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

Optimal. Leaf size=219 \[ \frac{a^4 \cosh \left (c-\frac{a d}{b}\right ) \text{Chi}\left (x d+\frac{a d}{b}\right )}{b^5}-\frac{a^2 \cosh (c+d x)}{b^3 d^2}+\frac{a^4 \sinh \left (c-\frac{a d}{b}\right ) \text{Shi}\left (x d+\frac{a d}{b}\right )}{b^5}-\frac{a^3 \sinh (c+d x)}{b^4 d}+\frac{a^2 x \sinh (c+d x)}{b^3 d}-\frac{2 a \sinh (c+d x)}{b^2 d^3}+\frac{2 a x \cosh (c+d x)}{b^2 d^2}-\frac{a x^2 \sinh (c+d x)}{b^2 d}-\frac{3 x^2 \cosh (c+d x)}{b d^2}+\frac{6 x \sinh (c+d x)}{b d^3}-\frac{6 \cosh (c+d x)}{b d^4}+\frac{x^3 \sinh (c+d x)}{b d} \]

[Out]

(-6*Cosh[c + d*x])/(b*d^4) - (a^2*Cosh[c + d*x])/(b^3*d^2) + (2*a*x*Cosh[c + d*x])/(b^2*d^2) - (3*x^2*Cosh[c +
 d*x])/(b*d^2) + (a^4*Cosh[c - (a*d)/b]*CoshIntegral[(a*d)/b + d*x])/b^5 - (2*a*Sinh[c + d*x])/(b^2*d^3) - (a^
3*Sinh[c + d*x])/(b^4*d) + (6*x*Sinh[c + d*x])/(b*d^3) + (a^2*x*Sinh[c + d*x])/(b^3*d) - (a*x^2*Sinh[c + d*x])
/(b^2*d) + (x^3*Sinh[c + d*x])/(b*d) + (a^4*Sinh[c - (a*d)/b]*SinhIntegral[(a*d)/b + d*x])/b^5

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Rubi [A]  time = 0.484027, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 15, 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} \[ \frac{a^4 \cosh \left (c-\frac{a d}{b}\right ) \text{Chi}\left (x d+\frac{a d}{b}\right )}{b^5}-\frac{a^2 \cosh (c+d x)}{b^3 d^2}+\frac{a^4 \sinh \left (c-\frac{a d}{b}\right ) \text{Shi}\left (x d+\frac{a d}{b}\right )}{b^5}-\frac{a^3 \sinh (c+d x)}{b^4 d}+\frac{a^2 x \sinh (c+d x)}{b^3 d}-\frac{2 a \sinh (c+d x)}{b^2 d^3}+\frac{2 a x \cosh (c+d x)}{b^2 d^2}-\frac{a x^2 \sinh (c+d x)}{b^2 d}-\frac{3 x^2 \cosh (c+d x)}{b d^2}+\frac{6 x \sinh (c+d x)}{b d^3}-\frac{6 \cosh (c+d x)}{b d^4}+\frac{x^3 \sinh (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*Cosh[c + d*x])/(a + b*x),x]

[Out]

(-6*Cosh[c + d*x])/(b*d^4) - (a^2*Cosh[c + d*x])/(b^3*d^2) + (2*a*x*Cosh[c + d*x])/(b^2*d^2) - (3*x^2*Cosh[c +
 d*x])/(b*d^2) + (a^4*Cosh[c - (a*d)/b]*CoshIntegral[(a*d)/b + d*x])/b^5 - (2*a*Sinh[c + d*x])/(b^2*d^3) - (a^
3*Sinh[c + d*x])/(b^4*d) + (6*x*Sinh[c + d*x])/(b*d^3) + (a^2*x*Sinh[c + d*x])/(b^3*d) - (a*x^2*Sinh[c + d*x])
/(b^2*d) + (x^3*Sinh[c + d*x])/(b*d) + (a^4*Sinh[c - (a*d)/b]*SinhIntegral[(a*d)/b + d*x])/b^5

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^4 \cosh (c+d x)}{a+b x} \, dx &=\int \left (-\frac{a^3 \cosh (c+d x)}{b^4}+\frac{a^2 x \cosh (c+d x)}{b^3}-\frac{a x^2 \cosh (c+d x)}{b^2}+\frac{x^3 \cosh (c+d x)}{b}+\frac{a^4 \cosh (c+d x)}{b^4 (a+b x)}\right ) \, dx\\ &=-\frac{a^3 \int \cosh (c+d x) \, dx}{b^4}+\frac{a^4 \int \frac{\cosh (c+d x)}{a+b x} \, dx}{b^4}+\frac{a^2 \int x \cosh (c+d x) \, dx}{b^3}-\frac{a \int x^2 \cosh (c+d x) \, dx}{b^2}+\frac{\int x^3 \cosh (c+d x) \, dx}{b}\\ &=-\frac{a^3 \sinh (c+d x)}{b^4 d}+\frac{a^2 x \sinh (c+d x)}{b^3 d}-\frac{a x^2 \sinh (c+d x)}{b^2 d}+\frac{x^3 \sinh (c+d x)}{b d}-\frac{a^2 \int \sinh (c+d x) \, dx}{b^3 d}+\frac{(2 a) \int x \sinh (c+d x) \, dx}{b^2 d}-\frac{3 \int x^2 \sinh (c+d x) \, dx}{b d}+\frac{\left (a^4 \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^4}+\frac{\left (a^4 \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^4}\\ &=-\frac{a^2 \cosh (c+d x)}{b^3 d^2}+\frac{2 a x \cosh (c+d x)}{b^2 d^2}-\frac{3 x^2 \cosh (c+d x)}{b d^2}+\frac{a^4 \cosh \left (c-\frac{a d}{b}\right ) \text{Chi}\left (\frac{a d}{b}+d x\right )}{b^5}-\frac{a^3 \sinh (c+d x)}{b^4 d}+\frac{a^2 x \sinh (c+d x)}{b^3 d}-\frac{a x^2 \sinh (c+d x)}{b^2 d}+\frac{x^3 \sinh (c+d x)}{b d}+\frac{a^4 \sinh \left (c-\frac{a d}{b}\right ) \text{Shi}\left (\frac{a d}{b}+d x\right )}{b^5}-\frac{(2 a) \int \cosh (c+d x) \, dx}{b^2 d^2}+\frac{6 \int x \cosh (c+d x) \, dx}{b d^2}\\ &=-\frac{a^2 \cosh (c+d x)}{b^3 d^2}+\frac{2 a x \cosh (c+d x)}{b^2 d^2}-\frac{3 x^2 \cosh (c+d x)}{b d^2}+\frac{a^4 \cosh \left (c-\frac{a d}{b}\right ) \text{Chi}\left (\frac{a d}{b}+d x\right )}{b^5}-\frac{2 a \sinh (c+d x)}{b^2 d^3}-\frac{a^3 \sinh (c+d x)}{b^4 d}+\frac{6 x \sinh (c+d x)}{b d^3}+\frac{a^2 x \sinh (c+d x)}{b^3 d}-\frac{a x^2 \sinh (c+d x)}{b^2 d}+\frac{x^3 \sinh (c+d x)}{b d}+\frac{a^4 \sinh \left (c-\frac{a d}{b}\right ) \text{Shi}\left (\frac{a d}{b}+d x\right )}{b^5}-\frac{6 \int \sinh (c+d x) \, dx}{b d^3}\\ &=-\frac{6 \cosh (c+d x)}{b d^4}-\frac{a^2 \cosh (c+d x)}{b^3 d^2}+\frac{2 a x \cosh (c+d x)}{b^2 d^2}-\frac{3 x^2 \cosh (c+d x)}{b d^2}+\frac{a^4 \cosh \left (c-\frac{a d}{b}\right ) \text{Chi}\left (\frac{a d}{b}+d x\right )}{b^5}-\frac{2 a \sinh (c+d x)}{b^2 d^3}-\frac{a^3 \sinh (c+d x)}{b^4 d}+\frac{6 x \sinh (c+d x)}{b d^3}+\frac{a^2 x \sinh (c+d x)}{b^3 d}-\frac{a x^2 \sinh (c+d x)}{b^2 d}+\frac{x^3 \sinh (c+d x)}{b d}+\frac{a^4 \sinh \left (c-\frac{a d}{b}\right ) \text{Shi}\left (\frac{a d}{b}+d x\right )}{b^5}\\ \end{align*}

Mathematica [A]  time = 0.631353, size = 159, normalized size = 0.73 \[ \frac{-b \left (d \left (-a^2 b d^2 x+a^3 d^2+a b^2 \left (d^2 x^2+2\right )-b^3 x \left (d^2 x^2+6\right )\right ) \sinh (c+d x)+b \left (a^2 d^2-2 a b d^2 x+3 b^2 \left (d^2 x^2+2\right )\right ) \cosh (c+d x)\right )+a^4 d^4 \cosh \left (c-\frac{a d}{b}\right ) \text{Chi}\left (d \left (\frac{a}{b}+x\right )\right )+a^4 d^4 \sinh \left (c-\frac{a d}{b}\right ) \text{Shi}\left (d \left (\frac{a}{b}+x\right )\right )}{b^5 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*d^4*Cosh[c - (a*d)/b]*CoshIntegral[d*(a/b + x)] - b*(b*(a^2*d^2 - 2*a*b*d^2*x + 3*b^2*(2 + d^2*x^2))*Cosh
[c + d*x] + d*(a^3*d^2 - a^2*b*d^2*x + a*b^2*(2 + d^2*x^2) - b^3*x*(6 + d^2*x^2))*Sinh[c + d*x]) + a^4*d^4*Sin
h[c - (a*d)/b]*SinhIntegral[d*(a/b + x)])/(b^5*d^4)

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Maple [A]  time = 0.133, size = 442, normalized size = 2. \begin{align*} -{\frac{{a}^{4}}{2\,{b}^{5}}{{\rm e}^{{\frac{da-cb}{b}}}}{\it Ei} \left ( 1,dx+c+{\frac{da-cb}{b}} \right ) }-{\frac{3\,{{\rm e}^{-dx-c}}{x}^{2}}{2\,b{d}^{2}}}-{\frac{{{\rm e}^{-dx-c}}{a}^{2}}{2\,{d}^{2}{b}^{3}}}-3\,{\frac{{{\rm e}^{-dx-c}}}{b{d}^{4}}}+{\frac{{{\rm e}^{-dx-c}}ax}{{d}^{2}{b}^{2}}}-{\frac{{{\rm e}^{-dx-c}}{x}^{3}}{2\,bd}}+{\frac{{{\rm e}^{-dx-c}}{a}^{3}}{2\,d{b}^{4}}}-3\,{\frac{{{\rm e}^{-dx-c}}x}{{d}^{3}b}}+{\frac{{{\rm e}^{-dx-c}}a}{{d}^{3}{b}^{2}}}+{\frac{{{\rm e}^{-dx-c}}a{x}^{2}}{2\,{b}^{2}d}}-{\frac{{{\rm e}^{-dx-c}}{a}^{2}x}{2\,{b}^{3}d}}-3\,{\frac{{{\rm e}^{dx+c}}}{b{d}^{4}}}-{\frac{a{{\rm e}^{dx+c}}{x}^{2}}{2\,{b}^{2}d}}+{\frac{a{{\rm e}^{dx+c}}x}{{d}^{2}{b}^{2}}}-{\frac{{a}^{4}}{2\,{b}^{5}}{{\rm e}^{-{\frac{da-cb}{b}}}}{\it Ei} \left ( 1,-dx-c-{\frac{da-cb}{b}} \right ) }+{\frac{{{\rm e}^{dx+c}}{x}^{3}}{2\,bd}}-{\frac{3\,{{\rm e}^{dx+c}}{x}^{2}}{2\,b{d}^{2}}}+3\,{\frac{{{\rm e}^{dx+c}}x}{{d}^{3}b}}-{\frac{{{\rm e}^{dx+c}}{a}^{3}}{2\,d{b}^{4}}}-{\frac{a{{\rm e}^{dx+c}}}{{d}^{3}{b}^{2}}}-{\frac{{{\rm e}^{dx+c}}{a}^{2}}{2\,{d}^{2}{b}^{3}}}+{\frac{{{\rm e}^{dx+c}}{a}^{2}x}{2\,{b}^{3}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/b^5*exp((a*d-b*c)/b)*Ei(1,d*x+c+(a*d-b*c)/b)*a^4-3/2/d^2*exp(-d*x-c)/b*x^2-1/2/d^2*exp(-d*x-c)/b^3*a^2-3/
d^4*exp(-d*x-c)/b+1/d^2*exp(-d*x-c)/b^2*a*x-1/2/d*exp(-d*x-c)/b*x^3+1/2/d*exp(-d*x-c)/b^4*a^3-3/d^3*exp(-d*x-c
)/b*x+1/d^3*exp(-d*x-c)/b^2*a+1/2/d*exp(-d*x-c)/b^2*a*x^2-1/2/d*exp(-d*x-c)/b^3*a^2*x-3/d^4/b*exp(d*x+c)-1/2/d
/b^2*a*exp(d*x+c)*x^2+1/d^2/b^2*a*exp(d*x+c)*x-1/2/b^5*exp(-(a*d-b*c)/b)*Ei(1,-d*x-c-(a*d-b*c)/b)*a^4+1/2/d/b*
exp(d*x+c)*x^3-3/2/d^2/b*exp(d*x+c)*x^2+3/d^3/b*exp(d*x+c)*x-1/2/d/b^4*a^3*exp(d*x+c)-1/d^3/b^2*a*exp(d*x+c)-1
/2/d^2/b^3*a^2*exp(d*x+c)+1/2/d/b^3*a^2*exp(d*x+c)*x

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Maxima [A]  time = 1.44775, size = 590, normalized size = 2.69 \begin{align*} -\frac{1}{24} \, d{\left (\frac{12 \, a^{4}{\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^{4} d} - \frac{12 \, a^{3}{\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^{4}} + \frac{6 \, a^{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 )}}{b^{3}} - \frac{4 \, a{\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^{2}} + \frac{3 \,{\left (\frac{{\left (d^{4} x^{4} e^{c} - 4 \, d^{3} x^{3} e^{c} + 12 \, d^{2} x^{2} e^{c} - 24 \, d x e^{c} + 24 \, e^{c}\right )} e^{\left (d x\right )}}{d^{5}} + \frac{{\left (d^{4} x^{4} + 4 \, d^{3} x^{3} + 12 \, d^{2} x^{2} + 24 \, d x + 24\right )} e^{\left (-d x - c\right )}}{d^{5}}\right )}}{b} + \frac{24 \, a^{4} \cosh \left (d x + c\right ) \log \left (b x + a\right )}{b^{5} d}\right )} + \frac{1}{12} \,{\left (\frac{12 \, a^{4} \log \left (b x + a\right )}{b^{5}} + \frac{3 \, b^{3} x^{4} - 4 \, a b^{2} x^{3} + 6 \, a^{2} b x^{2} - 12 \, a^{3} x}{b^{4}}\right )} \cosh \left (d x + c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*d*(12*a^4*(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^4*d) - 12*a^3*((d*x*e^c - e^c)*e^(d*x)/d^2 + (d*x + 1)*e^(-d*x - c)/d^2)/b^4 + 6*a^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^3 - 4*a*((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^2 + 3*((d^4*x^
4*e^c - 4*d^3*x^3*e^c + 12*d^2*x^2*e^c - 24*d*x*e^c + 24*e^c)*e^(d*x)/d^5 + (d^4*x^4 + 4*d^3*x^3 + 12*d^2*x^2
+ 24*d*x + 24)*e^(-d*x - c)/d^5)/b + 24*a^4*cosh(d*x + c)*log(b*x + a)/(b^5*d)) + 1/12*(12*a^4*log(b*x + a)/b^
5 + (3*b^3*x^4 - 4*a*b^2*x^3 + 6*a^2*b*x^2 - 12*a^3*x)/b^4)*cosh(d*x + c)

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Fricas [A]  time = 2.09321, size = 481, normalized size = 2.2 \begin{align*} -\frac{2 \,{\left (3 \, b^{4} d^{2} x^{2} - 2 \, a b^{3} d^{2} x + a^{2} b^{2} d^{2} + 6 \, b^{4}\right )} \cosh \left (d x + c\right ) -{\left (a^{4} d^{4}{\rm Ei}\left (\frac{b d x + a d}{b}\right ) + a^{4} d^{4}{\rm Ei}\left (-\frac{b d x + a d}{b}\right )\right )} \cosh \left (-\frac{b c - a d}{b}\right ) - 2 \,{\left (b^{4} d^{3} x^{3} - a b^{3} d^{3} x^{2} - a^{3} b d^{3} - 2 \, a b^{3} d +{\left (a^{2} b^{2} d^{3} + 6 \, b^{4} d\right )} x\right )} \sinh \left (d x + c\right ) +{\left (a^{4} d^{4}{\rm Ei}\left (\frac{b d x + a d}{b}\right ) - a^{4} d^{4}{\rm Ei}\left (-\frac{b d x + a d}{b}\right )\right )} \sinh \left (-\frac{b c - a d}{b}\right )}{2 \, b^{5} d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(3*b^4*d^2*x^2 - 2*a*b^3*d^2*x + a^2*b^2*d^2 + 6*b^4)*cosh(d*x + c) - (a^4*d^4*Ei((b*d*x + a*d)/b) + a
^4*d^4*Ei(-(b*d*x + a*d)/b))*cosh(-(b*c - a*d)/b) - 2*(b^4*d^3*x^3 - a*b^3*d^3*x^2 - a^3*b*d^3 - 2*a*b^3*d + (
a^2*b^2*d^3 + 6*b^4*d)*x)*sinh(d*x + c) + (a^4*d^4*Ei((b*d*x + a*d)/b) - a^4*d^4*Ei(-(b*d*x + a*d)/b))*sinh(-(
b*c - a*d)/b))/(b^5*d^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4} \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**4*cosh(d*x+c)/(b*x+a),x)

[Out]

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

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Giac [A]  time = 1.14898, size = 84, normalized size = 0.38 \begin{align*} \frac{a^{4}{\rm Ei}\left (\frac{b d x + a d}{b}\right ) e^{\left (c - \frac{a d}{b}\right )} + a^{4}{\rm Ei}\left (-\frac{b d x + a d}{b}\right ) e^{\left (-c + \frac{a d}{b}\right )}}{2 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a^4*Ei((b*d*x + a*d)/b)*e^(c - a*d/b) + a^4*Ei(-(b*d*x + a*d)/b)*e^(-c + a*d/b))/b^5