3.137 \(\int \frac{A+B \sinh (x)}{\sqrt{a+b \sinh (x)}} \, dx\)

Optimal. Leaf size=136 \[ \frac{2 i (A b-a B) \sqrt{\frac{a+b \sinh (x)}{a-i b}} \text{EllipticF}\left (\frac{\pi }{4}-\frac{i x}{2},\frac{2 b}{b+i a}\right )}{b \sqrt{a+b \sinh (x)}}+\frac{2 i B \sqrt{a+b \sinh (x)} E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{b \sqrt{\frac{a+b \sinh (x)}{a-i b}}} \]

[Out]

((2*I)*B*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*b - a*B)*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]])

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Rubi [A]  time = 0.120167, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {2752, 2663, 2661, 2655, 2653} \[ \frac{2 i (A b-a B) \sqrt{\frac{a+b \sinh (x)}{a-i b}} F\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{b \sqrt{a+b \sinh (x)}}+\frac{2 i B \sqrt{a+b \sinh (x)} E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{b \sqrt{\frac{a+b \sinh (x)}{a-i b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)*B*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*b - a*B)*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 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]

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{A+B \sinh (x)}{\sqrt{a+b \sinh (x)}} \, dx &=\frac{B \int \sqrt{a+b \sinh (x)} \, dx}{b}+\frac{(i (-i A b+i a B)) \int \frac{1}{\sqrt{a+b \sinh (x)}} \, dx}{b}\\ &=\frac{\left (B \sqrt{a+b \sinh (x)}\right ) \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 (i (-i A b+i a B) \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 B 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 b-a B) 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*}

Mathematica [A]  time = 0.503481, size = 109, normalized size = 0.8 \[ \frac{2 \sqrt{\frac{a+b \sinh (x)}{a-i b}} \left (i (A b-a B) \text{EllipticF}\left (\frac{1}{4} (\pi -2 i x),-\frac{2 i b}{a-i b}\right )+B (b+i a) E\left (\frac{1}{4} (\pi -2 i x)|-\frac{2 i b}{a-i b}\right )\right )}{b \sqrt{a+b \sinh (x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((I*a + b)*B*EllipticE[(Pi - (2*I)*x)/4, ((-2*I)*b)/(a - I*b)] + I*(A*b - a*B)*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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Maple [A]  time = 0.092, size = 266, normalized size = 2. \begin{align*} -2\,{\frac{ib-a}{{b}^{2}\cosh \left ( x \right ) \sqrt{a+b\sinh \left ( x \right ) }}\sqrt{-{\frac{a+b\sinh \left ( x \right ) }{ib-a}}}\sqrt{{\frac{ \left ( i-\sinh \left ( x \right ) \right ) b}{ib+a}}}\sqrt{{\frac{ \left ( i+\sinh \left ( x \right ) \right ) b}{ib-a}}} \left ( -iB{\it EllipticE} \left ( \sqrt{-{\frac{a+b\sinh \left ( x \right ) }{ib-a}}},\sqrt{-{\frac{ib-a}{ib+a}}} \right ) b+iB{\it EllipticF} \left ( \sqrt{-{\frac{a+b\sinh \left ( x \right ) }{ib-a}}},\sqrt{-{\frac{ib-a}{ib+a}}} \right ) b+A{\it EllipticF} \left ( \sqrt{-{\frac{a+b\sinh \left ( x \right ) }{ib-a}}},\sqrt{-{\frac{ib-a}{ib+a}}} \right ) b-B{\it EllipticE} \left ( \sqrt{-{\frac{a+b\sinh \left ( x \right ) }{ib-a}}},\sqrt{-{\frac{ib-a}{ib+a}}} \right ) a \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*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*B*El
lipticE((-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^(1/2))*b+I*B*EllipticF((-(a+b*sinh(x))/(I*b-a))^(1/2
),(-(I*b-a)/(I*b+a))^(1/2))*b+A*EllipticF((-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^(1/2))*b-B*Ellipti
cE((-(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., size = 0, normalized size = 0. \begin{align*} \int \frac{B \sinh \left (x\right ) + A}{\sqrt{b \sinh \left (x\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{B \sinh \left (x\right ) + A}{\sqrt{b \sinh \left (x\right ) + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B \sinh{\left (x \right )}}{\sqrt{a + b \sinh{\left (x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \sinh \left (x\right ) + A}{\sqrt{b \sinh \left (x\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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