3.2.1 \(\int \frac {a+i a \sinh (e+f x)}{(c+d x)^3} \, dx\) [101]

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

[Out]

-1/2*a/d/(d*x+c)^2-1/2*I*a*f*cosh(f*x+e)/d^2/(d*x+c)+1/2*I*a*f^2*cosh(-e+c*f/d)*Shi(c*f/d+f*x)/d^3-1/2*I*a*f^2
*Chi(c*f/d+f*x)*sinh(-e+c*f/d)/d^3-1/2*I*a*sinh(f*x+e)/d/(d*x+c)^2

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Rubi [A]
time = 0.17, antiderivative size = 131, normalized size of antiderivative = 1.00, 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 = {3398, 3378, 3384, 3379, 3382} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*a/(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]*
Sinh[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 3378

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 3379

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 3382

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]

Rule 3384

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 3398

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])

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*}

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Mathematica [A]
time = 0.39, size = 109, normalized size = 0.83 \begin {gather*} \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 )-d (f (c+d x) \cosh (e+f x)+d (-i+\sinh (e+f x)))+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 )\right )}{2 d^3 (c+d x)^2} \end {gather*}

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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 302 vs. \(2 (119 ) = 238\).
time = 0.50, size = 303, normalized size = 2.31

method	result	size
risch	\(-\frac {a}{2 d \left (d x +c \right )^{2}}-\frac {i a \,f^{3} {\mathrm e}^{-f x -e} x}{4 d \left (d^{2} x^{2} f^{2}+2 c d \,f^{2} x +c^{2} f^{2}\right )}-\frac {i a \,f^{3} {\mathrm e}^{-f x -e} c}{4 d^{2} \left (d^{2} x^{2} f^{2}+2 c d \,f^{2} x +c^{2} f^{2}\right )}+\frac {i a \,f^{2} {\mathrm e}^{-f x -e}}{4 d \left (d^{2} x^{2} f^{2}+2 c d \,f^{2} x +c^{2} f^{2}\right )}+\frac {i a \,f^{2} {\mathrm e}^{\frac {c f -d e}{d}} \expIntegral \left (1, f x +e +\frac {c f -d e}{d}\right )}{4 d^{3}}-\frac {i a \,f^{2} {\mathrm e}^{f x +e}}{4 d^{3} \left (\frac {c f}{d}+f x \right )^{2}}-\frac {i a \,f^{2} {\mathrm e}^{f x +e}}{4 d^{3} \left (\frac {c f}{d}+f x \right )}-\frac {i a \,f^{2} {\mathrm e}^{-\frac {c f -d e}{d}} \expIntegral \left (1, -f x -e -\frac {c f -d e}{d}\right )}{4 d^{3}}\)	\(303\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*sinh(f*x+e))/(d*x+c)^3,x,method=_RETURNVERBOSE)

[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 = 0.32, size = 101, normalized size = 0.77 \begin {gather*} \frac {1}{2} i \, a {\left (\frac {e^{\left (\frac {c f}{d} - e\right )} E_{3}\left (\frac {{\left (d x + c\right )} f}{d}\right )}{{\left (d x + c\right )}^{2} d} - \frac {e^{\left (-\frac {c f}{d} + e\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 {gather*}

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^(c*f/d - e)*exp_integral_e(3, (d*x + c)*f/d)/((d*x + c)^2*d) - e^(-c*f/d + e)*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 = 0.36, size = 227, normalized size = 1.73 \begin {gather*} \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 {c f - d e}{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 {c f - d e}{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 {gather*}

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^((c*f - d*e)/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^(-(c*f - d*e)/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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 322 vs. \(2 (115) = 230\).
time = 0.44, size = 322, normalized size = 2.46 \begin {gather*} \frac {i \, a d^{2} f^{2} x^{2} {\rm Ei}\left (\frac {d f x + c f}{d}\right ) e^{\left (e - \frac {c f}{d}\right )} - i \, a d^{2} f^{2} x^{2} {\rm Ei}\left (-\frac {d f x + c f}{d}\right ) e^{\left (-e + \frac {c f}{d}\right )} + 2 i \, a c d f^{2} x {\rm Ei}\left (\frac {d f x + c f}{d}\right ) e^{\left (e - \frac {c f}{d}\right )} - 2 i \, a c d f^{2} x {\rm Ei}\left (-\frac {d f x + c f}{d}\right ) e^{\left (-e + \frac {c f}{d}\right )} + i \, a c^{2} f^{2} {\rm Ei}\left (\frac {d f x + c f}{d}\right ) e^{\left (e - \frac {c f}{d}\right )} - i \, a c^{2} f^{2} {\rm Ei}\left (-\frac {d f x + c f}{d}\right ) e^{\left (-e + \frac {c f}{d}\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 {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+a\,\mathrm {sinh}\left (e+f\,x\right )\,1{}\mathrm {i}}{{\left (c+d\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sinh(e + f*x)*1i)/(c + d*x)^3,x)

[Out]

int((a + a*sinh(e + f*x)*1i)/(c + d*x)^3, x)

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