3.906 \(\int \frac {(12-3 e^2 x^2)^{3/2}}{(2+e x)^{5/2}} \, dx\)
Optimal. Leaf size=66 \[ \frac {2 \sqrt {3} (2-e x)^{3/2}}{e}+\frac {24 \sqrt {3} \sqrt {2-e x}}{e}-\frac {48 \sqrt {3} \tanh ^{-1}\left (\frac {1}{2} \sqrt {2-e x}\right )}{e} \]
[Out]
2*(-e*x+2)^(3/2)/e*3^(1/2)-48*arctanh(1/2*(-e*x+2)^(1/2))*3^(1/2)/e+24*3^(1/2)*(-e*x+2)^(1/2)/e
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Rubi [A] time = 0.03, antiderivative size = 66, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used =
{627, 50, 63, 206} \[ \frac {2 \sqrt {3} (2-e x)^{3/2}}{e}+\frac {24 \sqrt {3} \sqrt {2-e x}}{e}-\frac {48 \sqrt {3} \tanh ^{-1}\left (\frac {1}{2} \sqrt {2-e x}\right )}{e} \]
Antiderivative was successfully verified.
[In]
Int[(12 - 3*e^2*x^2)^(3/2)/(2 + e*x)^(5/2),x]
[Out]
(24*Sqrt[3]*Sqrt[2 - e*x])/e + (2*Sqrt[3]*(2 - e*x)^(3/2))/e - (48*Sqrt[3]*ArcTanh[Sqrt[2 - e*x]/2])/e
Rule 50
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]
Rule 63
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
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]))
Rubi steps
\begin {align*} \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{5/2}} \, dx &=\int \frac {(6-3 e x)^{3/2}}{2+e x} \, dx\\ &=\frac {2 \sqrt {3} (2-e x)^{3/2}}{e}+12 \int \frac {\sqrt {6-3 e x}}{2+e x} \, dx\\ &=\frac {24 \sqrt {3} \sqrt {2-e x}}{e}+\frac {2 \sqrt {3} (2-e x)^{3/2}}{e}+144 \int \frac {1}{\sqrt {6-3 e x} (2+e x)} \, dx\\ &=\frac {24 \sqrt {3} \sqrt {2-e x}}{e}+\frac {2 \sqrt {3} (2-e x)^{3/2}}{e}-\frac {96 \operatorname {Subst}\left (\int \frac {1}{4-\frac {x^2}{3}} \, dx,x,\sqrt {6-3 e x}\right )}{e}\\ &=\frac {24 \sqrt {3} \sqrt {2-e x}}{e}+\frac {2 \sqrt {3} (2-e x)^{3/2}}{e}-\frac {48 \sqrt {3} \tanh ^{-1}\left (\frac {1}{2} \sqrt {2-e x}\right )}{e}\\ \end {align*}
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Mathematica [A] time = 0.07, size = 69, normalized size = 1.05 \[ -\frac {2 \sqrt {12-3 e^2 x^2} \left (\sqrt {e x-2} (e x-14)+24 \tan ^{-1}\left (\frac {1}{2} \sqrt {e x-2}\right )\right )}{e \sqrt {e x-2} \sqrt {e x+2}} \]
Antiderivative was successfully verified.
[In]
Integrate[(12 - 3*e^2*x^2)^(3/2)/(2 + e*x)^(5/2),x]
[Out]
(-2*Sqrt[12 - 3*e^2*x^2]*((-14 + e*x)*Sqrt[-2 + e*x] + 24*ArcTan[Sqrt[-2 + e*x]/2]))/(e*Sqrt[-2 + e*x]*Sqrt[2
+ e*x])
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fricas [B] time = 0.93, size = 106, normalized size = 1.61 \[ \frac {2 \, {\left (12 \, \sqrt {3} {\left (e x + 2\right )} \log \left (-\frac {3 \, e^{2} x^{2} - 12 \, e x + 4 \, \sqrt {3} \sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2} - 36}{e^{2} x^{2} + 4 \, e x + 4}\right ) - \sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2} {\left (e x - 14\right )}\right )}}{e^{2} x + 2 \, e} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-3*e^2*x^2+12)^(3/2)/(e*x+2)^(5/2),x, algorithm="fricas")
[Out]
2*(12*sqrt(3)*(e*x + 2)*log(-(3*e^2*x^2 - 12*e*x + 4*sqrt(3)*sqrt(-3*e^2*x^2 + 12)*sqrt(e*x + 2) - 36)/(e^2*x^
2 + 4*e*x + 4)) - sqrt(-3*e^2*x^2 + 12)*sqrt(e*x + 2)*(e*x - 14))/(e^2*x + 2*e)
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-3*e^2*x^2+12)^(3/2)/(e*x+2)^(5/2),x, algorithm="giac")
[Out]
Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Warning, need to
choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was done assuming
[exp(1),exp(2)]=[-55,-49]Precision problem choosing root in common_EXT, current precision 14Warning, need to
choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was done assuming
[exp(1),exp(2)]=[-96,-85]Warning, need to choose a branch for the root of a polynomial with parameters. This
might be wrong.The choice was done assuming [exp(1),exp(2)]=[-7,4]Error: Bad Argument ValueError: Bad Argument
ValueError: Bad Argument ValueLimit probably undefined, algorithm unable to handle rootof([[1,2*t^2-2*exp(2),
-3*t^4+8*t^2*exp(2)-5*exp(2)^2,-18*t^6+34*t^4*exp(2)-14*t^2*exp(2)^2-2*exp(2)^3],[1,0,-6*t^4+16*t^2*exp(2)-10*
exp(2)^2,-16*t^6+16*t^4*exp(2)+16*t^2*exp(2)^2-16*exp(2)^3,41*t^8-112*t^6*exp(2)+94*t^4*exp(2)^2-16*t^2*exp(2)
^3-7*exp(2)^4]])Error: Bad Argument ValueError: Bad Argument ValueError: Bad Argument ValueLimit probably unde
fined, algorithm unable to handle rootof([[-1,-2*t^2+2*exp(2),3*t^4-8*t^2*exp(2)+5*exp(2)^2,18*t^6-34*t^4*exp(
2)+14*t^2*exp(2)^2+2*exp(2)^3],[1,0,-6*t^4+16*t^2*exp(2)-10*exp(2)^2,-16*t^6+16*t^4*exp(2)+16*t^2*exp(2)^2-16*
exp(2)^3,41*t^8-112*t^6*exp(2)+94*t^4*exp(2)^2-16*t^2*exp(2)^3-7*exp(2)^4]])Error: Bad Argument ValueError: Ba
d Argument ValueError: Bad Argument ValueLimit probably undefined, algorithm unable to handle rootof([[1,2*t^2
-2*exp(2),-3*t^4+8*t^2*exp(2)-5*exp(2)^2,-18*t^6+34*t^4*exp(2)-14*t^2*exp(2)^2-2*exp(2)^3],[1,0,-6*t^4+16*t^2*
exp(2)-10*exp(2)^2,-16*t^6+16*t^4*exp(2)+16*t^2*exp(2)^2-16*exp(2)^3,41*t^8-112*t^6*exp(2)+94*t^4*exp(2)^2-16*
t^2*exp(2)^3-7*exp(2)^4]])Error: Bad Argument ValueError: Bad Argument ValueError: Bad Argument ValueLimit pro
bably undefined, algorithm unable to handle rootof([[-1,-2*t^2+2*exp(2),3*t^4-8*t^2*exp(2)+5*exp(2)^2,18*t^6-3
4*t^4*exp(2)+14*t^2*exp(2)^2+2*exp(2)^3],[1,0,-6*t^4+16*t^2*exp(2)-10*exp(2)^2,-16*t^6+16*t^4*exp(2)+16*t^2*ex
p(2)^2-16*exp(2)^3,41*t^8-112*t^6*exp(2)+94*t^4*exp(2)^2-16*t^2*exp(2)^3-7*exp(2)^4]])Error: Bad Argument Valu
eError: Bad Argument ValueError: Bad Argument ValueLimit probably undefined, algorithm unable to handle rootof
([[1,2*t^2-2*exp(2),-3*t^4+8*t^2*exp(2)-5*exp(2)^2,-18*t^6+34*t^4*exp(2)-14*t^2*exp(2)^2-2*exp(2)^3],[1,0,-6*t
^4+16*t^2*exp(2)-10*exp(2)^2,-16*t^6+16*t^4*exp(2)+16*t^2*exp(2)^2-16*exp(2)^3,41*t^8-112*t^6*exp(2)+94*t^4*ex
p(2)^2-16*t^2*exp(2)^3-7*exp(2)^4]])Error: Bad Argument ValueError: Bad Argument ValueError: Bad Argument Valu
eLimit probably undefined, algorithm unable to handle rootof([[-1,-2*t^2+2*exp(2),3*t^4-8*t^2*exp(2)+5*exp(2)^
2,18*t^6-34*t^4*exp(2)+14*t^2*exp(2)^2+2*exp(2)^3],[1,0,-6*t^4+16*t^2*exp(2)-10*exp(2)^2,-16*t^6+16*t^4*exp(2)
+16*t^2*exp(2)^2-16*exp(2)^3,41*t^8-112*t^6*exp(2)+94*t^4*exp(2)^2-16*t^2*exp(2)^3-7*exp(2)^4]])Error: Bad Arg
ument ValueError: Bad Argument ValueError: Bad Argument ValueLimit probably undefined, algorithm unable to han
dle rootof([[1,2*t^2-2*exp(2),-3*t^4+8*t^2*exp(2)-5*exp(2)^2,-18*t^6+34*t^4*exp(2)-14*t^2*exp(2)^2-2*exp(2)^3]
,[1,0,-6*t^4+16*t^2*exp(2)-10*exp(2)^2,-16*t^6+16*t^4*exp(2)+16*t^2*exp(2)^2-16*exp(2)^3,41*t^8-112*t^6*exp(2)
+94*t^4*exp(2)^2-16*t^2*exp(2)^3-7*exp(2)^4]])Error: Bad Argument ValueError: Bad Argument ValueError: Bad Arg
ument ValueLimit probably undefined, algorithm unable to handle rootof([[-1,-2*t^2+2*exp(2),3*t^4-8*t^2*exp(2)
+5*exp(2)^2,18*t^6-34*t^4*exp(2)+14*t^2*exp(2)^2+2*exp(2)^3],[1,0,-6*t^4+16*t^2*exp(2)-10*exp(2)^2,-16*t^6+16*
t^4*exp(2)+16*t^2*exp(2)^2-16*exp(2)^3,41*t^8-112*t^6*exp(2)+94*t^4*exp(2)^2-16*t^2*exp(2)^3-7*exp(2)^4]])Erro
r: Bad Argument ValueError: Bad Argument ValueError: Bad Argument ValueLimit probably undefined, algorithm una
ble to handle rootof([[1,2*t^2-2*exp(2),-3*t^4+8*t^2*exp(2)-5*exp(2)^2,-18*t^6+34*t^4*exp(2)-14*t^2*exp(2)^2-2
*exp(2)^3],[1,0,-6*t^4+16*t^2*exp(2)-10*exp(2)^2,-16*t^6+16*t^4*exp(2)+16*t^2*exp(2)^2-16*exp(2)^3,41*t^8-112*
t^6*exp(2)+94*t^4*exp(2)^2-16*t^2*exp(2)^3-7*exp(2)^4]])Error: Bad Argument ValueError: Bad Argument ValueErro
r: Bad Argument ValueLimit probably undefined, algorithm unable to handle rootof([[-1,-2*t^2+2*exp(2),3*t^4-8*
t^2*exp(2)+5*exp(2)^2,18*t^6-34*t^4*exp(2)+14*t^2*exp(2)^2+2*exp(2)^3],[1,0,-6*t^4+16*t^2*exp(2)-10*exp(2)^2,-
16*t^6+16*t^4*exp(2)+16*t^2*exp(2)^2-16*exp(2)^3,41*t^8-112*t^6*exp(2)+94*t^4*exp(2)^2-16*t^2*exp(2)^3-7*exp(2
)^4]])Evaluation time: 1.68-sqrt(3)/exp(1)/2*exp(1)/2/exp(1)^2*C_0*sqrt(4/(x*exp(1)+2)^2*exp(1)^2-4/(x*exp(1)+
2)^2*exp(2)+4/(x*exp(1)+2)*exp(2)-exp(2))+integrate(-sqrt(3)/exp(1)/2*exp(1)*(2*C_0/sqrt(x*exp(1)+2)+((416*exp
(1)^10-704*exp(1)^8*exp(2)-96*exp(1)^6*exp(2)^2+384*exp(1)^2*exp(2)^4)/(x*exp(1)+2)^2+(-232*exp(1)^10+328*exp(
1)^8*exp(2)-96*exp(1)^4*exp(2)^3-192*exp(1)^2*exp(2)^4)/(x*exp(1)+2)+64*exp(1)^10)/((8*exp(1)^10-8*exp(1)^8*ex
p(2))/(x*exp(1)+2)^3+(-4*exp(1)^10+4*exp(1)^8*exp(2))/(x*exp(1)+2)^2+exp(1)^10/(x*exp(1)+2)))/sqrt(4/(x*exp(1)
+2)^2*exp(1)^2-4/(x*exp(1)+2)^2*exp(2)+4/(x*exp(1)+2)*exp(2)-exp(2))/sqrt(x*exp(1)+2)/sqrt(x*exp(1)+2)/2*exp(1
)*(sqrt(x*exp(1)+2))^-1,x)-2*sqrt(3)/exp(1)^2*sqrt(x*exp(1)+2)*sqrt(-(x*exp(1)+2)^2*exp(2)+4*(x*exp(1)+2)*exp(
2)+4*exp(1)^2-4*exp(2))
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maple [A] time = 0.06, size = 77, normalized size = 1.17 \[ -\frac {2 \sqrt {-e^{2} x^{2}+4}\, \left (\sqrt {-3 e x +6}\, e x +24 \sqrt {3}\, \arctanh \left (\frac {\sqrt {3}\, \sqrt {-3 e x +6}}{6}\right )-14 \sqrt {-3 e x +6}\right ) \sqrt {3}}{\sqrt {e x +2}\, \sqrt {-3 e x +6}\, e} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((-3*e^2*x^2+12)^(3/2)/(e*x+2)^(5/2),x)
[Out]
-2*(-e^2*x^2+4)^(1/2)*((-3*e*x+6)^(1/2)*e*x+24*3^(1/2)*arctanh(1/6*3^(1/2)*(-3*e*x+6)^(1/2))-14*(-3*e*x+6)^(1/
2))*3^(1/2)/(e*x+2)^(1/2)/(-3*e*x+6)^(1/2)/e
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {3}{2}}}{{\left (e x + 2\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-3*e^2*x^2+12)^(3/2)/(e*x+2)^(5/2),x, algorithm="maxima")
[Out]
integrate((-3*e^2*x^2 + 12)^(3/2)/(e*x + 2)^(5/2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (12-3\,e^2\,x^2\right )}^{3/2}}{{\left (e\,x+2\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((12 - 3*e^2*x^2)^(3/2)/(e*x + 2)^(5/2),x)
[Out]
int((12 - 3*e^2*x^2)^(3/2)/(e*x + 2)^(5/2), x)
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-3*e**2*x**2+12)**(3/2)/(e*x+2)**(5/2),x)
[Out]
Timed out
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