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

Optimal. Leaf size=141 \[ \frac{1}{2} a^2 d^2 \cosh (c) \text{Chi}(d x)+\frac{1}{2} a^2 d^2 \sinh (c) \text{Shi}(d x)-\frac{a^2 \cosh (c+d x)}{2 x^2}-\frac{a^2 d \sinh (c+d x)}{2 x}+\frac{2 a b \sinh (c+d x)}{d}-\frac{3 b^2 x^2 \cosh (c+d x)}{d^2}+\frac{6 b^2 x \sinh (c+d x)}{d^3}-\frac{6 b^2 \cosh (c+d x)}{d^4}+\frac{b^2 x^3 \sinh (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.232806, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.421, Rules used = {5287, 2637, 3297, 3303, 3298, 3301, 3296, 2638} \[ \frac{1}{2} a^2 d^2 \cosh (c) \text{Chi}(d x)+\frac{1}{2} a^2 d^2 \sinh (c) \text{Shi}(d x)-\frac{a^2 \cosh (c+d x)}{2 x^2}-\frac{a^2 d \sinh (c+d x)}{2 x}+\frac{2 a b \sinh (c+d x)}{d}-\frac{3 b^2 x^2 \cosh (c+d x)}{d^2}+\frac{6 b^2 x \sinh (c+d x)}{d^3}-\frac{6 b^2 \cosh (c+d x)}{d^4}+\frac{b^2 x^3 \sinh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2637

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

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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[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^3} \, dx &=\int \left (2 a b \cosh (c+d x)+\frac{a^2 \cosh (c+d x)}{x^3}+b^2 x^3 \cosh (c+d x)\right ) \, dx\\ &=a^2 \int \frac{\cosh (c+d x)}{x^3} \, dx+(2 a b) \int \cosh (c+d x) \, dx+b^2 \int x^3 \cosh (c+d x) \, dx\\ &=-\frac{a^2 \cosh (c+d x)}{2 x^2}+\frac{2 a b \sinh (c+d x)}{d}+\frac{b^2 x^3 \sinh (c+d x)}{d}-\frac{\left (3 b^2\right ) \int x^2 \sinh (c+d x) \, dx}{d}+\frac{1}{2} \left (a^2 d\right ) \int \frac{\sinh (c+d x)}{x^2} \, dx\\ &=-\frac{a^2 \cosh (c+d x)}{2 x^2}-\frac{3 b^2 x^2 \cosh (c+d x)}{d^2}+\frac{2 a b \sinh (c+d x)}{d}-\frac{a^2 d \sinh (c+d x)}{2 x}+\frac{b^2 x^3 \sinh (c+d x)}{d}+\frac{\left (6 b^2\right ) \int x \cosh (c+d x) \, dx}{d^2}+\frac{1}{2} \left (a^2 d^2\right ) \int \frac{\cosh (c+d x)}{x} \, dx\\ &=-\frac{a^2 \cosh (c+d x)}{2 x^2}-\frac{3 b^2 x^2 \cosh (c+d x)}{d^2}+\frac{2 a b \sinh (c+d x)}{d}-\frac{a^2 d \sinh (c+d x)}{2 x}+\frac{6 b^2 x \sinh (c+d x)}{d^3}+\frac{b^2 x^3 \sinh (c+d x)}{d}-\frac{\left (6 b^2\right ) \int \sinh (c+d x) \, dx}{d^3}+\frac{1}{2} \left (a^2 d^2 \cosh (c)\right ) \int \frac{\cosh (d x)}{x} \, dx+\frac{1}{2} \left (a^2 d^2 \sinh (c)\right ) \int \frac{\sinh (d x)}{x} \, dx\\ &=-\frac{6 b^2 \cosh (c+d x)}{d^4}-\frac{a^2 \cosh (c+d x)}{2 x^2}-\frac{3 b^2 x^2 \cosh (c+d x)}{d^2}+\frac{1}{2} a^2 d^2 \cosh (c) \text{Chi}(d x)+\frac{2 a b \sinh (c+d x)}{d}-\frac{a^2 d \sinh (c+d x)}{2 x}+\frac{6 b^2 x \sinh (c+d x)}{d^3}+\frac{b^2 x^3 \sinh (c+d x)}{d}+\frac{1}{2} a^2 d^2 \sinh (c) \text{Shi}(d x)\\ \end{align*}

Mathematica [A]  time = 0.347951, size = 136, normalized size = 0.96 \[ \frac{1}{2} \left (a^2 d^2 \cosh (c) \text{Chi}(d x)+a^2 d^2 \sinh (c) \text{Shi}(d x)-\frac{a^2 \cosh (c+d x)}{x^2}-\frac{a^2 d \sinh (c+d x)}{x}+\frac{4 a b \sinh (c+d x)}{d}-\frac{6 b^2 x^2 \cosh (c+d x)}{d^2}+\frac{12 b^2 x \sinh (c+d x)}{d^3}-\frac{12 b^2 \cosh (c+d x)}{d^4}+\frac{2 b^2 x^3 \sinh (c+d x)}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-12*b^2*Cosh[c + d*x])/d^4 - (a^2*Cosh[c + d*x])/x^2 - (6*b^2*x^2*Cosh[c + d*x])/d^2 + a^2*d^2*Cosh[c]*CoshI
ntegral[d*x] + (4*a*b*Sinh[c + d*x])/d - (a^2*d*Sinh[c + d*x])/x + (12*b^2*x*Sinh[c + d*x])/d^3 + (2*b^2*x^3*S
inh[c + d*x])/d + a^2*d^2*Sinh[c]*SinhIntegral[d*x])/2

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Maple [A]  time = 0.115, size = 265, normalized size = 1.9 \begin{align*} -{\frac{{b}^{2}{{\rm e}^{-dx-c}}{x}^{3}}{2\,d}}-{\frac{3\,{b}^{2}{{\rm e}^{-dx-c}}{x}^{2}}{2\,{d}^{2}}}-3\,{\frac{{b}^{2}{{\rm e}^{-dx-c}}x}{{d}^{3}}}-{\frac{{d}^{2}{a}^{2}{{\rm e}^{-c}}{\it Ei} \left ( 1,dx \right ) }{4}}-{\frac{ab{{\rm e}^{-dx-c}}}{d}}+{\frac{d{a}^{2}{{\rm e}^{-dx-c}}}{4\,x}}-{\frac{{a}^{2}{{\rm e}^{-dx-c}}}{4\,{x}^{2}}}-3\,{\frac{{b}^{2}{{\rm e}^{-dx-c}}}{{d}^{4}}}-{\frac{{d}^{2}{a}^{2}{{\rm e}^{c}}{\it Ei} \left ( 1,-dx \right ) }{4}}-3\,{\frac{{{\rm e}^{dx+c}}{b}^{2}}{{d}^{4}}}+{\frac{ab{{\rm e}^{dx+c}}}{d}}-{\frac{3\,{{\rm e}^{dx+c}}{b}^{2}{x}^{2}}{2\,{d}^{2}}}+3\,{\frac{{{\rm e}^{dx+c}}{b}^{2}x}{{d}^{3}}}-{\frac{{{\rm e}^{dx+c}}{a}^{2}}{4\,{x}^{2}}}-{\frac{d{a}^{2}{{\rm e}^{dx+c}}}{4\,x}}+{\frac{{{\rm e}^{dx+c}}{b}^{2}{x}^{3}}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.20039, size = 274, normalized size = 1.94 \begin{align*} \frac{1}{8} \,{\left (2 \, a^{2} d e^{\left (-c\right )} \Gamma \left (-1, d x\right ) + 2 \, a^{2} d e^{c} \Gamma \left (-1, -d x\right ) - \frac{8 \,{\left (d x e^{c} - e^{c}\right )} a b e^{\left (d x\right )}}{d^{2}} - \frac{8 \,{\left (d x + 1\right )} a b e^{\left (-d x - c\right )}}{d^{2}} - \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 )} b^{2} 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 )} b^{2} e^{\left (-d x - c\right )}}{d^{5}}\right )} d + \frac{1}{4} \,{\left (b^{2} x^{4} + 8 \, a b x - \frac{2 \, a^{2}}{x^{2}}\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^3,x, algorithm="maxima")

[Out]

1/8*(2*a^2*d*e^(-c)*gamma(-1, d*x) + 2*a^2*d*e^c*gamma(-1, -d*x) - 8*(d*x*e^c - e^c)*a*b*e^(d*x)/d^2 - 8*(d*x
+ 1)*a*b*e^(-d*x - c)/d^2 - (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)*b^2*e^(d*x)/d
^5 - (d^4*x^4 + 4*d^3*x^3 + 12*d^2*x^2 + 24*d*x + 24)*b^2*e^(-d*x - c)/d^5)*d + 1/4*(b^2*x^4 + 8*a*b*x - 2*a^2
/x^2)*cosh(d*x + c)

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Fricas [A]  time = 1.75178, size = 351, normalized size = 2.49 \begin{align*} -\frac{2 \,{\left (6 \, b^{2} d^{2} x^{4} + a^{2} d^{4} + 12 \, b^{2} x^{2}\right )} \cosh \left (d x + c\right ) -{\left (a^{2} d^{6} x^{2}{\rm Ei}\left (d x\right ) + a^{2} d^{6} x^{2}{\rm Ei}\left (-d x\right )\right )} \cosh \left (c\right ) - 2 \,{\left (2 \, b^{2} d^{3} x^{5} - a^{2} d^{5} x + 4 \, a b d^{3} x^{2} + 12 \, b^{2} d x^{3}\right )} \sinh \left (d x + c\right ) -{\left (a^{2} d^{6} x^{2}{\rm Ei}\left (d x\right ) - a^{2} d^{6} x^{2}{\rm Ei}\left (-d x\right )\right )} \sinh \left (c\right )}{4 \, d^{4} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^{3}}\, 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**3,x)

[Out]

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

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Giac [B]  time = 1.33646, size = 378, normalized size = 2.68 \begin{align*} \frac{a^{2} d^{6} x^{2}{\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + a^{2} d^{6} x^{2}{\rm Ei}\left (d x\right ) e^{c} + 2 \, b^{2} d^{3} x^{5} e^{\left (d x + c\right )} - 2 \, b^{2} d^{3} x^{5} e^{\left (-d x - c\right )} - a^{2} d^{5} x e^{\left (d x + c\right )} - 6 \, b^{2} d^{2} x^{4} e^{\left (d x + c\right )} + a^{2} d^{5} x e^{\left (-d x - c\right )} - 6 \, b^{2} d^{2} x^{4} e^{\left (-d x - c\right )} + 4 \, a b d^{3} x^{2} e^{\left (d x + c\right )} - 4 \, a b d^{3} x^{2} e^{\left (-d x - c\right )} - a^{2} d^{4} e^{\left (d x + c\right )} + 12 \, b^{2} d x^{3} e^{\left (d x + c\right )} - a^{2} d^{4} e^{\left (-d x - c\right )} - 12 \, b^{2} d x^{3} e^{\left (-d x - c\right )} - 12 \, b^{2} x^{2} e^{\left (d x + c\right )} - 12 \, b^{2} x^{2} e^{\left (-d x - c\right )}}{4 \, d^{4} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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