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

Optimal. Leaf size=149 \[ \frac{2 i a^2 \text{Chi}\left (x f+\frac{c f}{d}\right ) \sinh \left (e-\frac{c f}{d}\right )}{d}-\frac{a^2 \text{Chi}\left (2 x f+\frac{2 c f}{d}\right ) \cosh \left (2 e-\frac{2 c f}{d}\right )}{2 d}-\frac{a^2 \sinh \left (2 e-\frac{2 c f}{d}\right ) \text{Shi}\left (2 x f+\frac{2 c f}{d}\right )}{2 d}+\frac{2 i a^2 \cosh \left (e-\frac{c f}{d}\right ) \text{Shi}\left (x f+\frac{c f}{d}\right )}{d}+\frac{3 a^2 \log (c+d x)}{2 d} \]

[Out]

-(a^2*Cosh[2*e - (2*c*f)/d]*CoshIntegral[(2*c*f)/d + 2*f*x])/(2*d) + (3*a^2*Log[c + d*x])/(2*d) + ((2*I)*a^2*C
oshIntegral[(c*f)/d + f*x]*Sinh[e - (c*f)/d])/d + ((2*I)*a^2*Cosh[e - (c*f)/d]*SinhIntegral[(c*f)/d + f*x])/d
- (a^2*Sinh[2*e - (2*c*f)/d]*SinhIntegral[(2*c*f)/d + 2*f*x])/(2*d)

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Rubi [A]  time = 0.352419, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3318, 3312, 3303, 3298, 3301} \[ \frac{2 i a^2 \text{Chi}\left (x f+\frac{c f}{d}\right ) \sinh \left (e-\frac{c f}{d}\right )}{d}-\frac{a^2 \text{Chi}\left (2 x f+\frac{2 c f}{d}\right ) \cosh \left (2 e-\frac{2 c f}{d}\right )}{2 d}-\frac{a^2 \sinh \left (2 e-\frac{2 c f}{d}\right ) \text{Shi}\left (2 x f+\frac{2 c f}{d}\right )}{2 d}+\frac{2 i a^2 \cosh \left (e-\frac{c f}{d}\right ) \text{Shi}\left (x f+\frac{c f}{d}\right )}{d}+\frac{3 a^2 \log (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*Cosh[2*e - (2*c*f)/d]*CoshIntegral[(2*c*f)/d + 2*f*x])/(2*d) + (3*a^2*Log[c + d*x])/(2*d) + ((2*I)*a^2*C
oshIntegral[(c*f)/d + f*x]*Sinh[e - (c*f)/d])/d + ((2*I)*a^2*Cosh[e - (c*f)/d]*SinhIntegral[(c*f)/d + f*x])/d
- (a^2*Sinh[2*e - (2*c*f)/d]*SinhIntegral[(2*c*f)/d + 2*f*x])/(2*d)

Rule 3318

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1*(e + (Pi*a)/(2*b)))/2 + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 0])

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && 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 \frac{(a+i a \sinh (e+f x))^2}{c+d x} \, dx &=\left (4 a^2\right ) \int \frac{\sin ^4\left (\frac{1}{2} \left (i e+\frac{\pi }{2}\right )+\frac{i f x}{2}\right )}{c+d x} \, dx\\ &=\left (4 a^2\right ) \int \left (\frac{3}{8 (c+d x)}-\frac{\cosh (2 e+2 f x)}{8 (c+d x)}+\frac{i \sinh (e+f x)}{2 (c+d x)}\right ) \, dx\\ &=\frac{3 a^2 \log (c+d x)}{2 d}+\left (2 i a^2\right ) \int \frac{\sinh (e+f x)}{c+d x} \, dx-\frac{1}{2} a^2 \int \frac{\cosh (2 e+2 f x)}{c+d x} \, dx\\ &=\frac{3 a^2 \log (c+d x)}{2 d}-\frac{1}{2} \left (a^2 \cosh \left (2 e-\frac{2 c f}{d}\right )\right ) \int \frac{\cosh \left (\frac{2 c f}{d}+2 f x\right )}{c+d x} \, dx+\left (2 i a^2 \cosh \left (e-\frac{c f}{d}\right )\right ) \int \frac{\sinh \left (\frac{c f}{d}+f x\right )}{c+d x} \, dx-\frac{1}{2} \left (a^2 \sinh \left (2 e-\frac{2 c f}{d}\right )\right ) \int \frac{\sinh \left (\frac{2 c f}{d}+2 f x\right )}{c+d x} \, dx+\left (2 i a^2 \sinh \left (e-\frac{c f}{d}\right )\right ) \int \frac{\cosh \left (\frac{c f}{d}+f x\right )}{c+d x} \, dx\\ &=-\frac{a^2 \cosh \left (2 e-\frac{2 c f}{d}\right ) \text{Chi}\left (\frac{2 c f}{d}+2 f x\right )}{2 d}+\frac{3 a^2 \log (c+d x)}{2 d}+\frac{2 i a^2 \text{Chi}\left (\frac{c f}{d}+f x\right ) \sinh \left (e-\frac{c f}{d}\right )}{d}+\frac{2 i a^2 \cosh \left (e-\frac{c f}{d}\right ) \text{Shi}\left (\frac{c f}{d}+f x\right )}{d}-\frac{a^2 \sinh \left (2 e-\frac{2 c f}{d}\right ) \text{Shi}\left (\frac{2 c f}{d}+2 f x\right )}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.395477, size = 117, normalized size = 0.79 \[ -\frac{a^2 \left (-4 i \text{Chi}\left (f \left (\frac{c}{d}+x\right )\right ) \sinh \left (e-\frac{c f}{d}\right )+\text{Chi}\left (\frac{2 f (c+d x)}{d}\right ) \cosh \left (2 e-\frac{2 c f}{d}\right )+\sinh \left (2 e-\frac{2 c f}{d}\right ) \text{Shi}\left (\frac{2 f (c+d x)}{d}\right )-4 i \cosh \left (e-\frac{c f}{d}\right ) \text{Shi}\left (f \left (\frac{c}{d}+x\right )\right )-3 \log (c+d x)\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(Cosh[2*e - (2*c*f)/d]*CoshIntegral[(2*f*(c + d*x))/d] - 3*Log[c + d*x] - (4*I)*CoshIntegral[f*(c/d + x)
]*Sinh[e - (c*f)/d] - (4*I)*Cosh[e - (c*f)/d]*SinhIntegral[f*(c/d + x)] + Sinh[2*e - (2*c*f)/d]*SinhIntegral[(
2*f*(c + d*x))/d]))/(2*d)

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Maple [A]  time = 0.119, size = 193, normalized size = 1.3 \begin{align*}{\frac{-i{a}^{2}}{d}{{\rm e}^{-{\frac{cf-de}{d}}}}{\it Ei} \left ( 1,-fx-e-{\frac{cf-de}{d}} \right ) }+{\frac{3\,{a}^{2}\ln \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{2}}{4\,d}{{\rm e}^{2\,{\frac{cf-de}{d}}}}{\it Ei} \left ( 1,2\,fx+2\,e+2\,{\frac{cf-de}{d}} \right ) }+{\frac{{a}^{2}}{4\,d}{{\rm e}^{-2\,{\frac{cf-de}{d}}}}{\it Ei} \left ( 1,-2\,fx-2\,e-2\,{\frac{cf-de}{d}} \right ) }+{\frac{i{a}^{2}}{d}{{\rm e}^{{\frac{cf-de}{d}}}}{\it Ei} \left ( 1,fx+e+{\frac{cf-de}{d}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*sinh(f*x+e))^2/(d*x+c),x)

[Out]

-I*a^2/d*exp(-(c*f-d*e)/d)*Ei(1,-f*x-e-(c*f-d*e)/d)+3/2*a^2*ln(d*x+c)/d+1/4*a^2/d*exp(2*(c*f-d*e)/d)*Ei(1,2*f*
x+2*e+2*(c*f-d*e)/d)+1/4*a^2/d*exp(-2*(c*f-d*e)/d)*Ei(1,-2*f*x-2*e-2*(c*f-d*e)/d)+I*a^2/d*exp((c*f-d*e)/d)*Ei(
1,f*x+e+(c*f-d*e)/d)

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Maxima [A]  time = 1.38847, size = 203, normalized size = 1.36 \begin{align*} \frac{1}{4} \, a^{2}{\left (\frac{e^{\left (-2 \, e + \frac{2 \, c f}{d}\right )} E_{1}\left (\frac{2 \,{\left (d x + c\right )} f}{d}\right )}{d} + \frac{e^{\left (2 \, e - \frac{2 \, c f}{d}\right )} E_{1}\left (-\frac{2 \,{\left (d x + c\right )} f}{d}\right )}{d} + \frac{2 \, \log \left (d x + c\right )}{d}\right )} + i \, a^{2}{\left (\frac{e^{\left (-e + \frac{c f}{d}\right )} E_{1}\left (\frac{{\left (d x + c\right )} f}{d}\right )}{d} - \frac{e^{\left (e - \frac{c f}{d}\right )} E_{1}\left (-\frac{{\left (d x + c\right )} f}{d}\right )}{d}\right )} + \frac{a^{2} \log \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2*(e^(-2*e + 2*c*f/d)*exp_integral_e(1, 2*(d*x + c)*f/d)/d + e^(2*e - 2*c*f/d)*exp_integral_e(1, -2*(d*x
+ c)*f/d)/d + 2*log(d*x + c)/d) + I*a^2*(e^(-e + c*f/d)*exp_integral_e(1, (d*x + c)*f/d)/d - e^(e - c*f/d)*ex
p_integral_e(1, -(d*x + c)*f/d)/d) + a^2*log(d*x + c)/d

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Fricas [A]  time = 2.81089, size = 309, normalized size = 2.07 \begin{align*} -\frac{a^{2}{\rm Ei}\left (\frac{2 \,{\left (d f x + c f\right )}}{d}\right ) e^{\left (\frac{2 \,{\left (d e - c f\right )}}{d}\right )} - 4 i \, a^{2}{\rm Ei}\left (\frac{d f x + c f}{d}\right ) e^{\left (\frac{d e - c f}{d}\right )} + 4 i \, a^{2}{\rm Ei}\left (-\frac{d f x + c f}{d}\right ) e^{\left (-\frac{d e - c f}{d}\right )} + a^{2}{\rm Ei}\left (-\frac{2 \,{\left (d f x + c f\right )}}{d}\right ) e^{\left (-\frac{2 \,{\left (d e - c f\right )}}{d}\right )} - 6 \, a^{2} \log \left (\frac{d x + c}{d}\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a^2*Ei(2*(d*f*x + c*f)/d)*e^(2*(d*e - c*f)/d) - 4*I*a^2*Ei((d*f*x + c*f)/d)*e^((d*e - c*f)/d) + 4*I*a^2*
Ei(-(d*f*x + c*f)/d)*e^(-(d*e - c*f)/d) + a^2*Ei(-2*(d*f*x + c*f)/d)*e^(-2*(d*e - c*f)/d) - 6*a^2*log((d*x + c
)/d))/d

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int - \frac{\sinh ^{2}{\left (e + f x \right )}}{c + d x}\, dx + \int \frac{2 i \sinh{\left (e + f x \right )}}{c + d x}\, dx + \int \frac{1}{c + d x}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(-sinh(e + f*x)**2/(c + d*x), x) + Integral(2*I*sinh(e + f*x)/(c + d*x), x) + Integral(1/(c + d*
x), x))

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Giac [A]  time = 1.3336, size = 188, normalized size = 1.26 \begin{align*} -\frac{a^{2}{\rm Ei}\left (-\frac{2 \,{\left (d f x + c f\right )}}{d}\right ) e^{\left (\frac{2 \, c f}{d} - 2 \, e\right )} + 4 i \, a^{2}{\rm Ei}\left (-\frac{d f x + c f}{d}\right ) e^{\left (\frac{c f}{d} - e\right )} - 4 i \, a^{2}{\rm Ei}\left (\frac{d f x + c f}{d}\right ) e^{\left (-\frac{c f}{d} + e\right )} + a^{2}{\rm Ei}\left (\frac{2 \,{\left (d f x + c f\right )}}{d}\right ) e^{\left (-\frac{2 \, c f}{d} + 2 \, e\right )} - 6 \, a^{2} \log \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a^2*Ei(-2*(d*f*x + c*f)/d)*e^(2*c*f/d - 2*e) + 4*I*a^2*Ei(-(d*f*x + c*f)/d)*e^(c*f/d - e) - 4*I*a^2*Ei((
d*f*x + c*f)/d)*e^(-c*f/d + e) + a^2*Ei(2*(d*f*x + c*f)/d)*e^(-2*c*f/d + 2*e) - 6*a^2*log(d*x + c))/d