3.1.100 \(\int \frac {a+i a \sinh (e+f x)}{(c+d x)^2} \, dx\) [100]

Optimal. Leaf size=95 \[ -\frac {a}{d (c+d x)}+\frac {i a f \cosh \left (e-\frac {c f}{d}\right ) \text {Chi}\left (\frac {c f}{d}+f x\right )}{d^2}-\frac {i a \sinh (e+f x)}{d (c+d x)}+\frac {i a f \sinh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (\frac {c f}{d}+f x\right )}{d^2} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, 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 \text {Chi}\left (x f+\frac {c f}{d}\right ) \cosh \left (e-\frac {c f}{d}\right )}{d^2}+\frac {i a f \sinh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (x f+\frac {c f}{d}\right )}{d^2}-\frac {i a \sinh (e+f x)}{d (c+d x)}-\frac {a}{d (c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 0.28, size = 83, normalized size = 0.87 \begin {gather*} \frac {i a \left (f (c+d x) \cosh \left (e-\frac {c f}{d}\right ) \text {Chi}\left (f \left (\frac {c}{d}+x\right )\right )-d (-i+\sinh (e+f x))+f (c+d x) \sinh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (f \left (\frac {c}{d}+x\right )\right )\right )}{d^2 (c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.50, size = 153, normalized size = 1.61

method	result	size
risch	\(-\frac {a}{d \left (d x +c \right )}+\frac {i a f \,{\mathrm e}^{-f x -e}}{2 d \left (d x f +c f \right )}-\frac {i a f \,{\mathrm e}^{\frac {c f -d e}{d}} \expIntegral \left (1, f x +e +\frac {c f -d e}{d}\right )}{2 d^{2}}-\frac {i a f \,{\mathrm e}^{f x +e}}{2 d^{2} \left (\frac {c f}{d}+f x \right )}-\frac {i a f \,{\mathrm e}^{-\frac {c f -d e}{d}} \expIntegral \left (1, -f x -e -\frac {c f -d e}{d}\right )}{2 d^{2}}\)	\(153\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.31, size = 90, normalized size = 0.95 \begin {gather*} \frac {1}{2} i \, a {\left (\frac {e^{\left (\frac {c f}{d} - e\right )} E_{2}\left (\frac {{\left (d x + c\right )} f}{d}\right )}{{\left (d x + c\right )} d} - \frac {e^{\left (-\frac {c f}{d} + e\right )} E_{2}\left (-\frac {{\left (d x + c\right )} f}{d}\right )}{{\left (d x + c\right )} d}\right )} - \frac {a}{d^{2} x + c 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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 139, normalized size = 1.46 \begin {gather*} \frac {{\left (-i \, a d e^{\left (2 \, f x + 2 \, e\right )} + i \, a d + {\left ({\left (i \, a d f x + i \, a c f\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 f x + i \, a c f\right )} {\rm Ei}\left (\frac {d f x + c f}{d}\right ) e^{\left (-\frac {c f - d e}{d}\right )} - 2 \, a d\right )} e^{\left (f x + e\right )}\right )} e^{\left (-f x - e\right )}}{2 \, {\left (d^{3} x + c d^{2}\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)^2,x, algorithm="fricas")

[Out]

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

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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)**2,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. 630 vs. \(2 (91) = 182\).
time = 0.45, size = 630, normalized size = 6.63 \begin {gather*} \frac {1}{2} i \, a {\left (\frac {{\left ({\left (d x + c\right )} {\left (\frac {d e}{d x + c} - \frac {c f}{d x + c} + f\right )} f^{2} {\rm Ei}\left (\frac {{\left (d x + c\right )} {\left (\frac {d e}{d x + c} - \frac {c f}{d x + c} + f\right )} - d e + c f}{d}\right ) e^{\left (\frac {d e - c f}{d}\right )} - d e f^{2} {\rm Ei}\left (\frac {{\left (d x + c\right )} {\left (\frac {d e}{d x + c} - \frac {c f}{d x + c} + f\right )} - d e + c f}{d}\right ) e^{\left (\frac {d e - c f}{d}\right )} + c f^{3} {\rm Ei}\left (\frac {{\left (d x + c\right )} {\left (\frac {d e}{d x + c} - \frac {c f}{d x + c} + f\right )} - d e + c f}{d}\right ) e^{\left (\frac {d e - c f}{d}\right )} - d f^{2} e^{\left (\frac {{\left (d x + c\right )} {\left (\frac {d e}{d x + c} - \frac {c f}{d x + c} + f\right )}}{d}\right )}\right )} d^{2}}{{\left ({\left (d x + c\right )} d^{4} {\left (\frac {d e}{d x + c} - \frac {c f}{d x + c} + f\right )} - d^{5} e + c d^{4} f\right )} f} + \frac {{\left ({\left (d x + c\right )} {\left (\frac {d e}{d x + c} - \frac {c f}{d x + c} + f\right )} f^{2} {\rm Ei}\left (-\frac {{\left (d x + c\right )} {\left (\frac {d e}{d x + c} - \frac {c f}{d x + c} + f\right )} - d e + c f}{d}\right ) e^{\left (-\frac {d e - c f}{d}\right )} - d e f^{2} {\rm Ei}\left (-\frac {{\left (d x + c\right )} {\left (\frac {d e}{d x + c} - \frac {c f}{d x + c} + f\right )} - d e + c f}{d}\right ) e^{\left (-\frac {d e - c f}{d}\right )} + c f^{3} {\rm Ei}\left (-\frac {{\left (d x + c\right )} {\left (\frac {d e}{d x + c} - \frac {c f}{d x + c} + f\right )} - d e + c f}{d}\right ) e^{\left (-\frac {d e - c f}{d}\right )} + d f^{2} e^{\left (-\frac {{\left (d x + c\right )} {\left (\frac {d e}{d x + c} - \frac {c f}{d x + c} + f\right )}}{d}\right )}\right )} d^{2}}{{\left ({\left (d x + c\right )} d^{4} {\left (\frac {d e}{d x + c} - \frac {c f}{d x + c} + f\right )} - d^{5} e + c d^{4} f\right )} f}\right )} - \frac {a}{{\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)^2,x, algorithm="giac")

[Out]

1/2*I*a*(((d*x + c)*(d*e/(d*x + c) - c*f/(d*x + c) + f)*f^2*Ei(((d*x + c)*(d*e/(d*x + c) - c*f/(d*x + c) + f)
- d*e + c*f)/d)*e^((d*e - c*f)/d) - d*e*f^2*Ei(((d*x + c)*(d*e/(d*x + c) - c*f/(d*x + c) + f) - d*e + c*f)/d)*
e^((d*e - c*f)/d) + c*f^3*Ei(((d*x + c)*(d*e/(d*x + c) - c*f/(d*x + c) + f) - d*e + c*f)/d)*e^((d*e - c*f)/d)
- d*f^2*e^((d*x + c)*(d*e/(d*x + c) - c*f/(d*x + c) + f)/d))*d^2/(((d*x + c)*d^4*(d*e/(d*x + c) - c*f/(d*x + c
) + f) - d^5*e + c*d^4*f)*f) + ((d*x + c)*(d*e/(d*x + c) - c*f/(d*x + c) + f)*f^2*Ei(-((d*x + c)*(d*e/(d*x + c
) - c*f/(d*x + c) + f) - d*e + c*f)/d)*e^(-(d*e - c*f)/d) - d*e*f^2*Ei(-((d*x + c)*(d*e/(d*x + c) - c*f/(d*x +
c) + f) - d*e + c*f)/d)*e^(-(d*e - c*f)/d) + c*f^3*Ei(-((d*x + c)*(d*e/(d*x + c) - c*f/(d*x + c) + f) - d*e +
c*f)/d)*e^(-(d*e - c*f)/d) + d*f^2*e^(-(d*x + c)*(d*e/(d*x + c) - c*f/(d*x + c) + f)/d))*d^2/(((d*x + c)*d^4*
(d*e/(d*x + c) - c*f/(d*x + c) + f) - d^5*e + c*d^4*f)*f)) - a/((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}}{{\left (c+d\,x\right )}^2} \,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)^2,x)

[Out]

int((a + a*sinh(e + f*x)*1i)/(c + d*x)^2, x)

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