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

Optimal. Leaf size=56 \[ a \cosh (c) \text {Chi}(d x)+a \sinh (c) \text {Shi}(d x)+\frac {2 b \sinh (c+d x)}{d^3}-\frac {2 b x \cosh (c+d x)}{d^2}+\frac {b x^2 \sinh (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {5287, 3303, 3298, 3301, 3296, 2637} \[ a \cosh (c) \text {Chi}(d x)+a \sinh (c) \text {Shi}(d x)+\frac {2 b \sinh (c+d x)}{d^3}-\frac {2 b x \cosh (c+d x)}{d^2}+\frac {b x^2 \sinh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)*Cosh[c + d*x])/x,x]

[Out]

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

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Mathematica [A]  time = 0.18, size = 49, normalized size = 0.88 \[ a \cosh (c) \text {Chi}(d x)+a \sinh (c) \text {Shi}(d x)+\frac {b \left (\left (d^2 x^2+2\right ) \sinh (c+d x)-2 d x \cosh (c+d x)\right )}{d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.55, size = 86, normalized size = 1.54 \[ -\frac {4 \, b d x \cosh \left (d x + c\right ) - {\left (a d^{3} {\rm Ei}\left (d x\right ) + a d^{3} {\rm Ei}\left (-d x\right )\right )} \cosh \relax (c) - 2 \, {\left (b d^{2} x^{2} + 2 \, b\right )} \sinh \left (d x + c\right ) - {\left (a d^{3} {\rm Ei}\left (d x\right ) - a d^{3} {\rm Ei}\left (-d x\right )\right )} \sinh \relax (c)}{2 \, d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 109, normalized size = 1.95 \[ \frac {b d^{2} x^{2} e^{\left (d x + c\right )} - b d^{2} x^{2} e^{\left (-d x - c\right )} + a d^{3} {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + a d^{3} {\rm Ei}\left (d x\right ) e^{c} - 2 \, b d x e^{\left (d x + c\right )} - 2 \, b d x e^{\left (-d x - c\right )} + 2 \, b e^{\left (d x + c\right )} - 2 \, b e^{\left (-d x - c\right )}}{2 \, d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.12, size = 113, normalized size = 2.02 \[ -\frac {b \,{\mathrm e}^{d x +c} x}{d^{2}}+\frac {b \,{\mathrm e}^{d x +c} x^{2}}{2 d}-\frac {b \,{\mathrm e}^{-d x -c} x^{2}}{2 d}-\frac {b \,{\mathrm e}^{-d x -c} x}{d^{2}}-\frac {a \,{\mathrm e}^{c} \Ei \left (1, -d x \right )}{2}-\frac {b \,{\mathrm e}^{-d x -c}}{d^{3}}-\frac {a \,{\mathrm e}^{-c} \Ei \left (1, d x \right )}{2}+\frac {b \,{\mathrm e}^{d x +c}}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.40, size = 140, normalized size = 2.50 \[ -\frac {1}{6} \, {\left (b {\left (\frac {{\left (d^{3} x^{3} e^{c} - 3 \, d^{2} x^{2} e^{c} + 6 \, d x e^{c} - 6 \, e^{c}\right )} e^{\left (d x\right )}}{d^{4}} + \frac {{\left (d^{3} x^{3} + 3 \, d^{2} x^{2} + 6 \, d x + 6\right )} e^{\left (-d x - c\right )}}{d^{4}}\right )} + \frac {2 \, a \cosh \left (d x + c\right ) \log \left (x^{3}\right )}{d} - \frac {3 \, {\left ({\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + {\rm Ei}\left (d x\right ) e^{c}\right )} a}{d}\right )} d + \frac {1}{3} \, {\left (b x^{3} + a \log \left (x^{3}\right )\right )} \cosh \left (d x + c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(c + d*x)*(a + b*x^3))/x,x)

[Out]

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

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sympy [A]  time = 3.91, size = 66, normalized size = 1.18 \[ a \sinh {\relax (c )} \operatorname {Shi}{\left (d x \right )} + a \cosh {\relax (c )} \operatorname {Chi}\left (d x\right ) + b \left (\begin {cases} \frac {x^{2} \sinh {\left (c + d x \right )}}{d} - \frac {2 x \cosh {\left (c + d x \right )}}{d^{2}} + \frac {2 \sinh {\left (c + d x \right )}}{d^{3}} & \text {for}\: d \neq 0 \\\frac {x^{3} \cosh {\relax (c )}}{3} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*sinh(c)*Shi(d*x) + a*cosh(c)*Chi(d*x) + b*Piecewise((x**2*sinh(c + d*x)/d - 2*x*cosh(c + d*x)/d**2 + 2*sinh(
c + d*x)/d**3, Ne(d, 0)), (x**3*cosh(c)/3, True))

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