Integrand size = 25, antiderivative size = 128 \[ \int \frac {\sinh ^2(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx=-\frac {i E\left (i e+i f x\left |\frac {b}{a}\right .\right ) \sqrt {a+b \sinh ^2(e+f x)}}{b f \sqrt {1+\frac {b \sinh ^2(e+f x)}{a}}}+\frac {i a \operatorname {EllipticF}\left (i e+i f x,\frac {b}{a}\right ) \sqrt {1+\frac {b \sinh ^2(e+f x)}{a}}}{b f \sqrt {a+b \sinh ^2(e+f x)}} \]
-I*(cos(I*e+I*f*x)^2)^(1/2)/cos(I*e+I*f*x)*EllipticE(sin(I*e+I*f*x),(b/a)^ (1/2))*(a+b*sinh(f*x+e)^2)^(1/2)/b/f/(1+b*sinh(f*x+e)^2/a)^(1/2)+I*a*(cos( I*e+I*f*x)^2)^(1/2)/cos(I*e+I*f*x)*EllipticF(sin(I*e+I*f*x),(b/a)^(1/2))*( 1+b*sinh(f*x+e)^2/a)^(1/2)/b/f/(a+b*sinh(f*x+e)^2)^(1/2)
Time = 0.41 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.70 \[ \int \frac {\sinh ^2(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx=-\frac {i \sqrt {2 a-b+b \cosh (2 (e+f x))} \left (E\left (i (e+f x)\left |\frac {b}{a}\right .\right )-\operatorname {EllipticF}\left (i (e+f x),\frac {b}{a}\right )\right )}{b f \sqrt {\frac {2 a-b+b \cosh (2 (e+f x))}{a}}} \]
((-I)*Sqrt[2*a - b + b*Cosh[2*(e + f*x)]]*(EllipticE[I*(e + f*x), b/a] - E llipticF[I*(e + f*x), b/a]))/(b*f*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a])
Time = 0.64 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 25, 3651, 3042, 3657, 3042, 3656, 3662, 3042, 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.
| \(\displaystyle \int \frac {\sinh ^2(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\sin (i e+i f x)^2}{\sqrt {a-b \sin (i e+i f x)^2}}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\sin (i e+i f x)^2}{\sqrt {a-b \sin (i e+i f x)^2}}dx\) |
| \(\Big \downarrow \) 3651 |
| \(\displaystyle \frac {\int \sqrt {b \sinh ^2(e+f x)+a}dx}{b}-\frac {a \int \frac {1}{\sqrt {b \sinh ^2(e+f x)+a}}dx}{b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sqrt {a-b \sin (i e+i f x)^2}dx}{b}-\frac {a \int \frac {1}{\sqrt {a-b \sin (i e+i f x)^2}}dx}{b}\) |
| \(\Big \downarrow \) 3657 |
| \(\displaystyle \frac {\sqrt {a+b \sinh ^2(e+f x)} \int \sqrt {\frac {b \sinh ^2(e+f x)}{a}+1}dx}{b \sqrt {\frac {b \sinh ^2(e+f x)}{a}+1}}-\frac {a \int \frac {1}{\sqrt {a-b \sin (i e+i f x)^2}}dx}{b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {a+b \sinh ^2(e+f x)} \int \sqrt {1-\frac {b \sin (i e+i f x)^2}{a}}dx}{b \sqrt {\frac {b \sinh ^2(e+f x)}{a}+1}}-\frac {a \int \frac {1}{\sqrt {a-b \sin (i e+i f x)^2}}dx}{b}\) |
| \(\Big \downarrow \) 3656 |
| \(\displaystyle -\frac {a \int \frac {1}{\sqrt {a-b \sin (i e+i f x)^2}}dx}{b}-\frac {i \sqrt {a+b \sinh ^2(e+f x)} E\left (i e+i f x\left |\frac {b}{a}\right .\right )}{b f \sqrt {\frac {b \sinh ^2(e+f x)}{a}+1}}\) |
| \(\Big \downarrow \) 3662 |
| \(\displaystyle -\frac {a \sqrt {\frac {b \sinh ^2(e+f x)}{a}+1} \int \frac {1}{\sqrt {\frac {b \sinh ^2(e+f x)}{a}+1}}dx}{b \sqrt {a+b \sinh ^2(e+f x)}}-\frac {i \sqrt {a+b \sinh ^2(e+f x)} E\left (i e+i f x\left |\frac {b}{a}\right .\right )}{b f \sqrt {\frac {b \sinh ^2(e+f x)}{a}+1}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a \sqrt {\frac {b \sinh ^2(e+f x)}{a}+1} \int \frac {1}{\sqrt {1-\frac {b \sin (i e+i f x)^2}{a}}}dx}{b \sqrt {a+b \sinh ^2(e+f x)}}-\frac {i \sqrt {a+b \sinh ^2(e+f x)} E\left (i e+i f x\left |\frac {b}{a}\right .\right )}{b f \sqrt {\frac {b \sinh ^2(e+f x)}{a}+1}}\) |
| \(\Big \downarrow \) 3661 |
| \(\displaystyle \frac {i a \sqrt {\frac {b \sinh ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (i e+i f x,\frac {b}{a}\right )}{b f \sqrt {a+b \sinh ^2(e+f x)}}-\frac {i \sqrt {a+b \sinh ^2(e+f x)} E\left (i e+i f x\left |\frac {b}{a}\right .\right )}{b f \sqrt {\frac {b \sinh ^2(e+f x)}{a}+1}}\) |
((-I)*EllipticE[I*e + I*f*x, b/a]*Sqrt[a + b*Sinh[e + f*x]^2])/(b*f*Sqrt[1 + (b*Sinh[e + f*x]^2)/a]) + (I*a*EllipticF[I*e + I*f*x, b/a]*Sqrt[1 + (b* Sinh[e + f*x]^2)/a])/(b*f*Sqrt[a + b*Sinh[e + f*x]^2])
3.2.3.3.1 Defintions of rubi rules used
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[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[1 + b*(Sin[e + f*x]^2/a)] Int[Sqrt[1 + (b*Sin[e + f*x]^2)/a], x], x] /; FreeQ[{a, b, e, f}, x] && !GtQ[a, 0]
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]
Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[Sqrt[ 1 + b*(Sin[e + f*x]^2/a)]/Sqrt[a + b*Sin[e + f*x]^2] Int[1/Sqrt[1 + (b*Si n[e + f*x]^2)/a], x], x] /; FreeQ[{a, b, e, f}, x] && !GtQ[a, 0]
Time = 0.35 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.88
| method | result | size |
| default | \(-\frac {\sqrt {\frac {a +b \sinh \left (f x +e \right )^{2}}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \left (\operatorname {EllipticF}\left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )-\operatorname {EllipticE}\left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )\right )}{\sqrt {-\frac {b}{a}}\, \cosh \left (f x +e \right ) \sqrt {a +b \sinh \left (f x +e \right )^{2}}\, f}\) | \(113\) |
-1/(-b/a)^(1/2)*((a+b*sinh(f*x+e)^2)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*(Ellip ticF(sinh(f*x+e)*(-b/a)^(1/2),(a/b)^(1/2))-EllipticE(sinh(f*x+e)*(-b/a)^(1 /2),(a/b)^(1/2)))/cosh(f*x+e)/(a+b*sinh(f*x+e)^2)^(1/2)/f
\[ \int \frac {\sinh ^2(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx=\int { \frac {\sinh \left (f x + e\right )^{2}}{\sqrt {b \sinh \left (f x + e\right )^{2} + a}} \,d x } \]
\[ \int \frac {\sinh ^2(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx=\int \frac {\sinh ^{2}{\left (e + f x \right )}}{\sqrt {a + b \sinh ^{2}{\left (e + f x \right )}}}\, dx \]
\[ \int \frac {\sinh ^2(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx=\int { \frac {\sinh \left (f x + e\right )^{2}}{\sqrt {b \sinh \left (f x + e\right )^{2} + a}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 374 vs. \(2 (148) = 296\).
Time = 0.92 (sec) , antiderivative size = 374, normalized size of antiderivative = 2.92 \[ \int \frac {\sinh ^2(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx=-\frac {{\left (\frac {e^{\left (2 \, e\right )} \log \left ({\left | {\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )} \sqrt {b} + 2 \, a - b \right |}\right )}{\sqrt {b}} + \frac {2 \, {\left (2 \, a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} \arctan \left (-\frac {\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b} - \frac {2 \, {\left (2 \, {\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )} a e^{\left (2 \, e\right )} - {\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )} b e^{\left (2 \, e\right )} + b^{\frac {3}{2}} e^{\left (2 \, e\right )}\right )}}{{\left ({\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )}^{2} - b\right )} b}\right )} e^{\left (-e\right )}}{4 \, f^{2}} \]
-1/4*(e^(2*e)*log(abs((sqrt(b)*e^(2*f*x + 2*e) - sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))*sqrt(b) + 2*a - b))/sqrt(b ) + 2*(2*a*e^(2*e) + b*e^(2*e))*arctan(-(sqrt(b)*e^(2*f*x + 2*e) - sqrt(b* e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))/sqrt(-b) )/(sqrt(-b)*b) - 2*(2*(sqrt(b)*e^(2*f*x + 2*e) - sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))*a*e^(2*e) - (sqrt(b)*e^(2* f*x + 2*e) - sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))*b*e^(2*e) + b^(3/2)*e^(2*e))/(((sqrt(b)*e^(2*f*x + 2*e) - sqrt (b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))^2 - b )*b))*e^(-e)/f^2
Timed out. \[ \int \frac {\sinh ^2(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx=\int \frac {{\mathrm {sinh}\left (e+f\,x\right )}^2}{\sqrt {b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a}} \,d x \]