Integrand size = 28, antiderivative size = 99 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx=\frac {3 \left (b^2-4 a c\right ) \sqrt [3]{a+b x+c x^2} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},-\frac {7}{6},-\frac {1}{6},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{56 c^2 d (d (b+2 c x))^{7/3} \sqrt [3]{1-\frac {(b+2 c x)^2}{b^2-4 a c}}} \]
3/56*(-4*a*c+b^2)*(c*x^2+b*x+a)^(1/3)*hypergeom([-4/3, -7/6],[-1/6],(2*c*x +b)^2/(-4*a*c+b^2))/c^2/d/(d*(2*c*x+b))^(7/3)/(1-(2*c*x+b)^2/(-4*a*c+b^2)) ^(1/3)
Time = 10.06 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx=\frac {3 \left (b^2-4 a c\right ) \sqrt [3]{a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},-\frac {7}{6},-\frac {1}{6},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{56\ 2^{2/3} c^2 d (d (b+2 c x))^{7/3} \sqrt [3]{\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \]
(3*(b^2 - 4*a*c)*(a + x*(b + c*x))^(1/3)*Hypergeometric2F1[-4/3, -7/6, -1/ 6, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(56*2^(2/3)*c^2*d*(d*(b + 2*c*x))^(7/3)*( (c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(1/3))
Time = 0.24 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.42, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1118, 27, 279, 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.
\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx\) |
\(\Big \downarrow \) 1118 |
\(\displaystyle \frac {\int \frac {\left (-\frac {b^2}{c}+\frac {(b d+2 c x d)^2}{c d^2}+4 a\right )^{4/3}}{4\ 2^{2/3} (b d+2 c x d)^{10/3}}d(b d+2 c x d)}{2 c d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\left (-\frac {b^2}{c}+\frac {(b d+2 c x d)^2}{c d^2}+4 a\right )^{4/3}}{(b d+2 c x d)^{10/3}}d(b d+2 c x d)}{8\ 2^{2/3} c d}\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \frac {\left (4 a-\frac {b^2}{c}\right ) \sqrt [3]{4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}} \int \frac {\left (1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}\right )^{4/3}}{(b d+2 c x d)^{10/3}}d(b d+2 c x d)}{8\ 2^{2/3} c d \sqrt [3]{1-\frac {(b d+2 c d x)^2}{d^2 \left (b^2-4 a c\right )}}}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle -\frac {3 \left (4 a-\frac {b^2}{c}\right ) \sqrt [3]{4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},-\frac {7}{6},-\frac {1}{6},\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}\right )}{56\ 2^{2/3} c d (b d+2 c d x)^{7/3} \sqrt [3]{1-\frac {(b d+2 c d x)^2}{d^2 \left (b^2-4 a c\right )}}}\) |
(-3*(4*a - b^2/c)*(4*a - b^2/c + (b*d + 2*c*d*x)^2/(c*d^2))^(1/3)*Hypergeo metric2F1[-4/3, -7/6, -1/6, (b*d + 2*c*d*x)^2/((b^2 - 4*a*c)*d^2)])/(56*2^ (2/3)*c*d*(b*d + 2*c*d*x)^(7/3)*(1 - (b*d + 2*c*d*x)^2/((b^2 - 4*a*c)*d^2) )^(1/3))
3.15.21.3.1 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[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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(c*x)^m* (1 + b*(x^2/a))^p, x], 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]
\[\int \frac {\left (c \,x^{2}+b x +a \right )^{\frac {4}{3}}}{\left (2 c d x +b d \right )^{\frac {10}{3}}}d x\]
\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {10}{3}}} \,d x } \]
integral((2*c*d*x + b*d)^(2/3)*(c*x^2 + b*x + a)^(4/3)/(16*c^4*d^4*x^4 + 3 2*b*c^3*d^4*x^3 + 24*b^2*c^2*d^4*x^2 + 8*b^3*c*d^4*x + b^4*d^4), x)
\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx=\int \frac {\left (a + b x + c x^{2}\right )^{\frac {4}{3}}}{\left (d \left (b + 2 c x\right )\right )^{\frac {10}{3}}}\, dx \]
\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {10}{3}}} \,d x } \]
\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {10}{3}}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{4/3}}{{\left (b\,d+2\,c\,d\,x\right )}^{10/3}} \,d x \]