\(\int \frac {x \sinh (a+b x)}{\sqrt {\cosh (a+b x)}} \, dx\) [531]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [B] (verified)
   Fricas [F(-2)]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 18, antiderivative size = 37 \[ \int \frac {x \sinh (a+b x)}{\sqrt {\cosh (a+b x)}} \, dx=\frac {2 x \sqrt {\cosh (a+b x)}}{b}+\frac {4 i E\left (\left .\frac {1}{2} i (a+b x)\right |2\right )}{b^2} \]

[Out]

4*I*(cosh(1/2*a+1/2*b*x)^2)^(1/2)/cosh(1/2*a+1/2*b*x)*EllipticE(I*sinh(1/2*a+1/2*b*x),2^(1/2))/b^2+2*x*cosh(b*
x+a)^(1/2)/b

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {5481, 2719} \[ \int \frac {x \sinh (a+b x)}{\sqrt {\cosh (a+b x)}} \, dx=\frac {2 x \sqrt {\cosh (a+b x)}}{b}+\frac {4 i E\left (\left .\frac {1}{2} i (a+b x)\right |2\right )}{b^2} \]

[In]

Int[(x*Sinh[a + b*x])/Sqrt[Cosh[a + b*x]],x]

[Out]

(2*x*Sqrt[Cosh[a + b*x]])/b + ((4*I)*EllipticE[(I/2)*(a + b*x), 2])/b^2

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 5481

Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp[x^(m - n
 + 1)*(Cosh[a + b*x^n]^(p + 1)/(b*n*(p + 1))), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Cosh[a + b*x
^n]^(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 x \sqrt {\cosh (a+b x)}}{b}-\frac {2 \int \sqrt {\cosh (a+b x)} \, dx}{b} \\ & = \frac {2 x \sqrt {\cosh (a+b x)}}{b}+\frac {4 i E\left (\left .\frac {1}{2} i (a+b x)\right |2\right )}{b^2} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 1.29 (sec) , antiderivative size = 109, normalized size of antiderivative = 2.95 \[ \int \frac {x \sinh (a+b x)}{\sqrt {\cosh (a+b x)}} \, dx=\frac {(\cosh (a+b x)-\sinh (a+b x)) \left (4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-\cosh (2 (a+b x))-\sinh (2 (a+b x))\right ) \sqrt {1+\cosh (2 (a+b x))+\sinh (2 (a+b x))}+(-2+b x) (1+\cosh (2 (a+b x))+\sinh (2 (a+b x)))\right )}{b^2 \sqrt {\cosh (a+b x)}} \]

[In]

Integrate[(x*Sinh[a + b*x])/Sqrt[Cosh[a + b*x]],x]

[Out]

((Cosh[a + b*x] - Sinh[a + b*x])*(4*Hypergeometric2F1[-1/4, 1/2, 3/4, -Cosh[2*(a + b*x)] - Sinh[2*(a + b*x)]]*
Sqrt[1 + Cosh[2*(a + b*x)] + Sinh[2*(a + b*x)]] + (-2 + b*x)*(1 + Cosh[2*(a + b*x)] + Sinh[2*(a + b*x)])))/(b^
2*Sqrt[Cosh[a + b*x]])

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 249 vs. \(2 (61 ) = 122\).

Time = 0.26 (sec) , antiderivative size = 250, normalized size of antiderivative = 6.76

method result size
risch \(\frac {\left (b x -2\right ) \left (1+{\mathrm e}^{2 b x +2 a}\right ) \sqrt {2}\, {\mathrm e}^{-b x -a}}{b^{2} \sqrt {\left (1+{\mathrm e}^{2 b x +2 a}\right ) {\mathrm e}^{-b x -a}}}-\frac {2 \left (-\frac {2 \left (1+{\mathrm e}^{2 b x +2 a}\right )}{\sqrt {\left (1+{\mathrm e}^{2 b x +2 a}\right ) {\mathrm e}^{b x +a}}}+\frac {i \sqrt {-i \left ({\mathrm e}^{b x +a}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{b x +a}-i\right )}\, \sqrt {i {\mathrm e}^{b x +a}}\, \left (-2 i \operatorname {EllipticE}\left (\sqrt {-i \left ({\mathrm e}^{b x +a}+i\right )}, \frac {\sqrt {2}}{2}\right )+i \operatorname {EllipticF}\left (\sqrt {-i \left ({\mathrm e}^{b x +a}+i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {{\mathrm e}^{3 b x +3 a}+{\mathrm e}^{b x +a}}}\right ) \sqrt {2}\, \sqrt {\left (1+{\mathrm e}^{2 b x +2 a}\right ) {\mathrm e}^{b x +a}}\, {\mathrm e}^{-b x -a}}{b^{2} \sqrt {\left (1+{\mathrm e}^{2 b x +2 a}\right ) {\mathrm e}^{-b x -a}}}\) \(250\)

[In]

int(x*sinh(b*x+a)/cosh(b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

(b*x-2)*(exp(b*x+a)^2+1)/b^2*2^(1/2)/((exp(b*x+a)^2+1)/exp(b*x+a))^(1/2)/exp(b*x+a)-2/b^2*(-2*(exp(b*x+a)^2+1)
/((exp(b*x+a)^2+1)*exp(b*x+a))^(1/2)+I*(-I*(exp(b*x+a)+I))^(1/2)*2^(1/2)*(I*(exp(b*x+a)-I))^(1/2)*(I*exp(b*x+a
))^(1/2)/(exp(b*x+a)^3+exp(b*x+a))^(1/2)*(-2*I*EllipticE((-I*(exp(b*x+a)+I))^(1/2),1/2*2^(1/2))+I*EllipticF((-
I*(exp(b*x+a)+I))^(1/2),1/2*2^(1/2))))*2^(1/2)/((exp(b*x+a)^2+1)/exp(b*x+a))^(1/2)*((exp(b*x+a)^2+1)*exp(b*x+a
))^(1/2)/exp(b*x+a)

Fricas [F(-2)]

Exception generated. \[ \int \frac {x \sinh (a+b x)}{\sqrt {\cosh (a+b x)}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

Sympy [F]

\[ \int \frac {x \sinh (a+b x)}{\sqrt {\cosh (a+b x)}} \, dx=\int \frac {x \sinh {\left (a + b x \right )}}{\sqrt {\cosh {\left (a + b x \right )}}}\, dx \]

[In]

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

[Out]

Integral(x*sinh(a + b*x)/sqrt(cosh(a + b*x)), x)

Maxima [F]

\[ \int \frac {x \sinh (a+b x)}{\sqrt {\cosh (a+b x)}} \, dx=\int { \frac {x \sinh \left (b x + a\right )}{\sqrt {\cosh \left (b x + a\right )}} \,d x } \]

[In]

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

[Out]

integrate(x*sinh(b*x + a)/sqrt(cosh(b*x + a)), x)

Giac [F]

\[ \int \frac {x \sinh (a+b x)}{\sqrt {\cosh (a+b x)}} \, dx=\int { \frac {x \sinh \left (b x + a\right )}{\sqrt {\cosh \left (b x + a\right )}} \,d x } \]

[In]

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

[Out]

integrate(x*sinh(b*x + a)/sqrt(cosh(b*x + a)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x \sinh (a+b x)}{\sqrt {\cosh (a+b x)}} \, dx=\int \frac {x\,\mathrm {sinh}\left (a+b\,x\right )}{\sqrt {\mathrm {cosh}\left (a+b\,x\right )}} \,d x \]

[In]

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

[Out]

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