\(\int \frac {7}{20} \sqrt [5]{-2 e^3 x^2+3 e^6 x^2} \, dx\) [7561]

Optimal result

Integrand size = 25, antiderivative size = 26 \[ \int \frac {7}{20} \sqrt [5]{-2 e^3 x^2+3 e^6 x^2} \, dx=\frac {1}{4} e^{3/5} x \sqrt [5]{x \left (x+3 \left (-1+e^3\right ) x\right )} \]

[Out]

1/4*x*(x*exp(3)*(3*(exp(3)-1)*x+x))^(1/5)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6, 12, 15, 30} \[ \int \frac {7}{20} \sqrt [5]{-2 e^3 x^2+3 e^6 x^2} \, dx=\frac {1}{4} \sqrt [5]{3 e^6-2 e^3} x \sqrt [5]{x^2} \]

[In]

Int[(7*(-2*E^3*x^2 + 3*E^6*x^2)^(1/5))/20,x]

[Out]

((-2*E^3 + 3*E^6)^(1/5)*x*(x^2)^(1/5))/4

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \int \frac {7}{20} \sqrt [5]{-2 e^3+3 e^6} \sqrt [5]{x^2} \, dx \\ & = \frac {1}{20} \left (7 \sqrt [5]{-2 e^3+3 e^6}\right ) \int \sqrt [5]{x^2} \, dx \\ & = \frac {\left (7 \sqrt [5]{-2 e^3+3 e^6} \sqrt [5]{x^2}\right ) \int x^{2/5} \, dx}{20 x^{2/5}} \\ & = \frac {1}{4} \sqrt [5]{-2 e^3+3 e^6} x \sqrt [5]{x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {7}{20} \sqrt [5]{-2 e^3 x^2+3 e^6 x^2} \, dx=\frac {1}{4} x \sqrt [5]{-2 e^3 x^2+3 e^6 x^2} \]

[In]

Integrate[(7*(-2*E^3*x^2 + 3*E^6*x^2)^(1/5))/20,x]

[Out]

(x*(-2*E^3*x^2 + 3*E^6*x^2)^(1/5))/4

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69

method	result	size
risch	\(\frac {\left (x^{2} {\mathrm e}^{3} \left (3 \,{\mathrm e}^{3}-2\right )\right )^{\frac {1}{5}} x}{4}\)	\(18\)
gosper	\(\frac {x \left (3 x^{2} {\mathrm e}^{6}-2 x^{2} {\mathrm e}^{3}\right )^{\frac {1}{5}}}{4}\)	\(23\)
trager	\(\frac {x \left (3 x^{2} {\mathrm e}^{6}-2 x^{2} {\mathrm e}^{3}\right )^{\frac {1}{5}}}{4}\)	\(23\)

[In]

int(7/20*(3*x^2*exp(3)^2-2*x^2*exp(3))^(1/5),x,method=_RETURNVERBOSE)

[Out]

1/4*(x^2*exp(3)*(3*exp(3)-2))^(1/5)*x

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {7}{20} \sqrt [5]{-2 e^3 x^2+3 e^6 x^2} \, dx=\frac {1}{4} \, {\left (3 \, x^{2} e^{6} - 2 \, x^{2} e^{3}\right )}^{\frac {1}{5}} x \]

[In]

integrate(7/20*(3*x^2*exp(3)^2-2*x^2*exp(3))^(1/5),x, algorithm="fricas")

[Out]

1/4*(3*x^2*e^6 - 2*x^2*e^3)^(1/5)*x

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (22) = 44\).

Time = 1.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.08 \[ \int \frac {7}{20} \sqrt [5]{-2 e^3 x^2+3 e^6 x^2} \, dx=- \frac {x^{3} e^{\frac {3}{5}}}{2 \left (-2 + 3 e^{3}\right )^{\frac {4}{5}} \left (x^{2}\right )^{\frac {4}{5}}} + \frac {3 x^{3} e^{\frac {18}{5}}}{4 \left (-2 + 3 e^{3}\right )^{\frac {4}{5}} \left (x^{2}\right )^{\frac {4}{5}}} \]

[In]

integrate(7/20*(3*x**2*exp(3)**2-2*x**2*exp(3))**(1/5),x)

[Out]

-x**3*exp(3/5)/(2*(-2 + 3*exp(3))**(4/5)*(x**2)**(4/5)) + 3*x**3*exp(18/5)/(4*(-2 + 3*exp(3))**(4/5)*(x**2)**(
4/5))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.58 \[ \int \frac {7}{20} \sqrt [5]{-2 e^3 x^2+3 e^6 x^2} \, dx=\frac {1}{4} \, x^{\frac {7}{5}} {\left (3 \, e^{3} - 2\right )}^{\frac {1}{5}} e^{\frac {3}{5}} \]

[In]

integrate(7/20*(3*x^2*exp(3)^2-2*x^2*exp(3))^(1/5),x, algorithm="maxima")

[Out]

1/4*x^(7/5)*(3*e^3 - 2)^(1/5)*e^(3/5)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {7}{20} \sqrt [5]{-2 e^3 x^2+3 e^6 x^2} \, dx=\frac {1}{4} \, {\left (3 \, x^{2} e^{6} - 2 \, x^{2} e^{3}\right )}^{\frac {1}{5}} x \]

[In]

integrate(7/20*(3*x^2*exp(3)^2-2*x^2*exp(3))^(1/5),x, algorithm="giac")

[Out]

1/4*(3*x^2*e^6 - 2*x^2*e^3)^(1/5)*x

Mupad [B] (verification not implemented)

Time = 13.57 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {7}{20} \sqrt [5]{-2 e^3 x^2+3 e^6 x^2} \, dx=\frac {x\,{\left ({\mathrm {e}}^3\right )}^{1/5}\,{\left (3\,{\mathrm {e}}^3-2\right )}^{1/5}\,{\left (x^2\right )}^{1/5}}{4} \]

[In]

int((7*(3*x^2*exp(6) - 2*x^2*exp(3))^(1/5))/20,x)

[Out]

(x*exp(3)^(1/5)*(3*exp(3) - 2)^(1/5)*(x^2)^(1/5))/4