3.1.99 \(\int \frac {a+i a \sinh (e+f x)}{c+d x} \, dx\) [99]

Optimal. Leaf size=70 \[ \frac {a \log (c+d x)}{d}+\frac {i a \text {Chi}\left (\frac {c f}{d}+f x\right ) \sinh \left (e-\frac {c f}{d}\right )}{d}+\frac {i a \cosh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (\frac {c f}{d}+f x\right )}{d} \]

[Out]

a*ln(d*x+c)/d+I*a*cosh(-e+c*f/d)*Shi(c*f/d+f*x)/d-I*a*Chi(c*f/d+f*x)*sinh(-e+c*f/d)/d

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Rubi [A]
time = 0.12, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3398, 3384, 3379, 3382} \begin {gather*} \frac {i a \text {Chi}\left (x f+\frac {c f}{d}\right ) \sinh \left (e-\frac {c f}{d}\right )}{d}+\frac {i a \cosh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (x f+\frac {c f}{d}\right )}{d}+\frac {a \log (c+d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Log[c + d*x])/d + (I*a*CoshIntegral[(c*f)/d + f*x]*Sinh[e - (c*f)/d])/d + (I*a*Cosh[e - (c*f)/d]*SinhIntegr
al[(c*f)/d + f*x])/d

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

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Mathematica [A]
time = 0.17, size = 60, normalized size = 0.86 \begin {gather*} \frac {a \left (\log (c+d x)+i \text {Chi}\left (f \left (\frac {c}{d}+x\right )\right ) \sinh \left (e-\frac {c f}{d}\right )+i \cosh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (f \left (\frac {c}{d}+x\right )\right )\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.54, size = 96, normalized size = 1.37

method	result	size
risch	\(\frac {a \ln \left (d x +c \right )}{d}+\frac {i a \,{\mathrm e}^{\frac {c f -d e}{d}} \expIntegral \left (1, f x +e +\frac {c f -d e}{d}\right )}{2 d}-\frac {i a \,{\mathrm e}^{-\frac {c f -d e}{d}} \expIntegral \left (1, -f x -e -\frac {c f -d e}{d}\right )}{2 d}\)	\(96\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*ln(d*x+c)/d+1/2*I*a/d*exp((c*f-d*e)/d)*Ei(1,f*x+e+(c*f-d*e)/d)-1/2*I*a/d*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 = 73, normalized size = 1.04 \begin {gather*} \frac {1}{2} i \, a {\left (\frac {e^{\left (\frac {c f}{d} - e\right )} E_{1}\left (\frac {{\left (d x + c\right )} f}{d}\right )}{d} - \frac {e^{\left (-\frac {c f}{d} + e\right )} E_{1}\left (-\frac {{\left (d x + c\right )} f}{d}\right )}{d}\right )} + \frac {a \log \left (d x + c\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 81, normalized size = 1.16 \begin {gather*} \frac {-i \, a {\rm Ei}\left (-\frac {d f x + c f}{d}\right ) e^{\left (\frac {c f - d e}{d}\right )} + i \, a {\rm Ei}\left (\frac {d f x + c f}{d}\right ) e^{\left (-\frac {c f - d e}{d}\right )} + 2 \, a \log \left (\frac {d x + c}{d}\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} i a \left (\int \left (- \frac {i}{c + d x}\right )\, dx + \int \frac {\sinh {\left (e + f x \right )}}{c + d x}\, dx\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),x)

[Out]

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

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Giac [A]
time = 0.42, size = 69, normalized size = 0.99 \begin {gather*} -\frac {-i \, a {\rm Ei}\left (\frac {d f x + c f}{d}\right ) e^{\left (e - \frac {c f}{d}\right )} + i \, a {\rm Ei}\left (-\frac {d f x + c f}{d}\right ) e^{\left (-e + \frac {c f}{d}\right )} - 2 \, a \log \left (d x + c\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}}{c+d\,x} \,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),x)

[Out]

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

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