\frac {{\mathrm e}^{\frac {2 b x}{d x +c}} x}{4}+\frac {c \,{\mathrm e}^{\frac {2 b x}{d x +c}}}{4 d}-\frac {b c \,{\mathrm e}^{\frac {2 b}{d}} \expIntegral \left (1, \frac {2 b c}{d \left (d x +c \right )}\right )}{2 d^{2}}\)	\(120\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(b*x/(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-1/2*x+1/4/d*exp(-2*b*x/(d*x+c))*(d*x+c)+1/2*b*c/d^2*exp(-2*b/d)*Ei(1,-2*b*c/d/(d*x+c))+1/4*exp(2*b*x/(d*x+c))
*x+1/4*c/d*exp(2*b*x/(d*x+c))-1/2*b*c/d^2*exp(2*b/d)*Ei(1,2*b*c/d/(d*x+c))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (80) = 160\).
time = 0.52, size = 277, normalized size = 3.46 \begin {gather*} -\frac {d^{2} x - {\left (d^{2} x + c d\right )} \cosh \left (\frac {b x}{d x + c}\right )^{2} + {\left (b c {\rm Ei}\left (-\frac {2 \, b c}{d^{2} x + c d}\right ) \cosh \left (\frac {2 \, b}{d}\right ) - d^{2} x - c d\right )} \sinh \left (\frac {b x}{d x + c}\right )^{2} - {\left (b c {\rm Ei}\left (-\frac {2 \, b c}{d^{2} x + c d}\right ) \cosh \left (\frac {b x}{d x + c}\right )^{2} - b c {\rm Ei}\left (\frac {2 \, b c}{d^{2} x + c d}\right )\right )} \cosh \left (\frac {2 \, b}{d}\right ) - {\left (b c {\rm Ei}\left (-\frac {2 \, b c}{d^{2} x + c d}\right ) \cosh \left (\frac {b x}{d x + c}\right )^{2} - b c {\rm Ei}\left (-\frac {2 \, b c}{d^{2} x + c d}\right ) \sinh \left (\frac {b x}{d x + c}\right )^{2} + b c {\rm Ei}\left (\frac {2 \, b c}{d^{2} x + c d}\right )\right )} \sinh \left (\frac {2 \, b}{d}\right )}{2 \, {\left (d^{2} \cosh \left (\frac {b x}{d x + c}\right )^{2} - d^{2} \sinh \left (\frac {b x}{d x + c}\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(d^2*x - (d^2*x + c*d)*cosh(b*x/(d*x + c))^2 + (b*c*Ei(-2*b*c/(d^2*x + c*d))*cosh(2*b/d) - d^2*x - c*d)*s
inh(b*x/(d*x + c))^2 - (b*c*Ei(-2*b*c/(d^2*x + c*d))*cosh(b*x/(d*x + c))^2 - b*c*Ei(2*b*c/(d^2*x + c*d)))*cosh
(2*b/d) - (b*c*Ei(-2*b*c/(d^2*x + c*d))*cosh(b*x/(d*x + c))^2 - b*c*Ei(-2*b*c/(d^2*x + c*d))*sinh(b*x/(d*x + c
))^2 + b*c*Ei(2*b*c/(d^2*x + c*d)))*sinh(2*b/d))/(d^2*cosh(b*x/(d*x + c))^2 - d^2*sinh(b*x/(d*x + c))^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sinh ^{2}{\left (\frac {b x}{c + d x} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x/(d*x+c))**2,x)

[Out]

Integral(sinh(b*x/(c + d*x))**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(b*x/(d*x + c))^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {sinh}\left (\frac {b\,x}{c+d\,x}\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh((b*x)/(c + d*x))^2,x)

[Out]

int(sinh((b*x)/(c + d*x))^2, x)

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