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

Optimal. Leaf size=95 \[ a^2 d \sinh (c) \text{Chi}(d x)+a^2 d \cosh (c) \text{Shi}(d x)-\frac{a^2 \cosh (c+d x)}{x}+\frac{2 a b \sinh (c+d x)}{d}+\frac{2 b^2 \sinh (c+d x)}{d^3}-\frac{2 b^2 x \cosh (c+d x)}{d^2}+\frac{b^2 x^2 \sinh (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.176575, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {5287, 2637, 3297, 3303, 3298, 3301, 3296} \[ a^2 d \sinh (c) \text{Chi}(d x)+a^2 d \cosh (c) \text{Shi}(d x)-\frac{a^2 \cosh (c+d x)}{x}+\frac{2 a b \sinh (c+d x)}{d}+\frac{2 b^2 \sinh (c+d x)}{d^3}-\frac{2 b^2 x \cosh (c+d x)}{d^2}+\frac{b^2 x^2 \sinh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.262884, size = 95, normalized size = 1. \[ a^2 d \sinh (c) \text{Chi}(d x)+a^2 d \cosh (c) \text{Shi}(d x)-\frac{a^2 \cosh (c+d x)}{x}+\frac{2 a b \sinh (c+d x)}{d}+\frac{2 b^2 \sinh (c+d x)}{d^3}-\frac{2 b^2 x \cosh (c+d x)}{d^2}+\frac{b^2 x^2 \sinh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.069, size = 190, normalized size = 2. \begin{align*}{\frac{d{a}^{2}{{\rm e}^{-c}}{\it Ei} \left ( 1,dx \right ) }{2}}-{\frac{ab{{\rm e}^{-dx-c}}}{d}}-{\frac{{a}^{2}{{\rm e}^{-dx-c}}}{2\,x}}-{\frac{{b}^{2}{{\rm e}^{-dx-c}}}{{d}^{3}}}-{\frac{{b}^{2}{{\rm e}^{-dx-c}}{x}^{2}}{2\,d}}-{\frac{{b}^{2}{{\rm e}^{-dx-c}}x}{{d}^{2}}}+{\frac{{{\rm e}^{dx+c}}{b}^{2}}{{d}^{3}}}-{\frac{d{a}^{2}{{\rm e}^{c}}{\it Ei} \left ( 1,-dx \right ) }{2}}+{\frac{ab{{\rm e}^{dx+c}}}{d}}+{\frac{{{\rm e}^{dx+c}}{b}^{2}{x}^{2}}{2\,d}}-{\frac{{{\rm e}^{dx+c}}{b}^{2}x}{{d}^{2}}}-{\frac{{{\rm e}^{dx+c}}{a}^{2}}{2\,x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.20407, size = 242, normalized size = 2.55 \begin{align*} -\frac{1}{6} \,{\left (3 \, a^{2}{\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - 3 \, a^{2}{\rm Ei}\left (d x\right ) e^{c} + \frac{6 \,{\left (d x e^{c} - e^{c}\right )} a b e^{\left (d x\right )}}{d^{2}} + \frac{6 \,{\left (d x + 1\right )} a b e^{\left (-d x - c\right )}}{d^{2}} + \frac{{\left (d^{3} x^{3} e^{c} - 3 \, d^{2} x^{2} e^{c} + 6 \, d x e^{c} - 6 \, e^{c}\right )} b^{2} e^{\left (d x\right )}}{d^{4}} + \frac{{\left (d^{3} x^{3} + 3 \, d^{2} x^{2} + 6 \, d x + 6\right )} b^{2} e^{\left (-d x - c\right )}}{d^{4}}\right )} d + \frac{1}{3} \,{\left (b^{2} x^{3} + 6 \, a b x - \frac{3 \, a^{2}}{x}\right )} \cosh \left (d x + c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.0144, size = 286, normalized size = 3.01 \begin{align*} -\frac{2 \,{\left (a^{2} d^{3} + 2 \, b^{2} d x^{2}\right )} \cosh \left (d x + c\right ) -{\left (a^{2} d^{4} x{\rm Ei}\left (d x\right ) - a^{2} d^{4} x{\rm Ei}\left (-d x\right )\right )} \cosh \left (c\right ) - 2 \,{\left (b^{2} d^{2} x^{3} + 2 \,{\left (a b d^{2} + b^{2}\right )} x\right )} \sinh \left (d x + c\right ) -{\left (a^{2} d^{4} x{\rm Ei}\left (d x\right ) + a^{2} d^{4} x{\rm Ei}\left (-d x\right )\right )} \sinh \left (c\right )}{2 \, d^{3} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.19945, size = 266, normalized size = 2.8 \begin{align*} -\frac{a^{2} d^{4} x{\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - a^{2} d^{4} x{\rm Ei}\left (d x\right ) e^{c} - b^{2} d^{2} x^{3} e^{\left (d x + c\right )} + b^{2} d^{2} x^{3} e^{\left (-d x - c\right )} + a^{2} d^{3} e^{\left (d x + c\right )} - 2 \, a b d^{2} x e^{\left (d x + c\right )} + 2 \, b^{2} d x^{2} e^{\left (d x + c\right )} + a^{2} d^{3} e^{\left (-d x - c\right )} + 2 \, a b d^{2} x e^{\left (-d x - c\right )} + 2 \, b^{2} d x^{2} e^{\left (-d x - c\right )} - 2 \, b^{2} x e^{\left (d x + c\right )} + 2 \, b^{2} x e^{\left (-d x - c\right )}}{2 \, d^{3} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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