3.127 \(\int x^3 \sqrt {a+a \cosh (x)} \, dx\)

Optimal. Leaf size=68 \[ 2 x^3 \tanh \left (\frac {x}{2}\right ) \sqrt {a \cosh (x)+a}-12 x^2 \sqrt {a \cosh (x)+a}-96 \sqrt {a \cosh (x)+a}+48 x \tanh \left (\frac {x}{2}\right ) \sqrt {a \cosh (x)+a} \]

[Out]

-96*(a+a*cosh(x))^(1/2)-12*x^2*(a+a*cosh(x))^(1/2)+48*x*(a+a*cosh(x))^(1/2)*tanh(1/2*x)+2*x^3*(a+a*cosh(x))^(1
/2)*tanh(1/2*x)

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Rubi [A]  time = 0.12, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3319, 3296, 2638} \[ -12 x^2 \sqrt {a \cosh (x)+a}+2 x^3 \tanh \left (\frac {x}{2}\right ) \sqrt {a \cosh (x)+a}-96 \sqrt {a \cosh (x)+a}+48 x \tanh \left (\frac {x}{2}\right ) \sqrt {a \cosh (x)+a} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Sqrt[a + a*Cosh[x]],x]

[Out]

-96*Sqrt[a + a*Cosh[x]] - 12*x^2*Sqrt[a + a*Cosh[x]] + 48*x*Sqrt[a + a*Cosh[x]]*Tanh[x/2] + 2*x^3*Sqrt[a + a*C
osh[x]]*Tanh[x/2]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[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^3 \sqrt {a+a \cosh (x)} \, dx &=\left (\sqrt {a+a \cosh (x)} \text {sech}\left (\frac {x}{2}\right )\right ) \int x^3 \cosh \left (\frac {x}{2}\right ) \, dx\\ &=2 x^3 \sqrt {a+a \cosh (x)} \tanh \left (\frac {x}{2}\right )-\left (6 \sqrt {a+a \cosh (x)} \text {sech}\left (\frac {x}{2}\right )\right ) \int x^2 \sinh \left (\frac {x}{2}\right ) \, dx\\ &=-12 x^2 \sqrt {a+a \cosh (x)}+2 x^3 \sqrt {a+a \cosh (x)} \tanh \left (\frac {x}{2}\right )+\left (24 \sqrt {a+a \cosh (x)} \text {sech}\left (\frac {x}{2}\right )\right ) \int x \cosh \left (\frac {x}{2}\right ) \, dx\\ &=-12 x^2 \sqrt {a+a \cosh (x)}+48 x \sqrt {a+a \cosh (x)} \tanh \left (\frac {x}{2}\right )+2 x^3 \sqrt {a+a \cosh (x)} \tanh \left (\frac {x}{2}\right )-\left (48 \sqrt {a+a \cosh (x)} \text {sech}\left (\frac {x}{2}\right )\right ) \int \sinh \left (\frac {x}{2}\right ) \, dx\\ &=-96 \sqrt {a+a \cosh (x)}-12 x^2 \sqrt {a+a \cosh (x)}+48 x \sqrt {a+a \cosh (x)} \tanh \left (\frac {x}{2}\right )+2 x^3 \sqrt {a+a \cosh (x)} \tanh \left (\frac {x}{2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 33, normalized size = 0.49 \[ 2 \left (x \left (x^2+24\right ) \tanh \left (\frac {x}{2}\right )-6 \left (x^2+8\right )\right ) \sqrt {a (\cosh (x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sqrt[a + a*Cosh[x]],x]

[Out]

2*Sqrt[a*(1 + Cosh[x])]*(-6*(8 + x^2) + x*(24 + x^2)*Tanh[x/2])

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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^3*(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 \[ \int \sqrt {a \cosh \relax (x) + a} x^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+a*cosh(x))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(a*cosh(x) + a)*x^3, x)

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maple [A]  time = 0.10, size = 62, normalized size = 0.91 \[ \frac {\sqrt {2}\, \sqrt {a \left ({\mathrm e}^{x}+1\right )^{2} {\mathrm e}^{-x}}\, \left (x^{3} {\mathrm e}^{x}-x^{3}-6 x^{2} {\mathrm e}^{x}-6 x^{2}+24 x \,{\mathrm e}^{x}-24 x -48 \,{\mathrm e}^{x}-48\right )}{{\mathrm e}^{x}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+a*cosh(x))^(1/2),x)

[Out]

2^(1/2)*(a*(exp(x)+1)^2*exp(-x))^(1/2)/(exp(x)+1)*(x^3*exp(x)-x^3-6*x^2*exp(x)-6*x^2+24*x*exp(x)-24*x-48*exp(x
)-48)

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maxima [A]  time = 0.41, size = 88, normalized size = 1.29 \[ -{\left (\sqrt {2} \sqrt {a} x^{3} + 6 \, \sqrt {2} \sqrt {a} x^{2} + 24 \, \sqrt {2} \sqrt {a} x - {\left (\sqrt {2} \sqrt {a} x^{3} - 6 \, \sqrt {2} \sqrt {a} x^{2} + 24 \, \sqrt {2} \sqrt {a} x - 48 \, \sqrt {2} \sqrt {a}\right )} e^{x} + 48 \, \sqrt {2} \sqrt {a}\right )} e^{\left (-\frac {1}{2} \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+a*cosh(x))^(1/2),x, algorithm="maxima")

[Out]

-(sqrt(2)*sqrt(a)*x^3 + 6*sqrt(2)*sqrt(a)*x^2 + 24*sqrt(2)*sqrt(a)*x - (sqrt(2)*sqrt(a)*x^3 - 6*sqrt(2)*sqrt(a
)*x^2 + 24*sqrt(2)*sqrt(a)*x - 48*sqrt(2)*sqrt(a))*e^x + 48*sqrt(2)*sqrt(a))*e^(-1/2*x)

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mupad [B]  time = 0.91, size = 63, normalized size = 0.93 \[ -\frac {\sqrt {a+a\,\left (\frac {{\mathrm {e}}^{-x}}{2}+\frac {{\mathrm {e}}^x}{2}\right )}\,\left (48\,x+96\,{\mathrm {e}}^x+12\,x^2\,{\mathrm {e}}^x-2\,x^3\,{\mathrm {e}}^x-48\,x\,{\mathrm {e}}^x+12\,x^2+2\,x^3+96\right )}{{\mathrm {e}}^x+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + a*cosh(x))^(1/2),x)

[Out]

-((a + a*(exp(-x)/2 + exp(x)/2))^(1/2)*(48*x + 96*exp(x) + 12*x^2*exp(x) - 2*x^3*exp(x) - 48*x*exp(x) + 12*x^2
+ 2*x^3 + 96))/(exp(x) + 1)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \sqrt {a \left (\cosh {\relax (x )} + 1\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+a*cosh(x))**(1/2),x)

[Out]

Integral(x**3*sqrt(a*(cosh(x) + 1)), x)

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