Integrand size = 19, antiderivative size = 419 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{14/3}} \, dx=-\frac {24 b \sqrt [3]{a+b x^2}}{55 c^3 (c x)^{5/3}}-\frac {3 \left (a+b x^2\right )^{4/3}}{11 c (c x)^{11/3}}+\frac {8\ 3^{3/4} b^2 \sqrt [3]{c x} \sqrt [3]{a+b x^2} \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right ) \sqrt {\frac {c^{4/3}+\frac {b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac {\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{\left (c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {c^{2/3}-\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{55 a c^{17/3} \sqrt {-\frac {\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{\sqrt [3]{a+b x^2} \left (c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}}} \]
-24/55*b*(b*x^2+a)^(1/3)/c^3/(c*x)^(5/3)-3/11*(b*x^2+a)^(4/3)/c/(c*x)^(11/ 3)+8/55*3^(3/4)*b^2*(c*x)^(1/3)*(b*x^2+a)^(1/3)*(c^(2/3)-b^(1/3)*(c*x)^(2/ 3)/(b*x^2+a)^(1/3))*((c^(2/3)-b^(1/3)*(c*x)^(2/3)*(1-3^(1/2))/(b*x^2+a)^(1 /3))^2/(c^(2/3)-b^(1/3)*(c*x)^(2/3)*(1+3^(1/2))/(b*x^2+a)^(1/3))^2)^(1/2)/ (c^(2/3)-b^(1/3)*(c*x)^(2/3)*(1-3^(1/2))/(b*x^2+a)^(1/3))*(c^(2/3)-b^(1/3) *(c*x)^(2/3)*(1+3^(1/2))/(b*x^2+a)^(1/3))*EllipticF((1-(c^(2/3)-b^(1/3)*(c *x)^(2/3)*(1-3^(1/2))/(b*x^2+a)^(1/3))^2/(c^(2/3)-b^(1/3)*(c*x)^(2/3)*(1+3 ^(1/2))/(b*x^2+a)^(1/3))^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))*((c^(4/3)+b^(2/ 3)*(c*x)^(4/3)/(b*x^2+a)^(2/3)+b^(1/3)*c^(2/3)*(c*x)^(2/3)/(b*x^2+a)^(1/3) )/(c^(2/3)-b^(1/3)*(c*x)^(2/3)*(1+3^(1/2))/(b*x^2+a)^(1/3))^2)^(1/2)/a/c^( 17/3)/(-b^(1/3)*(c*x)^(2/3)*(c^(2/3)-b^(1/3)*(c*x)^(2/3)/(b*x^2+a)^(1/3))/ (b*x^2+a)^(1/3)/(c^(2/3)-b^(1/3)*(c*x)^(2/3)*(1+3^(1/2))/(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 = 10.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.14 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{14/3}} \, dx=-\frac {3 a x \sqrt [3]{a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {11}{6},-\frac {4}{3},-\frac {5}{6},-\frac {b x^2}{a}\right )}{11 (c x)^{14/3} \sqrt [3]{1+\frac {b x^2}{a}}} \]
(-3*a*x*(a + b*x^2)^(1/3)*Hypergeometric2F1[-11/6, -4/3, -5/6, -((b*x^2)/a )])/(11*(c*x)^(14/3)*(1 + (b*x^2)/a)^(1/3))
Time = 0.36 (sec) , antiderivative size = 360, normalized size of antiderivative = 0.86, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {247, 247, 266, 771, 766}
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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{14/3}} \, dx\) |
\(\Big \downarrow \) 247 |
\(\displaystyle \frac {8 b \int \frac {\sqrt [3]{b x^2+a}}{(c x)^{8/3}}dx}{11 c^2}-\frac {3 \left (a+b x^2\right )^{4/3}}{11 c (c x)^{11/3}}\) |
\(\Big \downarrow \) 247 |
\(\displaystyle \frac {8 b \left (\frac {2 b \int \frac {1}{(c x)^{2/3} \left (b x^2+a\right )^{2/3}}dx}{5 c^2}-\frac {3 \sqrt [3]{a+b x^2}}{5 c (c x)^{5/3}}\right )}{11 c^2}-\frac {3 \left (a+b x^2\right )^{4/3}}{11 c (c x)^{11/3}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {8 b \left (\frac {6 b \int \frac {1}{\left (b x^2+a\right )^{2/3}}d\sqrt [3]{c x}}{5 c^3}-\frac {3 \sqrt [3]{a+b x^2}}{5 c (c x)^{5/3}}\right )}{11 c^2}-\frac {3 \left (a+b x^2\right )^{4/3}}{11 c (c x)^{11/3}}\) |
\(\Big \downarrow \) 771 |
\(\displaystyle \frac {8 b \left (\frac {6 b \int \frac {1}{\sqrt {1-b x^2}}d\frac {\sqrt [3]{c x}}{\sqrt [6]{b x^2+a}}}{5 c^3 \sqrt {a+b x^2} \sqrt {\frac {a c^2}{a c^2+b c^2 x^2}}}-\frac {3 \sqrt [3]{a+b x^2}}{5 c (c x)^{5/3}}\right )}{11 c^2}-\frac {3 \left (a+b x^2\right )^{4/3}}{11 c (c x)^{11/3}}\) |
\(\Big \downarrow \) 766 |
\(\displaystyle \frac {8 b \left (\frac {3^{3/4} b \sqrt [3]{c x} \left (c^{2/3}-\sqrt [3]{b} (c x)^{2/3}\right ) \sqrt {\frac {b^{2/3} (c x)^{4/3}+\sqrt [3]{b} c^{2/3} (c x)^{2/3}+c^{4/3}}{\left (c^{2/3}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {c^{2/3}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{c^{2/3}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{5 c^{11/3} \sqrt {1-b x^2} \left (a+b x^2\right )^{2/3} \sqrt {-\frac {\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\sqrt [3]{b} (c x)^{2/3}\right )}{\left (c^{2/3}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}\right )^2}} \sqrt {\frac {a c^2}{a c^2+b c^2 x^2}}}-\frac {3 \sqrt [3]{a+b x^2}}{5 c (c x)^{5/3}}\right )}{11 c^2}-\frac {3 \left (a+b x^2\right )^{4/3}}{11 c (c x)^{11/3}}\) |
(-3*(a + b*x^2)^(4/3))/(11*c*(c*x)^(11/3)) + (8*b*((-3*(a + b*x^2)^(1/3))/ (5*c*(c*x)^(5/3)) + (3^(3/4)*b*(c*x)^(1/3)*(c^(2/3) - b^(1/3)*(c*x)^(2/3)) *Sqrt[(c^(4/3) + b^(1/3)*c^(2/3)*(c*x)^(2/3) + b^(2/3)*(c*x)^(4/3))/(c^(2/ 3) - (1 + Sqrt[3])*b^(1/3)*(c*x)^(2/3))^2]*EllipticF[ArcCos[(c^(2/3) - (1 - Sqrt[3])*b^(1/3)*(c*x)^(2/3))/(c^(2/3) - (1 + Sqrt[3])*b^(1/3)*(c*x)^(2/ 3))], (2 + Sqrt[3])/4])/(5*c^(11/3)*Sqrt[1 - b*x^2]*(a + b*x^2)^(2/3)*Sqrt [(a*c^2)/(a*c^2 + b*c^2*x^2)]*Sqrt[-((b^(1/3)*(c*x)^(2/3)*(c^(2/3) - b^(1/ 3)*(c*x)^(2/3)))/(c^(2/3) - (1 + Sqrt[3])*b^(1/3)*(c*x)^(2/3))^2)])))/(11* c^2)
3.8.66.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[x*(s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/ (s + (1 + Sqrt[3])*r*x^2)^2]/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[r*x^2*((s + r*x^2)/(s + (1 + Sqrt[3])*r*x^2)^2)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])* r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x ]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a/(a + b*x^n))^(p + 1 /n)*(a + b*x^n)^(p + 1/n) Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x /(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && LtQ[Denominator[p + 1/n], Denominator[p]]
\[\int \frac {\left (b \,x^{2}+a \right )^{\frac {4}{3}}}{\left (c x \right )^{\frac {14}{3}}}d x\]
\[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{14/3}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {4}{3}}}{\left (c x\right )^{\frac {14}{3}}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{14/3}} \, dx=\text {Timed out} \]
\[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{14/3}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {4}{3}}}{\left (c x\right )^{\frac {14}{3}}} \,d x } \]
\[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{14/3}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {4}{3}}}{\left (c x\right )^{\frac {14}{3}}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{14/3}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{4/3}}{{\left (c\,x\right )}^{14/3}} \,d x \]