\(\int \frac {(a+b x+c x^2)^{4/3}}{(b d+2 c d x)^{11/3}} \, dx\) [308]

Optimal result

Integrand size = 28, antiderivative size = 324 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{11/3}} \, dx=-\frac {3 (b d+2 c d x)^{4/3} \sqrt [3]{a+b x+c x^2}}{16 c^2 \left (b^2-4 a c\right ) d^5}+\frac {9 (b d+2 c d x)^{4/3} \left (a+b x+c x^2\right )^{4/3}}{16 c \left (b^2-4 a c\right )^2 d^5}+\frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right ) d (b d+2 c d x)^{8/3}}-\frac {9 \left (a+b x+c x^2\right )^{7/3}}{4 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{2/3}}-\frac {\sqrt {3} \arctan \left (\frac {1+\frac {\sqrt [3]{2} (b d+2 c d x)^{2/3}}{\sqrt [3]{c} d^{2/3} \sqrt [3]{a+b x+c x^2}}}{\sqrt {3}}\right )}{16\ 2^{2/3} c^{7/3} d^{11/3}}-\frac {3 \log \left ((b d+2 c d x)^{2/3}-2^{2/3} \sqrt [3]{c} d^{2/3} \sqrt [3]{a+b x+c x^2}\right )}{32\ 2^{2/3} c^{7/3} d^{11/3}} \] Output:

-3/16*(2*c*d*x+b*d)^(4/3)*(c*x^2+b*x+a)^(1/3)/c^2/(-4*a*c+b^2)/d^5+9/16*(2 
*c*d*x+b*d)^(4/3)*(c*x^2+b*x+a)^(4/3)/c/(-4*a*c+b^2)^2/d^5+3/4*(c*x^2+b*x+ 
a)^(7/3)/(-4*a*c+b^2)/d/(2*c*d*x+b*d)^(8/3)-9/4*(c*x^2+b*x+a)^(7/3)/(-4*a* 
c+b^2)^2/d^3/(2*c*d*x+b*d)^(2/3)-1/32*3^(1/2)*arctan(1/3*(1+2^(1/3)*(2*c*d 
*x+b*d)^(2/3)/c^(1/3)/d^(2/3)/(c*x^2+b*x+a)^(1/3))*3^(1/2))*2^(1/3)/c^(7/3 
)/d^(11/3)-3/64*ln((2*c*d*x+b*d)^(2/3)-2^(2/3)*c^(1/3)*d^(2/3)*(c*x^2+b*x+ 
a)^(1/3))*2^(1/3)/c^(7/3)/d^(11/3)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 10.10 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.32 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{11/3}} \, dx=\frac {3 \left (b^2-4 a c\right ) \sqrt [3]{a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},-\frac {4}{3},-\frac {1}{3},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{64\ 2^{2/3} c^2 d (d (b+2 c x))^{8/3} \sqrt [3]{\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \] Input:

Integrate[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(11/3),x]
 

Output:

(3*(b^2 - 4*a*c)*(a + x*(b + c*x))^(1/3)*Hypergeometric2F1[-4/3, -4/3, -1/ 
3, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(64*2^(2/3)*c^2*d*(d*(b + 2*c*x))^(8/3)*( 
(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(1/3))
 

Rubi [A] (warning: unable to verify)

Time = 0.55 (sec) , antiderivative size = 424, normalized size of antiderivative = 1.31, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {1117, 1117, 1118, 27, 248, 248, 266, 807, 853}

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.

\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{11/3}} \, dx\)

\(\Big \downarrow \) 1117

\(\displaystyle \frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 d \left (b^2-4 a c\right ) (b d+2 c d x)^{8/3}}-\frac {3 \int \frac {\left (c x^2+b x+a\right )^{4/3}}{(b d+2 c x d)^{5/3}}dx}{4 d^2 \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 1117

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

\(\Big \downarrow \) 1118

\(\displaystyle \frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 d \left (b^2-4 a c\right ) (b d+2 c d x)^{8/3}}-\frac {3 \left (\frac {3 \left (a+b x+c x^2\right )^{7/3}}{d \left (b^2-4 a c\right ) (b d+2 c d x)^{2/3}}-\frac {3 \int \frac {\sqrt [3]{b d+2 c x d} \left (-\frac {b^2}{c}+\frac {(b d+2 c x d)^2}{c d^2}+4 a\right )^{4/3}}{4\ 2^{2/3}}d(b d+2 c x d)}{c d^3 \left (b^2-4 a c\right )}\right )}{4 d^2 \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 d \left (b^2-4 a c\right ) (b d+2 c d x)^{8/3}}-\frac {3 \left (\frac {3 \left (a+b x+c x^2\right )^{7/3}}{d \left (b^2-4 a c\right ) (b d+2 c d x)^{2/3}}-\frac {3 \int \sqrt [3]{b d+2 c x d} \left (-\frac {b^2}{c}+\frac {(b d+2 c x d)^2}{c d^2}+4 a\right )^{4/3}d(b d+2 c x d)}{4\ 2^{2/3} c d^3 \left (b^2-4 a c\right )}\right )}{4 d^2 \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 248

\(\displaystyle \frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 d \left (b^2-4 a c\right ) (b d+2 c d x)^{8/3}}-\frac {3 \left (\frac {3 \left (a+b x+c x^2\right )^{7/3}}{d \left (b^2-4 a c\right ) (b d+2 c d x)^{2/3}}-\frac {3 \left (\frac {2}{3} \left (4 a-\frac {b^2}{c}\right ) \int \sqrt [3]{b d+2 c x d} \sqrt [3]{-\frac {b^2}{c}+\frac {(b d+2 c x d)^2}{c d^2}+4 a}d(b d+2 c x d)+\frac {1}{4} (b d+2 c d x)^{4/3} \left (4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}\right )^{4/3}\right )}{4\ 2^{2/3} c d^3 \left (b^2-4 a c\right )}\right )}{4 d^2 \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 248

\(\displaystyle \frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 d \left (b^2-4 a c\right ) (b d+2 c d x)^{8/3}}-\frac {3 \left (\frac {3 \left (a+b x+c x^2\right )^{7/3}}{d \left (b^2-4 a c\right ) (b d+2 c d x)^{2/3}}-\frac {3 \left (\frac {2}{3} \left (4 a-\frac {b^2}{c}\right ) \left (\frac {1}{3} \left (4 a-\frac {b^2}{c}\right ) \int \frac {\sqrt [3]{b d+2 c x d}}{\left (-\frac {b^2}{c}+\frac {(b d+2 c x d)^2}{c d^2}+4 a\right )^{2/3}}d(b d+2 c x d)+\frac {1}{2} (b d+2 c d x)^{4/3} \sqrt [3]{4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}}\right )+\frac {1}{4} (b d+2 c d x)^{4/3} \left (4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}\right )^{4/3}\right )}{4\ 2^{2/3} c d^3 \left (b^2-4 a c\right )}\right )}{4 d^2 \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 d \left (b^2-4 a c\right ) (b d+2 c d x)^{8/3}}-\frac {3 \left (\frac {3 \left (a+b x+c x^2\right )^{7/3}}{d \left (b^2-4 a c\right ) (b d+2 c d x)^{2/3}}-\frac {3 \left (\frac {2}{3} \left (4 a-\frac {b^2}{c}\right ) \left (\left (4 a-\frac {b^2}{c}\right ) \int \frac {b d+2 c x d}{\left (-\frac {b^2}{c}+\frac {(b d+2 c x d)^2}{c d^2}+4 a\right )^{2/3}}d\sqrt [3]{b d+2 c x d}+\frac {1}{2} (b d+2 c d x)^{4/3} \sqrt [3]{4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}}\right )+\frac {1}{4} (b d+2 c d x)^{4/3} \left (4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}\right )^{4/3}\right )}{4\ 2^{2/3} c d^3 \left (b^2-4 a c\right )}\right )}{4 d^2 \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 d \left (b^2-4 a c\right ) (b d+2 c d x)^{8/3}}-\frac {3 \left (\frac {3 \left (a+b x+c x^2\right )^{7/3}}{d \left (b^2-4 a c\right ) (b d+2 c d x)^{2/3}}-\frac {3 \left (\frac {2}{3} \left (4 a-\frac {b^2}{c}\right ) \left (\frac {1}{2} \left (4 a-\frac {b^2}{c}\right ) \int \frac {(b d+2 c x d)^{2/3}}{\left (-\frac {b^2}{c}+4 a+\frac {b d+2 c x d}{c d^2}\right )^{2/3}}d(b d+2 c x d)^{2/3}+\frac {1}{2} (b d+2 c d x)^{4/3} \sqrt [3]{4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}}\right )+\frac {1}{4} (b d+2 c d x)^{4/3} \left (4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}\right )^{4/3}\right )}{4\ 2^{2/3} c d^3 \left (b^2-4 a c\right )}\right )}{4 d^2 \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 853

\(\displaystyle \frac {3 \left (a+b x+c x^2\right )^{7/3}}{4 d \left (b^2-4 a c\right ) (b d+2 c d x)^{8/3}}-\frac {3 \left (\frac {3 \left (a+b x+c x^2\right )^{7/3}}{d \left (b^2-4 a c\right ) (b d+2 c d x)^{2/3}}-\frac {3 \left (\frac {2}{3} \left (4 a-\frac {b^2}{c}\right ) \left (\frac {1}{2} \left (4 a-\frac {b^2}{c}\right ) \left (-\frac {c^{2/3} d^{4/3} \arctan \left (\frac {\frac {2 (b d+2 c d x)^{2/3}}{\sqrt [3]{c} d^{2/3} \sqrt [3]{4 a-\frac {b^2}{c}+\frac {b d+2 c d x}{c d^2}}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} c^{2/3} d^{4/3} \log \left (\frac {(b d+2 c d x)^{2/3}}{\sqrt [3]{c} d^{2/3}}-\sqrt [3]{4 a-\frac {b^2}{c}+\frac {b d+2 c d x}{c d^2}}\right )\right )+\frac {1}{2} (b d+2 c d x)^{4/3} \sqrt [3]{4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}}\right )+\frac {1}{4} (b d+2 c d x)^{4/3} \left (4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}\right )^{4/3}\right )}{4\ 2^{2/3} c d^3 \left (b^2-4 a c\right )}\right )}{4 d^2 \left (b^2-4 a c\right )}\)

Input:

Int[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(11/3),x]
 

Output:

(3*(a + b*x + c*x^2)^(7/3))/(4*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(8/3)) - (3 
*((3*(a + b*x + c*x^2)^(7/3))/((b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(2/3)) - (3 
*(((b*d + 2*c*d*x)^(4/3)*(4*a - b^2/c + (b*d + 2*c*d*x)^2/(c*d^2))^(4/3))/ 
4 + (2*(4*a - b^2/c)*(((b*d + 2*c*d*x)^(4/3)*(4*a - b^2/c + (b*d + 2*c*d*x 
)^2/(c*d^2))^(1/3))/2 + ((4*a - b^2/c)*(-((c^(2/3)*d^(4/3)*ArcTan[(1 + (2* 
(b*d + 2*c*d*x)^(2/3))/(c^(1/3)*d^(2/3)*(4*a - b^2/c + (b*d + 2*c*d*x)/(c* 
d^2))^(1/3)))/Sqrt[3]])/Sqrt[3]) - (c^(2/3)*d^(4/3)*Log[(b*d + 2*c*d*x)^(2 
/3)/(c^(1/3)*d^(2/3)) - (4*a - b^2/c + (b*d + 2*c*d*x)/(c*d^2))^(1/3)])/2) 
)/2))/3))/(4*2^(2/3)*c*(b^2 - 4*a*c)*d^3)))/(4*(b^2 - 4*a*c)*d^2)
 

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 248

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]
 

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 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 853

Int[(x_)/((a_) + (b_.)*(x_)^3)^(2/3), x_Symbol] :> With[{q = Rt[b, 3]}, Sim 
p[-ArcTan[(1 + 2*q*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*q^2), x] - Simp 
[Log[q*x - (a + b*x^3)^(1/3)]/(2*q^2), x]] /; FreeQ[{a, b}, x]
 

rule 1117

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[-2*b*d*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(d^2*(m 
+ 1)*(b^2 - 4*a*c))), x] + Simp[b^2*((m + 2*p + 3)/(d^2*(m + 1)*(b^2 - 4*a* 
c)))   Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, 
 d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[m, -1] & 
& (IntegerQ[2*p] || (IntegerQ[m] && RationalQ[p]) || IntegerQ[(m + 2*p + 3) 
/2])
 

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]

\[\int \frac {\left (c \,x^{2}+b x +a \right )^{\frac {4}{3}}}{\left (2 c d x +b d \right )^{\frac {11}{3}}}d x\]

Input:

int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(11/3),x)
 

Output:

int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(11/3),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{11/3}} \, dx=\text {Timed out} \] Input:

integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(11/3),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{11/3}} \, dx=\int \frac {\left (a + b x + c x^{2}\right )^{\frac {4}{3}}}{\left (d \left (b + 2 c x\right )\right )^{\frac {11}{3}}}\, dx \] Input:

integrate((c*x**2+b*x+a)**(4/3)/(2*c*d*x+b*d)**(11/3),x)
 

Output:

Integral((a + b*x + c*x**2)**(4/3)/(d*(b + 2*c*x))**(11/3), x)
 

Maxima [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{11/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {11}{3}}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(11/3),x, algorithm="maxima")
 

Output:

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(11/3), x)
 

Giac [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{11/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {11}{3}}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(11/3),x, algorithm="giac")
 

Output:

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(11/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{11/3}} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{4/3}}{{\left (b\,d+2\,c\,d\,x\right )}^{11/3}} \,d x \] Input:

int((a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(11/3),x)
 

Output:

int((a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(11/3), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{11/3}} \, dx=\int \frac {\left (c \,x^{2}+b x +a \right )^{\frac {4}{3}}}{\left (2 c d x +b d \right )^{\frac {11}{3}}}d x \] Input:

int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(11/3),x)
 

Output:

int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(11/3),x)