Integrand size = 22, antiderivative size = 638 \[ \int (d+e x)^2 \left (a+b x+c x^2\right )^{4/3} \, dx=-\frac {3 \left (b^2-4 a c\right ) \left (17 c^2 d^2+5 b^2 e^2-c e (17 b d+3 a e)\right ) (b+2 c x) \sqrt [3]{a+b x+c x^2}}{935 c^4}+\frac {3 \left (17 c^2 d^2+5 b^2 e^2-c e (17 b d+3 a e)\right ) (b+2 c x) \left (a+b x+c x^2\right )^{4/3}}{374 c^3}+\frac {15 e (2 c d-b e) \left (a+b x+c x^2\right )^{7/3}}{119 c^2}+\frac {3 e (d+e x) \left (a+b x+c x^2\right )^{7/3}}{17 c}+\frac {\sqrt [3]{2} 3^{3/4} \sqrt {2+\sqrt {3}} \left (b^2-4 a c\right )^2 \left (17 c^2 d^2+5 b^2 e^2-c e (17 b d+3 a e)\right ) \left (\sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right ) \sqrt {\frac {\left (b^2-4 a c\right )^{2/3}-2^{2/3} \sqrt [3]{c} \sqrt [3]{b^2-4 a c} \sqrt [3]{a+b x+c x^2}+2 \sqrt [3]{2} c^{2/3} \left (a+b x+c x^2\right )^{2/3}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}}{\left (1+\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}}\right ),-7-4 \sqrt {3}\right )}{935 c^{13/3} (b+2 c x) \sqrt {\frac {\sqrt [3]{b^2-4 a c} \left (\sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right )^2}}} \]
-3/935*(-4*a*c+b^2)*(17*c^2*d^2+5*b^2*e^2-c*e*(3*a*e+17*b*d))*(2*c*x+b)*(c *x^2+b*x+a)^(1/3)/c^4+3/374*(17*c^2*d^2+5*b^2*e^2-c*e*(3*a*e+17*b*d))*(2*c *x+b)*(c*x^2+b*x+a)^(4/3)/c^3+15/119*e*(-b*e+2*c*d)*(c*x^2+b*x+a)^(7/3)/c^ 2+3/17*e*(e*x+d)*(c*x^2+b*x+a)^(7/3)/c+1/935*2^(1/3)*3^(3/4)*(-4*a*c+b^2)^ 2*(17*c^2*d^2+5*b^2*e^2-c*e*(3*a*e+17*b*d))*((-4*a*c+b^2)^(1/3)+2^(2/3)*c^ (1/3)*(c*x^2+b*x+a)^(1/3))*EllipticF((2^(2/3)*c^(1/3)*(c*x^2+b*x+a)^(1/3)+ (-4*a*c+b^2)^(1/3)*(1-3^(1/2)))/(2^(2/3)*c^(1/3)*(c*x^2+b*x+a)^(1/3)+(-4*a *c+b^2)^(1/3)*(1+3^(1/2))),I*3^(1/2)+2*I)*(1/2*6^(1/2)+1/2*2^(1/2))*(((-4* a*c+b^2)^(2/3)-2^(2/3)*c^(1/3)*(-4*a*c+b^2)^(1/3)*(c*x^2+b*x+a)^(1/3)+2*2^ (1/3)*c^(2/3)*(c*x^2+b*x+a)^(2/3))/(2^(2/3)*c^(1/3)*(c*x^2+b*x+a)^(1/3)+(- 4*a*c+b^2)^(1/3)*(1+3^(1/2)))^2)^(1/2)/c^(13/3)/(2*c*x+b)/((-4*a*c+b^2)^(1 /3)*((-4*a*c+b^2)^(1/3)+2^(2/3)*c^(1/3)*(c*x^2+b*x+a)^(1/3))/(2^(2/3)*c^(1 /3)*(c*x^2+b*x+a)^(1/3)+(-4*a*c+b^2)^(1/3)*(1+3^(1/2)))^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.28 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.26 \[ \int (d+e x)^2 \left (a+b x+c x^2\right )^{4/3} \, dx=\frac {3 (a+x (b+c x))^{4/3} \left (-\frac {160 e (-2 c d+b e) (a+x (b+c x))}{c}+224 e (d+e x) (a+x (b+c x))+\frac {14 \sqrt [3]{2} \left (17 c^2 d^2+5 b^2 e^2-c e (17 b d+3 a e)\right ) (b+2 c x) \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},\frac {1}{2},\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{3 c^2 \left (-\frac {c (a+x (b+c x))}{b^2-4 a c}\right )^{4/3}}\right )}{3808 c} \]
(3*(a + x*(b + c*x))^(4/3)*((-160*e*(-2*c*d + b*e)*(a + x*(b + c*x)))/c + 224*e*(d + e*x)*(a + x*(b + c*x)) + (14*2^(1/3)*(17*c^2*d^2 + 5*b^2*e^2 - c*e*(17*b*d + 3*a*e))*(b + 2*c*x)*Hypergeometric2F1[-4/3, 1/2, 3/2, (b + 2 *c*x)^2/(b^2 - 4*a*c)])/(3*c^2*(-((c*(a + x*(b + c*x)))/(b^2 - 4*a*c)))^(4 /3))))/(3808*c)
Time = 0.61 (sec) , antiderivative size = 634, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1166, 27, 1160, 1087, 1087, 1095, 759}
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.
| \(\displaystyle \int (d+e x)^2 \left (a+b x+c x^2\right )^{4/3} \, dx\) |
| \(\Big \downarrow \) 1166 |
| \(\displaystyle \frac {3 \int \frac {1}{3} \left (17 c d^2-7 b e d-3 a e^2+10 e (2 c d-b e) x\right ) \left (c x^2+b x+a\right )^{4/3}dx}{17 c}+\frac {3 e (d+e x) \left (a+b x+c x^2\right )^{7/3}}{17 c}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \left (17 c d^2-7 b e d-3 a e^2+10 e (2 c d-b e) x\right ) \left (c x^2+b x+a\right )^{4/3}dx}{17 c}+\frac {3 e (d+e x) \left (a+b x+c x^2\right )^{7/3}}{17 c}\) |
| \(\Big \downarrow \) 1160 |
| \(\displaystyle \frac {\frac {\left (-c e (3 a e+17 b d)+5 b^2 e^2+17 c^2 d^2\right ) \int \left (c x^2+b x+a\right )^{4/3}dx}{c}+\frac {15 e \left (a+b x+c x^2\right )^{7/3} (2 c d-b e)}{7 c}}{17 c}+\frac {3 e (d+e x) \left (a+b x+c x^2\right )^{7/3}}{17 c}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {\frac {\left (-c e (3 a e+17 b d)+5 b^2 e^2+17 c^2 d^2\right ) \left (\frac {3 (b+2 c x) \left (a+b x+c x^2\right )^{4/3}}{22 c}-\frac {2 \left (b^2-4 a c\right ) \int \sqrt [3]{c x^2+b x+a}dx}{11 c}\right )}{c}+\frac {15 e \left (a+b x+c x^2\right )^{7/3} (2 c d-b e)}{7 c}}{17 c}+\frac {3 e (d+e x) \left (a+b x+c x^2\right )^{7/3}}{17 c}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {\frac {\left (-c e (3 a e+17 b d)+5 b^2 e^2+17 c^2 d^2\right ) \left (\frac {3 (b+2 c x) \left (a+b x+c x^2\right )^{4/3}}{22 c}-\frac {2 \left (b^2-4 a c\right ) \left (\frac {3 (b+2 c x) \sqrt [3]{a+b x+c x^2}}{10 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\left (c x^2+b x+a\right )^{2/3}}dx}{10 c}\right )}{11 c}\right )}{c}+\frac {15 e \left (a+b x+c x^2\right )^{7/3} (2 c d-b e)}{7 c}}{17 c}+\frac {3 e (d+e x) \left (a+b x+c x^2\right )^{7/3}}{17 c}\) |
| \(\Big \downarrow \) 1095 |
| \(\displaystyle \frac {\frac {\left (-c e (3 a e+17 b d)+5 b^2 e^2+17 c^2 d^2\right ) \left (\frac {3 (b+2 c x) \left (a+b x+c x^2\right )^{4/3}}{22 c}-\frac {2 \left (b^2-4 a c\right ) \left (\frac {3 (b+2 c x) \sqrt [3]{a+b x+c x^2}}{10 c}-\frac {3 \left (b^2-4 a c\right ) \sqrt {(b+2 c x)^2} \int \frac {1}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [3]{c x^2+b x+a}}{10 c (b+2 c x)}\right )}{11 c}\right )}{c}+\frac {15 e \left (a+b x+c x^2\right )^{7/3} (2 c d-b e)}{7 c}}{17 c}+\frac {3 e (d+e x) \left (a+b x+c x^2\right )^{7/3}}{17 c}\) |
| \(\Big \downarrow \) 759 |
| \(\displaystyle \frac {\frac {\left (-c e (3 a e+17 b d)+5 b^2 e^2+17 c^2 d^2\right ) \left (\frac {3 (b+2 c x) \left (a+b x+c x^2\right )^{4/3}}{22 c}-\frac {2 \left (b^2-4 a c\right ) \left (\frac {3 (b+2 c x) \sqrt [3]{a+b x+c x^2}}{10 c}-\frac {3^{3/4} \sqrt {2+\sqrt {3}} \left (b^2-4 a c\right ) \sqrt {(b+2 c x)^2} \left (\sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right ) \sqrt {\frac {-2^{2/3} \sqrt [3]{c} \sqrt [3]{b^2-4 a c} \sqrt [3]{a+b x+c x^2}+\left (b^2-4 a c\right )^{2/3}+2 \sqrt [3]{2} c^{2/3} \left (a+b x+c x^2\right )^{2/3}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{c x^2+b x+a}}{\left (1+\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{c x^2+b x+a}}\right ),-7-4 \sqrt {3}\right )}{5\ 2^{2/3} c^{4/3} (b+2 c x) \sqrt {\frac {\sqrt [3]{b^2-4 a c} \left (\sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{b^2-4 a c}+2^{2/3} \sqrt [3]{c} \sqrt [3]{a+b x+c x^2}\right )^2}} \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}\right )}{11 c}\right )}{c}+\frac {15 e \left (a+b x+c x^2\right )^{7/3} (2 c d-b e)}{7 c}}{17 c}+\frac {3 e (d+e x) \left (a+b x+c x^2\right )^{7/3}}{17 c}\) |
(3*e*(d + e*x)*(a + b*x + c*x^2)^(7/3))/(17*c) + ((15*e*(2*c*d - b*e)*(a + b*x + c*x^2)^(7/3))/(7*c) + ((17*c^2*d^2 + 5*b^2*e^2 - c*e*(17*b*d + 3*a* e))*((3*(b + 2*c*x)*(a + b*x + c*x^2)^(4/3))/(22*c) - (2*(b^2 - 4*a*c)*((3 *(b + 2*c*x)*(a + b*x + c*x^2)^(1/3))/(10*c) - (3^(3/4)*Sqrt[2 + Sqrt[3]]* (b^2 - 4*a*c)*Sqrt[(b + 2*c*x)^2]*((b^2 - 4*a*c)^(1/3) + 2^(2/3)*c^(1/3)*( a + b*x + c*x^2)^(1/3))*Sqrt[((b^2 - 4*a*c)^(2/3) - 2^(2/3)*c^(1/3)*(b^2 - 4*a*c)^(1/3)*(a + b*x + c*x^2)^(1/3) + 2*2^(1/3)*c^(2/3)*(a + b*x + c*x^2 )^(2/3))/((1 + Sqrt[3])*(b^2 - 4*a*c)^(1/3) + 2^(2/3)*c^(1/3)*(a + b*x + c *x^2)^(1/3))^2]*EllipticF[ArcSin[((1 - Sqrt[3])*(b^2 - 4*a*c)^(1/3) + 2^(2 /3)*c^(1/3)*(a + b*x + c*x^2)^(1/3))/((1 + Sqrt[3])*(b^2 - 4*a*c)^(1/3) + 2^(2/3)*c^(1/3)*(a + b*x + c*x^2)^(1/3))], -7 - 4*Sqrt[3]])/(5*2^(2/3)*c^( 4/3)*(b + 2*c*x)*Sqrt[((b^2 - 4*a*c)^(1/3)*((b^2 - 4*a*c)^(1/3) + 2^(2/3)* c^(1/3)*(a + b*x + c*x^2)^(1/3)))/((1 + Sqrt[3])*(b^2 - 4*a*c)^(1/3) + 2^( 2/3)*c^(1/3)*(a + b*x + c*x^2)^(1/3))^2]*Sqrt[b^2 - 4*a*c + 4*c*(a + b*x + c*x^2)])))/(11*c)))/c)/(17*c)
3.25.83.3.1 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[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] & & PosQ[a]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[3*(Sqrt[(b + 2*c*x)^2]/(b + 2*c*x)) Subst[Int[x^(3*(p + 1) - 1)/Sqrt[b^2 - 4*a*c + 4 *c*x^3], x], x, (a + b*x + c*x^2)^(1/3)], x] /; FreeQ[{a, b, c}, x] && Inte gerQ[3*p]
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Simp[1/(c*(m + 2*p + 1)) Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]* (a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && If[Ration alQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadrat icQ[a, b, c, d, e, m, p, x]
\[\int \left (e x +d \right )^{2} \left (c \,x^{2}+b x +a \right )^{\frac {4}{3}}d x\]
\[ \int (d+e x)^2 \left (a+b x+c x^2\right )^{4/3} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {4}{3}} {\left (e x + d\right )}^{2} \,d x } \]
integral((c*e^2*x^4 + (2*c*d*e + b*e^2)*x^3 + a*d^2 + (c*d^2 + 2*b*d*e + a *e^2)*x^2 + (b*d^2 + 2*a*d*e)*x)*(c*x^2 + b*x + a)^(1/3), x)
\[ \int (d+e x)^2 \left (a+b x+c x^2\right )^{4/3} \, dx=\int \left (d + e x\right )^{2} \left (a + b x + c x^{2}\right )^{\frac {4}{3}}\, dx \]
\[ \int (d+e x)^2 \left (a+b x+c x^2\right )^{4/3} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {4}{3}} {\left (e x + d\right )}^{2} \,d x } \]
\[ \int (d+e x)^2 \left (a+b x+c x^2\right )^{4/3} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {4}{3}} {\left (e x + d\right )}^{2} \,d x } \]
Timed out. \[ \int (d+e x)^2 \left (a+b x+c x^2\right )^{4/3} \, dx=\int {\left (d+e\,x\right )}^2\,{\left (c\,x^2+b\,x+a\right )}^{4/3} \,d x \]