3.295 \(\int \sinh (\frac{a+b x}{c+d x}) \, dx\)

Optimal. Leaf size=101 \[ \frac{\cosh \left (\frac{b}{d}\right ) (b c-a d) \text{Chi}\left (\frac{b c-a d}{d (c+d x)}\right )}{d^2}-\frac{\sinh \left (\frac{b}{d}\right ) (b c-a d) \text{Shi}\left (\frac{b c-a d}{d (c+d x)}\right )}{d^2}+\frac{(c+d x) \sinh \left (\frac{a+b x}{c+d x}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.178836, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {5607, 3297, 3303, 3298, 3301} \[ \frac{\cosh \left (\frac{b}{d}\right ) (b c-a d) \text{Chi}\left (\frac{b c-a d}{d (c+d x)}\right )}{d^2}-\frac{\sinh \left (\frac{b}{d}\right ) (b c-a d) \text{Shi}\left (\frac{b c-a d}{d (c+d x)}\right )}{d^2}+\frac{(c+d x) \sinh \left (\frac{a+b x}{c+d x}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[(a + b*x)/(c + d*x)],x]

[Out]

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

Rule 5607

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]

Rule 3297

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 3303

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 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 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]

Rubi steps

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

Mathematica [B]  time = 0.706277, size = 373, normalized size = 3.69 \[ \frac{(b c-a d) \left (\cosh \left (\frac{b}{d}\right )-\sinh \left (\frac{b}{d}\right )\right ) \text{Chi}\left (\frac{b c-a d}{x d^2+c d}\right )+(b c-a d) \left (\sinh \left (\frac{b}{d}\right )+\cosh \left (\frac{b}{d}\right )\right ) \text{Chi}\left (\frac{a d-b c}{d (c+d x)}\right )+a d \sinh \left (\frac{b}{d}\right ) \text{Shi}\left (\frac{b c-a d}{x d^2+c d}\right )-b c \sinh \left (\frac{b}{d}\right ) \text{Shi}\left (\frac{b c-a d}{x d^2+c d}\right )-a d \cosh \left (\frac{b}{d}\right ) \text{Shi}\left (\frac{b c-a d}{x d^2+c d}\right )+b c \cosh \left (\frac{b}{d}\right ) \text{Shi}\left (\frac{b c-a d}{x d^2+c d}\right )+2 d^2 x \sinh \left (\frac{a+b x}{c+d x}\right )-a d \sinh \left (\frac{b}{d}\right ) \text{Shi}\left (\frac{a d-b c}{d (c+d x)}\right )+b c \sinh \left (\frac{b}{d}\right ) \text{Shi}\left (\frac{a d-b c}{d (c+d x)}\right )-a d \cosh \left (\frac{b}{d}\right ) \text{Shi}\left (\frac{a d-b c}{d (c+d x)}\right )+b c \cosh \left (\frac{b}{d}\right ) \text{Shi}\left (\frac{a d-b c}{d (c+d x)}\right )+2 c d \sinh \left (\frac{a+b x}{c+d x}\right )}{2 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[(a + b*x)/(c + d*x)],x]

[Out]

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

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Maple [B]  time = 0.028, size = 347, normalized size = 3.4 \begin{align*} -{\frac{a}{2}{{\rm e}^{-{\frac{bx+a}{dx+c}}}} \left ({\frac{da}{dx+c}}-{\frac{cb}{dx+c}} \right ) ^{-1}}+{\frac{cb}{2\,d}{{\rm e}^{-{\frac{bx+a}{dx+c}}}} \left ({\frac{da}{dx+c}}-{\frac{cb}{dx+c}} \right ) ^{-1}}+{\frac{a}{2\,d}{{\rm e}^{-{\frac{b}{d}}}}{\it Ei} \left ( 1,{\frac{da-cb}{d \left ( dx+c \right ) }} \right ) }-{\frac{cb}{2\,{d}^{2}}{{\rm e}^{-{\frac{b}{d}}}}{\it Ei} \left ( 1,{\frac{da-cb}{d \left ( dx+c \right ) }} \right ) }+{\frac{dxa}{2\,da-2\,cb}{{\rm e}^{{\frac{bx+a}{dx+c}}}}}-{\frac{bcx}{2\,da-2\,cb}{{\rm e}^{{\frac{bx+a}{dx+c}}}}}+{\frac{ac}{2\,da-2\,cb}{{\rm e}^{{\frac{bx+a}{dx+c}}}}}-{\frac{b{c}^{2}}{2\,d \left ( da-cb \right ) }{{\rm e}^{{\frac{bx+a}{dx+c}}}}}+{\frac{a}{2\,d}{{\rm e}^{{\frac{b}{d}}}}{\it Ei} \left ( 1,-{\frac{da-cb}{d \left ( dx+c \right ) }} \right ) }-{\frac{cb}{2\,{d}^{2}}{{\rm e}^{{\frac{b}{d}}}}{\it Ei} \left ( 1,-{\frac{da-cb}{d \left ( dx+c \right ) }} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.04179, size = 344, normalized size = 3.41 \begin{align*} \frac{{\left ({\left (b c - a d\right )}{\rm Ei}\left (\frac{b c - a d}{d^{2} x + c d}\right ) +{\left (b c - a d\right )}{\rm Ei}\left (-\frac{b c - a d}{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 + a}{d x + c}\right ) -{\left ({\left (b c - a d\right )}{\rm Ei}\left (\frac{b c - a d}{d^{2} x + c d}\right ) -{\left (b c - a d\right )}{\rm Ei}\left (-\frac{b c - a d}{d^{2} x + c d}\right )\right )} \sinh \left (\frac{b}{d}\right )}{2 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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