3.7 \(\int \frac{\sinh (a+b x)}{(c+d x)^3} \, dx\)

Optimal. Leaf size=104 \[ \frac{b^2 \sinh \left (a-\frac{b c}{d}\right ) \text{Chi}\left (\frac{b c}{d}+b x\right )}{2 d^3}+\frac{b^2 \cosh \left (a-\frac{b c}{d}\right ) \text{Shi}\left (\frac{b c}{d}+b x\right )}{2 d^3}-\frac{b \cosh (a+b x)}{2 d^2 (c+d x)}-\frac{\sinh (a+b x)}{2 d (c+d x)^2} \]

[Out]

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

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Rubi [A]  time = 0.166022, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3297, 3303, 3298, 3301} \[ \frac{b^2 \sinh \left (a-\frac{b c}{d}\right ) \text{Chi}\left (\frac{b c}{d}+b x\right )}{2 d^3}+\frac{b^2 \cosh \left (a-\frac{b c}{d}\right ) \text{Shi}\left (\frac{b c}{d}+b x\right )}{2 d^3}-\frac{b \cosh (a+b x)}{2 d^2 (c+d x)}-\frac{\sinh (a+b x)}{2 d (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]/(c + d*x)^3,x]

[Out]

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

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{\sinh (a+b x)}{(c+d x)^3} \, dx &=-\frac{\sinh (a+b x)}{2 d (c+d x)^2}+\frac{b \int \frac{\cosh (a+b x)}{(c+d x)^2} \, dx}{2 d}\\ &=-\frac{b \cosh (a+b x)}{2 d^2 (c+d x)}-\frac{\sinh (a+b x)}{2 d (c+d x)^2}+\frac{b^2 \int \frac{\sinh (a+b x)}{c+d x} \, dx}{2 d^2}\\ &=-\frac{b \cosh (a+b x)}{2 d^2 (c+d x)}-\frac{\sinh (a+b x)}{2 d (c+d x)^2}+\frac{\left (b^2 \cosh \left (a-\frac{b c}{d}\right )\right ) \int \frac{\sinh \left (\frac{b c}{d}+b x\right )}{c+d x} \, dx}{2 d^2}+\frac{\left (b^2 \sinh \left (a-\frac{b c}{d}\right )\right ) \int \frac{\cosh \left (\frac{b c}{d}+b x\right )}{c+d x} \, dx}{2 d^2}\\ &=-\frac{b \cosh (a+b x)}{2 d^2 (c+d x)}+\frac{b^2 \text{Chi}\left (\frac{b c}{d}+b x\right ) \sinh \left (a-\frac{b c}{d}\right )}{2 d^3}-\frac{\sinh (a+b x)}{2 d (c+d x)^2}+\frac{b^2 \cosh \left (a-\frac{b c}{d}\right ) \text{Shi}\left (\frac{b c}{d}+b x\right )}{2 d^3}\\ \end{align*}

Mathematica [A]  time = 0.558116, size = 88, normalized size = 0.85 \[ \frac{b^2 \sinh \left (a-\frac{b c}{d}\right ) \text{Chi}\left (b \left (\frac{c}{d}+x\right )\right )+b^2 \cosh \left (a-\frac{b c}{d}\right ) \text{Shi}\left (b \left (\frac{c}{d}+x\right )\right )-\frac{d (b (c+d x) \cosh (a+b x)+d \sinh (a+b x))}{(c+d x)^2}}{2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.039, size = 277, normalized size = 2.7 \begin{align*} -{\frac{{b}^{3}{{\rm e}^{-bx-a}}x}{4\,d \left ({b}^{2}{d}^{2}{x}^{2}+2\,{b}^{2}cdx+{c}^{2}{b}^{2} \right ) }}-{\frac{{b}^{3}{{\rm e}^{-bx-a}}c}{4\,{d}^{2} \left ({b}^{2}{d}^{2}{x}^{2}+2\,{b}^{2}cdx+{c}^{2}{b}^{2} \right ) }}+{\frac{{b}^{2}{{\rm e}^{-bx-a}}}{4\,d \left ({b}^{2}{d}^{2}{x}^{2}+2\,{b}^{2}cdx+{c}^{2}{b}^{2} \right ) }}+{\frac{{b}^{2}}{4\,{d}^{3}}{{\rm e}^{-{\frac{da-cb}{d}}}}{\it Ei} \left ( 1,bx+a-{\frac{da-cb}{d}} \right ) }-{\frac{{b}^{2}{{\rm e}^{bx+a}}}{4\,{d}^{3}} \left ({\frac{cb}{d}}+bx \right ) ^{-2}}-{\frac{{b}^{2}{{\rm e}^{bx+a}}}{4\,{d}^{3}} \left ({\frac{cb}{d}}+bx \right ) ^{-1}}-{\frac{{b}^{2}}{4\,{d}^{3}}{{\rm e}^{{\frac{da-cb}{d}}}}{\it Ei} \left ( 1,-bx-a-{\frac{-da+cb}{d}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(b*x+a)/(d*x+c)^3,x)

[Out]

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

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Maxima [A]  time = 1.38369, size = 127, normalized size = 1.22 \begin{align*} -\frac{b{\left (\frac{e^{\left (-a + \frac{b c}{d}\right )} E_{2}\left (\frac{{\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )} d} + \frac{e^{\left (a - \frac{b c}{d}\right )} E_{2}\left (-\frac{{\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )} d}\right )}}{4 \, d} - \frac{\sinh \left (b x + a\right )}{2 \,{\left (d x + c\right )}^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b*(e^(-a + b*c/d)*exp_integral_e(2, (d*x + c)*b/d)/((d*x + c)*d) + e^(a - b*c/d)*exp_integral_e(2, -(d*x
+ c)*b/d)/((d*x + c)*d))/d - 1/2*sinh(b*x + a)/((d*x + c)^2*d)

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Fricas [B]  time = 2.60996, size = 518, normalized size = 4.98 \begin{align*} -\frac{2 \, d^{2} \sinh \left (b x + a\right ) + 2 \,{\left (b d^{2} x + b c d\right )} \cosh \left (b x + a\right ) -{\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )}{\rm Ei}\left (\frac{b d x + b c}{d}\right ) -{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )}{\rm Ei}\left (-\frac{b d x + b c}{d}\right )\right )} \cosh \left (-\frac{b c - a d}{d}\right ) -{\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )}{\rm Ei}\left (\frac{b d x + b c}{d}\right ) +{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )}{\rm Ei}\left (-\frac{b d x + b c}{d}\right )\right )} \sinh \left (-\frac{b c - a d}{d}\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(sinh(b*x+a)/(d*x+c)^3,x, algorithm="fricas")

[Out]

-1/4*(2*d^2*sinh(b*x + a) + 2*(b*d^2*x + b*c*d)*cosh(b*x + a) - ((b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*Ei((b*d
*x + b*c)/d) - (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*Ei(-(b*d*x + b*c)/d))*cosh(-(b*c - a*d)/d) - ((b^2*d^2*x^
2 + 2*b^2*c*d*x + b^2*c^2)*Ei((b*d*x + b*c)/d) + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*Ei(-(b*d*x + b*c)/d))*s
inh(-(b*c - a*d)/d))/(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(sinh(b*x+a)/(d*x+c)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.23032, size = 406, normalized size = 3.9 \begin{align*} \frac{b^{2} d^{2} x^{2}{\rm Ei}\left (\frac{b d x + b c}{d}\right ) e^{\left (a - \frac{b c}{d}\right )} - b^{2} d^{2} x^{2}{\rm Ei}\left (-\frac{b d x + b c}{d}\right ) e^{\left (-a + \frac{b c}{d}\right )} + 2 \, b^{2} c d x{\rm Ei}\left (\frac{b d x + b c}{d}\right ) e^{\left (a - \frac{b c}{d}\right )} - 2 \, b^{2} c d x{\rm Ei}\left (-\frac{b d x + b c}{d}\right ) e^{\left (-a + \frac{b c}{d}\right )} + b^{2} c^{2}{\rm Ei}\left (\frac{b d x + b c}{d}\right ) e^{\left (a - \frac{b c}{d}\right )} - b^{2} c^{2}{\rm Ei}\left (-\frac{b d x + b c}{d}\right ) e^{\left (-a + \frac{b c}{d}\right )} - b d^{2} x e^{\left (b x + a\right )} - b d^{2} x e^{\left (-b x - a\right )} - b c d e^{\left (b x + a\right )} - b c d e^{\left (-b x - a\right )} - d^{2} e^{\left (b x + a\right )} + d^{2} e^{\left (-b x - a\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(sinh(b*x+a)/(d*x+c)^3,x, algorithm="giac")

[Out]

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