3.105 \(\int (a+b \sinh (x))^{5/2} \, dx\)

Optimal. Leaf size=179 \[ -\frac{16 i a \left (a^2+b^2\right ) \sqrt{\frac{a+b \sinh (x)}{a-i b}} \text{EllipticF}\left (\frac{\pi }{4}-\frac{i x}{2},\frac{2 b}{b+i a}\right )}{15 \sqrt{a+b \sinh (x)}}+\frac{2 i \left (23 a^2-9 b^2\right ) \sqrt{a+b \sinh (x)} E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{15 \sqrt{\frac{a+b \sinh (x)}{a-i b}}}+\frac{2}{5} b \cosh (x) (a+b \sinh (x))^{3/2}+\frac{16}{15} a b \cosh (x) \sqrt{a+b \sinh (x)} \]

[Out]

(16*a*b*Cosh[x]*Sqrt[a + b*Sinh[x]])/15 + (2*b*Cosh[x]*(a + b*Sinh[x])^(3/2))/5 + (((2*I)/15)*(23*a^2 - 9*b^2)
*EllipticE[Pi/4 - (I/2)*x, (2*b)/(I*a + b)]*Sqrt[a + b*Sinh[x]])/Sqrt[(a + b*Sinh[x])/(a - I*b)] - (((16*I)/15
)*a*(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 = 0.257808, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {2656, 2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac{16 i a \left (a^2+b^2\right ) \sqrt{\frac{a+b \sinh (x)}{a-i b}} F\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{15 \sqrt{a+b \sinh (x)}}+\frac{2 i \left (23 a^2-9 b^2\right ) \sqrt{a+b \sinh (x)} E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{15 \sqrt{\frac{a+b \sinh (x)}{a-i b}}}+\frac{2}{5} b \cosh (x) (a+b \sinh (x))^{3/2}+\frac{16}{15} a b \cosh (x) \sqrt{a+b \sinh (x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sinh[x])^(5/2),x]

[Out]

(16*a*b*Cosh[x]*Sqrt[a + b*Sinh[x]])/15 + (2*b*Cosh[x]*(a + b*Sinh[x])^(3/2))/5 + (((2*I)/15)*(23*a^2 - 9*b^2)
*EllipticE[Pi/4 - (I/2)*x, (2*b)/(I*a + b)]*Sqrt[a + b*Sinh[x]])/Sqrt[(a + b*Sinh[x])/(a - I*b)] - (((16*I)/15
)*a*(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 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 (a+b \sinh (x))^{5/2} \, dx &=\frac{2}{5} b \cosh (x) (a+b \sinh (x))^{3/2}+\frac{2}{5} \int \sqrt{a+b \sinh (x)} \left (\frac{1}{2} \left (5 a^2-3 b^2\right )+4 a b \sinh (x)\right ) \, dx\\ &=\frac{16}{15} a b \cosh (x) \sqrt{a+b \sinh (x)}+\frac{2}{5} b \cosh (x) (a+b \sinh (x))^{3/2}+\frac{4}{15} \int \frac{\frac{1}{4} a \left (15 a^2-17 b^2\right )+\frac{1}{4} b \left (23 a^2-9 b^2\right ) \sinh (x)}{\sqrt{a+b \sinh (x)}} \, dx\\ &=\frac{16}{15} a b \cosh (x) \sqrt{a+b \sinh (x)}+\frac{2}{5} b \cosh (x) (a+b \sinh (x))^{3/2}+\frac{1}{15} \left (23 a^2-9 b^2\right ) \int \sqrt{a+b \sinh (x)} \, dx-\frac{1}{15} \left (8 a \left (a^2+b^2\right )\right ) \int \frac{1}{\sqrt{a+b \sinh (x)}} \, dx\\ &=\frac{16}{15} a b \cosh (x) \sqrt{a+b \sinh (x)}+\frac{2}{5} b \cosh (x) (a+b \sinh (x))^{3/2}+\frac{\left (\left (23 a^2-9 b^2\right ) \sqrt{a+b \sinh (x)}\right ) \int \sqrt{\frac{a}{a-i b}+\frac{b \sinh (x)}{a-i b}} \, dx}{15 \sqrt{\frac{a+b \sinh (x)}{a-i b}}}-\frac{\left (8 a \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}{15 \sqrt{a+b \sinh (x)}}\\ &=\frac{16}{15} a b \cosh (x) \sqrt{a+b \sinh (x)}+\frac{2}{5} b \cosh (x) (a+b \sinh (x))^{3/2}+\frac{2 i \left (23 a^2-9 b^2\right ) E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right ) \sqrt{a+b \sinh (x)}}{15 \sqrt{\frac{a+b \sinh (x)}{a-i b}}}-\frac{16 i a \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}}}{15 \sqrt{a+b \sinh (x)}}\\ \end{align*}

Mathematica [A]  time = 0.413323, size = 178, normalized size = 0.99 \[ \frac{-16 i a \left (a^2+b^2\right ) \sqrt{\frac{a+b \sinh (x)}{a-i b}} \text{EllipticF}\left (\frac{1}{4} (\pi -2 i x),-\frac{2 i b}{a-i b}\right )+b \cosh (x) \left (22 a^2+28 a b \sinh (x)+3 b^2 \cosh (2 x)-3 b^2\right )+2 \left (23 a^2 b+23 i a^3-9 i a b^2-9 b^3\right ) \sqrt{\frac{a+b \sinh (x)}{a-i b}} E\left (\frac{1}{4} (\pi -2 i x)|-\frac{2 i b}{a-i b}\right )}{15 \sqrt{a+b \sinh (x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sinh[x])^(5/2),x]

[Out]

(2*((23*I)*a^3 + 23*a^2*b - (9*I)*a*b^2 - 9*b^3)*EllipticE[(Pi - (2*I)*x)/4, ((-2*I)*b)/(a - I*b)]*Sqrt[(a + b
*Sinh[x])/(a - I*b)] - (16*I)*a*(a^2 + b^2)*EllipticF[(Pi - (2*I)*x)/4, ((-2*I)*b)/(a - I*b)]*Sqrt[(a + b*Sinh
[x])/(a - I*b)] + b*Cosh[x]*(22*a^2 - 3*b^2 + 3*b^2*Cosh[2*x] + 28*a*b*Sinh[x]))/(15*Sqrt[a + b*Sinh[x]])

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Maple [B]  time = 0.155, size = 917, normalized size = 5.1 \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))^(5/2),x)

[Out]

2/15*(8*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^3*b+8*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*b^3+15*(-(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)*Elli
pticF((-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^(1/2))*a^4+6*(-(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^2-9*(-(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
lipticF((-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^(1/2))*b^4-23*(-(a+b*sinh(x))/(I*b-a))^(1/2)*((I-sin
h(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^4-14*(-(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^2+9*(-(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^4+3*b^4*sinh(x)^4+14*a*b^3*sinh(x)^3+11*a^2*b^2*sinh(x)^2+3*b^4*sinh(x)^2+14*a*b^3*sinh(x)+11*a
^2*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{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sinh(x))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*sinh(x) + a)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{2} \sinh \left (x\right )^{2} + 2 \, a b \sinh \left (x\right ) + a^{2}\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))^(5/2),x, algorithm="fricas")

[Out]

integral((b^2*sinh(x)^2 + 2*a*b*sinh(x) + a^2)*sqrt(b*sinh(x) + a), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sinh(x))**(5/2),x)

[Out]

Timed out

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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{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sinh(x))^(5/2),x, algorithm="giac")

[Out]

integrate((b*sinh(x) + a)^(5/2), x)