\(\int \frac {(a+b x+c x^2)^{4/3}}{(b d+2 c d x)^{14/3}} \, dx\) [314]

Optimal result

Integrand size = 28, antiderivative size = 643 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{14/3}} \, dx=-\frac {3 \sqrt [3]{a+b x+c x^2}}{55 c^2 d^3 (b d+2 c d x)^{5/3}}-\frac {3 \left (a+b x+c x^2\right )^{4/3}}{22 c d (b d+2 c d x)^{11/3}}+\frac {3^{3/4} \sqrt [3]{b d+2 c d x} \left (2 \sqrt [3]{c} d^{2/3}-\frac {\sqrt [3]{2} (b d+2 c d x)^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right ) \sqrt {\frac {2 \sqrt [3]{2} c^{2/3} d^{4/3}+\frac {(b d+2 c d x)^{4/3}}{\left (a+b x+c x^2\right )^{2/3}}+\frac {2^{2/3} \sqrt [3]{c} d^{2/3} (b d+2 c d x)^{2/3}}{\sqrt [3]{a+b x+c x^2}}}{\left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1+\sqrt {3}\right ) (b d+2 c d x)^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1-\sqrt {3}\right ) (b d+2 c d x)^{2/3}}{\sqrt [3]{a+b x+c x^2}}}{2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1+\sqrt {3}\right ) (b d+2 c d x)^{2/3}}{\sqrt [3]{a+b x+c x^2}}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{440 c^{10/3} d^{17/3} \left (a+b x+c x^2\right )^{2/3} \sqrt {\frac {b^2-4 a c}{b^2-4 a c-(b+2 c x)^2}} \sqrt {-\frac {(b d+2 c d x)^{2/3} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {(b d+2 c d x)^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )}{\sqrt [3]{a+b x+c x^2} \left (2^{2/3} \sqrt [3]{c} d^{2/3}-\frac {\left (1+\sqrt {3}\right ) (b d+2 c d x)^{2/3}}{\sqrt [3]{a+b x+c x^2}}\right )^2}} \sqrt {1+\frac {(b+2 c x)^2}{b^2-4 a c-(b+2 c x)^2}}} \] Output:

-3/55*(c*x^2+b*x+a)^(1/3)/c^2/d^3/(2*c*d*x+b*d)^(5/3)-3/22*(c*x^2+b*x+a)^( 
4/3)/c/d/(2*c*d*x+b*d)^(11/3)+1/440*3^(3/4)*(2*c*d*x+b*d)^(1/3)*(2*c^(1/3) 
*d^(2/3)-2^(1/3)*(2*c*d*x+b*d)^(2/3)/(c*x^2+b*x+a)^(1/3))*((2*2^(1/3)*c^(2 
/3)*d^(4/3)+(2*c*d*x+b*d)^(4/3)/(c*x^2+b*x+a)^(2/3)+2^(2/3)*c^(1/3)*d^(2/3 
)*(2*c*d*x+b*d)^(2/3)/(c*x^2+b*x+a)^(1/3))/(2^(2/3)*c^(1/3)*d^(2/3)-(1+3^( 
1/2))*(2*c*d*x+b*d)^(2/3)/(c*x^2+b*x+a)^(1/3))^2)^(1/2)*InverseJacobiAM(ar 
ccos((2^(2/3)*c^(1/3)*d^(2/3)-(1-3^(1/2))*(2*c*d*x+b*d)^(2/3)/(c*x^2+b*x+a 
)^(1/3))/(2^(2/3)*c^(1/3)*d^(2/3)-(1+3^(1/2))*(2*c*d*x+b*d)^(2/3)/(c*x^2+b 
*x+a)^(1/3))),1/4*6^(1/2)+1/4*2^(1/2))/c^(10/3)/d^(17/3)/(c*x^2+b*x+a)^(2/ 
3)/((-4*a*c+b^2)/(b^2-4*a*c-(2*c*x+b)^2))^(1/2)/(-(2*c*d*x+b*d)^(2/3)*(2^( 
2/3)*c^(1/3)*d^(2/3)-(2*c*d*x+b*d)^(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)-(1+3^(1/2))*(2*c*d*x+b*d)^(2/3)/(c*x^2+b* 
x+a)^(1/3))^2)^(1/2)/(1+(2*c*x+b)^2/(b^2-4*a*c-(2*c*x+b)^2))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.08 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.17 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{14/3}} \, dx=\frac {3 \left (b^2-4 a c\right ) \sqrt [3]{d (b+2 c x)} \sqrt [3]{a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {11}{6},-\frac {4}{3},-\frac {5}{6},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{88\ 2^{2/3} c^2 d^5 (b+2 c x)^4 \sqrt [3]{\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \] Input:

Integrate[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(14/3),x]
 

Output:

(3*(b^2 - 4*a*c)*(d*(b + 2*c*x))^(1/3)*(a + x*(b + c*x))^(1/3)*Hypergeomet 
ric2F1[-11/6, -4/3, -5/6, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(88*2^(2/3)*c^2*d^ 
5*(b + 2*c*x)^4*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(1/3))
 

Rubi [A] (warning: unable to verify)

Time = 0.45 (sec) , antiderivative size = 557, normalized size of antiderivative = 0.87, 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, 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.

\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{14/3}} \, dx\)

\(\Big \downarrow \) 1118

\(\displaystyle \frac {\int \frac {\left (-\frac {b^2}{c}+\frac {(b d+2 c x d)^2}{c d^2}+4 a\right )^{4/3}}{4\ 2^{2/3} (b d+2 c x d)^{14/3}}d(b d+2 c x d)}{2 c d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\left (-\frac {b^2}{c}+\frac {(b d+2 c x d)^2}{c d^2}+4 a\right )^{4/3}}{(b d+2 c x d)^{14/3}}d(b d+2 c x d)}{8\ 2^{2/3} c d}\)

\(\Big \downarrow \) 247

\(\displaystyle \frac {\frac {8 \int \frac {\sqrt [3]{-\frac {b^2}{c}+\frac {(b d+2 c x d)^2}{c d^2}+4 a}}{(b d+2 c x d)^{8/3}}d(b d+2 c x d)}{11 c d^2}-\frac {3 \left (4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}\right )^{4/3}}{11 (b d+2 c d x)^{11/3}}}{8\ 2^{2/3} c d}\)

\(\Big \downarrow \) 247

\(\displaystyle \frac {\frac {8 \left (\frac {2 \int \frac {1}{(b d+2 c x d)^{2/3} \left (-\frac {b^2}{c}+\frac {(b d+2 c x d)^2}{c d^2}+4 a\right )^{2/3}}d(b d+2 c x d)}{5 c d^2}-\frac {3 \sqrt [3]{4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}}}{5 (b d+2 c d x)^{5/3}}\right )}{11 c d^2}-\frac {3 \left (4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}\right )^{4/3}}{11 (b d+2 c d x)^{11/3}}}{8\ 2^{2/3} c d}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {\frac {8 \left (\frac {6 \int \frac {1}{\left (-\frac {b^2}{c}+\frac {(b d+2 c x d)^2}{c d^2}+4 a\right )^{2/3}}d\sqrt [3]{b d+2 c x d}}{5 c d^2}-\frac {3 \sqrt [3]{4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}}}{5 (b d+2 c d x)^{5/3}}\right )}{11 c d^2}-\frac {3 \left (4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}\right )^{4/3}}{11 (b d+2 c d x)^{11/3}}}{8\ 2^{2/3} c d}\)

\(\Big \downarrow \) 771

\(\displaystyle \frac {\frac {8 \left (\frac {6 \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{c d^2}}}d\frac {\sqrt [3]{b d+2 c x d}}{\sqrt [6]{-\frac {b^2}{c}+\frac {(b d+2 c x d)^2}{c d^2}+4 a}}}{5 c d^2 \sqrt {\frac {c d^2 \left (4 a-\frac {b^2}{c}\right )}{c d^2 \left (4 a-\frac {b^2}{c}\right )+(b d+2 c d x)^2}} \sqrt {4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}}}-\frac {3 \sqrt [3]{4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}}}{5 (b d+2 c d x)^{5/3}}\right )}{11 c d^2}-\frac {3 \left (4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}\right )^{4/3}}{11 (b d+2 c d x)^{11/3}}}{8\ 2^{2/3} c d}\)

\(\Big \downarrow \) 766

\(\displaystyle \frac {\frac {8 \left (\frac {3^{3/4} \sqrt [3]{b d+2 c d x} \left (\sqrt [3]{c} d^{2/3}-(b d+2 c d x)^{2/3}\right ) \sqrt {\frac {\sqrt [3]{c} d^{2/3} (b d+2 c d x)^{2/3}+(b d+2 c d x)^{4/3}+c^{2/3} d^{4/3}}{\left (\sqrt [3]{c} d^{2/3}-\left (1+\sqrt {3}\right ) (b d+2 c d x)^{2/3}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{c} d^{2/3}-\left (1-\sqrt {3}\right ) (b d+2 c x d)^{2/3}}{\sqrt [3]{c} d^{2/3}-\left (1+\sqrt {3}\right ) (b d+2 c x d)^{2/3}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{5 c^{4/3} d^{8/3} \sqrt {-\frac {(b d+2 c d x)^{2/3} \left (\sqrt [3]{c} d^{2/3}-(b d+2 c d x)^{2/3}\right )}{\left (\sqrt [3]{c} d^{2/3}-\left (1+\sqrt {3}\right ) (b d+2 c d x)^{2/3}\right )^2}} \sqrt {1-\frac {(b d+2 c d x)^2}{c d^2}} \sqrt {\frac {c d^2 \left (4 a-\frac {b^2}{c}\right )}{c d^2 \left (4 a-\frac {b^2}{c}\right )+(b d+2 c d x)^2}} \left (4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}\right )^{2/3}}-\frac {3 \sqrt [3]{4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}}}{5 (b d+2 c d x)^{5/3}}\right )}{11 c d^2}-\frac {3 \left (4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}\right )^{4/3}}{11 (b d+2 c d x)^{11/3}}}{8\ 2^{2/3} c d}\)

Input:

Int[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(14/3),x]
 

Output:

((-3*(4*a - b^2/c + (b*d + 2*c*d*x)^2/(c*d^2))^(4/3))/(11*(b*d + 2*c*d*x)^ 
(11/3)) + (8*((-3*(4*a - b^2/c + (b*d + 2*c*d*x)^2/(c*d^2))^(1/3))/(5*(b*d 
 + 2*c*d*x)^(5/3)) + (3^(3/4)*(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])/(5*c^(4/3)*d^(8/3)*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))))/(11*c*d^2))/(8*2^(2/3)*c*d)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 247

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]
 

rule 266

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]
 

rule 766

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 771

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]]
 

rule 1118

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]
 

Maple [F]

\[\int \frac {\left (c \,x^{2}+b x +a \right )^{\frac {4}{3}}}{\left (2 c d x +b d \right )^{\frac {14}{3}}}d x\]

Input:

int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(14/3),x)
 

Output:

int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(14/3),x)
 

Fricas [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{14/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {14}{3}}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(14/3),x, algorithm="fricas")
 

Output:

integral((2*c*d*x + b*d)^(1/3)*(c*x^2 + b*x + a)^(4/3)/(32*c^5*d^5*x^5 + 8 
0*b*c^4*d^5*x^4 + 80*b^2*c^3*d^5*x^3 + 40*b^3*c^2*d^5*x^2 + 10*b^4*c*d^5*x 
 + b^5*d^5), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{14/3}} \, dx=\text {Timed out} \] Input:

integrate((c*x**2+b*x+a)**(4/3)/(2*c*d*x+b*d)**(14/3),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{14/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {14}{3}}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(14/3),x, algorithm="maxima")
 

Output:

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(14/3), x)
 

Giac [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{14/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {14}{3}}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(14/3),x, algorithm="giac")
 

Output:

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(14/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{14/3}} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{4/3}}{{\left (b\,d+2\,c\,d\,x\right )}^{14/3}} \,d x \] Input:

int((a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(14/3),x)
 

Output:

int((a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(14/3), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{14/3}} \, dx=\int \frac {\left (c \,x^{2}+b x +a \right )^{\frac {4}{3}}}{\left (2 c d x +b d \right )^{\frac {14}{3}}}d x \] Input:

int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(14/3),x)
                                                                                    
                                                                                    
 

Output:

int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(14/3),x)