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

Optimal. Leaf size=164 \[ -\frac{2 i B \left (a^2+b^2\right ) \sqrt{\frac{a+b \sinh (x)}{a-i b}} \text{EllipticF}\left (\frac{\pi }{4}-\frac{i x}{2},\frac{2 b}{b+i a}\right )}{3 b \sqrt{a+b \sinh (x)}}+\frac{2 i (a B+3 A b) \sqrt{a+b \sinh (x)} E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{3 b \sqrt{\frac{a+b \sinh (x)}{a-i b}}}+\frac{2}{3} B \cosh (x) \sqrt{a+b \sinh (x)} \]

[Out]

(2*B*Cosh[x]*Sqrt[a + b*Sinh[x]])/3 + (((2*I)/3)*(3*A*b + a*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)/3)*(a^2 + b^2)*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 = 0.21153, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac{2 i B \left (a^2+b^2\right ) \sqrt{\frac{a+b \sinh (x)}{a-i b}} F\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{3 b \sqrt{a+b \sinh (x)}}+\frac{2 i (a B+3 A b) \sqrt{a+b \sinh (x)} E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{3 b \sqrt{\frac{a+b \sinh (x)}{a-i b}}}+\frac{2}{3} B \cosh (x) \sqrt{a+b \sinh (x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*B*Cosh[x]*Sqrt[a + b*Sinh[x]])/3 + (((2*I)/3)*(3*A*b + a*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)/3)*(a^2 + b^2)*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 2753

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

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 \sqrt{a+b \sinh (x)} (A+B \sinh (x)) \, dx &=\frac{2}{3} B \cosh (x) \sqrt{a+b \sinh (x)}+\frac{2}{3} \int \frac{\frac{1}{2} (3 a A-b B)+\frac{1}{2} (3 A b+a B) \sinh (x)}{\sqrt{a+b \sinh (x)}} \, dx\\ &=\frac{2}{3} B \cosh (x) \sqrt{a+b \sinh (x)}-\frac{\left (\left (a^2+b^2\right ) B\right ) \int \frac{1}{\sqrt{a+b \sinh (x)}} \, dx}{3 b}+\frac{(3 A b+a B) \int \sqrt{a+b \sinh (x)} \, dx}{3 b}\\ &=\frac{2}{3} B \cosh (x) \sqrt{a+b \sinh (x)}+\frac{\left ((3 A b+a B) \sqrt{a+b \sinh (x)}\right ) \int \sqrt{\frac{a}{a-i b}+\frac{b \sinh (x)}{a-i b}} \, dx}{3 b \sqrt{\frac{a+b \sinh (x)}{a-i b}}}-\frac{\left (\left (a^2+b^2\right ) 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}{3 b \sqrt{a+b \sinh (x)}}\\ &=\frac{2}{3} B \cosh (x) \sqrt{a+b \sinh (x)}+\frac{2 i (3 A b+a B) E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right ) \sqrt{a+b \sinh (x)}}{3 b \sqrt{\frac{a+b \sinh (x)}{a-i b}}}-\frac{2 i \left (a^2+b^2\right ) 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}}}{3 b \sqrt{a+b \sinh (x)}}\\ \end{align*}

Mathematica [A]  time = 0.401473, size = 151, normalized size = 0.92 \[ \frac{-2 i B \left (a^2+b^2\right ) \sqrt{\frac{a+b \sinh (x)}{a-i b}} \text{EllipticF}\left (\frac{1}{4} (\pi -2 i x),-\frac{2 i b}{a-i b}\right )+2 (b+i a) (a B+3 A b) \sqrt{\frac{a+b \sinh (x)}{a-i b}} E\left (\frac{1}{4} (\pi -2 i x)|-\frac{2 i b}{a-i b}\right )+2 b B \cosh (x) (a+b \sinh (x))}{3 b \sqrt{a+b \sinh (x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*B*Cosh[x]*(a + b*Sinh[x]) + 2*(I*a + b)*(3*A*b + a*B)*EllipticE[(Pi - (2*I)*x)/4, ((-2*I)*b)/(a - I*b)]*S
qrt[(a + b*Sinh[x])/(a - I*b)] - (2*I)*(a^2 + b^2)*B*EllipticF[(Pi - (2*I)*x)/4, ((-2*I)*b)/(a - I*b)]*Sqrt[(a
+ b*Sinh[x])/(a - I*b)])/(3*b*Sqrt[a + b*Sinh[x]])

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Maple [B]  time = 0.093, size = 897, normalized size = 5.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(I*B*(-(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*b+I*B*(-(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^3+3*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)*Ellipt
icF((-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^(1/2))*a^2*b+3*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)*EllipticF((-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a)
)^(1/2))*b^3-3*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)*El
lipticE((-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^(1/2))*a^2*b-3*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)*EllipticE((-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*
b+a))^(1/2))*b^3-B*(-(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^3-B*(-(a+b*sinh(x))/(I*b-a))^(1/2)*((I-si
nh(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*b^2+B*b^3*sinh(x)^3+B*a*b^2*sinh(x)^2+B*b^3*sinh(x)+B*a*b^2)/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{\left (B \sinh \left (x\right ) + A\right )} \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))^(1/2)*(A+B*sinh(x)),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 ({\left (B \sinh \left (x\right ) + A\right )} \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))^(1/2)*(A+B*sinh(x)),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 \left (A + B \sinh{\left (x \right )}\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))**(1/2)*(A+B*sinh(x)),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{\left (B \sinh \left (x\right ) + A\right )} \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))^(1/2)*(A+B*sinh(x)),x, algorithm="giac")

[Out]

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