3.111 \(\int \frac {\sinh (x)}{\sqrt {a+b \sinh (x)}} \, dx\)

Optimal. Leaf size=128 \[ \frac {2 i \sqrt {a+b \sinh (x)} E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {2 i a \sqrt {\frac {a+b \sinh (x)}{a-i b}} F\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{b \sqrt {a+b \sinh (x)}} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2752, 2663, 2661, 2655, 2653} \[ \frac {2 i \sqrt {a+b \sinh (x)} E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {2 i a \sqrt {\frac {a+b \sinh (x)}{a-i b}} F\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{b \sqrt {a+b \sinh (x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]/Sqrt[a + b*Sinh[x]],x]

[Out]

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

Rule 2653

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

Rule 2655

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*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

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

Rule 2663

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

Rule 2752

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {\sinh (x)}{\sqrt {a+b \sinh (x)}} \, dx &=\frac {\int \sqrt {a+b \sinh (x)} \, dx}{b}-\frac {a \int \frac {1}{\sqrt {a+b \sinh (x)}} \, dx}{b}\\ &=\frac {\sqrt {a+b \sinh (x)} \int \sqrt {\frac {a}{a-i b}+\frac {b \sinh (x)}{a-i b}} \, dx}{b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {\left (a \sqrt {\frac {a+b \sinh (x)}{a-i b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a-i b}+\frac {b \sinh (x)}{a-i b}}} \, dx}{b \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)}}{b \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {2 i a F\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}}}{b \sqrt {a+b \sinh (x)}}\\ \end {align*}

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Mathematica [A]  time = 0.36, size = 101, normalized size = 0.79 \[ \frac {2 \sqrt {\frac {a+b \sinh (x)}{a-i b}} \left ((b+i a) E\left (\frac {1}{4} (\pi -2 i x)|-\frac {2 i b}{a-i b}\right )-i a F\left (\frac {1}{4} (\pi -2 i x)|-\frac {2 i b}{a-i b}\right )\right )}{b \sqrt {a+b \sinh (x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sinh \relax (x)}{\sqrt {b \sinh \relax (x) + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sinh(x)/sqrt(b*sinh(x) + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh \relax (x)}{\sqrt {b \sinh \relax (x) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.12, size = 218, normalized size = 1.70 \[ \frac {2 \left (i b -a \right ) \sqrt {-\frac {a +b \sinh \relax (x )}{i b -a}}\, \sqrt {\frac {\left (i-\sinh \relax (x )\right ) b}{i b +a}}\, \sqrt {\frac {\left (i+\sinh \relax (x )\right ) b}{i b -a}}\, \left (i \EllipticE \left (\sqrt {-\frac {a +b \sinh \relax (x )}{i b -a}}, \sqrt {-\frac {i b -a}{i b +a}}\right ) b -i \EllipticF \left (\sqrt {-\frac {a +b \sinh \relax (x )}{i b -a}}, \sqrt {-\frac {i b -a}{i b +a}}\right ) b +\EllipticE \left (\sqrt {-\frac {a +b \sinh \relax (x )}{i b -a}}, \sqrt {-\frac {i b -a}{i b +a}}\right ) a \right )}{b^{2} \cosh \relax (x ) \sqrt {a +b \sinh \relax (x )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh \relax (x)}{\sqrt {b \sinh \relax (x) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {sinh}\relax (x)}{\sqrt {a+b\,\mathrm {sinh}\relax (x)}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)/(a + b*sinh(x))^(1/2),x)

[Out]

int(sinh(x)/(a + b*sinh(x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh {\relax (x )}}{\sqrt {a + b \sinh {\relax (x )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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