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

Optimal. Leaf size=153 \[ \frac{16 i a \left (a^2-b^2\right ) \sqrt{\frac{a+b \cosh (x)}{a+b}} \text{EllipticF}\left (\frac{i x}{2},\frac{2 b}{a+b}\right )}{15 \sqrt{a+b \cosh (x)}}-\frac{2 i \left (23 a^2+9 b^2\right ) \sqrt{a+b \cosh (x)} E\left (\frac{i x}{2}|\frac{2 b}{a+b}\right )}{15 \sqrt{\frac{a+b \cosh (x)}{a+b}}}+\frac{2}{5} b \sinh (x) (a+b \cosh (x))^{3/2}+\frac{16}{15} a b \sinh (x) \sqrt{a+b \cosh (x)} \]

[Out]

(((-2*I)/15)*(23*a^2 + 9*b^2)*Sqrt[a + b*Cosh[x]]*EllipticE[(I/2)*x, (2*b)/(a + b)])/Sqrt[(a + b*Cosh[x])/(a +
b)] + (((16*I)/15)*a*(a^2 - b^2)*Sqrt[(a + b*Cosh[x])/(a + b)]*EllipticF[(I/2)*x, (2*b)/(a + b)])/Sqrt[a + b*
Cosh[x]] + (16*a*b*Sqrt[a + b*Cosh[x]]*Sinh[x])/15 + (2*b*(a + b*Cosh[x])^(3/2)*Sinh[x])/5

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Rubi [A]  time = 0.24402, antiderivative size = 153, 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} \[ \frac{16 i a \left (a^2-b^2\right ) \sqrt{\frac{a+b \cosh (x)}{a+b}} F\left (\frac{i x}{2}|\frac{2 b}{a+b}\right )}{15 \sqrt{a+b \cosh (x)}}-\frac{2 i \left (23 a^2+9 b^2\right ) \sqrt{a+b \cosh (x)} E\left (\frac{i x}{2}|\frac{2 b}{a+b}\right )}{15 \sqrt{\frac{a+b \cosh (x)}{a+b}}}+\frac{2}{5} b \sinh (x) (a+b \cosh (x))^{3/2}+\frac{16}{15} a b \sinh (x) \sqrt{a+b \cosh (x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cosh[x])^(5/2),x]

[Out]

(((-2*I)/15)*(23*a^2 + 9*b^2)*Sqrt[a + b*Cosh[x]]*EllipticE[(I/2)*x, (2*b)/(a + b)])/Sqrt[(a + b*Cosh[x])/(a +
b)] + (((16*I)/15)*a*(a^2 - b^2)*Sqrt[(a + b*Cosh[x])/(a + b)]*EllipticF[(I/2)*x, (2*b)/(a + b)])/Sqrt[a + b*
Cosh[x]] + (16*a*b*Sqrt[a + b*Cosh[x]]*Sinh[x])/15 + (2*b*(a + b*Cosh[x])^(3/2)*Sinh[x])/5

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 \cosh (x))^{5/2} \, dx &=\frac{2}{5} b (a+b \cosh (x))^{3/2} \sinh (x)+\frac{2}{5} \int \sqrt{a+b \cosh (x)} \left (\frac{1}{2} \left (5 a^2+3 b^2\right )+4 a b \cosh (x)\right ) \, dx\\ &=\frac{16}{15} a b \sqrt{a+b \cosh (x)} \sinh (x)+\frac{2}{5} b (a+b \cosh (x))^{3/2} \sinh (x)+\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 ) \cosh (x)}{\sqrt{a+b \cosh (x)}} \, dx\\ &=\frac{16}{15} a b \sqrt{a+b \cosh (x)} \sinh (x)+\frac{2}{5} b (a+b \cosh (x))^{3/2} \sinh (x)-\frac{1}{15} \left (8 a \left (a^2-b^2\right )\right ) \int \frac{1}{\sqrt{a+b \cosh (x)}} \, dx+\frac{1}{15} \left (23 a^2+9 b^2\right ) \int \sqrt{a+b \cosh (x)} \, dx\\ &=\frac{16}{15} a b \sqrt{a+b \cosh (x)} \sinh (x)+\frac{2}{5} b (a+b \cosh (x))^{3/2} \sinh (x)+\frac{\left (\left (23 a^2+9 b^2\right ) \sqrt{a+b \cosh (x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \cosh (x)}{a+b}} \, dx}{15 \sqrt{\frac{a+b \cosh (x)}{a+b}}}-\frac{\left (8 a \left (a^2-b^2\right ) \sqrt{\frac{a+b \cosh (x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \cosh (x)}{a+b}}} \, dx}{15 \sqrt{a+b \cosh (x)}}\\ &=-\frac{2 i \left (23 a^2+9 b^2\right ) \sqrt{a+b \cosh (x)} E\left (\frac{i x}{2}|\frac{2 b}{a+b}\right )}{15 \sqrt{\frac{a+b \cosh (x)}{a+b}}}+\frac{16 i a \left (a^2-b^2\right ) \sqrt{\frac{a+b \cosh (x)}{a+b}} F\left (\frac{i x}{2}|\frac{2 b}{a+b}\right )}{15 \sqrt{a+b \cosh (x)}}+\frac{16}{15} a b \sqrt{a+b \cosh (x)} \sinh (x)+\frac{2}{5} b (a+b \cosh (x))^{3/2} \sinh (x)\\ \end{align*}

Mathematica [A]  time = 0.501181, size = 150, normalized size = 0.98 \[ \frac{16 i a \left (a^2-b^2\right ) \sqrt{\frac{a+b \cosh (x)}{a+b}} \text{EllipticF}\left (\frac{i x}{2},\frac{2 b}{a+b}\right )+b \sinh (x) \left (22 a^2+28 a b \cosh (x)+3 b^2 \cosh (2 x)+3 b^2\right )-2 i \left (23 a^2 b+23 a^3+9 a b^2+9 b^3\right ) \sqrt{\frac{a+b \cosh (x)}{a+b}} E\left (\frac{i x}{2}|\frac{2 b}{a+b}\right )}{15 \sqrt{a+b \cosh (x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cosh[x])^(5/2),x]

[Out]

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

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Maple [B]  time = 0.102, size = 685, 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*cosh(x))^(5/2),x)

[Out]

2/15*(24*(-2*b/(a-b))^(1/2)*b^3*cosh(1/2*x)*sinh(1/2*x)^6+(56*(-2*b/(a-b))^(1/2)*a*b^2+24*(-2*b/(a-b))^(1/2)*b
^3)*sinh(1/2*x)^4*cosh(1/2*x)+(22*(-2*b/(a-b))^(1/2)*a^2*b+28*(-2*b/(a-b))^(1/2)*a*b^2+6*(-2*b/(a-b))^(1/2)*b^
3)*sinh(1/2*x)^2*cosh(1/2*x)+15*a^3*(2*b/(a-b)*sinh(1/2*x)^2+(a+b)/(a-b))^(1/2)*(-sinh(1/2*x)^2)^(1/2)*Ellipti
cF(cosh(1/2*x)*(-2*b/(a-b))^(1/2),1/2*(-2*(a-b)/b)^(1/2))+23*a^2*b*(2*b/(a-b)*sinh(1/2*x)^2+(a+b)/(a-b))^(1/2)
*(-sinh(1/2*x)^2)^(1/2)*EllipticF(cosh(1/2*x)*(-2*b/(a-b))^(1/2),1/2*(-2*(a-b)/b)^(1/2))+17*a*b^2*(2*b/(a-b)*s
inh(1/2*x)^2+(a+b)/(a-b))^(1/2)*(-sinh(1/2*x)^2)^(1/2)*EllipticF(cosh(1/2*x)*(-2*b/(a-b))^(1/2),1/2*(-2*(a-b)/
b)^(1/2))+9*b^3*(2*b/(a-b)*sinh(1/2*x)^2+(a+b)/(a-b))^(1/2)*(-sinh(1/2*x)^2)^(1/2)*EllipticF(cosh(1/2*x)*(-2*b
/(a-b))^(1/2),1/2*(-2*(a-b)/b)^(1/2))-46*(2*b/(a-b)*sinh(1/2*x)^2+(a+b)/(a-b))^(1/2)*(-sinh(1/2*x)^2)^(1/2)*El
lipticE(cosh(1/2*x)*(-2*b/(a-b))^(1/2),1/2*(-2*(a-b)/b)^(1/2))*a^2*b-18*(2*b/(a-b)*sinh(1/2*x)^2+(a+b)/(a-b))^
(1/2)*(-sinh(1/2*x)^2)^(1/2)*EllipticE(cosh(1/2*x)*(-2*b/(a-b))^(1/2),1/2*(-2*(a-b)/b)^(1/2))*b^3)*((2*cosh(1/
2*x)^2*b+a-b)*sinh(1/2*x)^2)^(1/2)/(-2*b/(a-b))^(1/2)/(2*b*sinh(1/2*x)^4+(a+b)*sinh(1/2*x)^2)^(1/2)/sinh(1/2*x
)/(2*sinh(1/2*x)^2*b+a+b)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \cosh \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*cosh(x))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*cosh(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} \cosh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + a^{2}\right )} \sqrt{b \cosh \left (x\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*cosh(x)^2 + 2*a*b*cosh(x) + a^2)*sqrt(b*cosh(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*cosh(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 \cosh \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*cosh(x))^(5/2),x, algorithm="giac")

[Out]

integrate((b*cosh(x) + a)^(5/2), x)