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

Optimal. Leaf size=70 \[ a^2 d \sinh (c) \text {Chi}(d x)+a^2 d \cosh (c) \text {Shi}(d x)-\frac {a^2 \cosh (c+d x)}{x}+2 a b \cosh (c) \text {Chi}(d x)+2 a b \sinh (c) \text {Shi}(d x)+\frac {b^2 \sinh (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {6742, 2637, 3297, 3303, 3298, 3301} \[ a^2 d \sinh (c) \text {Chi}(d x)+a^2 d \cosh (c) \text {Shi}(d x)-\frac {a^2 \cosh (c+d x)}{x}+2 a b \cosh (c) \text {Chi}(d x)+2 a b \sinh (c) \text {Shi}(d x)+\frac {b^2 \sinh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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 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]

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 6742

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

Rubi steps

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

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Mathematica [A]  time = 0.25, size = 62, normalized size = 0.89 \[ -\frac {a^2 \cosh (c+d x)}{x}+a \text {Chi}(d x) (a d \sinh (c)+2 b \cosh (c))+a \text {Shi}(d x) (a d \cosh (c)+2 b \sinh (c))+\frac {b^2 \sinh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.54, size = 122, normalized size = 1.74 \[ -\frac {2 \, a^{2} d \cosh \left (d x + c\right ) - 2 \, b^{2} x \sinh \left (d x + c\right ) - {\left ({\left (a^{2} d^{2} + 2 \, a b d\right )} x {\rm Ei}\left (d x\right ) - {\left (a^{2} d^{2} - 2 \, a b d\right )} x {\rm Ei}\left (-d x\right )\right )} \cosh \relax (c) - {\left ({\left (a^{2} d^{2} + 2 \, a b d\right )} x {\rm Ei}\left (d x\right ) + {\left (a^{2} d^{2} - 2 \, a b d\right )} x {\rm Ei}\left (-d x\right )\right )} \sinh \relax (c)}{2 \, d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.13, size = 119, normalized size = 1.70 \[ -\frac {a^{2} d^{2} x {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - a^{2} d^{2} x {\rm Ei}\left (d x\right ) e^{c} - 2 \, a b d x {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - 2 \, a b d x {\rm Ei}\left (d x\right ) e^{c} + a^{2} d e^{\left (d x + c\right )} - b^{2} x e^{\left (d x + c\right )} + a^{2} d e^{\left (-d x - c\right )} + b^{2} x e^{\left (-d x - c\right )}}{2 \, d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.12, size = 118, normalized size = 1.69 \[ -\frac {a^{2} {\mathrm e}^{-d x -c}}{2 x}+\frac {d \,a^{2} {\mathrm e}^{-c} \Ei \left (1, d x \right )}{2}-\frac {b^{2} {\mathrm e}^{-d x -c}}{2 d}-a b \,{\mathrm e}^{-c} \Ei \left (1, d x \right )-\frac {a^{2} {\mathrm e}^{d x +c}}{2 x}-\frac {d \,a^{2} {\mathrm e}^{c} \Ei \left (1, -d x \right )}{2}+\frac {b^{2} {\mathrm e}^{d x +c}}{2 d}-a b \,{\mathrm e}^{c} \Ei \left (1, -d x \right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 136, normalized size = 1.94 \[ -\frac {1}{2} \, {\left ({\left ({\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - {\rm Ei}\left (d x\right ) e^{c}\right )} a^{2} + b^{2} {\left (\frac {{\left (d x e^{c} - e^{c}\right )} e^{\left (d x\right )}}{d^{2}} + \frac {{\left (d x + 1\right )} e^{\left (-d x - c\right )}}{d^{2}}\right )} + \frac {4 \, a b \cosh \left (d x + c\right ) \log \relax (x)}{d} - \frac {2 \, {\left ({\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + {\rm Ei}\left (d x\right ) e^{c}\right )} a b}{d}\right )} d + {\left (b^{2} x + 2 \, a b \log \relax (x) - \frac {a^{2}}{x}\right )} \cosh \left (d x + c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {cosh}\left (c+d\,x\right )\,{\left (a+b\,x\right )}^2}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x\right )^{2} \cosh {\left (c + d x \right )}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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