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

Optimal. Leaf size=58 \[ -\frac {\text {Chi}\left (\frac {2 \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{2 a}-\frac {\log \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6681, 3312, 3301} \[ -\frac {\text {Chi}\left (\frac {2 \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{2 a}-\frac {\log \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-CoshIntegral[(2*Sqrt[1 - a*x])/Sqrt[1 + a*x]]/(2*a) - Log[Sqrt[1 - a*x]/Sqrt[1 + a*x]]/(2*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 ^2\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\cosh ^2(x)}{x} \, dx,x,\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {1}{2 x}+\frac {\cosh (2 x)}{2 x}\right ) \, dx,x,\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a}\\ &=-\frac {\log \left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{2 a}-\frac {\operatorname {Subst}\left (\int \frac {\cosh (2 x)}{x} \, dx,x,\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{2 a}\\ &=-\frac {\text {Chi}\left (\frac {2 \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{2 a}-\frac {\log \left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{2 a}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 57, normalized size = 0.98 \[ -\frac {\text {Chi}\left (\frac {2 \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{2 a}-\frac {\log (1-a x)}{4 a}+\frac {\log (a x+1)}{4 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*CoshIntegral[(2*Sqrt[1 - a*x])/Sqrt[1 + a*x]]/a - Log[1 - a*x]/(4*a) + Log[1 + a*x]/(4*a)

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\cosh \left (\frac {\sqrt {-a x + 1}}{\sqrt {a x + 1}}\right )^{2}}{a^{2} x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.35, size = 0, normalized size = 0.00 \[ \int \frac {\cosh ^{2}\left (\frac {\sqrt {-a x +1}}{\sqrt {a x +1}}\right )}{-a^{2} x^{2}+1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\log \left (a x + 1\right )}{4 \, a} - \frac {\log \left (a x - 1\right )}{4 \, a} - \frac {1}{4} \, \int \frac {e^{\left (\frac {2 \, \sqrt {-a x + 1}}{\sqrt {a x + 1}}\right )}}{a^{2} x^{2} - 1}\,{d x} - \frac {1}{4} \, \int \frac {e^{\left (-\frac {2 \, \sqrt {-a x + 1}}{\sqrt {a x + 1}}\right )}}{a^{2} x^{2} - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*log(a*x + 1)/a - 1/4*log(a*x - 1)/a - 1/4*integrate(e^(2*sqrt(-a*x + 1)/sqrt(a*x + 1))/(a^2*x^2 - 1), x) -
1/4*integrate(e^(-2*sqrt(-a*x + 1)/sqrt(a*x + 1))/(a^2*x^2 - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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