\(\int \frac {(a+b x+c x^2)^{4/3}}{(b d+2 c d x)^{2/3}} \, dx\) [1416]

Optimal result

Integrand size = 28, antiderivative size = 597 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{2/3}} \, dx=-\frac {\left (b^2-4 a c\right ) \sqrt [3]{d (b+2 c x)} \sqrt [3]{a+b x+c x^2}}{9 c^2 d}+\frac {\sqrt [3]{d (b+2 c x)} \left (a+b x+c x^2\right )^{4/3}}{6 c d}+\frac {\left (b^2-4 a c\right ) \sqrt [3]{d (b+2 c x)} \left (b^2-4 a c-(b+2 c x)^2\right ) \left (2 \sqrt [3]{c} d^{2/3}-\frac {\sqrt [3]{2} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right ) \sqrt {\frac {2 \sqrt [3]{2} c^{2/3} d^{4/3}+\frac {(d (b+2 c x))^{4/3}}{\left (a+b x+c x^2\right )^{2/3}}+\frac {2^{2/3} \sqrt [3]{c} d^{2/3} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}}{\left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1+\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1-\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}}{2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1+\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{72 \sqrt [4]{3} c^{10/3} d^{5/3} \left (a+b x+c x^2\right )^{2/3} \sqrt {-\frac {(d (b+2 c x))^{2/3} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {(d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )}{\sqrt [3]{a+b x+c x^2} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1+\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )^2}}} \]

[Out]

-1/9*(-4*a*c+b^2)*(d*(2*c*x+b))^(1/3)*(c*x^2+b*x+a)^(1/3)/c^2/d+1/6*(d*(2*c*x+b))^(1/3)*(c*x^2+b*x+a)^(4/3)/c/
d+1/216*(-4*a*c+b^2)*(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^(5/3)/(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)

Rubi [A] (warning: unable to verify)

Time = 1.58 (sec) , antiderivative size = 597, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {708, 285, 335, 247, 231} \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{2/3}} \, dx=\frac {\left (b^2-4 a c\right ) \left (-4 a c+b^2-(b+2 c x)^2\right ) \sqrt [3]{d (b+2 c x)} \left (2 \sqrt [3]{c} d^{2/3}-\frac {\sqrt [3]{2} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right ) \sqrt {\frac {\frac {2^{2/3} \sqrt [3]{c} d^{2/3} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}+\frac {(d (b+2 c x))^{4/3}}{\left (a+b x+c x^2\right )^{2/3}}+2 \sqrt [3]{2} c^{2/3} d^{4/3}}{\left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1+\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1-\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{c x^2+b x+a}}}{2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1+\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{c x^2+b x+a}}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{72 \sqrt [4]{3} c^{10/3} d^{5/3} \left (a+b x+c x^2\right )^{2/3} \sqrt {-\frac {(d (b+2 c x))^{2/3} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {(d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )}{\sqrt [3]{a+b x+c x^2} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1+\sqrt {3}\right ) (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )^2}}}-\frac {\left (b^2-4 a c\right ) \sqrt [3]{a+b x+c x^2} \sqrt [3]{d (b+2 c x)}}{9 c^2 d}+\frac {\left (a+b x+c x^2\right )^{4/3} \sqrt [3]{d (b+2 c x)}}{6 c d} \]

[In]

Int[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(2/3),x]

[Out]

-1/9*((b^2 - 4*a*c)*(d*(b + 2*c*x))^(1/3)*(a + b*x + c*x^2)^(1/3))/(c^2*d) + ((d*(b + 2*c*x))^(1/3)*(a + b*x +
c*x^2)^(4/3))/(6*c*d) + ((b^2 - 4*a*c)*(d*(b + 2*c*x))^(1/3)*(b^2 - 4*a*c - (b + 2*c*x)^2)*(2*c^(1/3)*d^(2/3)
- (2^(1/3)*(d*(b + 2*c*x))^(2/3))/(a + b*x + c*x^2)^(1/3))*Sqrt[(2*2^(1/3)*c^(2/3)*d^(4/3) + (d*(b + 2*c*x))^
(4/3)/(a + b*x + c*x^2)^(2/3) + (2^(2/3)*c^(1/3)*d^(2/3)*(d*(b + 2*c*x))^(2/3))/(a + b*x + c*x^2)^(1/3))/(2^(2
/3)*c^(1/3)*d^(2/3) - ((1 + Sqrt[3])*(d*(b + 2*c*x))^(2/3))/(a + b*x + c*x^2)^(1/3))^2]*EllipticF[ArcCos[(2^(2
/3)*c^(1/3)*d^(2/3) - ((1 - Sqrt[3])*(d*(b + 2*c*x))^(2/3))/(a + b*x + c*x^2)^(1/3))/(2^(2/3)*c^(1/3)*d^(2/3)
- ((1 + Sqrt[3])*(d*(b + 2*c*x))^(2/3))/(a + b*x + c*x^2)^(1/3))], (2 + Sqrt[3])/4])/(72*3^(1/4)*c^(10/3)*d^(5
/3)*(a + b*x + c*x^2)^(2/3)*Sqrt[-(((d*(b + 2*c*x))^(2/3)*(2^(2/3)*c^(1/3)*d^(2/3) - (d*(b + 2*c*x))^(2/3)/(a
+ b*x + c*x^2)^(1/3)))/((a + b*x + c*x^2)^(1/3)*(2^(2/3)*c^(1/3)*d^(2/3) - ((1 + Sqrt[3])*(d*(b + 2*c*x))^(2/3
))/(a + b*x + c*x^2)^(1/3))^2))])

Rule 231

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]

Rule 247

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a/(a + b*x^n))^(p + 1/n)*(a + b*x^n)^(p + 1/n), Subst[In
t[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]]

Rule 285

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + n
*p + 1))), x] + Dist[a*n*(p/(m + n*p + 1)), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 708

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[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] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}\right )^{4/3}}{x^{2/3}} \, dx,x,b d+2 c d x\right )}{2 c d} \\ & = \frac {\sqrt [3]{d (b+2 c x)} \left (a+b x+c x^2\right )^{4/3}}{6 c d}-\frac {\left (b^2-4 a c\right ) \text {Subst}\left (\int \frac {\sqrt [3]{a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}}}{x^{2/3}} \, dx,x,b d+2 c d x\right )}{9 c^2 d} \\ & = -\frac {\left (b^2-4 a c\right ) \sqrt [3]{d (b+2 c x)} \sqrt [3]{a+b x+c x^2}}{9 c^2 d}+\frac {\sqrt [3]{d (b+2 c x)} \left (a+b x+c x^2\right )^{4/3}}{6 c d}+\frac {\left (b^2-4 a c\right )^2 \text {Subst}\left (\int \frac {1}{x^{2/3} \left (a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}\right )^{2/3}} \, dx,x,b d+2 c d x\right )}{54 c^3 d} \\ & = -\frac {\left (b^2-4 a c\right ) \sqrt [3]{d (b+2 c x)} \sqrt [3]{a+b x+c x^2}}{9 c^2 d}+\frac {\sqrt [3]{d (b+2 c x)} \left (a+b x+c x^2\right )^{4/3}}{6 c d}+\frac {\left (b^2-4 a c\right )^2 \text {Subst}\left (\int \frac {1}{\left (a-\frac {b^2}{4 c}+\frac {x^6}{4 c d^2}\right )^{2/3}} \, dx,x,\sqrt [3]{d (b+2 c x)}\right )}{18 c^3 d} \\ & = -\frac {\left (b^2-4 a c\right ) \sqrt [3]{d (b+2 c x)} \sqrt [3]{a+b x+c x^2}}{9 c^2 d}+\frac {\sqrt [3]{d (b+2 c x)} \left (a+b x+c x^2\right )^{4/3}}{6 c d}+\frac {\left (b^2-4 a c\right )^2 \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^6}{4 c d^2}}} \, dx,x,\frac {\sqrt [3]{d (b+2 c x)}}{\sqrt [6]{a+x (b+c x)}}\right )}{18 c^3 d \sqrt {\frac {a-\frac {b^2}{4 c}}{a+x (b+c x)}} \sqrt {a+x (b+c x)}} \\ & = -\frac {\left (b^2-4 a c\right ) \sqrt [3]{d (b+2 c x)} \sqrt [3]{a+b x+c x^2}}{9 c^2 d}+\frac {\sqrt [3]{d (b+2 c x)} \left (a+b x+c x^2\right )^{4/3}}{6 c d}+\frac {\left (b^2-4 a c\right )^2 \sqrt [3]{d (b+2 c x)} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {(d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}\right ) \sqrt {\frac {2 \sqrt [3]{2} c^{2/3} d^{4/3}+\frac {(d (b+2 c x))^{4/3}}{(a+x (b+c x))^{2/3}}+\frac {2^{2/3} \sqrt [3]{c} d^{2/3} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+x (b+c x)}}}{\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+x (b+c x)}}\right )^2}} F\left (\cos ^{-1}\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+x (b+c x)}}}{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+x (b+c x)}}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{9\ 2^{2/3} \sqrt [4]{3} c^{10/3} d^{5/3} \sqrt {\frac {4 a-\frac {b^2}{c}}{a+x (b+c x)}} (a+x (b+c x))^{2/3} \sqrt {4-\frac {(b+2 c x)^2}{c (a+x (b+c x))}} \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+x (b+c x)}}\right )}{\sqrt [3]{a+x (b+c x)} \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+x (b+c x)}}\right )^2}}} \\ \end{align*}

Mathematica [C] (verified)

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 = 0.17 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{2/3}} \, dx=-\frac {3 \left (b^2-4 a c\right ) \sqrt [3]{d (b+2 c x)} \sqrt [3]{a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},\frac {1}{6},\frac {7}{6},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{8\ 2^{2/3} c^2 d \sqrt [3]{\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \]

[In]

Integrate[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(2/3),x]

[Out]

(-3*(b^2 - 4*a*c)*(d*(b + 2*c*x))^(1/3)*(a + x*(b + c*x))^(1/3)*Hypergeometric2F1[-4/3, 1/6, 7/6, (b + 2*c*x)^
2/(b^2 - 4*a*c)])/(8*2^(2/3)*c^2*d*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(1/3))

Maple [F]

\[\int \frac {\left (c \,x^{2}+b x +a \right )^{\frac {4}{3}}}{\left (2 c d x +b d \right )^{\frac {2}{3}}}d x\]

[In]

int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(2/3),x)

[Out]

int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(2/3),x)

Fricas [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{2/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {2}{3}}} \,d x } \]

[In]

integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(2/3),x, algorithm="fricas")

[Out]

integral((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(2/3), x)

Sympy [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{2/3}} \, dx=\int \frac {\left (a + b x + c x^{2}\right )^{\frac {4}{3}}}{\left (d \left (b + 2 c x\right )\right )^{\frac {2}{3}}}\, dx \]

[In]

integrate((c*x**2+b*x+a)**(4/3)/(2*c*d*x+b*d)**(2/3),x)

[Out]

Integral((a + b*x + c*x**2)**(4/3)/(d*(b + 2*c*x))**(2/3), x)

Maxima [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{2/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {2}{3}}} \,d x } \]

[In]

integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(2/3),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(2/3), x)

Giac [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{2/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {2}{3}}} \,d x } \]

[In]

integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(2/3),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(2/3), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{2/3}} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{4/3}}{{\left (b\,d+2\,c\,d\,x\right )}^{2/3}} \,d x \]

[In]

int((a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(2/3),x)

[Out]

int((a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(2/3), x)