Integrand size = 32, antiderivative size = 65 \[ \int \frac {\sqrt {x^2} \left (2-6 x^2\right )^p}{\sqrt {2 x^2-3 x^4}} \, dx=-\frac {2^{-1+p} \sqrt {x^2} \left (1-3 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+p}{2},\frac {3+p}{2},\left (1-3 x^2\right )^2\right )}{\sqrt {3} (1+p) x} \] Output:
-1/3*2^(-1+p)*(x^2)^(1/2)*(-3*x^2+1)^(p+1)*hypergeom([1/2, 1/2*p+1/2],[3/2 +1/2*p],(-3*x^2+1)^2)*3^(1/2)/(p+1)/x
Time = 0.60 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.05 \[ \int \frac {\sqrt {x^2} \left (2-6 x^2\right )^p}{\sqrt {2 x^2-3 x^4}} \, dx=-\frac {\sqrt {x^2} \left (2-6 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+p}{2},1+\frac {1+p}{2},\frac {1}{4} \left (2-6 x^2\right )^2\right )}{4 \sqrt {3} (1+p) x} \] Input:
Integrate[(Sqrt[x^2]*(2 - 6*x^2)^p)/Sqrt[2*x^2 - 3*x^4],x]
Output:
-1/4*(Sqrt[x^2]*(2 - 6*x^2)^(1 + p)*Hypergeometric2F1[1/2, (1 + p)/2, 1 + (1 + p)/2, (2 - 6*x^2)^2/4])/(Sqrt[3]*(1 + p)*x)
Time = 0.21 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {34, 1940, 1118, 27, 278}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sqrt {x^2} \left (2-6 x^2\right )^p}{\sqrt {2 x^2-3 x^4}} \, dx\) |
\(\Big \downarrow \) 34 |
\(\displaystyle \frac {\sqrt {x^2} \int \frac {x \left (2-6 x^2\right )^p}{\sqrt {2 x^2-3 x^4}}dx}{x}\) |
\(\Big \downarrow \) 1940 |
\(\displaystyle \frac {\sqrt {x^2} \int \frac {\left (2-6 x^2\right )^p}{\sqrt {2 x^2-3 x^4}}dx^2}{2 x}\) |
\(\Big \downarrow \) 1118 |
\(\displaystyle -\frac {\sqrt {x^2} \int \frac {2 \sqrt {3} \left (2-6 x^2\right )^p}{\sqrt {4-x^4}}d\left (2-6 x^2\right )}{12 x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\sqrt {x^2} \int \frac {\left (2-6 x^2\right )^p}{\sqrt {4-x^4}}d\left (2-6 x^2\right )}{2 \sqrt {3} x}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle -\frac {\sqrt {x^2} \left (2-6 x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {p+1}{2},\frac {p+3}{2},\frac {x^4}{4}\right )}{4 \sqrt {3} (p+1) x}\) |
Input:
Int[(Sqrt[x^2]*(2 - 6*x^2)^p)/Sqrt[2*x^2 - 3*x^4],x]
Output:
-1/4*(Sqrt[x^2]*(2 - 6*x^2)^(1 + p)*Hypergeometric2F1[1/2, (1 + p)/2, (3 + p)/2, x^4/4])/(Sqrt[3]*(1 + p)*x)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_.)*((a_.)*(x_)^(m_))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a*x^m)^F racPart[p]/x^(m*FracPart[p])) Int[u*x^(m*p), x], x] /; FreeQ[{a, m, p}, x ] && !IntegerQ[p]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[1/e Subst[Int[x^m*(a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[2*c*d - b*e, 0]
Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_) ^(n_))^(q_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1) *(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] && !IntegerQ[p] && NeQ[k, j] && I ntegerQ[Simplify[j/n]] && IntegerQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]
\[\int \frac {\sqrt {x^{2}}\, \left (-6 x^{2}+2\right )^{p}}{\sqrt {-3 x^{4}+2 x^{2}}}d x\]
Input:
int((x^2)^(1/2)*(-6*x^2+2)^p/(-3*x^4+2*x^2)^(1/2),x)
Output:
int((x^2)^(1/2)*(-6*x^2+2)^p/(-3*x^4+2*x^2)^(1/2),x)
\[ \int \frac {\sqrt {x^2} \left (2-6 x^2\right )^p}{\sqrt {2 x^2-3 x^4}} \, dx=\int { \frac {{\left (-6 \, x^{2} + 2\right )}^{p} \sqrt {x^{2}}}{\sqrt {-3 \, x^{4} + 2 \, x^{2}}} \,d x } \] Input:
integrate((x^2)^(1/2)*(-6*x^2+2)^p/(-3*x^4+2*x^2)^(1/2),x, algorithm="fric as")
Output:
integral(-sqrt(-3*x^4 + 2*x^2)*(-6*x^2 + 2)^p*sqrt(x^2)/(3*x^4 - 2*x^2), s qrt(x^2))
\[ \int \frac {\sqrt {x^2} \left (2-6 x^2\right )^p}{\sqrt {2 x^2-3 x^4}} \, dx=\int \frac {\left (2 - 6 x^{2}\right )^{p} \sqrt {x^{2}}}{\sqrt {- x^{2} \cdot \left (3 x^{2} - 2\right )}}\, dx \] Input:
integrate((x**2)**(1/2)*(-6*x**2+2)**p/(-3*x**4+2*x**2)**(1/2),x)
Output:
Integral((2 - 6*x**2)**p*sqrt(x**2)/sqrt(-x**2*(3*x**2 - 2)), x)
\[ \int \frac {\sqrt {x^2} \left (2-6 x^2\right )^p}{\sqrt {2 x^2-3 x^4}} \, dx=\int { \frac {{\left (-6 \, x^{2} + 2\right )}^{p} \sqrt {x^{2}}}{\sqrt {-3 \, x^{4} + 2 \, x^{2}}} \,d x } \] Input:
integrate((x^2)^(1/2)*(-6*x^2+2)^p/(-3*x^4+2*x^2)^(1/2),x, algorithm="maxi ma")
Output:
integrate((-6*x^2 + 2)^p*sqrt(x^2)/sqrt(-3*x^4 + 2*x^2), x)
\[ \int \frac {\sqrt {x^2} \left (2-6 x^2\right )^p}{\sqrt {2 x^2-3 x^4}} \, dx=\int { \frac {{\left (-6 \, x^{2} + 2\right )}^{p} \sqrt {x^{2}}}{\sqrt {-3 \, x^{4} + 2 \, x^{2}}} \,d x } \] Input:
integrate((x^2)^(1/2)*(-6*x^2+2)^p/(-3*x^4+2*x^2)^(1/2),x, algorithm="giac ")
Output:
integrate((-6*x^2 + 2)^p*sqrt(x^2)/sqrt(-3*x^4 + 2*x^2), x)
Timed out. \[ \int \frac {\sqrt {x^2} \left (2-6 x^2\right )^p}{\sqrt {2 x^2-3 x^4}} \, dx=\int \frac {{\left (2-6\,x^2\right )}^p\,\sqrt {x^2}}{\sqrt {2\,x^2-3\,x^4}} \,d x \] Input:
int(((2 - 6*x^2)^p*(x^2)^(1/2))/(2*x^2 - 3*x^4)^(1/2),x)
Output:
int(((2 - 6*x^2)^p*(x^2)^(1/2))/(2*x^2 - 3*x^4)^(1/2), x)
\[ \int \frac {\sqrt {x^2} \left (2-6 x^2\right )^p}{\sqrt {2 x^2-3 x^4}} \, dx=\int \frac {\left (-6 x^{2}+2\right )^{p}}{\sqrt {-3 x^{2}+2}}d x \] Input:
int((x^2)^(1/2)*(-6*x^2+2)^p/(-3*x^4+2*x^2)^(1/2),x)
Output:
int(( - 6*x**2 + 2)**p/sqrt( - 3*x**2 + 2),x)