Integrand size = 11, antiderivative size = 285 \[ \int \left (a+b x^2\right )^{4/3} \, dx=\frac {24}{55} a x \sqrt [3]{a+b x^2}+\frac {3}{11} x \left (a+b x^2\right )^{4/3}-\frac {16\ 3^{3/4} \sqrt {2-\sqrt {3}} a^2 \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right ),-7+4 \sqrt {3}\right )}{55 b x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}} \] Output:
24/55*a*x*(b*x^2+a)^(1/3)+3/11*x*(b*x^2+a)^(4/3)-16/55*3^(3/4)*(1/2*6^(1/2 )-1/2*2^(1/2))*a^2*(a^(1/3)-(b*x^2+a)^(1/3))*((a^(2/3)+a^(1/3)*(b*x^2+a)^( 1/3)+(b*x^2+a)^(2/3))/((1-3^(1/2))*a^(1/3)-(b*x^2+a)^(1/3))^2)^(1/2)*Ellip ticF(((1+3^(1/2))*a^(1/3)-(b*x^2+a)^(1/3))/((1-3^(1/2))*a^(1/3)-(b*x^2+a)^ (1/3)),2*I-I*3^(1/2))/b/x/(-a^(1/3)*(a^(1/3)-(b*x^2+a)^(1/3))/((1-3^(1/2)) *a^(1/3)-(b*x^2+a)^(1/3))^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.00 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.16 \[ \int \left (a+b x^2\right )^{4/3} \, dx=\frac {a x \sqrt [3]{a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )}{\sqrt [3]{1+\frac {b x^2}{a}}} \] Input:
Integrate[(a + b*x^2)^(4/3),x]
Output:
(a*x*(a + b*x^2)^(1/3)*Hypergeometric2F1[-4/3, 1/2, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^(1/3)
Time = 0.26 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {211, 211, 234, 760}
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.
\(\displaystyle \int \left (a+b x^2\right )^{4/3} \, dx\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {8}{11} a \int \sqrt [3]{b x^2+a}dx+\frac {3}{11} x \left (a+b x^2\right )^{4/3}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {8}{11} a \left (\frac {2}{5} a \int \frac {1}{\left (b x^2+a\right )^{2/3}}dx+\frac {3}{5} x \sqrt [3]{a+b x^2}\right )+\frac {3}{11} x \left (a+b x^2\right )^{4/3}\) |
\(\Big \downarrow \) 234 |
\(\displaystyle \frac {8}{11} a \left (\frac {3 a \sqrt {b x^2} \int \frac {1}{\sqrt {b x^2}}d\sqrt [3]{b x^2+a}}{5 b x}+\frac {3}{5} x \sqrt [3]{a+b x^2}\right )+\frac {3}{11} x \left (a+b x^2\right )^{4/3}\) |
\(\Big \downarrow \) 760 |
\(\displaystyle \frac {8}{11} a \left (\frac {3}{5} x \sqrt [3]{a+b x^2}-\frac {2\ 3^{3/4} \sqrt {2-\sqrt {3}} a \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}\right ),-7+4 \sqrt {3}\right )}{5 b x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}\right )+\frac {3}{11} x \left (a+b x^2\right )^{4/3}\) |
Input:
Int[(a + b*x^2)^(4/3),x]
Output:
(3*x*(a + b*x^2)^(4/3))/11 + (8*a*((3*x*(a + b*x^2)^(1/3))/5 - (2*3^(3/4)* Sqrt[2 - Sqrt[3]]*a*(a^(1/3) - (a + b*x^2)^(1/3))*Sqrt[(a^(2/3) + a^(1/3)* (a + b*x^2)^(1/3) + (a + b*x^2)^(2/3))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2 )^(1/3))^2]*EllipticF[ArcSin[((1 + Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))/( (1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))], -7 + 4*Sqrt[3]])/(5*b*x*Sqrt[ -((a^(1/3)*(a^(1/3) - (a + b*x^2)^(1/3)))/((1 - Sqrt[3])*a^(1/3) - (a + b* x^2)^(1/3))^2)])))/11
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Simp[3*(Sqrt[b*x^2]/(2*b*x)) Subst[Int[1/Sqrt[-a + x^3], x], x, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b }, x]
Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s *x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(- s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 + Sqrt[3]) *s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x ] && NegQ[a]
\[\int \left (b \,x^{2}+a \right )^{\frac {4}{3}}d x\]
Input:
int((b*x^2+a)^(4/3),x)
Output:
int((b*x^2+a)^(4/3),x)
\[ \int \left (a+b x^2\right )^{4/3} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {4}{3}} \,d x } \] Input:
integrate((b*x^2+a)^(4/3),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(4/3), x)
Time = 0.56 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.09 \[ \int \left (a+b x^2\right )^{4/3} \, dx=a^{\frac {4}{3}} x {{}_{2}F_{1}\left (\begin {matrix} - \frac {4}{3}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} \] Input:
integrate((b*x**2+a)**(4/3),x)
Output:
a**(4/3)*x*hyper((-4/3, 1/2), (3/2,), b*x**2*exp_polar(I*pi)/a)
\[ \int \left (a+b x^2\right )^{4/3} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {4}{3}} \,d x } \] Input:
integrate((b*x^2+a)^(4/3),x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^(4/3), x)
\[ \int \left (a+b x^2\right )^{4/3} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {4}{3}} \,d x } \] Input:
integrate((b*x^2+a)^(4/3),x, algorithm="giac")
Output:
integrate((b*x^2 + a)^(4/3), x)
Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.13 \[ \int \left (a+b x^2\right )^{4/3} \, dx=\frac {x\,{\left (b\,x^2+a\right )}^{4/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {4}{3},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{4/3}} \] Input:
int((a + b*x^2)^(4/3),x)
Output:
(x*(a + b*x^2)^(4/3)*hypergeom([-4/3, 1/2], 3/2, -(b*x^2)/a))/((b*x^2)/a + 1)^(4/3)
\[ \int \left (a+b x^2\right )^{4/3} \, dx=\frac {39 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a x}{55}+\frac {3 \left (b \,x^{2}+a \right )^{\frac {1}{3}} b \,x^{3}}{11}+\frac {16 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {2}{3}}}d x \right ) a^{2}}{55} \] Input:
int((b*x^2+a)^(4/3),x)
Output:
(39*(a + b*x**2)**(1/3)*a*x + 15*(a + b*x**2)**(1/3)*b*x**3 + 16*int((a + b*x**2)**(1/3)/(a + b*x**2),x)*a**2)/55