3.901 \(\int (2+e x)^{5/2} (12-3 e^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=109 \[ -\frac{6 \sqrt{3} (2-e x)^{13/2}}{13 e}+\frac{96 \sqrt{3} (2-e x)^{11/2}}{11 e}-\frac{64 \sqrt{3} (2-e x)^{9/2}}{e}+\frac{1536 \sqrt{3} (2-e x)^{7/2}}{7 e}-\frac{1536 \sqrt{3} (2-e x)^{5/2}}{5 e} \]

[Out]

(-1536*Sqrt[3]*(2 - e*x)^(5/2))/(5*e) + (1536*Sqrt[3]*(2 - e*x)^(7/2))/(7*e) - (64*Sqrt[3]*(2 - e*x)^(9/2))/e
+ (96*Sqrt[3]*(2 - e*x)^(11/2))/(11*e) - (6*Sqrt[3]*(2 - e*x)^(13/2))/(13*e)

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Rubi [A]  time = 0.0274405, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {627, 43} \[ -\frac{6 \sqrt{3} (2-e x)^{13/2}}{13 e}+\frac{96 \sqrt{3} (2-e x)^{11/2}}{11 e}-\frac{64 \sqrt{3} (2-e x)^{9/2}}{e}+\frac{1536 \sqrt{3} (2-e x)^{7/2}}{7 e}-\frac{1536 \sqrt{3} (2-e x)^{5/2}}{5 e} \]

Antiderivative was successfully verified.

[In]

Int[(2 + e*x)^(5/2)*(12 - 3*e^2*x^2)^(3/2),x]

[Out]

(-1536*Sqrt[3]*(2 - e*x)^(5/2))/(5*e) + (1536*Sqrt[3]*(2 - e*x)^(7/2))/(7*e) - (64*Sqrt[3]*(2 - e*x)^(9/2))/e
+ (96*Sqrt[3]*(2 - e*x)^(11/2))/(11*e) - (6*Sqrt[3]*(2 - e*x)^(13/2))/(13*e)

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int (2+e x)^{5/2} \left (12-3 e^2 x^2\right )^{3/2} \, dx &=\int (6-3 e x)^{3/2} (2+e x)^4 \, dx\\ &=\int \left (256 (6-3 e x)^{3/2}-\frac{256}{3} (6-3 e x)^{5/2}+\frac{32}{3} (6-3 e x)^{7/2}-\frac{16}{27} (6-3 e x)^{9/2}+\frac{1}{81} (6-3 e x)^{11/2}\right ) \, dx\\ &=-\frac{1536 \sqrt{3} (2-e x)^{5/2}}{5 e}+\frac{1536 \sqrt{3} (2-e x)^{7/2}}{7 e}-\frac{64 \sqrt{3} (2-e x)^{9/2}}{e}+\frac{96 \sqrt{3} (2-e x)^{11/2}}{11 e}-\frac{6 \sqrt{3} (2-e x)^{13/2}}{13 e}\\ \end{align*}

Mathematica [A]  time = 0.0682521, size = 67, normalized size = 0.61 \[ -\frac{2 (e x-2)^2 \sqrt{12-3 e^2 x^2} \left (1155 e^4 x^4+12600 e^3 x^3+56840 e^2 x^2+133600 e x+154928\right )}{5005 e \sqrt{e x+2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + e*x)^(5/2)*(12 - 3*e^2*x^2)^(3/2),x]

[Out]

(-2*(-2 + e*x)^2*Sqrt[12 - 3*e^2*x^2]*(154928 + 133600*e*x + 56840*e^2*x^2 + 12600*e^3*x^3 + 1155*e^4*x^4))/(5
005*e*Sqrt[2 + e*x])

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Maple [A]  time = 0.043, size = 60, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2\,ex-4 \right ) \left ( 1155\,{e}^{4}{x}^{4}+12600\,{e}^{3}{x}^{3}+56840\,{e}^{2}{x}^{2}+133600\,ex+154928 \right ) }{15015\,e} \left ( -3\,{e}^{2}{x}^{2}+12 \right ) ^{{\frac{3}{2}}} \left ( ex+2 \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+2)^(5/2)*(-3*e^2*x^2+12)^(3/2),x)

[Out]

2/15015*(e*x-2)*(1155*e^4*x^4+12600*e^3*x^3+56840*e^2*x^2+133600*e*x+154928)*(-3*e^2*x^2+12)^(3/2)/e/(e*x+2)^(
3/2)

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Maxima [C]  time = 1.77502, size = 126, normalized size = 1.16 \begin{align*} -\frac{{\left (2310 i \, \sqrt{3} e^{6} x^{6} + 15960 i \, \sqrt{3} e^{5} x^{5} + 22120 i \, \sqrt{3} e^{4} x^{4} - 86720 i \, \sqrt{3} e^{3} x^{3} - 304224 i \, \sqrt{3} e^{2} x^{2} - 170624 i \, \sqrt{3} e x + 1239424 i \, \sqrt{3}\right )}{\left (e x + 2\right )} \sqrt{e x - 2}}{5005 \,{\left (e^{2} x + 2 \, e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+2)^(5/2)*(-3*e^2*x^2+12)^(3/2),x, algorithm="maxima")

[Out]

-1/5005*(2310*I*sqrt(3)*e^6*x^6 + 15960*I*sqrt(3)*e^5*x^5 + 22120*I*sqrt(3)*e^4*x^4 - 86720*I*sqrt(3)*e^3*x^3
- 304224*I*sqrt(3)*e^2*x^2 - 170624*I*sqrt(3)*e*x + 1239424*I*sqrt(3))*(e*x + 2)*sqrt(e*x - 2)/(e^2*x + 2*e)

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Fricas [A]  time = 1.84497, size = 215, normalized size = 1.97 \begin{align*} -\frac{2 \,{\left (1155 \, e^{6} x^{6} + 7980 \, e^{5} x^{5} + 11060 \, e^{4} x^{4} - 43360 \, e^{3} x^{3} - 152112 \, e^{2} x^{2} - 85312 \, e x + 619712\right )} \sqrt{-3 \, e^{2} x^{2} + 12} \sqrt{e x + 2}}{5005 \,{\left (e^{2} x + 2 \, e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+2)^(5/2)*(-3*e^2*x^2+12)^(3/2),x, algorithm="fricas")

[Out]

-2/5005*(1155*e^6*x^6 + 7980*e^5*x^5 + 11060*e^4*x^4 - 43360*e^3*x^3 - 152112*e^2*x^2 - 85312*e*x + 619712)*sq
rt(-3*e^2*x^2 + 12)*sqrt(e*x + 2)/(e^2*x + 2*e)

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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((e*x+2)**(5/2)*(-3*e**2*x**2+12)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac{3}{2}}{\left (e x + 2\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+2)^(5/2)*(-3*e^2*x^2+12)^(3/2),x, algorithm="giac")

[Out]

integrate((-3*e^2*x^2 + 12)^(3/2)*(e*x + 2)^(5/2), x)