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

Optimal. Leaf size=121 \[ \frac{f (b c-a d) \cosh \left (\frac{b f}{d}+e\right ) \text{Chi}\left (\frac{(b c-a d) f}{d (c+d x)}\right )}{d^2}-\frac{f (b c-a d) \sinh \left (\frac{b f}{d}+e\right ) \text{Shi}\left (\frac{(b c-a d) f}{d (c+d x)}\right )}{d^2}+\frac{(c+d x) \sinh \left (\frac{a f+b f x+c e+d e x}{c+d x}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.259482, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {5609, 5607, 3297, 3303, 3298, 3301} \[ \frac{f (b c-a d) \cosh \left (\frac{b f}{d}+e\right ) \text{Chi}\left (\frac{(b c-a d) f}{d (c+d x)}\right )}{d^2}-\frac{f (b c-a d) \sinh \left (\frac{b f}{d}+e\right ) \text{Shi}\left (\frac{(b c-a d) f}{d (c+d x)}\right )}{d^2}+\frac{(c+d x) \sinh \left (\frac{a f+b f x+c e+d e x}{c+d x}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[e + (f*(a + b*x))/(c + d*x)],x]

[Out]

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

Rule 5609

Int[Sinh[u_]^(n_.), x_Symbol] :> With[{lst = QuotientOfLinearsParts[u, x]}, Int[Sinh[(lst[[1]] + lst[[2]]*x)/(
lst[[3]] + lst[[4]]*x)]^n, x]] /; IGtQ[n, 0] && QuotientOfLinearsQ[u, x]

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

Mathematica [B]  time = 1.43234, size = 449, normalized size = 3.71 \[ \frac{f (b c-a d) \left (\cosh \left (\frac{b f}{d}+e\right )-\sinh \left (\frac{b f}{d}+e\right )\right ) \text{Chi}\left (\frac{(b c-a d) f}{d (c+d x)}\right )+f (b c-a d) \left (\sinh \left (\frac{b f}{d}+e\right )+\cosh \left (\frac{b f}{d}+e\right )\right ) \text{Chi}\left (\frac{a d f-b c f}{d (c+d x)}\right )+2 d^2 x \sinh \left (\frac{a f+b f x+c e+d e x}{c+d x}\right )+a d f \sinh \left (\frac{b f}{d}+e\right ) \text{Shi}\left (\frac{(b c-a d) f}{d (c+d x)}\right )-a d f \sinh \left (\frac{b f}{d}+e\right ) \text{Shi}\left (\frac{a d f-b c f}{d (c+d x)}\right )-b c f \sinh \left (\frac{b f}{d}+e\right ) \text{Shi}\left (\frac{(b c-a d) f}{d (c+d x)}\right )+b c f \sinh \left (\frac{b f}{d}+e\right ) \text{Shi}\left (\frac{a d f-b c f}{d (c+d x)}\right )-a d f \cosh \left (\frac{b f}{d}+e\right ) \text{Shi}\left (\frac{(b c-a d) f}{d (c+d x)}\right )-a d f \cosh \left (\frac{b f}{d}+e\right ) \text{Shi}\left (\frac{a d f-b c f}{d (c+d x)}\right )+b c f \cosh \left (\frac{b f}{d}+e\right ) \text{Shi}\left (\frac{(b c-a d) f}{d (c+d x)}\right )+b c f \cosh \left (\frac{b f}{d}+e\right ) \text{Shi}\left (\frac{a d f-b c f}{d (c+d x)}\right )+2 c d \sinh \left (\frac{a f+b f x+c e+d e x}{c+d x}\right )}{2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)*f*CoshIntegral[((b*c - a*d)*f)/(d*(c + d*x))]*(Cosh[e + (b*f)/d] - Sinh[e + (b*f)/d]) + (b*c - a*
d)*f*CoshIntegral[(-(b*c*f) + a*d*f)/(d*(c + d*x))]*(Cosh[e + (b*f)/d] + Sinh[e + (b*f)/d]) + 2*c*d*Sinh[(c*e
+ a*f + d*e*x + b*f*x)/(c + d*x)] + 2*d^2*x*Sinh[(c*e + a*f + d*e*x + b*f*x)/(c + d*x)] + b*c*f*Cosh[e + (b*f)
/d]*SinhIntegral[((b*c - a*d)*f)/(d*(c + d*x))] - a*d*f*Cosh[e + (b*f)/d]*SinhIntegral[((b*c - a*d)*f)/(d*(c +
d*x))] - b*c*f*Sinh[e + (b*f)/d]*SinhIntegral[((b*c - a*d)*f)/(d*(c + d*x))] + a*d*f*Sinh[e + (b*f)/d]*SinhIn
tegral[((b*c - a*d)*f)/(d*(c + d*x))] + b*c*f*Cosh[e + (b*f)/d]*SinhIntegral[(-(b*c*f) + a*d*f)/(d*(c + d*x))]
- a*d*f*Cosh[e + (b*f)/d]*SinhIntegral[(-(b*c*f) + a*d*f)/(d*(c + d*x))] + b*c*f*Sinh[e + (b*f)/d]*SinhIntegr
al[(-(b*c*f) + a*d*f)/(d*(c + d*x))] - a*d*f*Sinh[e + (b*f)/d]*SinhIntegral[(-(b*c*f) + a*d*f)/(d*(c + d*x))])
/(2*d^2)

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Maple [B]  time = 0.043, size = 459, normalized size = 3.8 \begin{align*} -{\frac{af}{2}{{\rm e}^{-{\frac{bfx+dex+af+ce}{dx+c}}}} \left ({\frac{adf}{dx+c}}-{\frac{bcf}{dx+c}} \right ) ^{-1}}+{\frac{bcf}{2\,d}{{\rm e}^{-{\frac{bfx+dex+af+ce}{dx+c}}}} \left ({\frac{adf}{dx+c}}-{\frac{bcf}{dx+c}} \right ) ^{-1}}+{\frac{af}{2\,d}{{\rm e}^{-{\frac{bf+de}{d}}}}{\it Ei} \left ( 1,{\frac{f \left ( da-cb \right ) }{d \left ( dx+c \right ) }} \right ) }-{\frac{bcf}{2\,{d}^{2}}{{\rm e}^{-{\frac{bf+de}{d}}}}{\it Ei} \left ( 1,{\frac{f \left ( da-cb \right ) }{d \left ( dx+c \right ) }} \right ) }+{\frac{af}{2\,d}{{\rm e}^{{\frac{bfx+dex+af+ce}{dx+c}}}} \left ({\frac{af}{dx+c}}-{\frac{bcf}{d \left ( dx+c \right ) }} \right ) ^{-1}}-{\frac{bcf}{2\,{d}^{2}}{{\rm e}^{{\frac{bfx+dex+af+ce}{dx+c}}}} \left ({\frac{af}{dx+c}}-{\frac{bcf}{d \left ( dx+c \right ) }} \right ) ^{-1}}+{\frac{af}{2\,d}{{\rm e}^{{\frac{bf+de}{d}}}}{\it Ei} \left ( 1,-{\frac{f \left ( da-cb \right ) }{d \left ( dx+c \right ) }}-{\frac{bf+de}{d}}-{\frac{-bf-de}{d}} \right ) }-{\frac{bcf}{2\,{d}^{2}}{{\rm e}^{{\frac{bf+de}{d}}}}{\it Ei} \left ( 1,-{\frac{f \left ( da-cb \right ) }{d \left ( dx+c \right ) }}-{\frac{bf+de}{d}}-{\frac{-bf-de}{d}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.04329, size = 417, normalized size = 3.45 \begin{align*} \frac{{\left ({\left (b c - a d\right )} f{\rm Ei}\left (\frac{{\left (b c - a d\right )} f}{d^{2} x + c d}\right ) +{\left (b c - a d\right )} f{\rm Ei}\left (-\frac{{\left (b c - a d\right )} f}{d^{2} x + c d}\right )\right )} \cosh \left (\frac{d e + b f}{d}\right ) + 2 \,{\left (d^{2} x + c d\right )} \sinh \left (\frac{c e + a f +{\left (d e + b f\right )} x}{d x + c}\right ) -{\left ({\left (b c - a d\right )} f{\rm Ei}\left (\frac{{\left (b c - a d\right )} f}{d^{2} x + c d}\right ) -{\left (b c - a d\right )} f{\rm Ei}\left (-\frac{{\left (b c - a d\right )} f}{d^{2} x + c d}\right )\right )} \sinh \left (\frac{d e + b f}{d}\right )}{2 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(((b*c - a*d)*f*Ei((b*c - a*d)*f/(d^2*x + c*d)) + (b*c - a*d)*f*Ei(-(b*c - a*d)*f/(d^2*x + c*d)))*cosh((d*
e + b*f)/d) + 2*(d^2*x + c*d)*sinh((c*e + a*f + (d*e + b*f)*x)/(d*x + c)) - ((b*c - a*d)*f*Ei((b*c - a*d)*f/(d
^2*x + c*d)) - (b*c - a*d)*f*Ei(-(b*c - a*d)*f/(d^2*x + c*d)))*sinh((d*e + b*f)/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(e+f*(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 (e + \frac{{\left (b x + a\right )} f}{d x + c}\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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