3.106 \(\int (a+b \sinh (x))^{3/2} \, dx\)

Optimal. Leaf size=150 \[ -\frac{2 i \left (a^2+b^2\right ) \sqrt{\frac{a+b \sinh (x)}{a-i b}} \text{EllipticF}\left (\frac{\pi }{4}-\frac{i x}{2},\frac{2 b}{b+i a}\right )}{3 \sqrt{a+b \sinh (x)}}+\frac{2}{3} b \cosh (x) \sqrt{a+b \sinh (x)}+\frac{8 i a \sqrt{a+b \sinh (x)} E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{3 \sqrt{\frac{a+b \sinh (x)}{a-i b}}} \]

[Out]

(2*b*Cosh[x]*Sqrt[a + b*Sinh[x]])/3 + (((8*I)/3)*a*EllipticE[Pi/4 - (I/2)*x, (2*b)/(I*a + b)]*Sqrt[a + b*Sinh[
x]])/Sqrt[(a + b*Sinh[x])/(a - I*b)] - (((2*I)/3)*(a^2 + b^2)*EllipticF[Pi/4 - (I/2)*x, (2*b)/(I*a + b)]*Sqrt[
(a + b*Sinh[x])/(a - I*b)])/Sqrt[a + b*Sinh[x]]

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Rubi [A]  time = 0.165804, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {2656, 2752, 2663, 2661, 2655, 2653} \[ -\frac{2 i \left (a^2+b^2\right ) \sqrt{\frac{a+b \sinh (x)}{a-i b}} F\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{3 \sqrt{a+b \sinh (x)}}+\frac{2}{3} b \cosh (x) \sqrt{a+b \sinh (x)}+\frac{8 i a \sqrt{a+b \sinh (x)} E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{3 \sqrt{\frac{a+b \sinh (x)}{a-i b}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sinh[x])^(3/2),x]

[Out]

(2*b*Cosh[x]*Sqrt[a + b*Sinh[x]])/3 + (((8*I)/3)*a*EllipticE[Pi/4 - (I/2)*x, (2*b)/(I*a + b)]*Sqrt[a + b*Sinh[
x]])/Sqrt[(a + b*Sinh[x])/(a - I*b)] - (((2*I)/3)*(a^2 + b^2)*EllipticF[Pi/4 - (I/2)*x, (2*b)/(I*a + b)]*Sqrt[
(a + b*Sinh[x])/(a - I*b)])/Sqrt[a + b*Sinh[x]]

Rule 2656

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), x] + Dist[1/n, Int[(a + b*Sin[c + d*x])^(n - 2)*Simp[a^2*n + b^2*(n - 1) + a*b*(2*n - 1)*Sin[c + d*
x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

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 (a+b \sinh (x))^{3/2} \, dx &=\frac{2}{3} b \cosh (x) \sqrt{a+b \sinh (x)}+\frac{2}{3} \int \frac{\frac{1}{2} \left (3 a^2-b^2\right )+2 a b \sinh (x)}{\sqrt{a+b \sinh (x)}} \, dx\\ &=\frac{2}{3} b \cosh (x) \sqrt{a+b \sinh (x)}+\frac{1}{3} (4 a) \int \sqrt{a+b \sinh (x)} \, dx+\frac{1}{3} \left (-a^2-b^2\right ) \int \frac{1}{\sqrt{a+b \sinh (x)}} \, dx\\ &=\frac{2}{3} b \cosh (x) \sqrt{a+b \sinh (x)}+\frac{\left (4 a \sqrt{a+b \sinh (x)}\right ) \int \sqrt{\frac{a}{a-i b}+\frac{b \sinh (x)}{a-i b}} \, dx}{3 \sqrt{\frac{a+b \sinh (x)}{a-i b}}}+\frac{\left (\left (-a^2-b^2\right ) \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 \sqrt{a+b \sinh (x)}}\\ &=\frac{2}{3} b \cosh (x) \sqrt{a+b \sinh (x)}+\frac{8 i a E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right ) \sqrt{a+b \sinh (x)}}{3 \sqrt{\frac{a+b \sinh (x)}{a-i b}}}-\frac{2 i \left (a^2+b^2\right ) 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 \sqrt{a+b \sinh (x)}}\\ \end{align*}

Mathematica [A]  time = 0.378487, size = 139, normalized size = 0.93 \[ \frac{-2 i \left (a^2+b^2\right ) \sqrt{\frac{a+b \sinh (x)}{a-i b}} \text{EllipticF}\left (\frac{1}{4} (\pi -2 i x),-\frac{2 i b}{a-i b}\right )+2 b \cosh (x) (a+b \sinh (x))+8 a (b+i a) \sqrt{\frac{a+b \sinh (x)}{a-i b}} E\left (\frac{1}{4} (\pi -2 i x)|-\frac{2 i b}{a-i b}\right )}{3 \sqrt{a+b \sinh (x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sinh[x])^(3/2),x]

[Out]

(2*b*Cosh[x]*(a + b*Sinh[x]) + 8*a*(I*a + b)*EllipticE[(Pi - (2*I)*x)/4, ((-2*I)*b)/(a - I*b)]*Sqrt[(a + b*Sin
h[x])/(a - I*b)] - (2*I)*(a^2 + b^2)*EllipticF[(Pi - (2*I)*x)/4, ((-2*I)*b)/(a - I*b)]*Sqrt[(a + b*Sinh[x])/(a
- I*b)])/(3*Sqrt[a + b*Sinh[x]])

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Maple [B]  time = 0.079, size = 676, normalized size = 4.5 \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))^(3/2),x)

[Out]

2/3*(I*(-(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*(-(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+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^3+3*(-(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
*b^2-4*(-(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^3-4*(-(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
*b^2+b^3*sinh(x)^3+a*b^2*sinh(x)^2+b^3*sinh(x)+a*b^2)/b/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 )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sinh(x) + a)^(3/2), 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 )}^{\frac{3}{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*sinh(x) + a)^(3/2), 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 )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*sinh(x))**(3/2), 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 )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sinh(x) + a)^(3/2), x)