Integrand size = 15, antiderivative size = 119 \[ \int x^2 \left (a+b x^2\right )^{3/4} \, dx=-\frac {4 a^2 x}{15 b \sqrt [4]{a+b x^2}}+\frac {2 a x \left (a+b x^2\right )^{3/4}}{15 b}+\frac {2}{9} x^3 \left (a+b x^2\right )^{3/4}+\frac {4 a^{5/2} \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{15 b^{3/2} \sqrt [4]{a+b x^2}} \] Output:
-4/15*a^2*x/b/(b*x^2+a)^(1/4)+2/15*a*x*(b*x^2+a)^(3/4)/b+2/9*x^3*(b*x^2+a) ^(3/4)+4/15*a^(5/2)*(1+b*x^2/a)^(1/4)*EllipticE(sin(1/2*arctan(b^(1/2)*x/a ^(1/2))),2^(1/2))/b^(3/2)/(b*x^2+a)^(1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 6.87 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.52 \[ \int x^2 \left (a+b x^2\right )^{3/4} \, dx=\frac {2 x \left (a+b x^2\right )^{3/4} \left (a+b x^2-\frac {a \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )}{\left (1+\frac {b x^2}{a}\right )^{3/4}}\right )}{9 b} \] Input:
Integrate[x^2*(a + b*x^2)^(3/4),x]
Output:
(2*x*(a + b*x^2)^(3/4)*(a + b*x^2 - (a*Hypergeometric2F1[-3/4, 1/2, 3/2, - ((b*x^2)/a)])/(1 + (b*x^2)/a)^(3/4)))/(9*b)
Time = 0.20 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {248, 262, 227, 225, 212}
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.
\(\displaystyle \int x^2 \left (a+b x^2\right )^{3/4} \, dx\) |
\(\Big \downarrow \) 248 |
\(\displaystyle \frac {1}{3} a \int \frac {x^2}{\sqrt [4]{b x^2+a}}dx+\frac {2}{9} x^3 \left (a+b x^2\right )^{3/4}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {1}{3} a \left (\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac {2 a \int \frac {1}{\sqrt [4]{b x^2+a}}dx}{5 b}\right )+\frac {2}{9} x^3 \left (a+b x^2\right )^{3/4}\) |
\(\Big \downarrow \) 227 |
\(\displaystyle \frac {1}{3} a \left (\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^2}{a}+1} \int \frac {1}{\sqrt [4]{\frac {b x^2}{a}+1}}dx}{5 b \sqrt [4]{a+b x^2}}\right )+\frac {2}{9} x^3 \left (a+b x^2\right )^{3/4}\) |
\(\Big \downarrow \) 225 |
\(\displaystyle \frac {1}{3} a \left (\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx\right )}{5 b \sqrt [4]{a+b x^2}}\right )+\frac {2}{9} x^3 \left (a+b x^2\right )^{3/4}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle \frac {1}{3} a \left (\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\frac {2 \sqrt {a} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b}}\right )}{5 b \sqrt [4]{a+b x^2}}\right )+\frac {2}{9} x^3 \left (a+b x^2\right )^{3/4}\) |
Input:
Int[x^2*(a + b*x^2)^(3/4),x]
Output:
(2*x^3*(a + b*x^2)^(3/4))/9 + (a*((2*x*(a + b*x^2)^(3/4))/(5*b) - (2*a*(1 + (b*x^2)/a)^(1/4)*((2*x)/(1 + (b*x^2)/a)^(1/4) - (2*Sqrt[a]*EllipticE[Arc Tan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/Sqrt[b]))/(5*b*(a + b*x^2)^(1/4))))/3
Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) )*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a , 0] && PosQ[b/a]
Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)) , x] - Simp[a Int[1/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b}, x] && GtQ[ a, 0] && PosQ[b/a]
Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(1/4)/( a + b*x^2)^(1/4) Int[1/(1 + b*(x^2/a))^(1/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 2*p + 1))), x] + Simp[2*a*(p/(m + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && GtQ[ p, 0] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
\[\int x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{4}}d x\]
Input:
int(x^2*(b*x^2+a)^(3/4),x)
Output:
int(x^2*(b*x^2+a)^(3/4),x)
\[ \int x^2 \left (a+b x^2\right )^{3/4} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {3}{4}} x^{2} \,d x } \] Input:
integrate(x^2*(b*x^2+a)^(3/4),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(3/4)*x^2, x)
Result contains complex when optimal does not.
Time = 0.54 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.24 \[ \int x^2 \left (a+b x^2\right )^{3/4} \, dx=\frac {a^{\frac {3}{4}} x^{3} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} \] Input:
integrate(x**2*(b*x**2+a)**(3/4),x)
Output:
a**(3/4)*x**3*hyper((-3/4, 3/2), (5/2,), b*x**2*exp_polar(I*pi)/a)/3
\[ \int x^2 \left (a+b x^2\right )^{3/4} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {3}{4}} x^{2} \,d x } \] Input:
integrate(x^2*(b*x^2+a)^(3/4),x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^(3/4)*x^2, x)
\[ \int x^2 \left (a+b x^2\right )^{3/4} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {3}{4}} x^{2} \,d x } \] Input:
integrate(x^2*(b*x^2+a)^(3/4),x, algorithm="giac")
Output:
integrate((b*x^2 + a)^(3/4)*x^2, x)
Timed out. \[ \int x^2 \left (a+b x^2\right )^{3/4} \, dx=\int x^2\,{\left (b\,x^2+a\right )}^{3/4} \,d x \] Input:
int(x^2*(a + b*x^2)^(3/4),x)
Output:
int(x^2*(a + b*x^2)^(3/4), x)
\[ \int x^2 \left (a+b x^2\right )^{3/4} \, dx=\frac {\frac {2 \left (b \,x^{2}+a \right )^{\frac {3}{4}} a x}{15}+\frac {2 \left (b \,x^{2}+a \right )^{\frac {3}{4}} b \,x^{3}}{9}-\frac {2 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}}d x \right ) a^{2}}{15}}{b} \] Input:
int(x^2*(b*x^2+a)^(3/4),x)
Output:
(2*(3*(a + b*x**2)**(3/4)*a*x + 5*(a + b*x**2)**(3/4)*b*x**3 - 3*int((a + b*x**2)**(3/4)/(a + b*x**2),x)*a**2))/(45*b)