\(\int \frac {1}{(a+b \sinh (x))^{3/2}} \, dx\) [109]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [C] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 10, antiderivative size = 94 \[ \int \frac {1}{(a+b \sinh (x))^{3/2}} \, dx=-\frac {2 b \cosh (x)}{\left (a^2+b^2\right ) \sqrt {a+b \sinh (x)}}+\frac {2 i E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right ) \sqrt {a+b \sinh (x)}}{\left (a^2+b^2\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2743, 21, 2734, 2732} \[ \int \frac {1}{(a+b \sinh (x))^{3/2}} \, dx=-\frac {2 b \cosh (x)}{\left (a^2+b^2\right ) \sqrt {a+b \sinh (x)}}+\frac {2 i \sqrt {a+b \sinh (x)} E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{\left (a^2+b^2\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}}} \]

[In]

Int[(a + b*Sinh[x])^(-3/2),x]

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
 n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
 d*x, a + b*x])

Rule 2732

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2
+ d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2734

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
 d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
 b^2, 0] &&  !GtQ[a + b, 0]

Rule 2743

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((a + b*Sin[c + d*x])^(n
+ 1)/(d*(n + 1)*(a^2 - b^2))), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n +
 1) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integ
erQ[2*n]

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.86 \[ \int \frac {1}{(a+b \sinh (x))^{3/2}} \, dx=\frac {-2 b \cosh (x)+2 (i a+b) E\left (\frac {1}{4} (\pi -2 i x)|-\frac {2 i b}{a-i b}\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}}}{\left (a^2+b^2\right ) \sqrt {a+b \sinh (x)}} \]

[In]

Integrate[(a + b*Sinh[x])^(-3/2),x]

[Out]

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 455 vs. \(2 (110 ) = 220\).

Time = 1.30 (sec) , antiderivative size = 456, normalized size of antiderivative = 4.85

method result size
default \(\frac {2 \sqrt {-\frac {a +b \sinh \left (x \right )}{i b -a}}\, \sqrt {\frac {\left (i-\sinh \left (x \right )\right ) b}{i b +a}}\, \sqrt {\frac {\left (i+\sinh \left (x \right )\right ) b}{i b -a}}\, \operatorname {EllipticF}\left (\sqrt {-\frac {a +b \sinh \left (x \right )}{i b -a}}, \sqrt {-\frac {i b -a}{i b +a}}\right ) a^{2}+2 \sqrt {-\frac {a +b \sinh \left (x \right )}{i b -a}}\, \sqrt {\frac {\left (i-\sinh \left (x \right )\right ) b}{i b +a}}\, \sqrt {\frac {\left (i+\sinh \left (x \right )\right ) b}{i b -a}}\, \operatorname {EllipticF}\left (\sqrt {-\frac {a +b \sinh \left (x \right )}{i b -a}}, \sqrt {-\frac {i b -a}{i b +a}}\right ) b^{2}-2 \sqrt {-\frac {a +b \sinh \left (x \right )}{i b -a}}\, \sqrt {\frac {\left (i-\sinh \left (x \right )\right ) b}{i b +a}}\, \sqrt {\frac {\left (i+\sinh \left (x \right )\right ) b}{i b -a}}\, \operatorname {EllipticE}\left (\sqrt {-\frac {a +b \sinh \left (x \right )}{i b -a}}, \sqrt {-\frac {i b -a}{i b +a}}\right ) a^{2}-2 \sqrt {-\frac {a +b \sinh \left (x \right )}{i b -a}}\, \sqrt {\frac {\left (i-\sinh \left (x \right )\right ) b}{i b +a}}\, \sqrt {\frac {\left (i+\sinh \left (x \right )\right ) b}{i b -a}}\, \operatorname {EllipticE}\left (\sqrt {-\frac {a +b \sinh \left (x \right )}{i b -a}}, \sqrt {-\frac {i b -a}{i b +a}}\right ) b^{2}-2 b^{2} \sinh \left (x \right )^{2}-2 b^{2}}{\left (a^{2}+b^{2}\right ) b \cosh \left (x \right ) \sqrt {a +b \sinh \left (x \right )}}\) \(456\)

[In]

int(1/(a+b*sinh(x))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 407, normalized size of antiderivative = 4.33 \[ \int \frac {1}{(a+b \sinh (x))^{3/2}} \, dx=-\frac {2 \, {\left ({\left (\sqrt {2} a b \cosh \left (x\right )^{2} + \sqrt {2} a b \sinh \left (x\right )^{2} + 2 \, \sqrt {2} a^{2} \cosh \left (x\right ) - \sqrt {2} a b + 2 \, {\left (\sqrt {2} a b \cosh \left (x\right ) + \sqrt {2} a^{2}\right )} \sinh \left (x\right )\right )} \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} + 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} + 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cosh \left (x\right ) + 3 \, b \sinh \left (x\right ) + 2 \, a}{3 \, b}\right ) - 3 \, {\left (\sqrt {2} b^{2} \cosh \left (x\right )^{2} + \sqrt {2} b^{2} \sinh \left (x\right )^{2} + 2 \, \sqrt {2} a b \cosh \left (x\right ) - \sqrt {2} b^{2} + 2 \, {\left (\sqrt {2} b^{2} \cosh \left (x\right ) + \sqrt {2} a b\right )} \sinh \left (x\right )\right )} \sqrt {b} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} + 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} + 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} + 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} + 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cosh \left (x\right ) + 3 \, b \sinh \left (x\right ) + 2 \, a}{3 \, b}\right )\right ) - 6 \, {\left (b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + a b \cosh \left (x\right ) + {\left (2 \, b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right )\right )} \sqrt {b \sinh \left (x\right ) + a}\right )}}{3 \, {\left (a^{2} b^{2} + b^{4} - {\left (a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )^{2} - {\left (a^{2} b^{2} + b^{4}\right )} \sinh \left (x\right )^{2} - 2 \, {\left (a^{3} b + a b^{3}\right )} \cosh \left (x\right ) - 2 \, {\left (a^{3} b + a b^{3} + {\left (a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )\right )}} \]

[In]

integrate(1/(a+b*sinh(x))^(3/2),x, algorithm="fricas")

[Out]

-2/3*((sqrt(2)*a*b*cosh(x)^2 + sqrt(2)*a*b*sinh(x)^2 + 2*sqrt(2)*a^2*cosh(x) - sqrt(2)*a*b + 2*(sqrt(2)*a*b*co
sh(x) + sqrt(2)*a^2)*sinh(x))*sqrt(b)*weierstrassPInverse(4/3*(4*a^2 + 3*b^2)/b^2, -8/27*(8*a^3 + 9*a*b^2)/b^3
, 1/3*(3*b*cosh(x) + 3*b*sinh(x) + 2*a)/b) - 3*(sqrt(2)*b^2*cosh(x)^2 + sqrt(2)*b^2*sinh(x)^2 + 2*sqrt(2)*a*b*
cosh(x) - sqrt(2)*b^2 + 2*(sqrt(2)*b^2*cosh(x) + sqrt(2)*a*b)*sinh(x))*sqrt(b)*weierstrassZeta(4/3*(4*a^2 + 3*
b^2)/b^2, -8/27*(8*a^3 + 9*a*b^2)/b^3, weierstrassPInverse(4/3*(4*a^2 + 3*b^2)/b^2, -8/27*(8*a^3 + 9*a*b^2)/b^
3, 1/3*(3*b*cosh(x) + 3*b*sinh(x) + 2*a)/b)) - 6*(b^2*cosh(x)^2 + b^2*sinh(x)^2 + a*b*cosh(x) + (2*b^2*cosh(x)
 + a*b)*sinh(x))*sqrt(b*sinh(x) + a))/(a^2*b^2 + b^4 - (a^2*b^2 + b^4)*cosh(x)^2 - (a^2*b^2 + b^4)*sinh(x)^2 -
 2*(a^3*b + a*b^3)*cosh(x) - 2*(a^3*b + a*b^3 + (a^2*b^2 + b^4)*cosh(x))*sinh(x))

Sympy [F]

\[ \int \frac {1}{(a+b \sinh (x))^{3/2}} \, dx=\int \frac {1}{\left (a + b \sinh {\left (x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/(a+b*sinh(x))**(3/2),x)

[Out]

Integral((a + b*sinh(x))**(-3/2), x)

Maxima [F]

\[ \int \frac {1}{(a+b \sinh (x))^{3/2}} \, dx=\int { \frac {1}{{\left (b \sinh \left (x\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/(a+b*sinh(x))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sinh(x) + a)^(-3/2), x)

Giac [F]

\[ \int \frac {1}{(a+b \sinh (x))^{3/2}} \, dx=\int { \frac {1}{{\left (b \sinh \left (x\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/(a+b*sinh(x))^(3/2),x, algorithm="giac")

[Out]

integrate((b*sinh(x) + a)^(-3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \sinh (x))^{3/2}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {sinh}\left (x\right )\right )}^{3/2}} \,d x \]

[In]

int(1/(a + b*sinh(x))^(3/2),x)

[Out]

int(1/(a + b*sinh(x))^(3/2), x)