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

Optimal. Leaf size=42 \[ a d \sinh (c) \text {Chi}(d x)+a d \cosh (c) \text {Shi}(d x)-\frac {a \cosh (c+d x)}{x}+\frac {b \sinh (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {5287, 2637, 3297, 3303, 3298, 3301} \[ a d \sinh (c) \text {Chi}(d x)+a d \cosh (c) \text {Shi}(d x)-\frac {a \cosh (c+d x)}{x}+\frac {b \sinh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.09, size = 42, normalized size = 1.00 \[ a d \sinh (c) \text {Chi}(d x)+a d \cosh (c) \text {Shi}(d x)-\frac {a \cosh (c+d x)}{x}+\frac {b \sinh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.58, size = 82, normalized size = 1.95 \[ -\frac {2 \, a d \cosh \left (d x + c\right ) - 2 \, b x \sinh \left (d x + c\right ) - {\left (a d^{2} x {\rm Ei}\left (d x\right ) - a d^{2} x {\rm Ei}\left (-d x\right )\right )} \cosh \relax (c) - {\left (a d^{2} x {\rm Ei}\left (d x\right ) + a d^{2} x {\rm Ei}\left (-d x\right )\right )} \sinh \relax (c)}{2 \, d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.12, size = 80, normalized size = 1.90 \[ -\frac {a d^{2} x {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - a d^{2} x {\rm Ei}\left (d x\right ) e^{c} + a d e^{\left (d x + c\right )} - b x e^{\left (d x + c\right )} + a d e^{\left (-d x - c\right )} + b x e^{\left (-d x - c\right )}}{2 \, d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.11, size = 81, normalized size = 1.93 \[ -\frac {a \,{\mathrm e}^{-d x -c}}{2 x}+\frac {d a \,{\mathrm e}^{-c} \Ei \left (1, d x \right )}{2}-\frac {b \,{\mathrm e}^{-d x -c}}{2 d}-\frac {a \,{\mathrm e}^{d x +c}}{2 x}-\frac {d a \,{\mathrm e}^{c} \Ei \left (1, -d x \right )}{2}+\frac {b \,{\mathrm e}^{d x +c}}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+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)-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)

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maxima [A]  time = 0.38, size = 80, normalized size = 1.90 \[ -\frac {1}{2} \, {\left (a {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - a {\rm Ei}\left (d x\right ) e^{c} + \frac {{\left (d x e^{c} - e^{c}\right )} b e^{\left (d x\right )}}{d^{2}} + \frac {{\left (d x + 1\right )} b e^{\left (-d x - c\right )}}{d^{2}}\right )} d + {\left (b x - \frac {a}{x}\right )} \cosh \left (d x + c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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