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

Optimal. Leaf size=150 \[ -\frac{a^3 \cosh \left (c-\frac{a d}{b}\right ) \text{Chi}\left (x d+\frac{a d}{b}\right )}{b^4}-\frac{a^3 \sinh \left (c-\frac{a d}{b}\right ) \text{Shi}\left (x d+\frac{a d}{b}\right )}{b^4}+\frac{a^2 \sinh (c+d x)}{b^3 d}+\frac{a \cosh (c+d x)}{b^2 d^2}-\frac{a x \sinh (c+d x)}{b^2 d}+\frac{2 \sinh (c+d x)}{b d^3}-\frac{2 x \cosh (c+d x)}{b d^2}+\frac{x^2 \sinh (c+d x)}{b d} \]

[Out]

(a*Cosh[c + d*x])/(b^2*d^2) - (2*x*Cosh[c + d*x])/(b*d^2) - (a^3*Cosh[c - (a*d)/b]*CoshIntegral[(a*d)/b + d*x]
)/b^4 + (2*Sinh[c + d*x])/(b*d^3) + (a^2*Sinh[c + d*x])/(b^3*d) - (a*x*Sinh[c + d*x])/(b^2*d) + (x^2*Sinh[c +
d*x])/(b*d) - (a^3*Sinh[c - (a*d)/b]*SinhIntegral[(a*d)/b + d*x])/b^4

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Rubi [A]  time = 0.328539, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 11, 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} \[ -\frac{a^3 \cosh \left (c-\frac{a d}{b}\right ) \text{Chi}\left (x d+\frac{a d}{b}\right )}{b^4}-\frac{a^3 \sinh \left (c-\frac{a d}{b}\right ) \text{Shi}\left (x d+\frac{a d}{b}\right )}{b^4}+\frac{a^2 \sinh (c+d x)}{b^3 d}+\frac{a \cosh (c+d x)}{b^2 d^2}-\frac{a x \sinh (c+d x)}{b^2 d}+\frac{2 \sinh (c+d x)}{b d^3}-\frac{2 x \cosh (c+d x)}{b d^2}+\frac{x^2 \sinh (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*Cosh[c + d*x])/(a + b*x),x]

[Out]

(a*Cosh[c + d*x])/(b^2*d^2) - (2*x*Cosh[c + d*x])/(b*d^2) - (a^3*Cosh[c - (a*d)/b]*CoshIntegral[(a*d)/b + d*x]
)/b^4 + (2*Sinh[c + d*x])/(b*d^3) + (a^2*Sinh[c + d*x])/(b^3*d) - (a*x*Sinh[c + d*x])/(b^2*d) + (x^2*Sinh[c +
d*x])/(b*d) - (a^3*Sinh[c - (a*d)/b]*SinhIntegral[(a*d)/b + d*x])/b^4

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

Mathematica [A]  time = 0.430323, size = 118, normalized size = 0.79 \[ \frac{b \left (\left (a^2 d^2-a b d^2 x+b^2 \left (d^2 x^2+2\right )\right ) \sinh (c+d x)+b d (a-2 b x) \cosh (c+d x)\right )-a^3 d^3 \cosh \left (c-\frac{a d}{b}\right ) \text{Chi}\left (d \left (\frac{a}{b}+x\right )\right )-a^3 d^3 \sinh \left (c-\frac{a d}{b}\right ) \text{Shi}\left (d \left (\frac{a}{b}+x\right )\right )}{b^4 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a^3*d^3*Cosh[c - (a*d)/b]*CoshIntegral[d*(a/b + x)]) + b*(b*d*(a - 2*b*x)*Cosh[c + d*x] + (a^2*d^2 - a*b*d^
2*x + b^2*(2 + d^2*x^2))*Sinh[c + d*x]) - a^3*d^3*Sinh[c - (a*d)/b]*SinhIntegral[d*(a/b + x)])/(b^4*d^3)

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Maple [A]  time = 0.034, size = 292, normalized size = 2. \begin{align*} -{\frac{{{\rm e}^{-dx-c}}{x}^{2}}{2\,bd}}+{\frac{{{\rm e}^{-dx-c}}ax}{2\,d{b}^{2}}}-{\frac{{{\rm e}^{-dx-c}}{a}^{2}}{2\,d{b}^{3}}}-{\frac{{{\rm e}^{-dx-c}}x}{b{d}^{2}}}+{\frac{{{\rm e}^{-dx-c}}a}{2\,{d}^{2}{b}^{2}}}-{\frac{{{\rm e}^{-dx-c}}}{{d}^{3}b}}+{\frac{{a}^{3}}{2\,{b}^{4}}{{\rm e}^{{\frac{da-cb}{b}}}}{\it Ei} \left ( 1,dx+c+{\frac{da-cb}{b}} \right ) }-{\frac{{{\rm e}^{dx+c}}x}{b{d}^{2}}}+{\frac{{{\rm e}^{dx+c}}{x}^{2}}{2\,bd}}+{\frac{{{\rm e}^{dx+c}}}{{d}^{3}b}}+{\frac{{a}^{3}}{2\,{b}^{4}}{{\rm e}^{-{\frac{da-cb}{b}}}}{\it Ei} \left ( 1,-dx-c-{\frac{da-cb}{b}} \right ) }-{\frac{a{{\rm e}^{dx+c}}x}{2\,d{b}^{2}}}+{\frac{a{{\rm e}^{dx+c}}}{2\,{d}^{2}{b}^{2}}}+{\frac{{{\rm e}^{dx+c}}{a}^{2}}{2\,d{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d*exp(-d*x-c)/b*x^2+1/2/d*exp(-d*x-c)/b^2*a*x-1/2/d*exp(-d*x-c)/b^3*a^2-1/d^2*exp(-d*x-c)/b*x+1/2/d^2*exp
(-d*x-c)/b^2*a-1/d^3*exp(-d*x-c)/b+1/2/b^4*exp((a*d-b*c)/b)*Ei(1,d*x+c+(a*d-b*c)/b)*a^3-1/d^2/b*exp(d*x+c)*x+1
/2/d/b*exp(d*x+c)*x^2+1/d^3/b*exp(d*x+c)+1/2/b^4*exp(-(a*d-b*c)/b)*Ei(1,-d*x-c-(a*d-b*c)/b)*a^3-1/2/d/b^2*a*ex
p(d*x+c)*x+1/2/d^2/b^2*a*exp(d*x+c)+1/2/d/b^3*a^2*exp(d*x+c)

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Maxima [B]  time = 1.27553, size = 443, normalized size = 2.95 \begin{align*} \frac{1}{12} \, d{\left (\frac{6 \, a^{3}{\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^{3} d} - \frac{6 \, a^{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 )}}{b^{3}} + \frac{3 \, a{\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 )}}{b^{2}} - \frac{2 \,{\left (\frac{{\left (d^{3} x^{3} e^{c} - 3 \, d^{2} x^{2} e^{c} + 6 \, d x e^{c} - 6 \, e^{c}\right )} e^{\left (d x\right )}}{d^{4}} + \frac{{\left (d^{3} x^{3} + 3 \, d^{2} x^{2} + 6 \, d x + 6\right )} e^{\left (-d x - c\right )}}{d^{4}}\right )}}{b} + \frac{12 \, a^{3} \cosh \left (d x + c\right ) \log \left (b x + a\right )}{b^{4} d}\right )} - \frac{1}{6} \,{\left (\frac{6 \, a^{3} \log \left (b x + a\right )}{b^{4}} - \frac{2 \, b^{2} x^{3} - 3 \, a b x^{2} + 6 \, a^{2} x}{b^{3}}\right )} \cosh \left (d x + c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*d*(6*a^3*(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^3*d) - 6*a^2*((d*x*e^c - e^c)*e^(d*x)/d^2 + (d*x + 1)*e^(-d*x - c)/d^2)/b^3 + 3*a*((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^2 - 2*((d^3*x^3*e^c - 3*d^2*x^2*e^c
+ 6*d*x*e^c - 6*e^c)*e^(d*x)/d^4 + (d^3*x^3 + 3*d^2*x^2 + 6*d*x + 6)*e^(-d*x - c)/d^4)/b + 12*a^3*cosh(d*x + c
)*log(b*x + a)/(b^4*d)) - 1/6*(6*a^3*log(b*x + a)/b^4 - (2*b^2*x^3 - 3*a*b*x^2 + 6*a^2*x)/b^3)*cosh(d*x + c)

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Fricas [A]  time = 2.09382, size = 392, normalized size = 2.61 \begin{align*} -\frac{2 \,{\left (2 \, b^{3} d x - a b^{2} d\right )} \cosh \left (d x + c\right ) +{\left (a^{3} d^{3}{\rm Ei}\left (\frac{b d x + a d}{b}\right ) + a^{3} d^{3}{\rm Ei}\left (-\frac{b d x + a d}{b}\right )\right )} \cosh \left (-\frac{b c - a d}{b}\right ) - 2 \,{\left (b^{3} d^{2} x^{2} - a b^{2} d^{2} x + a^{2} b d^{2} + 2 \, b^{3}\right )} \sinh \left (d x + c\right ) -{\left (a^{3} d^{3}{\rm Ei}\left (\frac{b d x + a d}{b}\right ) - a^{3} d^{3}{\rm Ei}\left (-\frac{b d x + a d}{b}\right )\right )} \sinh \left (-\frac{b c - a d}{b}\right )}{2 \, b^{4} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(2*b^3*d*x - a*b^2*d)*cosh(d*x + c) + (a^3*d^3*Ei((b*d*x + a*d)/b) + a^3*d^3*Ei(-(b*d*x + a*d)/b))*cos
h(-(b*c - a*d)/b) - 2*(b^3*d^2*x^2 - a*b^2*d^2*x + a^2*b*d^2 + 2*b^3)*sinh(d*x + c) - (a^3*d^3*Ei((b*d*x + a*d
)/b) - a^3*d^3*Ei(-(b*d*x + a*d)/b))*sinh(-(b*c - a*d)/b))/(b^4*d^3)

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

[Out]

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

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Giac [A]  time = 1.1756, size = 84, normalized size = 0.56 \begin{align*} -\frac{a^{3}{\rm Ei}\left (\frac{b d x + a d}{b}\right ) e^{\left (c - \frac{a d}{b}\right )} + a^{3}{\rm Ei}\left (-\frac{b d x + a d}{b}\right ) e^{\left (-c + \frac{a d}{b}\right )}}{2 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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