Integrand size = 36, antiderivative size = 58 \[ \int \frac {\cosh ^3\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=-\frac {3 \text {Chi}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{4 a}-\frac {\text {Chi}\left (\frac {3 \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{4 a} \] Output:
-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
Time = 0.05 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.95 \[ \int \frac {\cosh ^3\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=\frac {-3 \text {Chi}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )-\text {Chi}\left (\frac {3 \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{4 a} \] Input:
Integrate[Cosh[Sqrt[1 - a*x]/Sqrt[1 + a*x]]^3/(1 - a^2*x^2),x]
Output:
(-3*CoshIntegral[Sqrt[1 - a*x]/Sqrt[1 + a*x]] - CoshIntegral[(3*Sqrt[1 - a *x])/Sqrt[1 + a*x]])/(4*a)
Time = 0.42 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {7232, 3042, 3793, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\cosh ^3\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )}{1-a^2 x^2} \, dx\) |
\(\Big \downarrow \) 7232 |
\(\displaystyle -\frac {\int \frac {\sqrt {a x+1} \cosh ^3\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )}{\sqrt {1-a x}}d\frac {\sqrt {1-a x}}{\sqrt {a x+1}}}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {\sqrt {a x+1} \sin \left (\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}+\frac {\pi }{2}\right )^3}{\sqrt {1-a x}}d\frac {\sqrt {1-a x}}{\sqrt {a x+1}}}{a}\) |
\(\Big \downarrow \) 3793 |
\(\displaystyle -\frac {\int \left (\frac {3 \sqrt {a x+1} \cosh \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )}{4 \sqrt {1-a x}}+\frac {\sqrt {a x+1} \cosh \left (\frac {3 \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{4 \sqrt {1-a x}}\right )d\frac {\sqrt {1-a x}}{\sqrt {a x+1}}}{a}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {3}{4} \text {Chi}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )+\frac {1}{4} \text {Chi}\left (\frac {3 \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a}\) |
Input:
Int[Cosh[Sqrt[1 - a*x]/Sqrt[1 + a*x]]^3/(1 - a^2*x^2),x]
Output:
-(((3*CoshIntegral[Sqrt[1 - a*x]/Sqrt[1 + a*x]])/4 + CoshIntegral[(3*Sqrt[ 1 - a*x])/Sqrt[1 + a*x]]/4)/a)
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In t[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]))
Int[((a_.) + (b_.)*(F_)[((c_.)*Sqrt[(d_.) + (e_.)*(x_)])/Sqrt[(f_.) + (g_.) *(x_)]])^(n_.)/((A_.) + (C_.)*(x_)^2), x_Symbol] :> Simp[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]
\[\int \frac {\cosh \left (\frac {\sqrt {-a x +1}}{\sqrt {a x +1}}\right )^{3}}{-a^{2} x^{2}+1}d x\]
Input:
int(cosh((-a*x+1)^(1/2)/(a*x+1)^(1/2))^3/(-a^2*x^2+1),x)
Output:
int(cosh((-a*x+1)^(1/2)/(a*x+1)^(1/2))^3/(-a^2*x^2+1),x)
\[ \int \frac {\cosh ^3\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=\int { -\frac {\cosh \left (\frac {\sqrt {-a x + 1}}{\sqrt {a x + 1}}\right )^{3}}{a^{2} x^{2} - 1} \,d x } \] Input:
integrate(cosh((-a*x+1)^(1/2)/(a*x+1)^(1/2))^3/(-a^2*x^2+1),x, algorithm=" fricas")
Output:
integral(-cosh(sqrt(-a*x + 1)/sqrt(a*x + 1))^3/(a^2*x^2 - 1), x)
\[ \int \frac {\cosh ^3\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=- \int \frac {\cosh ^{3}{\left (\frac {\sqrt {- a x + 1}}{\sqrt {a x + 1}} \right )}}{a^{2} x^{2} - 1}\, dx \] Input:
integrate(cosh((-a*x+1)**(1/2)/(a*x+1)**(1/2))**3/(-a**2*x**2+1),x)
Output:
-Integral(cosh(sqrt(-a*x + 1)/sqrt(a*x + 1))**3/(a**2*x**2 - 1), x)
\[ \int \frac {\cosh ^3\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=\int { -\frac {\cosh \left (\frac {\sqrt {-a x + 1}}{\sqrt {a x + 1}}\right )^{3}}{a^{2} x^{2} - 1} \,d x } \] Input:
integrate(cosh((-a*x+1)^(1/2)/(a*x+1)^(1/2))^3/(-a^2*x^2+1),x, algorithm=" maxima")
Output:
-integrate(cosh(sqrt(-a*x + 1)/sqrt(a*x + 1))^3/(a^2*x^2 - 1), x)
\[ \int \frac {\cosh ^3\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=\int { -\frac {\cosh \left (\frac {\sqrt {-a x + 1}}{\sqrt {a x + 1}}\right )^{3}}{a^{2} x^{2} - 1} \,d x } \] Input:
integrate(cosh((-a*x+1)^(1/2)/(a*x+1)^(1/2))^3/(-a^2*x^2+1),x, algorithm=" giac")
Output:
integrate(-cosh(sqrt(-a*x + 1)/sqrt(a*x + 1))^3/(a^2*x^2 - 1), x)
Timed out. \[ \int \frac {\cosh ^3\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=-\int \frac {{\mathrm {cosh}\left (\frac {\sqrt {1-a\,x}}{\sqrt {a\,x+1}}\right )}^3}{a^2\,x^2-1} \,d x \] Input:
int(-cosh((1 - a*x)^(1/2)/(a*x + 1)^(1/2))^3/(a^2*x^2 - 1),x)
Output:
-int(cosh((1 - a*x)^(1/2)/(a*x + 1)^(1/2))^3/(a^2*x^2 - 1), x)
\[ \int \frac {\cosh ^3\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=-\left (\int \frac {\cosh \left (\frac {\sqrt {-a x +1}}{\sqrt {a x +1}}\right )^{3}}{a^{2} x^{2}-1}d x \right ) \] Input:
int(cosh((-a*x+1)^(1/2)/(a*x+1)^(1/2))^3/(-a^2*x^2+1),x)
Output:
- int(cosh(sqrt( - a*x + 1)/sqrt(a*x + 1))**3/(a**2*x**2 - 1),x)