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

Optimal. Leaf size=125 \[ -\frac{a d \sinh \left (c-\frac{a d}{b}\right ) \text{Chi}\left (x d+\frac{a d}{b}\right )}{b^3}+\frac{\cosh \left (c-\frac{a d}{b}\right ) \text{Chi}\left (x d+\frac{a d}{b}\right )}{b^2}+\frac{\sinh \left (c-\frac{a d}{b}\right ) \text{Shi}\left (x d+\frac{a d}{b}\right )}{b^2}-\frac{a d \cosh \left (c-\frac{a d}{b}\right ) \text{Shi}\left (x d+\frac{a d}{b}\right )}{b^3}+\frac{a \cosh (c+d x)}{b^2 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.297248, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6742, 3297, 3303, 3298, 3301} \[ -\frac{a d \sinh \left (c-\frac{a d}{b}\right ) \text{Chi}\left (x d+\frac{a d}{b}\right )}{b^3}+\frac{\cosh \left (c-\frac{a d}{b}\right ) \text{Chi}\left (x d+\frac{a d}{b}\right )}{b^2}+\frac{\sinh \left (c-\frac{a d}{b}\right ) \text{Shi}\left (x d+\frac{a d}{b}\right )}{b^2}-\frac{a d \cosh \left (c-\frac{a d}{b}\right ) \text{Shi}\left (x d+\frac{a d}{b}\right )}{b^3}+\frac{a \cosh (c+d x)}{b^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[(x*Cosh[c + d*x])/(a + b*x)^2,x]

[Out]

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

Rule 6742

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

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

Mathematica [A]  time = 0.41379, size = 97, normalized size = 0.78 \[ \frac{\text{Chi}\left (d \left (\frac{a}{b}+x\right )\right ) \left (b \cosh \left (c-\frac{a d}{b}\right )-a d \sinh \left (c-\frac{a d}{b}\right )\right )+\text{Shi}\left (d \left (\frac{a}{b}+x\right )\right ) \left (b \sinh \left (c-\frac{a d}{b}\right )-a d \cosh \left (c-\frac{a d}{b}\right )\right )+\frac{a b \cosh (c+d x)}{a+b x}}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.027, size = 215, normalized size = 1.7 \begin{align*}{\frac{d{{\rm e}^{-dx-c}}a}{2\,{b}^{2} \left ( bdx+da \right ) }}-{\frac{da}{2\,{b}^{3}}{{\rm e}^{{\frac{da-cb}{b}}}}{\it Ei} \left ( 1,dx+c+{\frac{da-cb}{b}} \right ) }-{\frac{1}{2\,{b}^{2}}{{\rm e}^{{\frac{da-cb}{b}}}}{\it Ei} \left ( 1,dx+c+{\frac{da-cb}{b}} \right ) }+{\frac{d{{\rm e}^{dx+c}}a}{2\,{b}^{3}} \left ({\frac{da}{b}}+dx \right ) ^{-1}}+{\frac{da}{2\,{b}^{3}}{{\rm e}^{-{\frac{da-cb}{b}}}}{\it Ei} \left ( 1,-dx-c-{\frac{da-cb}{b}} \right ) }-{\frac{1}{2\,{b}^{2}}{{\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*cosh(d*x+c)/(b*x+a)^2,x)

[Out]

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

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Maxima [A]  time = 1.36264, size = 240, normalized size = 1.92 \begin{align*} -\frac{1}{2} \,{\left (a{\left (\frac{e^{\left (-c + \frac{a d}{b}\right )} E_{1}\left (\frac{{\left (b x + a\right )} d}{b}\right )}{b^{3}} - \frac{e^{\left (c - \frac{a d}{b}\right )} E_{1}\left (-\frac{{\left (b x + a\right )} d}{b}\right )}{b^{3}}\right )} + \frac{\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}}{b d} + \frac{2 \, \cosh \left (d x + c\right ) \log \left (b x + a\right )}{b^{2} d}\right )} d +{\left (\frac{a}{b^{3} x + a b^{2}} + \frac{\log \left (b x + a\right )}{b^{2}}\right )} \cosh \left (d x + c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.03375, size = 414, normalized size = 3.31 \begin{align*} \frac{2 \, a b \cosh \left (d x + c\right ) -{\left ({\left (a^{2} d - a b +{\left (a b d - b^{2}\right )} x\right )}{\rm Ei}\left (\frac{b d x + a d}{b}\right ) -{\left (a^{2} d + a b +{\left (a b d + b^{2}\right )} x\right )}{\rm Ei}\left (-\frac{b d x + a d}{b}\right )\right )} \cosh \left (-\frac{b c - a d}{b}\right ) +{\left ({\left (a^{2} d - a b +{\left (a b d - b^{2}\right )} x\right )}{\rm Ei}\left (\frac{b d x + a d}{b}\right ) +{\left (a^{2} d + a b +{\left (a b d + b^{2}\right )} x\right )}{\rm Ei}\left (-\frac{b d x + a d}{b}\right )\right )} \sinh \left (-\frac{b c - a d}{b}\right )}{2 \,{\left (b^{4} x + a b^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.24044, size = 373, normalized size = 2.98 \begin{align*} -\frac{a b d x{\rm Ei}\left (\frac{b d x + a d}{b}\right ) e^{\left (c - \frac{a d}{b}\right )} - a b d x{\rm Ei}\left (-\frac{b d x + a d}{b}\right ) e^{\left (-c + \frac{a d}{b}\right )} + a^{2} d{\rm Ei}\left (\frac{b d x + a d}{b}\right ) e^{\left (c - \frac{a d}{b}\right )} - b^{2} x{\rm Ei}\left (\frac{b d x + a d}{b}\right ) e^{\left (c - \frac{a d}{b}\right )} - a^{2} d{\rm Ei}\left (-\frac{b d x + a d}{b}\right ) e^{\left (-c + \frac{a d}{b}\right )} - b^{2} x{\rm Ei}\left (-\frac{b d x + a d}{b}\right ) e^{\left (-c + \frac{a d}{b}\right )} - a b{\rm Ei}\left (\frac{b d x + a d}{b}\right ) e^{\left (c - \frac{a d}{b}\right )} - a b{\rm Ei}\left (-\frac{b d x + a d}{b}\right ) e^{\left (-c + \frac{a d}{b}\right )} - a b e^{\left (d x + c\right )} - a b e^{\left (-d x - c\right )}}{2 \,{\left (b^{3} x + a b^{2}\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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