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

Optimal. Leaf size=55 \[ a d \sinh (c) \text {Chi}(d x)+a d \cosh (c) \text {Shi}(d x)-\frac {a \cosh (c+d x)}{x}-\frac {b \cosh (c+d x)}{d^2}+\frac {b x \sinh (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {5287, 3297, 3303, 3298, 3301, 3296, 2638} \[ a d \sinh (c) \text {Chi}(d x)+a d \cosh (c) \text {Shi}(d x)-\frac {a \cosh (c+d x)}{x}-\frac {b \cosh (c+d x)}{d^2}+\frac {b x \sinh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.14, size = 55, normalized size = 1.00 \[ a d \sinh (c) \text {Chi}(d x)+a d \cosh (c) \text {Shi}(d x)-\frac {a \cosh (c+d x)}{x}-\frac {b \cosh (c+d x)}{d^2}+\frac {b x \sinh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.59, size = 90, normalized size = 1.64 \[ \frac {2 \, b d x^{2} \sinh \left (d x + c\right ) - 2 \, {\left (a d^{2} + b x\right )} \cosh \left (d x + c\right ) + {\left (a d^{3} x {\rm Ei}\left (d x\right ) - a d^{3} x {\rm Ei}\left (-d x\right )\right )} \cosh \relax (c) + {\left (a d^{3} x {\rm Ei}\left (d x\right ) + a d^{3} x {\rm Ei}\left (-d x\right )\right )} \sinh \relax (c)}{2 \, d^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.14, size = 111, normalized size = 2.02 \[ -\frac {a d^{3} x {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - a d^{3} x {\rm Ei}\left (d x\right ) e^{c} - b d x^{2} e^{\left (d x + c\right )} + b d x^{2} e^{\left (-d x - c\right )} + a d^{2} e^{\left (d x + c\right )} + a d^{2} e^{\left (-d x - c\right )} + b x e^{\left (d x + c\right )} + b x e^{\left (-d x - c\right )}}{2 \, d^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.14, size = 110, normalized size = 2.00 \[ -\frac {a \,{\mathrm e}^{-d x -c}}{2 x}+\frac {d a \,{\mathrm e}^{-c} \Ei \left (1, d x \right )}{2}-\frac {b \,{\mathrm e}^{-d x -c} x}{2 d}-\frac {b \,{\mathrm e}^{-d x -c}}{2 d^{2}}-\frac {a \,{\mathrm e}^{d x +c}}{2 x}-\frac {d a \,{\mathrm e}^{c} \Ei \left (1, -d x \right )}{2}+\frac {b \,{\mathrm e}^{d x +c} x}{2 d}-\frac {b \,{\mathrm e}^{d x +c}}{2 d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.40, size = 102, normalized size = 1.85 \[ -\frac {1}{4} \, {\left (2 \, a {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - 2 \, a {\rm Ei}\left (d x\right ) e^{c} + \frac {{\left (d^{2} x^{2} e^{c} - 2 \, d x e^{c} + 2 \, e^{c}\right )} b e^{\left (d x\right )}}{d^{3}} + \frac {{\left (d^{2} x^{2} + 2 \, d x + 2\right )} b e^{\left (-d x - c\right )}}{d^{3}}\right )} d + \frac {1}{2} \, {\left (b x^{2} - \frac {2 \, a}{x}\right )} \cosh \left (d x + c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {cosh}\left (c+d\,x\right )\,\left (b\,x^3+a\right )}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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