3.15 \(\int \frac{(a+b x)^2 \cosh (c+d x)}{x^3} \, dx\)

Optimal. Leaf size=121 \[ \frac{1}{2} a^2 d^2 \cosh (c) \text{Chi}(d x)+\frac{1}{2} a^2 d^2 \sinh (c) \text{Shi}(d x)-\frac{a^2 \cosh (c+d x)}{2 x^2}-\frac{a^2 d \sinh (c+d x)}{2 x}+2 a b d \sinh (c) \text{Chi}(d x)+2 a b d \cosh (c) \text{Shi}(d x)-\frac{2 a b \cosh (c+d x)}{x}+b^2 \cosh (c) \text{Chi}(d x)+b^2 \sinh (c) \text{Shi}(d x) \]

[Out]

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

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Rubi [A]  time = 0.338688, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {6742, 3297, 3303, 3298, 3301} \[ \frac{1}{2} a^2 d^2 \cosh (c) \text{Chi}(d x)+\frac{1}{2} a^2 d^2 \sinh (c) \text{Shi}(d x)-\frac{a^2 \cosh (c+d x)}{2 x^2}-\frac{a^2 d \sinh (c+d x)}{2 x}+2 a b d \sinh (c) \text{Chi}(d x)+2 a b d \cosh (c) \text{Shi}(d x)-\frac{2 a b \cosh (c+d x)}{x}+b^2 \cosh (c) \text{Chi}(d x)+b^2 \sinh (c) \text{Shi}(d x) \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^2*Cosh[c + d*x])/x^3,x]

[Out]

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

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

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+b x)^2 \cosh (c+d x)}{x^3} \, dx &=\int \left (\frac{a^2 \cosh (c+d x)}{x^3}+\frac{2 a b \cosh (c+d x)}{x^2}+\frac{b^2 \cosh (c+d x)}{x}\right ) \, dx\\ &=a^2 \int \frac{\cosh (c+d x)}{x^3} \, dx+(2 a b) \int \frac{\cosh (c+d x)}{x^2} \, dx+b^2 \int \frac{\cosh (c+d x)}{x} \, dx\\ &=-\frac{a^2 \cosh (c+d x)}{2 x^2}-\frac{2 a b \cosh (c+d x)}{x}+\frac{1}{2} \left (a^2 d\right ) \int \frac{\sinh (c+d x)}{x^2} \, dx+(2 a b d) \int \frac{\sinh (c+d x)}{x} \, dx+\left (b^2 \cosh (c)\right ) \int \frac{\cosh (d x)}{x} \, dx+\left (b^2 \sinh (c)\right ) \int \frac{\sinh (d x)}{x} \, dx\\ &=-\frac{a^2 \cosh (c+d x)}{2 x^2}-\frac{2 a b \cosh (c+d x)}{x}+b^2 \cosh (c) \text{Chi}(d x)-\frac{a^2 d \sinh (c+d x)}{2 x}+b^2 \sinh (c) \text{Shi}(d x)+\frac{1}{2} \left (a^2 d^2\right ) \int \frac{\cosh (c+d x)}{x} \, dx+(2 a b d \cosh (c)) \int \frac{\sinh (d x)}{x} \, dx+(2 a b d \sinh (c)) \int \frac{\cosh (d x)}{x} \, dx\\ &=-\frac{a^2 \cosh (c+d x)}{2 x^2}-\frac{2 a b \cosh (c+d x)}{x}+b^2 \cosh (c) \text{Chi}(d x)+2 a b d \text{Chi}(d x) \sinh (c)-\frac{a^2 d \sinh (c+d x)}{2 x}+2 a b d \cosh (c) \text{Shi}(d x)+b^2 \sinh (c) \text{Shi}(d x)+\frac{1}{2} \left (a^2 d^2 \cosh (c)\right ) \int \frac{\cosh (d x)}{x} \, dx+\frac{1}{2} \left (a^2 d^2 \sinh (c)\right ) \int \frac{\sinh (d x)}{x} \, dx\\ &=-\frac{a^2 \cosh (c+d x)}{2 x^2}-\frac{2 a b \cosh (c+d x)}{x}+b^2 \cosh (c) \text{Chi}(d x)+\frac{1}{2} a^2 d^2 \cosh (c) \text{Chi}(d x)+2 a b d \text{Chi}(d x) \sinh (c)-\frac{a^2 d \sinh (c+d x)}{2 x}+2 a b d \cosh (c) \text{Shi}(d x)+b^2 \sinh (c) \text{Shi}(d x)+\frac{1}{2} a^2 d^2 \sinh (c) \text{Shi}(d x)\\ \end{align*}

Mathematica [A]  time = 0.373391, size = 93, normalized size = 0.77 \[ \frac{1}{2} \left (\text{Chi}(d x) \left (\cosh (c) \left (a^2 d^2+2 b^2\right )+4 a b d \sinh (c)\right )+\text{Shi}(d x) \left (\sinh (c) \left (a^2 d^2+2 b^2\right )+4 a b d \cosh (c)\right )-\frac{a ((a+4 b x) \cosh (c+d x)+a d x \sinh (c+d x))}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^2*Cosh[c + d*x])/x^3,x]

[Out]

(CoshIntegral[d*x]*((2*b^2 + a^2*d^2)*Cosh[c] + 4*a*b*d*Sinh[c]) - (a*((a + 4*b*x)*Cosh[c + d*x] + a*d*x*Sinh[
c + d*x]))/x^2 + (4*a*b*d*Cosh[c] + (2*b^2 + a^2*d^2)*Sinh[c])*SinhIntegral[d*x])/2

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Maple [A]  time = 0.052, size = 181, normalized size = 1.5 \begin{align*} -{\frac{{d}^{2}{a}^{2}{{\rm e}^{-c}}{\it Ei} \left ( 1,dx \right ) }{4}}-{\frac{ab{{\rm e}^{-dx-c}}}{x}}+dab{{\rm e}^{-c}}{\it Ei} \left ( 1,dx \right ) -{\frac{{b}^{2}{{\rm e}^{-c}}{\it Ei} \left ( 1,dx \right ) }{2}}+{\frac{d{a}^{2}{{\rm e}^{-dx-c}}}{4\,x}}-{\frac{{a}^{2}{{\rm e}^{-dx-c}}}{4\,{x}^{2}}}-{\frac{ab{{\rm e}^{dx+c}}}{x}}-dab{{\rm e}^{c}}{\it Ei} \left ( 1,-dx \right ) -{\frac{{b}^{2}{{\rm e}^{c}}{\it Ei} \left ( 1,-dx \right ) }{2}}-{\frac{{d}^{2}{a}^{2}{{\rm e}^{c}}{\it Ei} \left ( 1,-dx \right ) }{4}}-{\frac{{{\rm e}^{dx+c}}{a}^{2}}{4\,{x}^{2}}}-{\frac{d{a}^{2}{{\rm e}^{dx+c}}}{4\,x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2*cosh(d*x+c)/x^3,x)

[Out]

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

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Maxima [A]  time = 1.37234, size = 170, normalized size = 1.4 \begin{align*} \frac{1}{4} \,{\left ({\left (d e^{\left (-c\right )} \Gamma \left (-1, d x\right ) + d e^{c} \Gamma \left (-1, -d x\right )\right )} a^{2} - 4 \,{\left ({\rm Ei}\left (-d x\right ) e^{\left (-c\right )} -{\rm Ei}\left (d x\right ) e^{c}\right )} a b - \frac{4 \, b^{2} \cosh \left (d x + c\right ) \log \left (x\right )}{d} + \frac{2 \,{\left ({\rm Ei}\left (-d x\right ) e^{\left (-c\right )} +{\rm Ei}\left (d x\right ) e^{c}\right )} b^{2}}{d}\right )} d + \frac{1}{2} \,{\left (2 \, b^{2} \log \left (x\right ) - \frac{4 \, a b x + a^{2}}{x^{2}}\right )} \cosh \left (d x + c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2*cosh(d*x+c)/x^3,x, algorithm="maxima")

[Out]

1/4*((d*e^(-c)*gamma(-1, d*x) + d*e^c*gamma(-1, -d*x))*a^2 - 4*(Ei(-d*x)*e^(-c) - Ei(d*x)*e^c)*a*b - 4*b^2*cos
h(d*x + c)*log(x)/d + 2*(Ei(-d*x)*e^(-c) + Ei(d*x)*e^c)*b^2/d)*d + 1/2*(2*b^2*log(x) - (4*a*b*x + a^2)/x^2)*co
sh(d*x + c)

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Fricas [A]  time = 1.94994, size = 351, normalized size = 2.9 \begin{align*} -\frac{2 \, a^{2} d x \sinh \left (d x + c\right ) + 2 \,{\left (4 \, a b x + a^{2}\right )} \cosh \left (d x + c\right ) -{\left ({\left (a^{2} d^{2} + 4 \, a b d + 2 \, b^{2}\right )} x^{2}{\rm Ei}\left (d x\right ) +{\left (a^{2} d^{2} - 4 \, a b d + 2 \, b^{2}\right )} x^{2}{\rm Ei}\left (-d x\right )\right )} \cosh \left (c\right ) -{\left ({\left (a^{2} d^{2} + 4 \, a b d + 2 \, b^{2}\right )} x^{2}{\rm Ei}\left (d x\right ) -{\left (a^{2} d^{2} - 4 \, a b d + 2 \, b^{2}\right )} x^{2}{\rm Ei}\left (-d x\right )\right )} \sinh \left (c\right )}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2*cosh(d*x+c)/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right )^{2} \cosh{\left (c + d x \right )}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**2*cosh(d*x+c)/x**3,x)

[Out]

Integral((a + b*x)**2*cosh(c + d*x)/x**3, x)

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Giac [A]  time = 1.17773, size = 244, normalized size = 2.02 \begin{align*} \frac{a^{2} d^{2} x^{2}{\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + a^{2} d^{2} x^{2}{\rm Ei}\left (d x\right ) e^{c} - 4 \, a b d x^{2}{\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + 4 \, a b d x^{2}{\rm Ei}\left (d x\right ) e^{c} + 2 \, b^{2} x^{2}{\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + 2 \, b^{2} x^{2}{\rm Ei}\left (d x\right ) e^{c} - a^{2} d x e^{\left (d x + c\right )} + a^{2} d x e^{\left (-d x - c\right )} - 4 \, a b x e^{\left (d x + c\right )} - 4 \, a b x e^{\left (-d x - c\right )} - a^{2} e^{\left (d x + c\right )} - a^{2} e^{\left (-d x - c\right )}}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2*cosh(d*x+c)/x^3,x, algorithm="giac")

[Out]

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