3.101 \(\int \frac{a+i a \sinh (e+f x)}{(c+d x)^3} \, dx\)

Optimal. Leaf size=131 \[ \frac{i a f^2 \text{Chi}\left (x f+\frac{c f}{d}\right ) \sinh \left (e-\frac{c f}{d}\right )}{2 d^3}+\frac{i a f^2 \cosh \left (e-\frac{c f}{d}\right ) \text{Shi}\left (x f+\frac{c f}{d}\right )}{2 d^3}-\frac{i a f \cosh (e+f x)}{2 d^2 (c+d x)}-\frac{i a \sinh (e+f x)}{2 d (c+d x)^2}-\frac{a}{2 d (c+d x)^2} \]

[Out]

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

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Rubi [A]  time = 0.232228, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3317, 3297, 3303, 3298, 3301} \[ \frac{i a f^2 \text{Chi}\left (x f+\frac{c f}{d}\right ) \sinh \left (e-\frac{c f}{d}\right )}{2 d^3}+\frac{i a f^2 \cosh \left (e-\frac{c f}{d}\right ) \text{Shi}\left (x f+\frac{c f}{d}\right )}{2 d^3}-\frac{i a f \cosh (e+f x)}{2 d^2 (c+d x)}-\frac{i a \sinh (e+f x)}{2 d (c+d x)^2}-\frac{a}{2 d (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Sinh[e + f*x])/(c + d*x)^3,x]

[Out]

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

Rule 3317

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] ||
IGtQ[m, 0] || NeQ[a^2 - b^2, 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 \frac{a+i a \sinh (e+f x)}{(c+d x)^3} \, dx &=\int \left (\frac{a}{(c+d x)^3}+\frac{i a \sinh (e+f x)}{(c+d x)^3}\right ) \, dx\\ &=-\frac{a}{2 d (c+d x)^2}+(i a) \int \frac{\sinh (e+f x)}{(c+d x)^3} \, dx\\ &=-\frac{a}{2 d (c+d x)^2}-\frac{i a \sinh (e+f x)}{2 d (c+d x)^2}+\frac{(i a f) \int \frac{\cosh (e+f x)}{(c+d x)^2} \, dx}{2 d}\\ &=-\frac{a}{2 d (c+d x)^2}-\frac{i a f \cosh (e+f x)}{2 d^2 (c+d x)}-\frac{i a \sinh (e+f x)}{2 d (c+d x)^2}+\frac{\left (i a f^2\right ) \int \frac{\sinh (e+f x)}{c+d x} \, dx}{2 d^2}\\ &=-\frac{a}{2 d (c+d x)^2}-\frac{i a f \cosh (e+f x)}{2 d^2 (c+d x)}-\frac{i a \sinh (e+f x)}{2 d (c+d x)^2}+\frac{\left (i a f^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}{2 d^2}+\frac{\left (i a f^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}{2 d^2}\\ &=-\frac{a}{2 d (c+d x)^2}-\frac{i a f \cosh (e+f x)}{2 d^2 (c+d x)}+\frac{i a f^2 \text{Chi}\left (\frac{c f}{d}+f x\right ) \sinh \left (e-\frac{c f}{d}\right )}{2 d^3}-\frac{i a \sinh (e+f x)}{2 d (c+d x)^2}+\frac{i a f^2 \cosh \left (e-\frac{c f}{d}\right ) \text{Shi}\left (\frac{c f}{d}+f x\right )}{2 d^3}\\ \end{align*}

Mathematica [A]  time = 0.669285, size = 109, normalized size = 0.83 \[ \frac{i a \left (f^2 (c+d x)^2 \text{Chi}\left (f \left (\frac{c}{d}+x\right )\right ) \sinh \left (e-\frac{c f}{d}\right )+f^2 (c+d x)^2 \cosh \left (e-\frac{c f}{d}\right ) \text{Shi}\left (f \left (\frac{c}{d}+x\right )\right )-d (f (c+d x) \cosh (e+f x)+d (\sinh (e+f x)-i))\right )}{2 d^3 (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Sinh[e + f*x])/(c + d*x)^3,x]

[Out]

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

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Maple [B]  time = 0.06, size = 303, normalized size = 2.3 \begin{align*} -{\frac{a}{2\,d \left ( dx+c \right ) ^{2}}}-{\frac{{\frac{i}{4}}a{f}^{3}{{\rm e}^{-fx-e}}x}{d \left ({d}^{2}{f}^{2}{x}^{2}+2\,cd{f}^{2}x+{c}^{2}{f}^{2} \right ) }}-{\frac{{\frac{i}{4}}a{f}^{3}{{\rm e}^{-fx-e}}c}{{d}^{2} \left ({d}^{2}{f}^{2}{x}^{2}+2\,cd{f}^{2}x+{c}^{2}{f}^{2} \right ) }}+{\frac{{\frac{i}{4}}a{f}^{2}{{\rm e}^{-fx-e}}}{d \left ({d}^{2}{f}^{2}{x}^{2}+2\,cd{f}^{2}x+{c}^{2}{f}^{2} \right ) }}+{\frac{{\frac{i}{4}}a{f}^{2}}{{d}^{3}}{{\rm e}^{{\frac{cf-de}{d}}}}{\it Ei} \left ( 1,fx+e+{\frac{cf-de}{d}} \right ) }-{\frac{{\frac{i}{4}}a{f}^{2}{{\rm e}^{fx+e}}}{{d}^{3}} \left ({\frac{cf}{d}}+fx \right ) ^{-2}}-{\frac{{\frac{i}{4}}a{f}^{2}{{\rm e}^{fx+e}}}{{d}^{3}} \left ({\frac{cf}{d}}+fx \right ) ^{-1}}-{\frac{{\frac{i}{4}}a{f}^{2}}{{d}^{3}}{{\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))/(d*x+c)^3,x)

[Out]

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

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Maxima [A]  time = 1.21992, size = 134, normalized size = 1.02 \begin{align*} \frac{1}{2} i \, a{\left (\frac{e^{\left (-e + \frac{c f}{d}\right )} E_{3}\left (\frac{{\left (d x + c\right )} f}{d}\right )}{{\left (d x + c\right )}^{2} d} - \frac{e^{\left (e - \frac{c f}{d}\right )} E_{3}\left (-\frac{{\left (d x + c\right )} f}{d}\right )}{{\left (d x + c\right )}^{2} d}\right )} - \frac{a}{2 \,{\left (d^{3} x^{2} + 2 \, c d^{2} x + c^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*sinh(f*x+e))/(d*x+c)^3,x, algorithm="maxima")

[Out]

1/2*I*a*(e^(-e + c*f/d)*exp_integral_e(3, (d*x + c)*f/d)/((d*x + c)^2*d) - e^(e - c*f/d)*exp_integral_e(3, -(d
*x + c)*f/d)/((d*x + c)^2*d)) - 1/2*a/(d^3*x^2 + 2*c*d^2*x + c^2*d)

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Fricas [A]  time = 2.72999, size = 471, normalized size = 3.6 \begin{align*} \frac{{\left (-i \, a d^{2} f x - i \, a c d f + i \, a d^{2} +{\left (-i \, a d^{2} f x - i \, a c d f - i \, a d^{2}\right )} e^{\left (2 \, f x + 2 \, e\right )} -{\left (2 \, a d^{2} -{\left (i \, a d^{2} f^{2} x^{2} + 2 i \, a c d f^{2} x + i \, a c^{2} f^{2}\right )}{\rm Ei}\left (\frac{d f x + c f}{d}\right ) e^{\left (\frac{d e - c f}{d}\right )} -{\left (-i \, a d^{2} f^{2} x^{2} - 2 i \, a c d f^{2} x - i \, a c^{2} f^{2}\right )}{\rm Ei}\left (-\frac{d f x + c f}{d}\right ) e^{\left (-\frac{d e - c f}{d}\right )}\right )} e^{\left (f x + e\right )}\right )} e^{\left (-f x - e\right )}}{4 \,{\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*sinh(f*x+e))/(d*x+c)^3,x, algorithm="fricas")

[Out]

1/4*(-I*a*d^2*f*x - I*a*c*d*f + I*a*d^2 + (-I*a*d^2*f*x - I*a*c*d*f - I*a*d^2)*e^(2*f*x + 2*e) - (2*a*d^2 - (I
*a*d^2*f^2*x^2 + 2*I*a*c*d*f^2*x + I*a*c^2*f^2)*Ei((d*f*x + c*f)/d)*e^((d*e - c*f)/d) - (-I*a*d^2*f^2*x^2 - 2*
I*a*c*d*f^2*x - I*a*c^2*f^2)*Ei(-(d*f*x + c*f)/d)*e^(-(d*e - c*f)/d))*e^(f*x + e))*e^(-f*x - e)/(d^5*x^2 + 2*c
*d^4*x + c^2*d^3)

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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((a+I*a*sinh(f*x+e))/(d*x+c)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.34846, size = 451, normalized size = 3.44 \begin{align*} \frac{-i \, a d^{2} f^{2} x^{2}{\rm Ei}\left (-\frac{d f x + c f}{d}\right ) e^{\left (\frac{c f}{d} - e\right )} + i \, a d^{2} f^{2} x^{2}{\rm Ei}\left (\frac{d f x + c f}{d}\right ) e^{\left (-\frac{c f}{d} + e\right )} - 2 i \, a c d f^{2} x{\rm Ei}\left (-\frac{d f x + c f}{d}\right ) e^{\left (\frac{c f}{d} - e\right )} + 2 i \, a c d f^{2} x{\rm Ei}\left (\frac{d f x + c f}{d}\right ) e^{\left (-\frac{c f}{d} + e\right )} - i \, a c^{2} f^{2}{\rm Ei}\left (-\frac{d f x + c f}{d}\right ) e^{\left (\frac{c f}{d} - e\right )} + i \, a c^{2} f^{2}{\rm Ei}\left (\frac{d f x + c f}{d}\right ) e^{\left (-\frac{c f}{d} + e\right )} - i \, a d^{2} f x e^{\left (f x + e\right )} - i \, a d^{2} f x e^{\left (-f x - e\right )} - i \, a c d f e^{\left (f x + e\right )} - i \, a c d f e^{\left (-f x - e\right )} - i \, a d^{2} e^{\left (f x + e\right )} + i \, a d^{2} e^{\left (-f x - e\right )} - 2 \, a d^{2}}{4 \,{\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*sinh(f*x+e))/(d*x+c)^3,x, algorithm="giac")

[Out]

1/4*(-I*a*d^2*f^2*x^2*Ei(-(d*f*x + c*f)/d)*e^(c*f/d - e) + I*a*d^2*f^2*x^2*Ei((d*f*x + c*f)/d)*e^(-c*f/d + e)
- 2*I*a*c*d*f^2*x*Ei(-(d*f*x + c*f)/d)*e^(c*f/d - e) + 2*I*a*c*d*f^2*x*Ei((d*f*x + c*f)/d)*e^(-c*f/d + e) - I*
a*c^2*f^2*Ei(-(d*f*x + c*f)/d)*e^(c*f/d - e) + I*a*c^2*f^2*Ei((d*f*x + c*f)/d)*e^(-c*f/d + e) - I*a*d^2*f*x*e^
(f*x + e) - I*a*d^2*f*x*e^(-f*x - e) - I*a*c*d*f*e^(f*x + e) - I*a*c*d*f*e^(-f*x - e) - I*a*d^2*e^(f*x + e) +
I*a*d^2*e^(-f*x - e) - 2*a*d^2)/(d^5*x^2 + 2*c*d^4*x + c^2*d^3)