\(\int \frac {\cosh ^3(a+b x^2)}{x} \, dx\) [19]

Optimal result

Integrand size = 14, antiderivative size = 55 \[ \int \frac {\cosh ^3\left (a+b x^2\right )}{x} \, dx=\frac {3}{8} \cosh (a) \text {Chi}\left (b x^2\right )+\frac {1}{8} \cosh (3 a) \text {Chi}\left (3 b x^2\right )+\frac {3}{8} \sinh (a) \text {Shi}\left (b x^2\right )+\frac {1}{8} \sinh (3 a) \text {Shi}\left (3 b x^2\right ) \]

[Out]

3/8*Chi(b*x^2)*cosh(a)+1/8*Chi(3*b*x^2)*cosh(3*a)+3/8*Shi(b*x^2)*sinh(a)+1/8*Shi(3*b*x^2)*sinh(3*a)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5449, 5427, 5425, 5424} \[ \int \frac {\cosh ^3\left (a+b x^2\right )}{x} \, dx=\frac {3}{8} \cosh (a) \text {Chi}\left (b x^2\right )+\frac {1}{8} \cosh (3 a) \text {Chi}\left (3 b x^2\right )+\frac {3}{8} \sinh (a) \text {Shi}\left (b x^2\right )+\frac {1}{8} \sinh (3 a) \text {Shi}\left (3 b x^2\right ) \]

[In]

Int[Cosh[a + b*x^2]^3/x,x]

[Out]

(3*Cosh[a]*CoshIntegral[b*x^2])/8 + (Cosh[3*a]*CoshIntegral[3*b*x^2])/8 + (3*Sinh[a]*SinhIntegral[b*x^2])/8 +
(Sinh[3*a]*SinhIntegral[3*b*x^2])/8

Rule 5424

Int[Sinh[(d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Simp[SinhIntegral[d*x^n]/n, x] /; FreeQ[{d, n}, x]

Rule 5425

Int[Cosh[(d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Simp[CoshIntegral[d*x^n]/n, x] /; FreeQ[{d, n}, x]

Rule 5427

Int[Cosh[(c_) + (d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Dist[Cosh[c], Int[Cosh[d*x^n]/x, x], x] + Dist[Sinh[c], In
t[Sinh[d*x^n]/x, x], x] /; FreeQ[{c, d, n}, x]

Rule 5449

Int[((a_.) + Cosh[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandTrigReduce[(
e*x)^m, (a + b*Cosh[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {3 \cosh \left (a+b x^2\right )}{4 x}+\frac {\cosh \left (3 a+3 b x^2\right )}{4 x}\right ) \, dx \\ & = \frac {1}{4} \int \frac {\cosh \left (3 a+3 b x^2\right )}{x} \, dx+\frac {3}{4} \int \frac {\cosh \left (a+b x^2\right )}{x} \, dx \\ & = \frac {1}{4} (3 \cosh (a)) \int \frac {\cosh \left (b x^2\right )}{x} \, dx+\frac {1}{4} \cosh (3 a) \int \frac {\cosh \left (3 b x^2\right )}{x} \, dx+\frac {1}{4} (3 \sinh (a)) \int \frac {\sinh \left (b x^2\right )}{x} \, dx+\frac {1}{4} \sinh (3 a) \int \frac {\sinh \left (3 b x^2\right )}{x} \, dx \\ & = \frac {3}{8} \cosh (a) \text {Chi}\left (b x^2\right )+\frac {1}{8} \cosh (3 a) \text {Chi}\left (3 b x^2\right )+\frac {3}{8} \sinh (a) \text {Shi}\left (b x^2\right )+\frac {1}{8} \sinh (3 a) \text {Shi}\left (3 b x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89 \[ \int \frac {\cosh ^3\left (a+b x^2\right )}{x} \, dx=\frac {1}{8} \left (3 \cosh (a) \text {Chi}\left (b x^2\right )+\cosh (3 a) \text {Chi}\left (3 b x^2\right )+3 \sinh (a) \text {Shi}\left (b x^2\right )+\sinh (3 a) \text {Shi}\left (3 b x^2\right )\right ) \]

[In]

Integrate[Cosh[a + b*x^2]^3/x,x]

[Out]

(3*Cosh[a]*CoshIntegral[b*x^2] + Cosh[3*a]*CoshIntegral[3*b*x^2] + 3*Sinh[a]*SinhIntegral[b*x^2] + Sinh[3*a]*S
inhIntegral[3*b*x^2])/8

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.25

method	result	size
risch	\(-\frac {{\mathrm e}^{6 a} {\mathrm e}^{-3 a} \operatorname {Ei}_{1}\left (-3 b \,x^{2}\right )}{16}-\frac {3 \,{\mathrm e}^{4 a} {\mathrm e}^{-3 a} \operatorname {Ei}_{1}\left (-b \,x^{2}\right )}{16}-\frac {3 \,{\mathrm e}^{2 a} {\mathrm e}^{-3 a} \operatorname {Ei}_{1}\left (b \,x^{2}\right )}{16}-\frac {{\mathrm e}^{-3 a} \operatorname {Ei}_{1}\left (3 b \,x^{2}\right )}{16}\)	\(69\)

[In]

int(cosh(b*x^2+a)^3/x,x,method=_RETURNVERBOSE)

[Out]

-1/16*exp(6*a)*exp(-3*a)*Ei(1,-3*b*x^2)-3/16*exp(4*a)*exp(-3*a)*Ei(1,-b*x^2)-3/16*exp(2*a)*exp(-3*a)*Ei(1,b*x^
2)-1/16*exp(-3*a)*Ei(1,3*b*x^2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.51 \[ \int \frac {\cosh ^3\left (a+b x^2\right )}{x} \, dx=\frac {1}{16} \, {\left ({\rm Ei}\left (3 \, b x^{2}\right ) + {\rm Ei}\left (-3 \, b x^{2}\right )\right )} \cosh \left (3 \, a\right ) + \frac {3}{16} \, {\left ({\rm Ei}\left (b x^{2}\right ) + {\rm Ei}\left (-b x^{2}\right )\right )} \cosh \left (a\right ) + \frac {1}{16} \, {\left ({\rm Ei}\left (3 \, b x^{2}\right ) - {\rm Ei}\left (-3 \, b x^{2}\right )\right )} \sinh \left (3 \, a\right ) + \frac {3}{16} \, {\left ({\rm Ei}\left (b x^{2}\right ) - {\rm Ei}\left (-b x^{2}\right )\right )} \sinh \left (a\right ) \]

[In]

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

[Out]

1/16*(Ei(3*b*x^2) + Ei(-3*b*x^2))*cosh(3*a) + 3/16*(Ei(b*x^2) + Ei(-b*x^2))*cosh(a) + 1/16*(Ei(3*b*x^2) - Ei(-
3*b*x^2))*sinh(3*a) + 3/16*(Ei(b*x^2) - Ei(-b*x^2))*sinh(a)

Sympy [F]

\[ \int \frac {\cosh ^3\left (a+b x^2\right )}{x} \, dx=\int \frac {\cosh ^{3}{\left (a + b x^{2} \right )}}{x}\, dx \]

[In]

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

[Out]

Integral(cosh(a + b*x**2)**3/x, x)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.91 \[ \int \frac {\cosh ^3\left (a+b x^2\right )}{x} \, dx=\frac {1}{16} \, {\rm Ei}\left (3 \, b x^{2}\right ) e^{\left (3 \, a\right )} + \frac {3}{16} \, {\rm Ei}\left (-b x^{2}\right ) e^{\left (-a\right )} + \frac {1}{16} \, {\rm Ei}\left (-3 \, b x^{2}\right ) e^{\left (-3 \, a\right )} + \frac {3}{16} \, {\rm Ei}\left (b x^{2}\right ) e^{a} \]

[In]

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

[Out]

1/16*Ei(3*b*x^2)*e^(3*a) + 3/16*Ei(-b*x^2)*e^(-a) + 1/16*Ei(-3*b*x^2)*e^(-3*a) + 3/16*Ei(b*x^2)*e^a

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.91 \[ \int \frac {\cosh ^3\left (a+b x^2\right )}{x} \, dx=\frac {1}{16} \, {\rm Ei}\left (3 \, b x^{2}\right ) e^{\left (3 \, a\right )} + \frac {3}{16} \, {\rm Ei}\left (-b x^{2}\right ) e^{\left (-a\right )} + \frac {1}{16} \, {\rm Ei}\left (-3 \, b x^{2}\right ) e^{\left (-3 \, a\right )} + \frac {3}{16} \, {\rm Ei}\left (b x^{2}\right ) e^{a} \]

[In]

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

[Out]

1/16*Ei(3*b*x^2)*e^(3*a) + 3/16*Ei(-b*x^2)*e^(-a) + 1/16*Ei(-3*b*x^2)*e^(-3*a) + 3/16*Ei(b*x^2)*e^a

Mupad [F(-1)]

Timed out. \[ \int \frac {\cosh ^3\left (a+b x^2\right )}{x} \, dx=\int \frac {{\mathrm {cosh}\left (b\,x^2+a\right )}^3}{x} \,d x \]

[In]

int(cosh(a + b*x^2)^3/x,x)

[Out]

int(cosh(a + b*x^2)^3/x, x)