3.128 \(\int x^2 \sqrt {a+a \cosh (x)} \, dx\)

Optimal. Leaf size=53 \[ 2 x^2 \tanh \left (\frac {x}{2}\right ) \sqrt {a \cosh (x)+a}-8 x \sqrt {a \cosh (x)+a}+16 \tanh \left (\frac {x}{2}\right ) \sqrt {a \cosh (x)+a} \]

[Out]

-8*x*(a+a*cosh(x))^(1/2)+16*(a+a*cosh(x))^(1/2)*tanh(1/2*x)+2*x^2*(a+a*cosh(x))^(1/2)*tanh(1/2*x)

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Rubi [A]  time = 0.10, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3319, 3296, 2637} \[ 2 x^2 \tanh \left (\frac {x}{2}\right ) \sqrt {a \cosh (x)+a}-8 x \sqrt {a \cosh (x)+a}+16 \tanh \left (\frac {x}{2}\right ) \sqrt {a \cosh (x)+a} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[a + a*Cosh[x]],x]

[Out]

-8*x*Sqrt[a + a*Cosh[x]] + 16*Sqrt[a + a*Cosh[x]]*Tanh[x/2] + 2*x^2*Sqrt[a + a*Cosh[x]]*Tanh[x/2]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

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

Rule 3319

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[((2*a)^IntPart[n
]*(a + b*Sin[e + f*x])^FracPart[n])/Sin[e/2 + (a*Pi)/(4*b) + (f*x)/2]^(2*FracPart[n]), Int[(c + d*x)^m*Sin[e/2
+ (a*Pi)/(4*b) + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[n
+ 1/2] && (GtQ[n, 0] || IGtQ[m, 0])

Rubi steps

\begin {align*} \int x^2 \sqrt {a+a \cosh (x)} \, dx &=\left (\sqrt {a+a \cosh (x)} \text {sech}\left (\frac {x}{2}\right )\right ) \int x^2 \cosh \left (\frac {x}{2}\right ) \, dx\\ &=2 x^2 \sqrt {a+a \cosh (x)} \tanh \left (\frac {x}{2}\right )-\left (4 \sqrt {a+a \cosh (x)} \text {sech}\left (\frac {x}{2}\right )\right ) \int x \sinh \left (\frac {x}{2}\right ) \, dx\\ &=-8 x \sqrt {a+a \cosh (x)}+2 x^2 \sqrt {a+a \cosh (x)} \tanh \left (\frac {x}{2}\right )+\left (8 \sqrt {a+a \cosh (x)} \text {sech}\left (\frac {x}{2}\right )\right ) \int \cosh \left (\frac {x}{2}\right ) \, dx\\ &=-8 x \sqrt {a+a \cosh (x)}+16 \sqrt {a+a \cosh (x)} \tanh \left (\frac {x}{2}\right )+2 x^2 \sqrt {a+a \cosh (x)} \tanh \left (\frac {x}{2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 31, normalized size = 0.58 \[ 8 \left (\frac {1}{4} \left (x^2+8\right ) \tanh \left (\frac {x}{2}\right )-x\right ) \sqrt {a (\cosh (x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sqrt[a + a*Cosh[x]],x]

[Out]

8*Sqrt[a*(1 + Cosh[x])]*(-x + ((8 + x^2)*Tanh[x/2])/4)

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+a*cosh(x))^(1/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \cosh \relax (x) + a} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+a*cosh(x))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(a*cosh(x) + a)*x^2, x)

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maple [A]  time = 0.08, size = 50, normalized size = 0.94 \[ \frac {\sqrt {2}\, \sqrt {a \left ({\mathrm e}^{x}+1\right )^{2} {\mathrm e}^{-x}}\, \left (x^{2} {\mathrm e}^{x}-x^{2}-4 x \,{\mathrm e}^{x}-4 x +8 \,{\mathrm e}^{x}-8\right )}{{\mathrm e}^{x}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+a*cosh(x))^(1/2),x)

[Out]

2^(1/2)*(a*(exp(x)+1)^2*exp(-x))^(1/2)/(exp(x)+1)*(x^2*exp(x)-x^2-4*x*exp(x)-4*x+8*exp(x)-8)

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maxima [A]  time = 0.42, size = 66, normalized size = 1.25 \[ -{\left (\sqrt {2} \sqrt {a} x^{2} + 4 \, \sqrt {2} \sqrt {a} x - {\left (\sqrt {2} \sqrt {a} x^{2} - 4 \, \sqrt {2} \sqrt {a} x + 8 \, \sqrt {2} \sqrt {a}\right )} e^{x} + 8 \, \sqrt {2} \sqrt {a}\right )} e^{\left (-\frac {1}{2} \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+a*cosh(x))^(1/2),x, algorithm="maxima")

[Out]

-(sqrt(2)*sqrt(a)*x^2 + 4*sqrt(2)*sqrt(a)*x - (sqrt(2)*sqrt(a)*x^2 - 4*sqrt(2)*sqrt(a)*x + 8*sqrt(2)*sqrt(a))*
e^x + 8*sqrt(2)*sqrt(a))*e^(-1/2*x)

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mupad [B]  time = 0.07, size = 51, normalized size = 0.96 \[ -\frac {\sqrt {a+a\,\left (\frac {{\mathrm {e}}^{-x}}{2}+\frac {{\mathrm {e}}^x}{2}\right )}\,\left (8\,x-16\,{\mathrm {e}}^x-2\,x^2\,{\mathrm {e}}^x+8\,x\,{\mathrm {e}}^x+2\,x^2+16\right )}{{\mathrm {e}}^x+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + a*cosh(x))^(1/2),x)

[Out]

-((a + a*(exp(-x)/2 + exp(x)/2))^(1/2)*(8*x - 16*exp(x) - 2*x^2*exp(x) + 8*x*exp(x) + 2*x^2 + 16))/(exp(x) + 1
)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \sqrt {a \left (\cosh {\relax (x )} + 1\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+a*cosh(x))**(1/2),x)

[Out]

Integral(x**2*sqrt(a*(cosh(x) + 1)), x)

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