\(\int \sinh (\frac {b x}{c+d x}) \, dx\) [292]

Optimal result

Integrand size = 11, antiderivative size = 74 \[ \int \sinh \left (\frac {b x}{c+d x}\right ) \, dx=\frac {b c \cosh \left (\frac {b}{d}\right ) \text {Chi}\left (\frac {b c}{d (c+d x)}\right )}{d^2}+\frac {(c+d x) \sinh \left (\frac {b x}{c+d x}\right )}{d}-\frac {b c \sinh \left (\frac {b}{d}\right ) \text {Shi}\left (\frac {b c}{d (c+d x)}\right )}{d^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {5726, 3378, 3384, 3379, 3382} \[ \int \sinh \left (\frac {b x}{c+d x}\right ) \, dx=\frac {b c \cosh \left (\frac {b}{d}\right ) \text {Chi}\left (\frac {b c}{d (c+d x)}\right )}{d^2}-\frac {b c \sinh \left (\frac {b}{d}\right ) \text {Shi}\left (\frac {b c}{d (c+d x)}\right )}{d^2}+\frac {(c+d x) \sinh \left (\frac {b x}{c+d x}\right )}{d} \]

[In]

Int[Sinh[(b*x)/(c + d*x)],x]

[Out]

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

Rule 3378

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

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.26 \[ \int \sinh \left (\frac {b x}{c+d x}\right ) \, dx=\frac {d e^{-\frac {b x}{c+d x}} \left (-1+e^{\frac {2 b x}{c+d x}}\right ) (c+d x)+2 b c \cosh \left (\frac {b}{d}\right ) \text {Chi}\left (\frac {b c}{c d+d^2 x}\right )-2 b c \sinh \left (\frac {b}{d}\right ) \text {Shi}\left (\frac {b c}{c d+d^2 x}\right )}{2 d^2} \]

[In]

Integrate[Sinh[(b*x)/(c + d*x)],x]

[Out]

((d*(-1 + E^((2*b*x)/(c + d*x)))*(c + d*x))/E^((b*x)/(c + d*x)) + 2*b*c*Cosh[b/d]*CoshIntegral[(b*c)/(c*d + d^
2*x)] - 2*b*c*Sinh[b/d]*SinhIntegral[(b*c)/(c*d + d^2*x)])/(2*d^2)

Maple [A] (verified)

Time = 1.09 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.53

method	result	size
risch	\(-\frac {{\mathrm e}^{-\frac {b x}{d x +c}} \left (d x +c \right )}{2 d}-\frac {b c \,{\mathrm e}^{-\frac {b}{d}} \operatorname {Ei}_{1}\left (-\frac {b c}{d \left (d x +c \right )}\right )}{2 d^{2}}+\frac {{\mathrm e}^{\frac {b x}{d x +c}} x}{2}+\frac {c \,{\mathrm e}^{\frac {b x}{d x +c}}}{2 d}-\frac {b c \,{\mathrm e}^{\frac {b}{d}} \operatorname {Ei}_{1}\left (\frac {b c}{d \left (d x +c \right )}\right )}{2 d^{2}}\)	\(113\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (74) = 148\).

Time = 0.27 (sec) , antiderivative size = 253, normalized size of antiderivative = 3.42 \[ \int \sinh \left (\frac {b x}{c+d x}\right ) \, dx=-\frac {b c {\rm Ei}\left (-\frac {b c}{d^{2} x + c d}\right ) \cosh \left (\frac {b}{d}\right ) \sinh \left (\frac {b x}{d x + c}\right )^{2} - {\left (b c {\rm Ei}\left (-\frac {b c}{d^{2} x + c d}\right ) \cosh \left (\frac {b x}{d x + c}\right )^{2} + b c {\rm Ei}\left (\frac {b c}{d^{2} x + c d}\right )\right )} \cosh \left (\frac {b}{d}\right ) - 2 \, {\left (d^{2} x + c d\right )} \sinh \left (\frac {b x}{d x + c}\right ) - {\left (b c {\rm Ei}\left (-\frac {b c}{d^{2} x + c d}\right ) \cosh \left (\frac {b x}{d x + c}\right )^{2} - b c {\rm Ei}\left (-\frac {b c}{d^{2} x + c d}\right ) \sinh \left (\frac {b x}{d x + c}\right )^{2} - b c {\rm Ei}\left (\frac {b c}{d^{2} x + c d}\right )\right )} \sinh \left (\frac {b}{d}\right )}{2 \, {\left (d^{2} \cosh \left (\frac {b x}{d x + c}\right )^{2} - d^{2} \sinh \left (\frac {b x}{d x + c}\right )^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \sinh \left (\frac {b x}{c+d x}\right ) \, dx=\int \sinh {\left (\frac {b x}{c + d x} \right )}\, dx \]

[In]

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

[Out]

Integral(sinh(b*x/(c + d*x)), x)

Maxima [F]

\[ \int \sinh \left (\frac {b x}{c+d x}\right ) \, dx=\int { \sinh \left (\frac {b x}{d x + c}\right ) \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \sinh \left (\frac {b x}{c+d x}\right ) \, dx=\int { \sinh \left (\frac {b x}{d x + c}\right ) \,d x } \]

[In]

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

[Out]

integrate(sinh(b*x/(d*x + c)), x)

Mupad [F(-1)]

Timed out. \[ \int \sinh \left (\frac {b x}{c+d x}\right ) \, dx=\int \mathrm {sinh}\left (\frac {b\,x}{c+d\,x}\right ) \,d x \]

[In]

int(sinh((b*x)/(c + d*x)),x)

[Out]

int(sinh((b*x)/(c + d*x)), x)