3.213 \(\int \frac {\cosh ^3(\frac {\sqrt {1-a x}}{\sqrt {1+a x}})}{1-a^2 x^2} \, dx\)

Optimal. Leaf size=58 \[ -\frac {3 \text {Chi}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )}{4 a}-\frac {\text {Chi}\left (\frac {3 \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{4 a} \]

[Out]

-3/4*Chi((-a*x+1)^(1/2)/(a*x+1)^(1/2))/a-1/4*Chi(3*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/a

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Rubi [A]  time = 0.12, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6681, 3312, 3301} \[ -\frac {3 \text {Chi}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )}{4 a}-\frac {\text {Chi}\left (\frac {3 \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{4 a} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[Sqrt[1 - a*x]/Sqrt[1 + a*x]]^3/(1 - a^2*x^2),x]

[Out]

(-3*CoshIntegral[Sqrt[1 - a*x]/Sqrt[1 + a*x]])/(4*a) - CoshIntegral[(3*Sqrt[1 - a*x])/Sqrt[1 + a*x]]/(4*a)

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 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 6681

Int[((a_.) + (b_.)*(F_)[((c_.)*Sqrt[(d_.) + (e_.)*(x_)])/Sqrt[(f_.) + (g_.)*(x_)]])^(n_.)/((A_.) + (C_.)*(x_)^
2), x_Symbol] :> Dist[(2*e*g)/(C*(e*f - d*g)), Subst[Int[(a + b*F[c*x])^n/x, x], x, Sqrt[d + e*x]/Sqrt[f + g*x
]], x] /; FreeQ[{a, b, c, d, e, f, g, A, C, F}, x] && EqQ[C*d*f - A*e*g, 0] && EqQ[e*f + d*g, 0] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\cosh ^3\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\cosh ^3(x)}{x} \, dx,x,\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {3 \cosh (x)}{4 x}+\frac {\cosh (3 x)}{4 x}\right ) \, dx,x,\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {\cosh (3 x)}{x} \, dx,x,\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{4 a}-\frac {3 \operatorname {Subst}\left (\int \frac {\cosh (x)}{x} \, dx,x,\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{4 a}\\ &=-\frac {3 \text {Chi}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{4 a}-\frac {\text {Chi}\left (\frac {3 \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{4 a}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 55, normalized size = 0.95 \[ \frac {-3 \text {Chi}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-\text {Chi}\left (\frac {3 \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{4 a} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[Sqrt[1 - a*x]/Sqrt[1 + a*x]]^3/(1 - a^2*x^2),x]

[Out]

(-3*CoshIntegral[Sqrt[1 - a*x]/Sqrt[1 + a*x]] - CoshIntegral[(3*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/(4*a)

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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\cosh \left (\frac {\sqrt {-a x + 1}}{\sqrt {a x + 1}}\right )^{3}}{a^{2} x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh((-a*x+1)^(1/2)/(a*x+1)^(1/2))^3/(-a^2*x^2+1),x, algorithm="fricas")

[Out]

integral(-cosh(sqrt(-a*x + 1)/sqrt(a*x + 1))^3/(a^2*x^2 - 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {\cosh \left (\frac {\sqrt {-a x + 1}}{\sqrt {a x + 1}}\right )^{3}}{a^{2} x^{2} - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh((-a*x+1)^(1/2)/(a*x+1)^(1/2))^3/(-a^2*x^2+1),x, algorithm="giac")

[Out]

integrate(-cosh(sqrt(-a*x + 1)/sqrt(a*x + 1))^3/(a^2*x^2 - 1), x)

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maple [F]  time = 0.60, size = 0, normalized size = 0.00 \[ \int \frac {\cosh ^{3}\left (\frac {\sqrt {-a x +1}}{\sqrt {a x +1}}\right )}{-a^{2} x^{2}+1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh((-a*x+1)^(1/2)/(a*x+1)^(1/2))^3/(-a^2*x^2+1),x)

[Out]

int(cosh((-a*x+1)^(1/2)/(a*x+1)^(1/2))^3/(-a^2*x^2+1),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {\cosh \left (\frac {\sqrt {-a x + 1}}{\sqrt {a x + 1}}\right )^{3}}{a^{2} x^{2} - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh((-a*x+1)^(1/2)/(a*x+1)^(1/2))^3/(-a^2*x^2+1),x, algorithm="maxima")

[Out]

-integrate(cosh(sqrt(-a*x + 1)/sqrt(a*x + 1))^3/(a^2*x^2 - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-cosh((1 - a*x)^(1/2)/(a*x + 1)^(1/2))^3/(a^2*x^2 - 1),x)

[Out]

-int(cosh((1 - a*x)^(1/2)/(a*x + 1)^(1/2))^3/(a^2*x^2 - 1), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {\cosh ^{3}{\left (\frac {\sqrt {- a x + 1}}{\sqrt {a x + 1}} \right )}}{a^{2} x^{2} - 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh((-a*x+1)**(1/2)/(a*x+1)**(1/2))**3/(-a**2*x**2+1),x)

[Out]

-Integral(cosh(sqrt(-a*x + 1)/sqrt(a*x + 1))**3/(a**2*x**2 - 1), x)

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