3.92 \(\int \frac{(a+b x^3)^2 \cosh (c+d x)}{x^4} \, dx\)

Optimal. Leaf size=150 \[ \frac{1}{6} a^2 d^3 \sinh (c) \text{Chi}(d x)+\frac{1}{6} a^2 d^3 \cosh (c) \text{Shi}(d x)-\frac{a^2 d^2 \cosh (c+d x)}{6 x}-\frac{a^2 d \sinh (c+d x)}{6 x^2}-\frac{a^2 \cosh (c+d x)}{3 x^3}+2 a b \cosh (c) \text{Chi}(d x)+2 a b \sinh (c) \text{Shi}(d x)+\frac{2 b^2 \sinh (c+d x)}{d^3}-\frac{2 b^2 x \cosh (c+d x)}{d^2}+\frac{b^2 x^2 \sinh (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.278846, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {5287, 3297, 3303, 3298, 3301, 3296, 2637} \[ \frac{1}{6} a^2 d^3 \sinh (c) \text{Chi}(d x)+\frac{1}{6} a^2 d^3 \cosh (c) \text{Shi}(d x)-\frac{a^2 d^2 \cosh (c+d x)}{6 x}-\frac{a^2 d \sinh (c+d x)}{6 x^2}-\frac{a^2 \cosh (c+d x)}{3 x^3}+2 a b \cosh (c) \text{Chi}(d x)+2 a b \sinh (c) \text{Shi}(d x)+\frac{2 b^2 \sinh (c+d x)}{d^3}-\frac{2 b^2 x \cosh (c+d x)}{d^2}+\frac{b^2 x^2 \sinh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)^2*Cosh[c + d*x])/x^4,x]

[Out]

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

Rule 5287

Int[Cosh[(c_.) + (d_.)*(x_)]*((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[Cosh[c + d*x], (e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]

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]

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 2637

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

Rubi steps

\begin{align*} \int \frac{\left (a+b x^3\right )^2 \cosh (c+d x)}{x^4} \, dx &=\int \left (\frac{a^2 \cosh (c+d x)}{x^4}+\frac{2 a b \cosh (c+d x)}{x}+b^2 x^2 \cosh (c+d x)\right ) \, dx\\ &=a^2 \int \frac{\cosh (c+d x)}{x^4} \, dx+(2 a b) \int \frac{\cosh (c+d x)}{x} \, dx+b^2 \int x^2 \cosh (c+d x) \, dx\\ &=-\frac{a^2 \cosh (c+d x)}{3 x^3}+\frac{b^2 x^2 \sinh (c+d x)}{d}-\frac{\left (2 b^2\right ) \int x \sinh (c+d x) \, dx}{d}+\frac{1}{3} \left (a^2 d\right ) \int \frac{\sinh (c+d x)}{x^3} \, dx+(2 a b \cosh (c)) \int \frac{\cosh (d x)}{x} \, dx+(2 a b \sinh (c)) \int \frac{\sinh (d x)}{x} \, dx\\ &=-\frac{a^2 \cosh (c+d x)}{3 x^3}-\frac{2 b^2 x \cosh (c+d x)}{d^2}+2 a b \cosh (c) \text{Chi}(d x)-\frac{a^2 d \sinh (c+d x)}{6 x^2}+\frac{b^2 x^2 \sinh (c+d x)}{d}+2 a b \sinh (c) \text{Shi}(d x)+\frac{\left (2 b^2\right ) \int \cosh (c+d x) \, dx}{d^2}+\frac{1}{6} \left (a^2 d^2\right ) \int \frac{\cosh (c+d x)}{x^2} \, dx\\ &=-\frac{a^2 \cosh (c+d x)}{3 x^3}-\frac{a^2 d^2 \cosh (c+d x)}{6 x}-\frac{2 b^2 x \cosh (c+d x)}{d^2}+2 a b \cosh (c) \text{Chi}(d x)+\frac{2 b^2 \sinh (c+d x)}{d^3}-\frac{a^2 d \sinh (c+d x)}{6 x^2}+\frac{b^2 x^2 \sinh (c+d x)}{d}+2 a b \sinh (c) \text{Shi}(d x)+\frac{1}{6} \left (a^2 d^3\right ) \int \frac{\sinh (c+d x)}{x} \, dx\\ &=-\frac{a^2 \cosh (c+d x)}{3 x^3}-\frac{a^2 d^2 \cosh (c+d x)}{6 x}-\frac{2 b^2 x \cosh (c+d x)}{d^2}+2 a b \cosh (c) \text{Chi}(d x)+\frac{2 b^2 \sinh (c+d x)}{d^3}-\frac{a^2 d \sinh (c+d x)}{6 x^2}+\frac{b^2 x^2 \sinh (c+d x)}{d}+2 a b \sinh (c) \text{Shi}(d x)+\frac{1}{6} \left (a^2 d^3 \cosh (c)\right ) \int \frac{\sinh (d x)}{x} \, dx+\frac{1}{6} \left (a^2 d^3 \sinh (c)\right ) \int \frac{\cosh (d x)}{x} \, dx\\ &=-\frac{a^2 \cosh (c+d x)}{3 x^3}-\frac{a^2 d^2 \cosh (c+d x)}{6 x}-\frac{2 b^2 x \cosh (c+d x)}{d^2}+2 a b \cosh (c) \text{Chi}(d x)+\frac{1}{6} a^2 d^3 \text{Chi}(d x) \sinh (c)+\frac{2 b^2 \sinh (c+d x)}{d^3}-\frac{a^2 d \sinh (c+d x)}{6 x^2}+\frac{b^2 x^2 \sinh (c+d x)}{d}+\frac{1}{6} a^2 d^3 \cosh (c) \text{Shi}(d x)+2 a b \sinh (c) \text{Shi}(d x)\\ \end{align*}

Mathematica [A]  time = 0.569738, size = 135, normalized size = 0.9 \[ \frac{1}{6} \left (-\frac{a^2 d^2 \cosh (c+d x)}{x}-\frac{a^2 d \sinh (c+d x)}{x^2}-\frac{2 a^2 \cosh (c+d x)}{x^3}+a \text{Chi}(d x) \left (a d^3 \sinh (c)+12 b \cosh (c)\right )+a \text{Shi}(d x) \left (a d^3 \cosh (c)+12 b \sinh (c)\right )+\frac{12 b^2 \sinh (c+d x)}{d^3}-\frac{12 b^2 x \cosh (c+d x)}{d^2}+\frac{6 b^2 x^2 \sinh (c+d x)}{d}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^3)^2*Cosh[c + d*x])/x^4,x]

[Out]

((-2*a^2*Cosh[c + d*x])/x^3 - (a^2*d^2*Cosh[c + d*x])/x - (12*b^2*x*Cosh[c + d*x])/d^2 + a*CoshIntegral[d*x]*(
12*b*Cosh[c] + a*d^3*Sinh[c]) + (12*b^2*Sinh[c + d*x])/d^3 - (a^2*d*Sinh[c + d*x])/x^2 + (6*b^2*x^2*Sinh[c + d
*x])/d + a*(a*d^3*Cosh[c] + 12*b*Sinh[c])*SinhIntegral[d*x])/6

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Maple [A]  time = 0.128, size = 261, normalized size = 1.7 \begin{align*} -{\frac{{b}^{2}{{\rm e}^{-dx-c}}{x}^{2}}{2\,d}}-{\frac{{b}^{2}{{\rm e}^{-dx-c}}x}{{d}^{2}}}+{\frac{{d}^{3}{a}^{2}{{\rm e}^{-c}}{\it Ei} \left ( 1,dx \right ) }{12}}-{\frac{{a}^{2}{d}^{2}{{\rm e}^{-dx-c}}}{12\,x}}+{\frac{d{a}^{2}{{\rm e}^{-dx-c}}}{12\,{x}^{2}}}-{\frac{{a}^{2}{{\rm e}^{-dx-c}}}{6\,{x}^{3}}}-ab{{\rm e}^{-c}}{\it Ei} \left ( 1,dx \right ) -{\frac{{b}^{2}{{\rm e}^{-dx-c}}}{{d}^{3}}}-{\frac{{d}^{3}{a}^{2}{{\rm e}^{c}}{\it Ei} \left ( 1,-dx \right ) }{12}}+{\frac{{{\rm e}^{dx+c}}{b}^{2}}{{d}^{3}}}-{\frac{{{\rm e}^{dx+c}}{a}^{2}}{6\,{x}^{3}}}-{\frac{d{a}^{2}{{\rm e}^{dx+c}}}{12\,{x}^{2}}}-{\frac{{a}^{2}{d}^{2}{{\rm e}^{dx+c}}}{12\,x}}-ab{{\rm e}^{c}}{\it Ei} \left ( 1,-dx \right ) +{\frac{{{\rm e}^{dx+c}}{b}^{2}{x}^{2}}{2\,d}}-{\frac{{{\rm e}^{dx+c}}{b}^{2}x}{{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^2*cosh(d*x+c)/x^4,x)

[Out]

-1/2/d*b^2*exp(-d*x-c)*x^2-1/d^2*b^2*exp(-d*x-c)*x+1/12*d^3*a^2*exp(-c)*Ei(1,d*x)-1/12*d^2*a^2*exp(-d*x-c)/x+1
/12*d*a^2*exp(-d*x-c)/x^2-1/6*a^2*exp(-d*x-c)/x^3-a*b*exp(-c)*Ei(1,d*x)-1/d^3*b^2*exp(-d*x-c)-1/12*d^3*a^2*exp
(c)*Ei(1,-d*x)+1/d^3*b^2*exp(d*x+c)-1/6*a^2/x^3*exp(d*x+c)-1/12*d*a^2/x^2*exp(d*x+c)-1/12*d^2*a^2/x*exp(d*x+c)
-a*b*exp(c)*Ei(1,-d*x)+1/2/d*b^2*exp(d*x+c)*x^2-1/d^2*b^2*exp(d*x+c)*x

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Maxima [A]  time = 1.25149, size = 254, normalized size = 1.69 \begin{align*} \frac{1}{6} \,{\left ({\left (d^{2} e^{\left (-c\right )} \Gamma \left (-2, d x\right ) - d^{2} e^{c} \Gamma \left (-2, -d x\right )\right )} a^{2} - b^{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 )} - \frac{4 \, a b \cosh \left (d x + c\right ) \log \left (x^{3}\right )}{d} + \frac{6 \,{\left ({\rm Ei}\left (-d x\right ) e^{\left (-c\right )} +{\rm Ei}\left (d x\right ) e^{c}\right )} a b}{d}\right )} d + \frac{1}{3} \,{\left (b^{2} x^{3} + 2 \, a b \log \left (x^{3}\right ) - \frac{a^{2}}{x^{3}}\right )} \cosh \left (d x + c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*cosh(d*x+c)/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.75958, size = 412, normalized size = 2.75 \begin{align*} -\frac{2 \,{\left (a^{2} d^{5} x^{2} + 12 \, b^{2} d x^{4} + 2 \, a^{2} d^{3}\right )} \cosh \left (d x + c\right ) -{\left ({\left (a^{2} d^{6} + 12 \, a b d^{3}\right )} x^{3}{\rm Ei}\left (d x\right ) -{\left (a^{2} d^{6} - 12 \, a b d^{3}\right )} x^{3}{\rm Ei}\left (-d x\right )\right )} \cosh \left (c\right ) - 2 \,{\left (6 \, b^{2} d^{2} x^{5} - a^{2} d^{4} x + 12 \, b^{2} x^{3}\right )} \sinh \left (d x + c\right ) -{\left ({\left (a^{2} d^{6} + 12 \, a b d^{3}\right )} x^{3}{\rm Ei}\left (d x\right ) +{\left (a^{2} d^{6} - 12 \, a b d^{3}\right )} x^{3}{\rm Ei}\left (-d x\right )\right )} \sinh \left (c\right )}{12 \, d^{3} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*cosh(d*x+c)/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{3}\right )^{2} \cosh{\left (c + d x \right )}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**2*cosh(d*x+c)/x**4,x)

[Out]

Integral((a + b*x**3)**2*cosh(c + d*x)/x**4, x)

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Giac [A]  time = 1.33523, size = 377, normalized size = 2.51 \begin{align*} -\frac{a^{2} d^{6} x^{3}{\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - a^{2} d^{6} x^{3}{\rm Ei}\left (d x\right ) e^{c} + a^{2} d^{5} x^{2} e^{\left (d x + c\right )} - 6 \, b^{2} d^{2} x^{5} e^{\left (d x + c\right )} + a^{2} d^{5} x^{2} e^{\left (-d x - c\right )} + 6 \, b^{2} d^{2} x^{5} e^{\left (-d x - c\right )} - 12 \, a b d^{3} x^{3}{\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - 12 \, a b d^{3} x^{3}{\rm Ei}\left (d x\right ) e^{c} + a^{2} d^{4} x e^{\left (d x + c\right )} + 12 \, b^{2} d x^{4} e^{\left (d x + c\right )} - a^{2} d^{4} x e^{\left (-d x - c\right )} + 12 \, b^{2} d x^{4} e^{\left (-d x - c\right )} + 2 \, a^{2} d^{3} e^{\left (d x + c\right )} - 12 \, b^{2} x^{3} e^{\left (d x + c\right )} + 2 \, a^{2} d^{3} e^{\left (-d x - c\right )} + 12 \, b^{2} x^{3} e^{\left (-d x - c\right )}}{12 \, d^{3} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*cosh(d*x+c)/x^4,x, algorithm="giac")

[Out]

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