3.299 \(\int \sinh ^2(e+\frac{f (a+b x)}{c+d x}) \, dx\)

Optimal. Leaf size=129 \[ \frac{f (b c-a d) \sinh \left (2 \left (\frac{b f}{d}+e\right )\right ) \text{Chi}\left (\frac{2 (b c-a d) f}{d (c+d x)}\right )}{d^2}-\frac{f (b c-a d) \cosh \left (2 \left (\frac{b f}{d}+e\right )\right ) \text{Shi}\left (\frac{2 (b c-a d) f}{d (c+d x)}\right )}{d^2}+\frac{(c+d x) \sinh ^2\left (\frac{a f+b f x+c e+d e x}{c+d x}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.279059, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {5609, 5607, 3313, 12, 3303, 3298, 3301} \[ \frac{f (b c-a d) \sinh \left (2 \left (\frac{b f}{d}+e\right )\right ) \text{Chi}\left (\frac{2 (b c-a d) f}{d (c+d x)}\right )}{d^2}-\frac{f (b c-a d) \cosh \left (2 \left (\frac{b f}{d}+e\right )\right ) \text{Shi}\left (\frac{2 (b c-a d) f}{d (c+d x)}\right )}{d^2}+\frac{(c+d x) \sinh ^2\left (\frac{a f+b f x+c e+d e x}{c+d x}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[e + (f*(a + b*x))/(c + d*x)]^2,x]

[Out]

((b*c - a*d)*f*CoshIntegral[(2*(b*c - a*d)*f)/(d*(c + d*x))]*Sinh[2*(e + (b*f)/d)])/d^2 + ((c + d*x)*Sinh[(c*e
+ a*f + d*e*x + b*f*x)/(c + d*x)]^2)/d - ((b*c - a*d)*f*Cosh[2*(e + (b*f)/d)]*SinhIntegral[(2*(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 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 12

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

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

Mathematica [A]  time = 2.00019, size = 136, normalized size = 1.05 \[ \frac{2 f (b c-a d) \sinh \left (2 \left (\frac{b f}{d}+e\right )\right ) \text{Chi}\left (\frac{2 (a d f-b c f)}{d (c+d x)}\right )+2 f (b c-a d) \cosh \left (2 \left (\frac{b f}{d}+e\right )\right ) \text{Shi}\left (\frac{2 (a d f-b c f)}{d (c+d x)}\right )+d \left ((c+d x) \cosh \left (\frac{2 (a f+b f x+c e+d e x)}{c+d x}\right )-d x\right )}{2 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[e + (f*(a + b*x))/(c + d*x)]^2,x]

[Out]

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

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Maple [B]  time = 0.097, size = 468, normalized size = 3.6 \begin{align*} -{\frac{x}{2}}+{\frac{af}{4}{{\rm e}^{-2\,{\frac{bfx+dex+af+ce}{dx+c}}}} \left ({\frac{adf}{dx+c}}-{\frac{bcf}{dx+c}} \right ) ^{-1}}-{\frac{bcf}{4\,d}{{\rm e}^{-2\,{\frac{bfx+dex+af+ce}{dx+c}}}} \left ({\frac{adf}{dx+c}}-{\frac{bcf}{dx+c}} \right ) ^{-1}}-{\frac{af}{2\,d}{{\rm e}^{-2\,{\frac{bf+de}{d}}}}{\it Ei} \left ( 1,2\,{\frac{f \left ( da-cb \right ) }{d \left ( dx+c \right ) }} \right ) }+{\frac{bcf}{2\,{d}^{2}}{{\rm e}^{-2\,{\frac{bf+de}{d}}}}{\it Ei} \left ( 1,2\,{\frac{f \left ( da-cb \right ) }{d \left ( dx+c \right ) }} \right ) }+{\frac{af}{4\,d}{{\rm e}^{2\,{\frac{bfx+dex+af+ce}{dx+c}}}} \left ({\frac{af}{dx+c}}-{\frac{bcf}{d \left ( dx+c \right ) }} \right ) ^{-1}}-{\frac{bcf}{4\,{d}^{2}}{{\rm e}^{2\,{\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}^{2\,{\frac{bf+de}{d}}}}{\it Ei} \left ( 1,-2\,{\frac{f \left ( da-cb \right ) }{d \left ( dx+c \right ) }}-2\,{\frac{bf+de}{d}}-2\,{\frac{-bf-de}{d}} \right ) }-{\frac{bcf}{2\,{d}^{2}}{{\rm e}^{2\,{\frac{bf+de}{d}}}}{\it Ei} \left ( 1,-2\,{\frac{f \left ( da-cb \right ) }{d \left ( dx+c \right ) }}-2\,{\frac{bf+de}{d}}-2\,{\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))^2,x)

[Out]

-1/2*x+1/4*f*exp(-2*(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-1/4/d*f*exp(-2*(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(-2*(b*f+d*e)/d)*Ei(1,2*(a*d-b*c)*f/d/(d*x+c))
*a+1/2/d^2*f*exp(-2*(b*f+d*e)/d)*Ei(1,2*(a*d-b*c)*f/d/(d*x+c))*c*b+1/4/d*f*exp(2*(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-1/4/d^2*f*exp(2*(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(2*(b*f+d*e)/d)*Ei(1,-2*(a*d-b*c)*f/d/(d*x+c)-2*(b*f+d*e)/d-2*(-b*f-d*e)/d)*a-1/2/d^2*f*ex
p(2*(b*f+d*e)/d)*Ei(1,-2*(a*d-b*c)*f/d/(d*x+c)-2*(b*f+d*e)/d-2*(-b*f-d*e)/d)*c*b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2} \, x + \frac{1}{4} \, \int e^{\left (\frac{2 \, b c f}{d^{2} x + c d} - 2 \, e - \frac{2 \, a f}{d x + c} - \frac{2 \, b f}{d}\right )}\,{d x} + \frac{1}{4} \, \int e^{\left (-\frac{2 \, b c f}{d^{2} x + c d} + 2 \, e + \frac{2 \, a f}{d x + c} + \frac{2 \, b f}{d}\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))^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.31664, size = 1018, normalized size = 7.89 \begin{align*} -\frac{d^{2} x -{\left (d^{2} x + c d\right )} \cosh \left (\frac{c e + a f +{\left (d e + b f\right )} x}{d x + c}\right )^{2} +{\left ({\left (b c - a d\right )} f{\rm Ei}\left (-\frac{2 \,{\left (b c - a d\right )} f}{d^{2} x + c d}\right ) \cosh \left (\frac{2 \,{\left (d e + b f\right )}}{d}\right ) - d^{2} x - c d\right )} \sinh \left (\frac{c e + a f +{\left (d e + b f\right )} x}{d x + c}\right )^{2} -{\left ({\left (b c - a d\right )} f{\rm Ei}\left (-\frac{2 \,{\left (b c - a d\right )} f}{d^{2} x + c d}\right ) \cosh \left (\frac{c e + a f +{\left (d e + b f\right )} x}{d x + c}\right )^{2} -{\left (b c - a d\right )} f{\rm Ei}\left (\frac{2 \,{\left (b c - a d\right )} f}{d^{2} x + c d}\right )\right )} \cosh \left (\frac{2 \,{\left (d e + b f\right )}}{d}\right ) -{\left ({\left (b c - a d\right )} f{\rm Ei}\left (-\frac{2 \,{\left (b c - a d\right )} f}{d^{2} x + c d}\right ) \cosh \left (\frac{c e + a f +{\left (d e + b f\right )} x}{d x + c}\right )^{2} -{\left (b c - a d\right )} f{\rm Ei}\left (-\frac{2 \,{\left (b c - a d\right )} f}{d^{2} x + c d}\right ) \sinh \left (\frac{c e + a f +{\left (d e + b f\right )} x}{d x + c}\right )^{2} +{\left (b c - a d\right )} f{\rm Ei}\left (\frac{2 \,{\left (b c - a d\right )} f}{d^{2} x + c d}\right )\right )} \sinh \left (\frac{2 \,{\left (d e + b f\right )}}{d}\right )}{2 \,{\left (d^{2} \cosh \left (\frac{c e + a f +{\left (d e + b f\right )} x}{d x + c}\right )^{2} - d^{2} \sinh \left (\frac{c e + a f +{\left (d e + b f\right )} x}{d x + c}\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(d^2*x - (d^2*x + c*d)*cosh((c*e + a*f + (d*e + b*f)*x)/(d*x + c))^2 + ((b*c - a*d)*f*Ei(-2*(b*c - a*d)*f
/(d^2*x + c*d))*cosh(2*(d*e + b*f)/d) - d^2*x - c*d)*sinh((c*e + a*f + (d*e + b*f)*x)/(d*x + c))^2 - ((b*c - a
*d)*f*Ei(-2*(b*c - a*d)*f/(d^2*x + c*d))*cosh((c*e + a*f + (d*e + b*f)*x)/(d*x + c))^2 - (b*c - a*d)*f*Ei(2*(b
*c - a*d)*f/(d^2*x + c*d)))*cosh(2*(d*e + b*f)/d) - ((b*c - a*d)*f*Ei(-2*(b*c - a*d)*f/(d^2*x + c*d))*cosh((c*
e + a*f + (d*e + b*f)*x)/(d*x + c))^2 - (b*c - a*d)*f*Ei(-2*(b*c - a*d)*f/(d^2*x + c*d))*sinh((c*e + a*f + (d*
e + b*f)*x)/(d*x + c))^2 + (b*c - a*d)*f*Ei(2*(b*c - a*d)*f/(d^2*x + c*d)))*sinh(2*(d*e + b*f)/d))/(d^2*cosh((
c*e + a*f + (d*e + b*f)*x)/(d*x + c))^2 - d^2*sinh((c*e + a*f + (d*e + b*f)*x)/(d*x + c))^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))**2,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 )^{2}\,{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))^2,x, algorithm="giac")

[Out]

integrate(sinh(e + (b*x + a)*f/(d*x + c))^2, x)