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

Optimal. Leaf size=100 \[ \frac{a^2 \cosh \left (c-\frac{a d}{b}\right ) \text{Chi}\left (x d+\frac{a d}{b}\right )}{b^3}+\frac{a^2 \sinh \left (c-\frac{a d}{b}\right ) \text{Shi}\left (x d+\frac{a d}{b}\right )}{b^3}-\frac{a \sinh (c+d x)}{b^2 d}-\frac{\cosh (c+d x)}{b d^2}+\frac{x \sinh (c+d x)}{b d} \]

[Out]

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

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Rubi [A]  time = 0.26226, antiderivative size = 100, 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, 3296, 2638, 3303, 3298, 3301} \[ \frac{a^2 \cosh \left (c-\frac{a d}{b}\right ) \text{Chi}\left (x d+\frac{a d}{b}\right )}{b^3}+\frac{a^2 \sinh \left (c-\frac{a d}{b}\right ) \text{Shi}\left (x d+\frac{a d}{b}\right )}{b^3}-\frac{a \sinh (c+d x)}{b^2 d}-\frac{\cosh (c+d x)}{b d^2}+\frac{x \sinh (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*Cosh[c + d*x])/(a + b*x),x]

[Out]

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

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 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]

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

Mathematica [A]  time = 0.283274, size = 89, normalized size = 0.89 \[ \frac{a^2 d^2 \cosh \left (c-\frac{a d}{b}\right ) \text{Chi}\left (d \left (\frac{a}{b}+x\right )\right )+a^2 d^2 \sinh \left (c-\frac{a d}{b}\right ) \text{Shi}\left (d \left (\frac{a}{b}+x\right )\right )+b (d (b x-a) \sinh (c+d x)-b \cosh (c+d x))}{b^3 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.03, size = 184, normalized size = 1.8 \begin{align*} -{\frac{{{\rm e}^{-dx-c}}x}{2\,bd}}+{\frac{{{\rm e}^{-dx-c}}a}{2\,d{b}^{2}}}-{\frac{{{\rm e}^{-dx-c}}}{2\,b{d}^{2}}}-{\frac{{a}^{2}}{2\,{b}^{3}}{{\rm e}^{{\frac{da-cb}{b}}}}{\it Ei} \left ( 1,dx+c+{\frac{da-cb}{b}} \right ) }+{\frac{{{\rm e}^{dx+c}}x}{2\,bd}}-{\frac{{{\rm e}^{dx+c}}}{2\,b{d}^{2}}}-{\frac{a{{\rm e}^{dx+c}}}{2\,d{b}^{2}}}-{\frac{{a}^{2}}{2\,{b}^{3}}{{\rm e}^{-{\frac{da-cb}{b}}}}{\it Ei} \left ( 1,-dx-c-{\frac{da-cb}{b}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.37828, size = 315, normalized size = 3.15 \begin{align*} -\frac{1}{4} \, d{\left (\frac{2 \, a^{2}{\left (\frac{e^{\left (-c + \frac{a d}{b}\right )} E_{1}\left (\frac{{\left (b x + a\right )} d}{b}\right )}{b} + \frac{e^{\left (c - \frac{a d}{b}\right )} E_{1}\left (-\frac{{\left (b x + a\right )} d}{b}\right )}{b}\right )}}{b^{2} d} - \frac{2 \, a{\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}} + \frac{\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}}}{b} + \frac{4 \, a^{2} \cosh \left (d x + c\right ) \log \left (b x + a\right )}{b^{3} d}\right )} + \frac{1}{2} \,{\left (\frac{2 \, a^{2} \log \left (b x + a\right )}{b^{3}} + \frac{b x^{2} - 2 \, a x}{b^{2}}\right )} \cosh \left (d x + c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.15575, size = 327, normalized size = 3.27 \begin{align*} -\frac{2 \, b^{2} \cosh \left (d x + c\right ) -{\left (a^{2} d^{2}{\rm Ei}\left (\frac{b d x + a d}{b}\right ) + a^{2} d^{2}{\rm Ei}\left (-\frac{b d x + a d}{b}\right )\right )} \cosh \left (-\frac{b c - a d}{b}\right ) - 2 \,{\left (b^{2} d x - a b d\right )} \sinh \left (d x + c\right ) +{\left (a^{2} d^{2}{\rm Ei}\left (\frac{b d x + a d}{b}\right ) - a^{2} d^{2}{\rm Ei}\left (-\frac{b d x + a d}{b}\right )\right )} \sinh \left (-\frac{b c - a d}{b}\right )}{2 \, b^{3} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.20668, size = 84, normalized size = 0.84 \begin{align*} \frac{a^{2}{\rm Ei}\left (\frac{b d x + a d}{b}\right ) e^{\left (c - \frac{a d}{b}\right )} + a^{2}{\rm Ei}\left (-\frac{b d x + a d}{b}\right ) e^{\left (-c + \frac{a d}{b}\right )}}{2 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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