3.42 \(\int (a+a \cosh (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=89 \[ \frac{64 a^3 \sinh (c+d x)}{15 d \sqrt{a \cosh (c+d x)+a}}+\frac{16 a^2 \sinh (c+d x) \sqrt{a \cosh (c+d x)+a}}{15 d}+\frac{2 a \sinh (c+d x) (a \cosh (c+d x)+a)^{3/2}}{5 d} \]

[Out]

(64*a^3*Sinh[c + d*x])/(15*d*Sqrt[a + a*Cosh[c + d*x]]) + (16*a^2*Sqrt[a + a*Cosh[c + d*x]]*Sinh[c + d*x])/(15
*d) + (2*a*(a + a*Cosh[c + d*x])^(3/2)*Sinh[c + d*x])/(5*d)

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Rubi [A]  time = 0.0468435, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2647, 2646} \[ \frac{64 a^3 \sinh (c+d x)}{15 d \sqrt{a \cosh (c+d x)+a}}+\frac{16 a^2 \sinh (c+d x) \sqrt{a \cosh (c+d x)+a}}{15 d}+\frac{2 a \sinh (c+d x) (a \cosh (c+d x)+a)^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cosh[c + d*x])^(5/2),x]

[Out]

(64*a^3*Sinh[c + d*x])/(15*d*Sqrt[a + a*Cosh[c + d*x]]) + (16*a^2*Sqrt[a + a*Cosh[c + d*x]]*Sinh[c + d*x])/(15
*d) + (2*a*(a + a*Cosh[c + d*x])^(3/2)*Sinh[c + d*x])/(5*d)

Rule 2647

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[(a*(2*n - 1))/n, Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && Eq
Q[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

Rule 2646

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(-2*b*Cos[c + d*x])/(d*Sqrt[a + b*Sin[c + d*
x]]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int (a+a \cosh (c+d x))^{5/2} \, dx &=\frac{2 a (a+a \cosh (c+d x))^{3/2} \sinh (c+d x)}{5 d}+\frac{1}{5} (8 a) \int (a+a \cosh (c+d x))^{3/2} \, dx\\ &=\frac{16 a^2 \sqrt{a+a \cosh (c+d x)} \sinh (c+d x)}{15 d}+\frac{2 a (a+a \cosh (c+d x))^{3/2} \sinh (c+d x)}{5 d}+\frac{1}{15} \left (32 a^2\right ) \int \sqrt{a+a \cosh (c+d x)} \, dx\\ &=\frac{64 a^3 \sinh (c+d x)}{15 d \sqrt{a+a \cosh (c+d x)}}+\frac{16 a^2 \sqrt{a+a \cosh (c+d x)} \sinh (c+d x)}{15 d}+\frac{2 a (a+a \cosh (c+d x))^{3/2} \sinh (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.12481, size = 71, normalized size = 0.8 \[ \frac{a^2 \left (150 \sinh \left (\frac{1}{2} (c+d x)\right )+25 \sinh \left (\frac{3}{2} (c+d x)\right )+3 \sinh \left (\frac{5}{2} (c+d x)\right )\right ) \text{sech}\left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cosh (c+d x)+1)}}{30 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cosh[c + d*x])^(5/2),x]

[Out]

(a^2*Sqrt[a*(1 + Cosh[c + d*x])]*Sech[(c + d*x)/2]*(150*Sinh[(c + d*x)/2] + 25*Sinh[(3*(c + d*x))/2] + 3*Sinh[
(5*(c + d*x))/2]))/(30*d)

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Maple [A]  time = 0.042, size = 73, normalized size = 0.8 \begin{align*}{\frac{8\,{a}^{3}\sqrt{2}}{15\,d}\cosh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sinh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 3\, \left ( \cosh \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+4\, \left ( \cosh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+8 \right ){\frac{1}{\sqrt{a \left ( \cosh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cosh(d*x+c))^(5/2),x)

[Out]

8/15*a^3*cosh(1/2*d*x+1/2*c)*sinh(1/2*d*x+1/2*c)*(3*cosh(1/2*d*x+1/2*c)^4+4*cosh(1/2*d*x+1/2*c)^2+8)*2^(1/2)/(
a*cosh(1/2*d*x+1/2*c)^2)^(1/2)/d

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Maxima [A]  time = 1.59718, size = 163, normalized size = 1.83 \begin{align*} \frac{\sqrt{2} a^{\frac{5}{2}} e^{\left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )}}{20 \, d} + \frac{5 \, \sqrt{2} a^{\frac{5}{2}} e^{\left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right )}}{12 \, d} + \frac{5 \, \sqrt{2} a^{\frac{5}{2}} e^{\left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}}{2 \, d} - \frac{5 \, \sqrt{2} a^{\frac{5}{2}} e^{\left (-\frac{1}{2} \, d x - \frac{1}{2} \, c\right )}}{2 \, d} - \frac{5 \, \sqrt{2} a^{\frac{5}{2}} e^{\left (-\frac{3}{2} \, d x - \frac{3}{2} \, c\right )}}{12 \, d} - \frac{\sqrt{2} a^{\frac{5}{2}} e^{\left (-\frac{5}{2} \, d x - \frac{5}{2} \, c\right )}}{20 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cosh(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/20*sqrt(2)*a^(5/2)*e^(5/2*d*x + 5/2*c)/d + 5/12*sqrt(2)*a^(5/2)*e^(3/2*d*x + 3/2*c)/d + 5/2*sqrt(2)*a^(5/2)*
e^(1/2*d*x + 1/2*c)/d - 5/2*sqrt(2)*a^(5/2)*e^(-1/2*d*x - 1/2*c)/d - 5/12*sqrt(2)*a^(5/2)*e^(-3/2*d*x - 3/2*c)
/d - 1/20*sqrt(2)*a^(5/2)*e^(-5/2*d*x - 5/2*c)/d

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Fricas [B]  time = 1.8515, size = 856, normalized size = 9.62 \begin{align*} \frac{\sqrt{\frac{1}{2}}{\left (3 \, a^{2} \cosh \left (d x + c\right )^{5} + 3 \, a^{2} \sinh \left (d x + c\right )^{5} + 25 \, a^{2} \cosh \left (d x + c\right )^{4} + 150 \, a^{2} \cosh \left (d x + c\right )^{3} + 5 \,{\left (3 \, a^{2} \cosh \left (d x + c\right ) + 5 \, a^{2}\right )} \sinh \left (d x + c\right )^{4} - 150 \, a^{2} \cosh \left (d x + c\right )^{2} + 10 \,{\left (3 \, a^{2} \cosh \left (d x + c\right )^{2} + 10 \, a^{2} \cosh \left (d x + c\right ) + 15 \, a^{2}\right )} \sinh \left (d x + c\right )^{3} - 25 \, a^{2} \cosh \left (d x + c\right ) + 30 \,{\left (a^{2} \cosh \left (d x + c\right )^{3} + 5 \, a^{2} \cosh \left (d x + c\right )^{2} + 15 \, a^{2} \cosh \left (d x + c\right ) - 5 \, a^{2}\right )} \sinh \left (d x + c\right )^{2} - 3 \, a^{2} + 5 \,{\left (3 \, a^{2} \cosh \left (d x + c\right )^{4} + 20 \, a^{2} \cosh \left (d x + c\right )^{3} + 90 \, a^{2} \cosh \left (d x + c\right )^{2} - 60 \, a^{2} \cosh \left (d x + c\right ) - 5 \, a^{2}\right )} \sinh \left (d x + c\right )\right )} \sqrt{\frac{a}{\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )}}}{30 \,{\left (d \cosh \left (d x + c\right )^{2} + 2 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + d \sinh \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*sqrt(1/2)*(3*a^2*cosh(d*x + c)^5 + 3*a^2*sinh(d*x + c)^5 + 25*a^2*cosh(d*x + c)^4 + 150*a^2*cosh(d*x + c)
^3 + 5*(3*a^2*cosh(d*x + c) + 5*a^2)*sinh(d*x + c)^4 - 150*a^2*cosh(d*x + c)^2 + 10*(3*a^2*cosh(d*x + c)^2 + 1
0*a^2*cosh(d*x + c) + 15*a^2)*sinh(d*x + c)^3 - 25*a^2*cosh(d*x + c) + 30*(a^2*cosh(d*x + c)^3 + 5*a^2*cosh(d*
x + c)^2 + 15*a^2*cosh(d*x + c) - 5*a^2)*sinh(d*x + c)^2 - 3*a^2 + 5*(3*a^2*cosh(d*x + c)^4 + 20*a^2*cosh(d*x
+ c)^3 + 90*a^2*cosh(d*x + c)^2 - 60*a^2*cosh(d*x + c) - 5*a^2)*sinh(d*x + c))*sqrt(a/(cosh(d*x + c) + sinh(d*
x + c)))/(d*cosh(d*x + c)^2 + 2*d*cosh(d*x + c)*sinh(d*x + c) + d*sinh(d*x + c)^2)

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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+a*cosh(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.19484, size = 142, normalized size = 1.6 \begin{align*} -\frac{\sqrt{2}{\left ({\left (150 \, a^{\frac{5}{2}} e^{\left (2 \, d x + \frac{5}{2} \, c\right )} + 25 \, a^{\frac{5}{2}} e^{\left (d x + \frac{3}{2} \, c\right )} + 3 \, a^{\frac{5}{2}} e^{\left (\frac{1}{2} \, c\right )}\right )} e^{\left (-\frac{5}{2} \, d x - 3 \, c\right )} -{\left (3 \, a^{\frac{5}{2}} e^{\left (\frac{5}{2} \, d x + \frac{35}{2} \, c\right )} + 25 \, a^{\frac{5}{2}} e^{\left (\frac{3}{2} \, d x + \frac{33}{2} \, c\right )} + 150 \, a^{\frac{5}{2}} e^{\left (\frac{1}{2} \, d x + \frac{31}{2} \, c\right )}\right )} e^{\left (-15 \, c\right )}\right )}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cosh(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-1/60*sqrt(2)*((150*a^(5/2)*e^(2*d*x + 5/2*c) + 25*a^(5/2)*e^(d*x + 3/2*c) + 3*a^(5/2)*e^(1/2*c))*e^(-5/2*d*x
- 3*c) - (3*a^(5/2)*e^(5/2*d*x + 35/2*c) + 25*a^(5/2)*e^(3/2*d*x + 33/2*c) + 150*a^(5/2)*e^(1/2*d*x + 31/2*c))
*e^(-15*c))/d