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

Optimal. Leaf size=62 \[ a^2 \cosh (c) \text{Chi}(d x)+a^2 \sinh (c) \text{Shi}(d x)+\frac{2 a b \sinh (c+d x)}{d}-\frac{b^2 \cosh (c+d x)}{d^2}+\frac{b^2 x \sinh (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.184752, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.412, Rules used = {6742, 2637, 3303, 3298, 3301, 3296, 2638} \[ a^2 \cosh (c) \text{Chi}(d x)+a^2 \sinh (c) \text{Shi}(d x)+\frac{2 a b \sinh (c+d x)}{d}-\frac{b^2 \cosh (c+d x)}{d^2}+\frac{b^2 x \sinh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^2*Cosh[c + d*x])/x,x]

[Out]

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

Rule 6742

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

Rule 2637

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

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]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2638

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

Rubi steps

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

Mathematica [A]  time = 0.234957, size = 51, normalized size = 0.82 \[ a^2 \cosh (c) \text{Chi}(d x)+a^2 \sinh (c) \text{Shi}(d x)+\frac{b (d (2 a+b x) \sinh (c+d x)-b \cosh (c+d x))}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.036, size = 121, normalized size = 2. \begin{align*} -{\frac{{a}^{2}{{\rm e}^{-c}}{\it Ei} \left ( 1,dx \right ) }{2}}-{\frac{{b}^{2}{{\rm e}^{-dx-c}}x}{2\,d}}-{\frac{{b}^{2}{{\rm e}^{-dx-c}}}{2\,{d}^{2}}}-{\frac{ab{{\rm e}^{-dx-c}}}{d}}-{\frac{{a}^{2}{{\rm e}^{c}}{\it Ei} \left ( 1,-dx \right ) }{2}}+{\frac{{{\rm e}^{dx+c}}{b}^{2}x}{2\,d}}-{\frac{{{\rm e}^{dx+c}}{b}^{2}}{2\,{d}^{2}}}+{\frac{ab{{\rm e}^{dx+c}}}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.34019, size = 236, normalized size = 3.81 \begin{align*} -\frac{1}{4} \,{\left (4 \, a b{\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 )} + b^{2}{\left (\frac{{\left (d^{2} x^{2} e^{c} - 2 \, d x e^{c} + 2 \, e^{c}\right )} e^{\left (d x\right )}}{d^{3}} + \frac{{\left (d^{2} x^{2} + 2 \, d x + 2\right )} e^{\left (-d x - c\right )}}{d^{3}}\right )} + \frac{4 \, a^{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 )} a^{2}}{d}\right )} d + \frac{1}{2} \,{\left (b^{2} x^{2} + 4 \, a b x + 2 \, a^{2} \log \left (x\right )\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,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.97115, size = 221, normalized size = 3.56 \begin{align*} -\frac{2 \, b^{2} \cosh \left (d x + c\right ) -{\left (a^{2} d^{2}{\rm Ei}\left (d x\right ) + a^{2} d^{2}{\rm Ei}\left (-d x\right )\right )} \cosh \left (c\right ) - 2 \,{\left (b^{2} d x + 2 \, a b d\right )} \sinh \left (d x + c\right ) -{\left (a^{2} d^{2}{\rm Ei}\left (d x\right ) - a^{2} d^{2}{\rm Ei}\left (-d x\right )\right )} \sinh \left (c\right )}{2 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 5.07679, size = 73, normalized size = 1.18 \begin{align*} a^{2} \sinh{\left (c \right )} \operatorname{Shi}{\left (d x \right )} + a^{2} \cosh{\left (c \right )} \operatorname{Chi}\left (d x\right ) + 2 a b \left (\begin{cases} \frac{\sinh{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \cosh{\left (c \right )} & \text{otherwise} \end{cases}\right ) + b^{2} \left (\begin{cases} \frac{x \sinh{\left (c + d x \right )}}{d} - \frac{\cosh{\left (c + d x \right )}}{d^{2}} & \text{for}\: d \neq 0 \\\frac{x^{2} \cosh{\left (c \right )}}{2} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*sinh(c)*Shi(d*x) + a**2*cosh(c)*Chi(d*x) + 2*a*b*Piecewise((sinh(c + d*x)/d, Ne(d, 0)), (x*cosh(c), True)
) + b**2*Piecewise((x*sinh(c + d*x)/d - cosh(c + d*x)/d**2, Ne(d, 0)), (x**2*cosh(c)/2, True))

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Giac [A]  time = 1.26265, size = 153, normalized size = 2.47 \begin{align*} \frac{a^{2} d^{2}{\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + a^{2} d^{2}{\rm Ei}\left (d x\right ) e^{c} + b^{2} d x e^{\left (d x + c\right )} - b^{2} d x e^{\left (-d x - c\right )} + 2 \, a b d e^{\left (d x + c\right )} - 2 \, a b d e^{\left (-d x - c\right )} - b^{2} e^{\left (d x + c\right )} - b^{2} e^{\left (-d x - c\right )}}{2 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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