Integrand size = 28, antiderivative size = 581 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{8/3}} \, dx=\frac {\sqrt [3]{d (b+2 c x)} \sqrt [3]{a+b x+c x^2}}{5 c^2 d^3}-\frac {3 \left (a+b x+c x^2\right )^{4/3}}{10 c d (d (b+2 c x))^{5/3}}-\frac {\sqrt [3]{d (b+2 c x)} \left (b^2-4 a c-(b+2 c x)^2\right ) \left (2 \sqrt [3]{c} d^{2/3}-\frac {\sqrt [3]{2} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right ) \sqrt {\frac {2 \sqrt [3]{2} c^{2/3} d^{4/3}+\frac {(d (b+2 c x))^{4/3}}{\left (a+b x+c x^2\right )^{2/3}}+\frac {2^{2/3} \sqrt [3]{c} d^{2/3} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}}{\left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1+\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1-\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}}{2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1+\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{40 \sqrt [4]{3} c^{10/3} d^{11/3} \left (a+b x+c x^2\right )^{2/3} \sqrt {-\frac {(d (b+2 c x))^{2/3} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {(d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )}{\sqrt [3]{a+b x+c x^2} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1+\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )^2}}} \]
1/5*(d*(2*c*x+b))^(1/3)*(c*x^2+b*x+a)^(1/3)/c^2/d^3-3/10*(c*x^2+b*x+a)^(4/ 3)/c/d/(d*(2*c*x+b))^(5/3)-1/120*(d*(2*c*x+b))^(1/3)*(b^2-4*a*c-(2*c*x+b)^ 2)*(2*c^(1/3)*d^(2/3)-2^(1/3)*(d*(2*c*x+b))^(2/3)/(c*x^2+b*x+a)^(1/3))*((2 ^(2/3)*c^(1/3)*d^(2/3)-(d*(2*c*x+b))^(2/3)*(1-3^(1/2))/(c*x^2+b*x+a)^(1/3) )^2/(2^(2/3)*c^(1/3)*d^(2/3)-(d*(2*c*x+b))^(2/3)*(1+3^(1/2))/(c*x^2+b*x+a) ^(1/3))^2)^(1/2)/(2^(2/3)*c^(1/3)*d^(2/3)-(d*(2*c*x+b))^(2/3)*(1-3^(1/2))/ (c*x^2+b*x+a)^(1/3))*(2^(2/3)*c^(1/3)*d^(2/3)-(d*(2*c*x+b))^(2/3)*(1+3^(1/ 2))/(c*x^2+b*x+a)^(1/3))*EllipticF((1-(2^(2/3)*c^(1/3)*d^(2/3)-(d*(2*c*x+b ))^(2/3)*(1-3^(1/2))/(c*x^2+b*x+a)^(1/3))^2/(2^(2/3)*c^(1/3)*d^(2/3)-(d*(2 *c*x+b))^(2/3)*(1+3^(1/2))/(c*x^2+b*x+a)^(1/3))^2)^(1/2),1/4*6^(1/2)+1/4*2 ^(1/2))*((2*2^(1/3)*c^(2/3)*d^(4/3)+(d*(2*c*x+b))^(4/3)/(c*x^2+b*x+a)^(2/3 )+2^(2/3)*c^(1/3)*d^(2/3)*(d*(2*c*x+b))^(2/3)/(c*x^2+b*x+a)^(1/3))/(2^(2/3 )*c^(1/3)*d^(2/3)-(d*(2*c*x+b))^(2/3)*(1+3^(1/2))/(c*x^2+b*x+a)^(1/3))^2)^ (1/2)*3^(3/4)/c^(10/3)/d^(11/3)/(c*x^2+b*x+a)^(2/3)/(-(d*(2*c*x+b))^(2/3)* (2^(2/3)*c^(1/3)*d^(2/3)-(d*(2*c*x+b))^(2/3)/(c*x^2+b*x+a)^(1/3))/(c*x^2+b *x+a)^(1/3)/(2^(2/3)*c^(1/3)*d^(2/3)-(d*(2*c*x+b))^(2/3)*(1+3^(1/2))/(c*x^ 2+b*x+a)^(1/3))^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.05 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.18 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{8/3}} \, dx=\frac {3 \left (b^2-4 a c\right ) \sqrt [3]{a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},-\frac {5}{6},\frac {1}{6},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{40\ 2^{2/3} c^2 d (d (b+2 c x))^{5/3} \sqrt [3]{\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \]
(3*(b^2 - 4*a*c)*(a + x*(b + c*x))^(1/3)*Hypergeometric2F1[-4/3, -5/6, 1/6 , (b + 2*c*x)^2/(b^2 - 4*a*c)])/(40*2^(2/3)*c^2*d*(d*(b + 2*c*x))^(5/3)*(( c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(1/3))
Time = 0.49 (sec) , antiderivative size = 563, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1118, 27, 247, 248, 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.
| \(\displaystyle \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{8/3}} \, dx\) |
| \(\Big \downarrow \) 1118 |
| \(\displaystyle \frac {\int \frac {\left (-\frac {b^2}{c}+\frac {(b d+2 c x d)^2}{c d^2}+4 a\right )^{4/3}}{4\ 2^{2/3} (b d+2 c x d)^{8/3}}d(b d+2 c x d)}{2 c d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\left (-\frac {b^2}{c}+\frac {(b d+2 c x d)^2}{c d^2}+4 a\right )^{4/3}}{(b d+2 c x d)^{8/3}}d(b d+2 c x d)}{8\ 2^{2/3} c d}\) |
| \(\Big \downarrow \) 247 |
| \(\displaystyle \frac {\frac {8 \int \frac {\sqrt [3]{-\frac {b^2}{c}+\frac {(b d+2 c x d)^2}{c d^2}+4 a}}{(b d+2 c x d)^{2/3}}d(b d+2 c x d)}{5 c d^2}-\frac {3 \left (4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}\right )^{4/3}}{5 (b d+2 c d x)^{5/3}}}{8\ 2^{2/3} c d}\) |
| \(\Big \downarrow \) 248 |
| \(\displaystyle \frac {\frac {8 \left (\frac {2}{3} \left (4 a-\frac {b^2}{c}\right ) \int \frac {1}{(b d+2 c x d)^{2/3} \left (-\frac {b^2}{c}+\frac {(b d+2 c x d)^2}{c d^2}+4 a\right )^{2/3}}d(b d+2 c x d)+\sqrt [3]{b d+2 c d x} \sqrt [3]{4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}}\right )}{5 c d^2}-\frac {3 \left (4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}\right )^{4/3}}{5 (b d+2 c d x)^{5/3}}}{8\ 2^{2/3} c d}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {\frac {8 \left (2 \left (4 a-\frac {b^2}{c}\right ) \int \frac {1}{\left (-\frac {b^2}{c}+\frac {(b d+2 c x d)^2}{c d^2}+4 a\right )^{2/3}}d\sqrt [3]{b d+2 c x d}+\sqrt [3]{b d+2 c d x} \sqrt [3]{4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}}\right )}{5 c d^2}-\frac {3 \left (4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}\right )^{4/3}}{5 (b d+2 c d x)^{5/3}}}{8\ 2^{2/3} c d}\) |
| \(\Big \downarrow \) 771 |
| \(\displaystyle \frac {\frac {8 \left (\frac {2 \left (4 a-\frac {b^2}{c}\right ) \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{c d^2}}}d\frac {\sqrt [3]{b d+2 c x d}}{\sqrt [6]{-\frac {b^2}{c}+\frac {(b d+2 c x d)^2}{c d^2}+4 a}}}{\sqrt {\frac {c d^2 \left (4 a-\frac {b^2}{c}\right )}{c d^2 \left (4 a-\frac {b^2}{c}\right )+(b d+2 c d x)^2}} \sqrt {4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}}}+\sqrt [3]{b d+2 c d x} \sqrt [3]{4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}}\right )}{5 c d^2}-\frac {3 \left (4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}\right )^{4/3}}{5 (b d+2 c d x)^{5/3}}}{8\ 2^{2/3} c d}\) |
| \(\Big \downarrow \) 766 |
| \(\displaystyle \frac {\frac {8 \left (\frac {\left (4 a-\frac {b^2}{c}\right ) \sqrt [3]{b d+2 c d x} \left (\sqrt [3]{c} d^{2/3}-(b d+2 c d x)^{2/3}\right ) \sqrt {\frac {\sqrt [3]{c} d^{2/3} (b d+2 c d x)^{2/3}+(b d+2 c d x)^{4/3}+c^{2/3} d^{4/3}}{\left (\sqrt [3]{c} d^{2/3}-\left (1+\sqrt {3}\right ) (b d+2 c d x)^{2/3}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{c} d^{2/3}-\left (1-\sqrt {3}\right ) (b d+2 c x d)^{2/3}}{\sqrt [3]{c} d^{2/3}-\left (1+\sqrt {3}\right ) (b d+2 c x d)^{2/3}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt [4]{3} \sqrt [3]{c} d^{2/3} \sqrt {-\frac {(b d+2 c d x)^{2/3} \left (\sqrt [3]{c} d^{2/3}-(b d+2 c d x)^{2/3}\right )}{\left (\sqrt [3]{c} d^{2/3}-\left (1+\sqrt {3}\right ) (b d+2 c d x)^{2/3}\right )^2}} \sqrt {1-\frac {(b d+2 c d x)^2}{c d^2}} \sqrt {\frac {c d^2 \left (4 a-\frac {b^2}{c}\right )}{c d^2 \left (4 a-\frac {b^2}{c}\right )+(b d+2 c d x)^2}} \left (4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}\right )^{2/3}}+\sqrt [3]{b d+2 c d x} \sqrt [3]{4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}}\right )}{5 c d^2}-\frac {3 \left (4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}\right )^{4/3}}{5 (b d+2 c d x)^{5/3}}}{8\ 2^{2/3} c d}\) |
((-3*(4*a - b^2/c + (b*d + 2*c*d*x)^2/(c*d^2))^(4/3))/(5*(b*d + 2*c*d*x)^( 5/3)) + (8*((b*d + 2*c*d*x)^(1/3)*(4*a - b^2/c + (b*d + 2*c*d*x)^2/(c*d^2) )^(1/3) + ((4*a - b^2/c)*(b*d + 2*c*d*x)^(1/3)*(c^(1/3)*d^(2/3) - (b*d + 2 *c*d*x)^(2/3))*Sqrt[(c^(2/3)*d^(4/3) + c^(1/3)*d^(2/3)*(b*d + 2*c*d*x)^(2/ 3) + (b*d + 2*c*d*x)^(4/3))/(c^(1/3)*d^(2/3) - (1 + Sqrt[3])*(b*d + 2*c*d* x)^(2/3))^2]*EllipticF[ArcCos[(c^(1/3)*d^(2/3) - (1 - Sqrt[3])*(b*d + 2*c* d*x)^(2/3))/(c^(1/3)*d^(2/3) - (1 + Sqrt[3])*(b*d + 2*c*d*x)^(2/3))], (2 + Sqrt[3])/4])/(3^(1/4)*c^(1/3)*d^(2/3)*Sqrt[-(((b*d + 2*c*d*x)^(2/3)*(c^(1 /3)*d^(2/3) - (b*d + 2*c*d*x)^(2/3)))/(c^(1/3)*d^(2/3) - (1 + Sqrt[3])*(b* d + 2*c*d*x)^(2/3))^2)]*Sqrt[((4*a - b^2/c)*c*d^2)/((4*a - b^2/c)*c*d^2 + (b*d + 2*c*d*x)^2)]*Sqrt[1 - (b*d + 2*c*d*x)^2/(c*d^2)]*(4*a - b^2/c + (b* d + 2*c*d*x)^2/(c*d^2))^(2/3))))/(5*c*d^2))/(8*2^(2/3)*c*d)
3.15.17.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] :> 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] :> 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[((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]
\[\int \frac {\left (c \,x^{2}+b x +a \right )^{\frac {4}{3}}}{\left (2 c d x +b d \right )^{\frac {8}{3}}}d x\]
\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{8/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {8}{3}}} \,d x } \]
integral((2*c*d*x + b*d)^(1/3)*(c*x^2 + b*x + a)^(4/3)/(8*c^3*d^3*x^3 + 12 *b*c^2*d^3*x^2 + 6*b^2*c*d^3*x + b^3*d^3), x)
\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{8/3}} \, dx=\int \frac {\left (a + b x + c x^{2}\right )^{\frac {4}{3}}}{\left (d \left (b + 2 c x\right )\right )^{\frac {8}{3}}}\, dx \]
\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{8/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {8}{3}}} \,d x } \]
\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{8/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {8}{3}}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{8/3}} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{4/3}}{{\left (b\,d+2\,c\,d\,x\right )}^{8/3}} \,d x \]