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

Optimal. Leaf size=114 \[ \frac {1}{2} a^2 d^2 \cosh (c) \text {Chi}(d x)+\frac {1}{2} a^2 d^2 \sinh (c) \text {Shi}(d x)-\frac {a^2 \cosh (c+d x)}{2 x^2}-\frac {a^2 d \sinh (c+d x)}{2 x}+2 a b \cosh (c) \text {Chi}(d x)+2 a b \sinh (c) \text {Shi}(d x)-\frac {b^2 \cosh (c+d x)}{d^2}+\frac {b^2 x \sinh (c+d x)}{d} \]

[Out]

2*a*b*Chi(d*x)*cosh(c)+1/2*a^2*d^2*Chi(d*x)*cosh(c)-b^2*cosh(d*x+c)/d^2-1/2*a^2*cosh(d*x+c)/x^2+2*a*b*Shi(d*x)
*sinh(c)+1/2*a^2*d^2*Shi(d*x)*sinh(c)-1/2*a^2*d*sinh(d*x+c)/x+b^2*x*sinh(d*x+c)/d

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Rubi [A]  time = 0.23, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 12, 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, 2638} \[ \frac {1}{2} a^2 d^2 \cosh (c) \text {Chi}(d x)+\frac {1}{2} a^2 d^2 \sinh (c) \text {Shi}(d x)-\frac {a^2 \cosh (c+d x)}{2 x^2}-\frac {a^2 d \sinh (c+d x)}{2 x}+2 a b \cosh (c) \text {Chi}(d x)+2 a b \sinh (c) \text {Shi}(d x)-\frac {b^2 \cosh (c+d x)}{d^2}+\frac {b^2 x \sinh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[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 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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.40, size = 97, normalized size = 0.85 \[ \frac {1}{2} \left (-\frac {a^2 \cosh (c+d x)}{x^2}-\frac {a^2 d \sinh (c+d x)}{x}+a \cosh (c) \left (a d^2+4 b\right ) \text {Chi}(d x)+a \sinh (c) \left (a d^2+4 b\right ) \text {Shi}(d x)-\frac {2 b^2 \cosh (c+d x)}{d^2}+\frac {2 b^2 x \sinh (c+d x)}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.89, size = 164, normalized size = 1.44 \[ -\frac {2 \, {\left (a^{2} d^{2} + 2 \, b^{2} x^{2}\right )} \cosh \left (d x + c\right ) - {\left ({\left (a^{2} d^{4} + 4 \, a b d^{2}\right )} x^{2} {\rm Ei}\left (d x\right ) + {\left (a^{2} d^{4} + 4 \, a b d^{2}\right )} x^{2} {\rm Ei}\left (-d x\right )\right )} \cosh \relax (c) + 2 \, {\left (a^{2} d^{3} x - 2 \, b^{2} d x^{3}\right )} \sinh \left (d x + c\right ) - {\left ({\left (a^{2} d^{4} + 4 \, a b d^{2}\right )} x^{2} {\rm Ei}\left (d x\right ) - {\left (a^{2} d^{4} + 4 \, a b d^{2}\right )} x^{2} {\rm Ei}\left (-d x\right )\right )} \sinh \relax (c)}{4 \, d^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.13, size = 206, normalized size = 1.81 \[ \frac {a^{2} d^{4} x^{2} {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + a^{2} d^{4} x^{2} {\rm Ei}\left (d x\right ) e^{c} + 4 \, a b d^{2} x^{2} {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + 4 \, a b d^{2} x^{2} {\rm Ei}\left (d x\right ) e^{c} - a^{2} d^{3} x e^{\left (d x + c\right )} + 2 \, b^{2} d x^{3} e^{\left (d x + c\right )} + a^{2} d^{3} x e^{\left (-d x - c\right )} - 2 \, b^{2} d x^{3} e^{\left (-d x - c\right )} - a^{2} d^{2} e^{\left (d x + c\right )} - 2 \, b^{2} x^{2} e^{\left (d x + c\right )} - a^{2} d^{2} e^{\left (-d x - c\right )} - 2 \, b^{2} x^{2} e^{\left (-d x - c\right )}}{4 \, d^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.17, size = 188, normalized size = 1.65 \[ -\frac {b^{2} {\mathrm e}^{-d x -c} x}{2 d}-a b \,{\mathrm e}^{-c} \Ei \left (1, d x \right )+\frac {d \,a^{2} {\mathrm e}^{-d x -c}}{4 x}-\frac {a^{2} {\mathrm e}^{-d x -c}}{4 x^{2}}-\frac {d^{2} a^{2} {\mathrm e}^{-c} \Ei \left (1, d x \right )}{4}-\frac {b^{2} {\mathrm e}^{-d x -c}}{2 d^{2}}+\frac {b^{2} {\mathrm e}^{d x +c} x}{2 d}-\frac {d^{2} a^{2} {\mathrm e}^{c} \Ei \left (1, -d x \right )}{4}-a b \,{\mathrm e}^{c} \Ei \left (1, -d x \right )-\frac {a^{2} {\mathrm e}^{d x +c}}{4 x^{2}}-\frac {d \,a^{2} {\mathrm e}^{d x +c}}{4 x}-\frac {b^{2} {\mathrm e}^{d x +c}}{2 d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 165, normalized size = 1.45 \[ \frac {1}{4} \, {\left ({\left (d e^{\left (-c\right )} \Gamma \left (-1, d x\right ) + d e^{c} \Gamma \left (-1, -d x\right )\right )} a^{2} - b^{2} {\left (\frac {{\left (d^{2} x^{2} e^{c} - 2 \, d x e^{c} + 2 \, e^{c}\right )} e^{\left (d x\right )}}{d^{3}} + \frac {{\left (d^{2} x^{2} + 2 \, d x + 2\right )} e^{\left (-d x - c\right )}}{d^{3}}\right )} - \frac {4 \, a b \cosh \left (d x + c\right ) \log \left (x^{2}\right )}{d} + \frac {4 \, {\left ({\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + {\rm Ei}\left (d x\right ) e^{c}\right )} a b}{d}\right )} d + \frac {1}{2} \, {\left (b^{2} x^{2} + 2 \, a b \log \left (x^{2}\right ) - \frac {a^{2}}{x^{2}}\right )} \cosh \left (d x + c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {cosh}\left (c+d\,x\right )\,{\left (b\,x^2+a\right )}^2}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x^{2}\right )^{2} \cosh {\left (c + d x \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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