Integrand size = 10, antiderivative size = 55 \[ \int \left (1+\cosh ^2(x)\right )^{3/2} \, dx=-2 i E\left (\left .\frac {\pi }{2}+i x\right |-1\right )+\frac {2}{3} i \operatorname {EllipticF}\left (\frac {\pi }{2}+i x,-1\right )+\frac {1}{3} \cosh (x) \sqrt {1+\cosh ^2(x)} \sinh (x) \]
2*(-sinh(x)^2)^(1/2)/sinh(x)*EllipticE(cosh(x),I)-2/3*(-sinh(x)^2)^(1/2)/s inh(x)*EllipticF(cosh(x),I)+1/3*cosh(x)*sinh(x)*(1+cosh(x)^2)^(1/2)
Time = 0.04 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int \left (1+\cosh ^2(x)\right )^{3/2} \, dx=\frac {-24 i E\left (i x\left |\frac {1}{2}\right .\right )+4 i \operatorname {EllipticF}\left (i x,\frac {1}{2}\right )+\sqrt {3+\cosh (2 x)} \sinh (2 x)}{6 \sqrt {2}} \]
((-24*I)*EllipticE[I*x, 1/2] + (4*I)*EllipticF[I*x, 1/2] + Sqrt[3 + Cosh[2 *x]]*Sinh[2*x])/(6*Sqrt[2])
Time = 0.42 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {3042, 3659, 27, 3042, 3651, 3042, 3656, 3661}
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.
\(\displaystyle \int \left (\cosh ^2(x)+1\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (1+\sin \left (\frac {\pi }{2}+i x\right )^2\right )^{3/2}dx\) |
\(\Big \downarrow \) 3659 |
\(\displaystyle \frac {1}{3} \int \frac {2 \left (3 \cosh ^2(x)+2\right )}{\sqrt {\cosh ^2(x)+1}}dx+\frac {1}{3} \sinh (x) \cosh (x) \sqrt {\cosh ^2(x)+1}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{3} \int \frac {3 \cosh ^2(x)+2}{\sqrt {\cosh ^2(x)+1}}dx+\frac {1}{3} \sinh (x) \cosh (x) \sqrt {\cosh ^2(x)+1}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \sinh (x) \cosh (x) \sqrt {\cosh ^2(x)+1}+\frac {2}{3} \int \frac {3 \sin \left (i x+\frac {\pi }{2}\right )^2+2}{\sqrt {\sin \left (i x+\frac {\pi }{2}\right )^2+1}}dx\) |
\(\Big \downarrow \) 3651 |
\(\displaystyle \frac {2}{3} \left (3 \int \sqrt {\cosh ^2(x)+1}dx-\int \frac {1}{\sqrt {\cosh ^2(x)+1}}dx\right )+\frac {1}{3} \sinh (x) \cosh (x) \sqrt {\cosh ^2(x)+1}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \sinh (x) \cosh (x) \sqrt {\cosh ^2(x)+1}+\frac {2}{3} \left (3 \int \sqrt {\sin \left (i x+\frac {\pi }{2}\right )^2+1}dx-\int \frac {1}{\sqrt {\sin \left (i x+\frac {\pi }{2}\right )^2+1}}dx\right )\) |
\(\Big \downarrow \) 3656 |
\(\displaystyle \frac {1}{3} \sinh (x) \cosh (x) \sqrt {\cosh ^2(x)+1}+\frac {2}{3} \left (-\int \frac {1}{\sqrt {\sin \left (i x+\frac {\pi }{2}\right )^2+1}}dx-3 i E\left (\left .i x+\frac {\pi }{2}\right |-1\right )\right )\) |
\(\Big \downarrow \) 3661 |
\(\displaystyle \frac {1}{3} \sinh (x) \cosh (x) \sqrt {\cosh ^2(x)+1}+\frac {2}{3} \left (i \operatorname {EllipticF}\left (i x+\frac {\pi }{2},-1\right )-3 i E\left (\left .i x+\frac {\pi }{2}\right |-1\right )\right )\) |
(2*((-3*I)*EllipticE[Pi/2 + I*x, -1] + I*EllipticF[Pi/2 + I*x, -1]))/3 + ( Cosh[x]*Sqrt[1 + Cosh[x]^2]*Sinh[x])/3
3.1.47.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[B/b Int[Sqrt[a + b*Sin[e + f*x]^2], x] , x] + Simp[(A*b - a*B)/b Int[1/Sqrt[a + b*Sin[e + f*x]^2], x], x] /; Fre eQ[{a, b, e, f, A, B}, x]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(Sqrt[a ]/f)*EllipticE[e + f*x, -b/a], x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(-b)*C os[e + f*x]*Sin[e + f*x]*((a + b*Sin[e + f*x]^2)^(p - 1)/(2*f*p)), x] + Sim p[1/(2*p) Int[(a + b*Sin[e + f*x]^2)^(p - 2)*Simp[a*(b + 2*a*p) + b*(2*a + b)*(2*p - 1)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[ a + b, 0] && GtQ[p, 1]
Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(1/(S qrt[a]*f))*EllipticF[e + f*x, -b/a], x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]
Time = 0.50 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.80
method | result | size |
default | \(-\frac {\sqrt {\left (1+\cosh \left (x \right )^{2}\right ) \sinh \left (x \right )^{2}}\, \left (-\cosh \left (x \right )^{5}+10 i \sqrt {1+\cosh \left (x \right )^{2}}\, \sqrt {-\sinh \left (x \right )^{2}}\, \operatorname {EllipticF}\left (i \cosh \left (x \right ), i\right )-6 i \sqrt {1+\cosh \left (x \right )^{2}}\, \sqrt {-\sinh \left (x \right )^{2}}\, \operatorname {EllipticE}\left (i \cosh \left (x \right ), i\right )+\cosh \left (x \right )\right )}{3 \sqrt {\cosh \left (x \right )^{4}-1}\, \sinh \left (x \right ) \sqrt {1+\cosh \left (x \right )^{2}}}\) | \(99\) |
-1/3*((1+cosh(x)^2)*sinh(x)^2)^(1/2)*(-cosh(x)^5+10*I*(1+cosh(x)^2)^(1/2)* (-sinh(x)^2)^(1/2)*EllipticF(I*cosh(x),I)-6*I*(1+cosh(x)^2)^(1/2)*(-sinh(x )^2)^(1/2)*EllipticE(I*cosh(x),I)+cosh(x))/(cosh(x)^4-1)^(1/2)/sinh(x)/(1+ cosh(x)^2)^(1/2)
\[ \int \left (1+\cosh ^2(x)\right )^{3/2} \, dx=\int { {\left (\cosh \left (x\right )^{2} + 1\right )}^{\frac {3}{2}} \,d x } \]
Timed out. \[ \int \left (1+\cosh ^2(x)\right )^{3/2} \, dx=\text {Timed out} \]
\[ \int \left (1+\cosh ^2(x)\right )^{3/2} \, dx=\int { {\left (\cosh \left (x\right )^{2} + 1\right )}^{\frac {3}{2}} \,d x } \]
\[ \int \left (1+\cosh ^2(x)\right )^{3/2} \, dx=\int { {\left (\cosh \left (x\right )^{2} + 1\right )}^{\frac {3}{2}} \,d x } \]
Timed out. \[ \int \left (1+\cosh ^2(x)\right )^{3/2} \, dx=\int {\left ({\mathrm {cosh}\left (x\right )}^2+1\right )}^{3/2} \,d x \]