3.16 \(\int \frac{(a+b x)^2 \cosh (c+d x)}{x^4} \, dx\)

Optimal. Leaf size=172 \[ \frac{1}{6} a^2 d^3 \sinh (c) \text{Chi}(d x)+\frac{1}{6} a^2 d^3 \cosh (c) \text{Shi}(d x)-\frac{a^2 d^2 \cosh (c+d x)}{6 x}-\frac{a^2 d \sinh (c+d x)}{6 x^2}-\frac{a^2 \cosh (c+d x)}{3 x^3}+a b d^2 \cosh (c) \text{Chi}(d x)+a b d^2 \sinh (c) \text{Shi}(d x)-\frac{a b \cosh (c+d x)}{x^2}-\frac{a b d \sinh (c+d x)}{x}+b^2 d \sinh (c) \text{Chi}(d x)+b^2 d \cosh (c) \text{Shi}(d x)-\frac{b^2 \cosh (c+d x)}{x} \]

[Out]

-(a^2*Cosh[c + d*x])/(3*x^3) - (a*b*Cosh[c + d*x])/x^2 - (b^2*Cosh[c + d*x])/x - (a^2*d^2*Cosh[c + d*x])/(6*x)
 + a*b*d^2*Cosh[c]*CoshIntegral[d*x] + b^2*d*CoshIntegral[d*x]*Sinh[c] + (a^2*d^3*CoshIntegral[d*x]*Sinh[c])/6
 - (a^2*d*Sinh[c + d*x])/(6*x^2) - (a*b*d*Sinh[c + d*x])/x + b^2*d*Cosh[c]*SinhIntegral[d*x] + (a^2*d^3*Cosh[c
]*SinhIntegral[d*x])/6 + a*b*d^2*Sinh[c]*SinhIntegral[d*x]

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Rubi [A]  time = 0.42641, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 17, 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} \[ \frac{1}{6} a^2 d^3 \sinh (c) \text{Chi}(d x)+\frac{1}{6} a^2 d^3 \cosh (c) \text{Shi}(d x)-\frac{a^2 d^2 \cosh (c+d x)}{6 x}-\frac{a^2 d \sinh (c+d x)}{6 x^2}-\frac{a^2 \cosh (c+d x)}{3 x^3}+a b d^2 \cosh (c) \text{Chi}(d x)+a b d^2 \sinh (c) \text{Shi}(d x)-\frac{a b \cosh (c+d x)}{x^2}-\frac{a b d \sinh (c+d x)}{x}+b^2 d \sinh (c) \text{Chi}(d x)+b^2 d \cosh (c) \text{Shi}(d x)-\frac{b^2 \cosh (c+d x)}{x} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^2*Cosh[c + d*x])/x^4,x]

[Out]

-(a^2*Cosh[c + d*x])/(3*x^3) - (a*b*Cosh[c + d*x])/x^2 - (b^2*Cosh[c + d*x])/x - (a^2*d^2*Cosh[c + d*x])/(6*x)
 + a*b*d^2*Cosh[c]*CoshIntegral[d*x] + b^2*d*CoshIntegral[d*x]*Sinh[c] + (a^2*d^3*CoshIntegral[d*x]*Sinh[c])/6
 - (a^2*d*Sinh[c + d*x])/(6*x^2) - (a*b*d*Sinh[c + d*x])/x + b^2*d*Cosh[c]*SinhIntegral[d*x] + (a^2*d^3*Cosh[c
]*SinhIntegral[d*x])/6 + a*b*d^2*Sinh[c]*SinhIntegral[d*x]

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^4} \, dx &=\int \left (\frac{a^2 \cosh (c+d x)}{x^4}+\frac{2 a b \cosh (c+d x)}{x^3}+\frac{b^2 \cosh (c+d x)}{x^2}\right ) \, dx\\ &=a^2 \int \frac{\cosh (c+d x)}{x^4} \, dx+(2 a b) \int \frac{\cosh (c+d x)}{x^3} \, dx+b^2 \int \frac{\cosh (c+d x)}{x^2} \, dx\\ &=-\frac{a^2 \cosh (c+d x)}{3 x^3}-\frac{a b \cosh (c+d x)}{x^2}-\frac{b^2 \cosh (c+d x)}{x}+\frac{1}{3} \left (a^2 d\right ) \int \frac{\sinh (c+d x)}{x^3} \, dx+(a b d) \int \frac{\sinh (c+d x)}{x^2} \, dx+\left (b^2 d\right ) \int \frac{\sinh (c+d x)}{x} \, dx\\ &=-\frac{a^2 \cosh (c+d x)}{3 x^3}-\frac{a b \cosh (c+d x)}{x^2}-\frac{b^2 \cosh (c+d x)}{x}-\frac{a^2 d \sinh (c+d x)}{6 x^2}-\frac{a b d \sinh (c+d x)}{x}+\frac{1}{6} \left (a^2 d^2\right ) \int \frac{\cosh (c+d x)}{x^2} \, dx+\left (a b d^2\right ) \int \frac{\cosh (c+d x)}{x} \, dx+\left (b^2 d \cosh (c)\right ) \int \frac{\sinh (d x)}{x} \, dx+\left (b^2 d \sinh (c)\right ) \int \frac{\cosh (d x)}{x} \, dx\\ &=-\frac{a^2 \cosh (c+d x)}{3 x^3}-\frac{a b \cosh (c+d x)}{x^2}-\frac{b^2 \cosh (c+d x)}{x}-\frac{a^2 d^2 \cosh (c+d x)}{6 x}+b^2 d \text{Chi}(d x) \sinh (c)-\frac{a^2 d \sinh (c+d x)}{6 x^2}-\frac{a b d \sinh (c+d x)}{x}+b^2 d \cosh (c) \text{Shi}(d x)+\frac{1}{6} \left (a^2 d^3\right ) \int \frac{\sinh (c+d x)}{x} \, dx+\left (a b d^2 \cosh (c)\right ) \int \frac{\cosh (d x)}{x} \, dx+\left (a b d^2 \sinh (c)\right ) \int \frac{\sinh (d x)}{x} \, dx\\ &=-\frac{a^2 \cosh (c+d x)}{3 x^3}-\frac{a b \cosh (c+d x)}{x^2}-\frac{b^2 \cosh (c+d x)}{x}-\frac{a^2 d^2 \cosh (c+d x)}{6 x}+a b d^2 \cosh (c) \text{Chi}(d x)+b^2 d \text{Chi}(d x) \sinh (c)-\frac{a^2 d \sinh (c+d x)}{6 x^2}-\frac{a b d \sinh (c+d x)}{x}+b^2 d \cosh (c) \text{Shi}(d x)+a b d^2 \sinh (c) \text{Shi}(d x)+\frac{1}{6} \left (a^2 d^3 \cosh (c)\right ) \int \frac{\sinh (d x)}{x} \, dx+\frac{1}{6} \left (a^2 d^3 \sinh (c)\right ) \int \frac{\cosh (d x)}{x} \, dx\\ &=-\frac{a^2 \cosh (c+d x)}{3 x^3}-\frac{a b \cosh (c+d x)}{x^2}-\frac{b^2 \cosh (c+d x)}{x}-\frac{a^2 d^2 \cosh (c+d x)}{6 x}+a b d^2 \cosh (c) \text{Chi}(d x)+b^2 d \text{Chi}(d x) \sinh (c)+\frac{1}{6} a^2 d^3 \text{Chi}(d x) \sinh (c)-\frac{a^2 d \sinh (c+d x)}{6 x^2}-\frac{a b d \sinh (c+d x)}{x}+b^2 d \cosh (c) \text{Shi}(d x)+\frac{1}{6} a^2 d^3 \cosh (c) \text{Shi}(d x)+a b d^2 \sinh (c) \text{Shi}(d x)\\ \end{align*}

Mathematica [A]  time = 0.44756, size = 154, normalized size = 0.9 \[ -\frac{-d x^3 \text{Chi}(d x) \left (\sinh (c) \left (a^2 d^2+6 b^2\right )+6 a b d \cosh (c)\right )-d x^3 \text{Shi}(d x) \left (a^2 d^2 \cosh (c)+6 a b d \sinh (c)+6 b^2 \cosh (c)\right )+a^2 d^2 x^2 \cosh (c+d x)+a^2 d x \sinh (c+d x)+2 a^2 \cosh (c+d x)+6 a b d x^2 \sinh (c+d x)+6 a b x \cosh (c+d x)+6 b^2 x^2 \cosh (c+d x)}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^2*Cosh[c + d*x])/x^4,x]

[Out]

-(2*a^2*Cosh[c + d*x] + 6*a*b*x*Cosh[c + d*x] + 6*b^2*x^2*Cosh[c + d*x] + a^2*d^2*x^2*Cosh[c + d*x] - d*x^3*Co
shIntegral[d*x]*(6*a*b*d*Cosh[c] + (6*b^2 + a^2*d^2)*Sinh[c]) + a^2*d*x*Sinh[c + d*x] + 6*a*b*d*x^2*Sinh[c + d
*x] - d*x^3*(6*b^2*Cosh[c] + a^2*d^2*Cosh[c] + 6*a*b*d*Sinh[c])*SinhIntegral[d*x])/(6*x^3)

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Maple [A]  time = 0.06, size = 287, normalized size = 1.7 \begin{align*}{\frac{{d}^{3}{a}^{2}{{\rm e}^{-c}}{\it Ei} \left ( 1,dx \right ) }{12}}-{\frac{{a}^{2}{d}^{2}{{\rm e}^{-dx-c}}}{12\,x}}-{\frac{{a}^{2}{{\rm e}^{-dx-c}}}{6\,{x}^{3}}}+{\frac{d{a}^{2}{{\rm e}^{-dx-c}}}{12\,{x}^{2}}}-{\frac{{b}^{2}{{\rm e}^{-dx-c}}}{2\,x}}+{\frac{bda{{\rm e}^{-dx-c}}}{2\,x}}-{\frac{ab{{\rm e}^{-dx-c}}}{2\,{x}^{2}}}-{\frac{{d}^{2}ab{{\rm e}^{-c}}{\it Ei} \left ( 1,dx \right ) }{2}}+{\frac{d{b}^{2}{{\rm e}^{-c}}{\it Ei} \left ( 1,dx \right ) }{2}}-{\frac{{{\rm e}^{dx+c}}{a}^{2}}{6\,{x}^{3}}}-{\frac{d{a}^{2}{{\rm e}^{dx+c}}}{12\,{x}^{2}}}-{\frac{{a}^{2}{d}^{2}{{\rm e}^{dx+c}}}{12\,x}}-{\frac{{d}^{2}ab{{\rm e}^{c}}{\it Ei} \left ( 1,-dx \right ) }{2}}-{\frac{ab{{\rm e}^{dx+c}}}{2\,{x}^{2}}}-{\frac{bda{{\rm e}^{dx+c}}}{2\,x}}-{\frac{{d}^{3}{a}^{2}{{\rm e}^{c}}{\it Ei} \left ( 1,-dx \right ) }{12}}-{\frac{{{\rm e}^{dx+c}}{b}^{2}}{2\,x}}-{\frac{d{b}^{2}{{\rm e}^{c}}{\it Ei} \left ( 1,-dx \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2*cosh(d*x+c)/x^4,x)

[Out]

1/12*d^3*a^2*exp(-c)*Ei(1,d*x)-1/12*d^2*a^2*exp(-d*x-c)/x-1/6*a^2*exp(-d*x-c)/x^3+1/12*d*a^2*exp(-d*x-c)/x^2-1
/2*b^2*exp(-d*x-c)/x+1/2*d*a*b*exp(-d*x-c)/x-1/2*a*b*exp(-d*x-c)/x^2-1/2*d^2*a*b*exp(-c)*Ei(1,d*x)+1/2*d*b^2*e
xp(-c)*Ei(1,d*x)-1/6*a^2/x^3*exp(d*x+c)-1/12*d*a^2/x^2*exp(d*x+c)-1/12*d^2*a^2/x*exp(d*x+c)-1/2*d^2*a*b*exp(c)
*Ei(1,-d*x)-1/2*a*b/x^2*exp(d*x+c)-1/2*d*a*b/x*exp(d*x+c)-1/12*d^3*a^2*exp(c)*Ei(1,-d*x)-1/2*b^2/x*exp(d*x+c)-
1/2*d*b^2*exp(c)*Ei(1,-d*x)

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Maxima [A]  time = 1.43103, size = 158, normalized size = 0.92 \begin{align*} \frac{1}{6} \,{\left (a^{2} d^{2} e^{\left (-c\right )} \Gamma \left (-2, d x\right ) - a^{2} d^{2} e^{c} \Gamma \left (-2, -d x\right ) + 3 \, a b d e^{\left (-c\right )} \Gamma \left (-1, d x\right ) + 3 \, a b d e^{c} \Gamma \left (-1, -d x\right ) - 3 \, b^{2}{\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + 3 \, b^{2}{\rm Ei}\left (d x\right ) e^{c}\right )} d - \frac{{\left (3 \, b^{2} x^{2} + 3 \, a b x + a^{2}\right )} \cosh \left (d x + c\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2*cosh(d*x+c)/x^4,x, algorithm="maxima")

[Out]

1/6*(a^2*d^2*e^(-c)*gamma(-2, d*x) - a^2*d^2*e^c*gamma(-2, -d*x) + 3*a*b*d*e^(-c)*gamma(-1, d*x) + 3*a*b*d*e^c
*gamma(-1, -d*x) - 3*b^2*Ei(-d*x)*e^(-c) + 3*b^2*Ei(d*x)*e^c)*d - 1/3*(3*b^2*x^2 + 3*a*b*x + a^2)*cosh(d*x + c
)/x^3

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Fricas [A]  time = 2.01559, size = 431, normalized size = 2.51 \begin{align*} -\frac{2 \,{\left (6 \, a b x +{\left (a^{2} d^{2} + 6 \, b^{2}\right )} x^{2} + 2 \, a^{2}\right )} \cosh \left (d x + c\right ) -{\left ({\left (a^{2} d^{3} + 6 \, a b d^{2} + 6 \, b^{2} d\right )} x^{3}{\rm Ei}\left (d x\right ) -{\left (a^{2} d^{3} - 6 \, a b d^{2} + 6 \, b^{2} d\right )} x^{3}{\rm Ei}\left (-d x\right )\right )} \cosh \left (c\right ) + 2 \,{\left (6 \, a b d x^{2} + a^{2} d x\right )} \sinh \left (d x + c\right ) -{\left ({\left (a^{2} d^{3} + 6 \, a b d^{2} + 6 \, b^{2} d\right )} x^{3}{\rm Ei}\left (d x\right ) +{\left (a^{2} d^{3} - 6 \, a b d^{2} + 6 \, b^{2} d\right )} x^{3}{\rm Ei}\left (-d x\right )\right )} \sinh \left (c\right )}{12 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2*cosh(d*x+c)/x^4,x, algorithm="fricas")

[Out]

-1/12*(2*(6*a*b*x + (a^2*d^2 + 6*b^2)*x^2 + 2*a^2)*cosh(d*x + c) - ((a^2*d^3 + 6*a*b*d^2 + 6*b^2*d)*x^3*Ei(d*x
) - (a^2*d^3 - 6*a*b*d^2 + 6*b^2*d)*x^3*Ei(-d*x))*cosh(c) + 2*(6*a*b*d*x^2 + a^2*d*x)*sinh(d*x + c) - ((a^2*d^
3 + 6*a*b*d^2 + 6*b^2*d)*x^3*Ei(d*x) + (a^2*d^3 - 6*a*b*d^2 + 6*b^2*d)*x^3*Ei(-d*x))*sinh(c))/x^3

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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^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**2*cosh(d*x+c)/x**4,x)

[Out]

Integral((a + b*x)**2*cosh(c + d*x)/x**4, x)

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Giac [A]  time = 1.15162, size = 385, normalized size = 2.24 \begin{align*} -\frac{a^{2} d^{3} x^{3}{\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - a^{2} d^{3} x^{3}{\rm Ei}\left (d x\right ) e^{c} - 6 \, a b d^{2} x^{3}{\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - 6 \, a b d^{2} x^{3}{\rm Ei}\left (d x\right ) e^{c} + 6 \, b^{2} d x^{3}{\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - 6 \, b^{2} d x^{3}{\rm Ei}\left (d x\right ) e^{c} + a^{2} d^{2} x^{2} e^{\left (d x + c\right )} + a^{2} d^{2} x^{2} e^{\left (-d x - c\right )} + 6 \, a b d x^{2} e^{\left (d x + c\right )} - 6 \, a b d x^{2} e^{\left (-d x - c\right )} + a^{2} d x e^{\left (d x + c\right )} + 6 \, b^{2} x^{2} e^{\left (d x + c\right )} - a^{2} d x e^{\left (-d x - c\right )} + 6 \, b^{2} x^{2} e^{\left (-d x - c\right )} + 6 \, a b x e^{\left (d x + c\right )} + 6 \, a b x e^{\left (-d x - c\right )} + 2 \, a^{2} e^{\left (d x + c\right )} + 2 \, a^{2} e^{\left (-d x - c\right )}}{12 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2*cosh(d*x+c)/x^4,x, algorithm="giac")

[Out]

-1/12*(a^2*d^3*x^3*Ei(-d*x)*e^(-c) - a^2*d^3*x^3*Ei(d*x)*e^c - 6*a*b*d^2*x^3*Ei(-d*x)*e^(-c) - 6*a*b*d^2*x^3*E
i(d*x)*e^c + 6*b^2*d*x^3*Ei(-d*x)*e^(-c) - 6*b^2*d*x^3*Ei(d*x)*e^c + a^2*d^2*x^2*e^(d*x + c) + a^2*d^2*x^2*e^(
-d*x - c) + 6*a*b*d*x^2*e^(d*x + c) - 6*a*b*d*x^2*e^(-d*x - c) + a^2*d*x*e^(d*x + c) + 6*b^2*x^2*e^(d*x + c) -
 a^2*d*x*e^(-d*x - c) + 6*b^2*x^2*e^(-d*x - c) + 6*a*b*x*e^(d*x + c) + 6*a*b*x*e^(-d*x - c) + 2*a^2*e^(d*x + c
) + 2*a^2*e^(-d*x - c))/x^3