\(\int \frac {(a+b x+c x^2)^{4/3}}{(b d+2 c d x)^{10/3}} \, dx\) [317]

Optimal result

Integrand size = 28, antiderivative size = 100 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx=\frac {3 \left (b^2-4 a c\right ) \sqrt [3]{a+b x+c x^2} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},-\frac {7}{6},-\frac {1}{6},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{56 c^2 d (b d+2 c d x)^{7/3} \sqrt [3]{1-\frac {(b+2 c x)^2}{b^2-4 a c}}} \] Output:

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/(2*c*d*x+b*d)^(7/3)/(1-(2*c*x+b)^2/(-4*a*c+b^2)) 
^(1/3)
 

Mathematica [A] (verified)

Time = 10.07 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx=\frac {3 \left (b^2-4 a c\right ) \sqrt [3]{a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},-\frac {7}{6},-\frac {1}{6},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{56\ 2^{2/3} c^2 d (d (b+2 c x))^{7/3} \sqrt [3]{\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \] Input:

Integrate[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(10/3),x]
 

Output:

(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))
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.41, 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

\(\displaystyle -\frac {3 \left (4 a-\frac {b^2}{c}\right ) \sqrt [3]{4 a-\frac {b^2}{c}+\frac {(b d+2 c d x)^2}{c d^2}} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},-\frac {7}{6},-\frac {1}{6},\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}\right )}{56\ 2^{2/3} c d (b d+2 c d x)^{7/3} \sqrt [3]{1-\frac {(b d+2 c d x)^2}{d^2 \left (b^2-4 a c\right )}}}\)

Input:

Int[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(10/3),x]
 

Output:

(-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))
 

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 278

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])
 

rule 279

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])
 

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]

\[\int \frac {\left (c \,x^{2}+b x +a \right )^{\frac {4}{3}}}{\left (2 c d x +b d \right )^{\frac {10}{3}}}d x\]

Input:

int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(10/3),x)
 

Output:

int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(10/3),x)
 

Fricas [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {10}{3}}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(10/3),x, algorithm="fricas")
 

Output:

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)
 

Sympy [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx=\int \frac {\left (a + b x + c x^{2}\right )^{\frac {4}{3}}}{\left (d \left (b + 2 c x\right )\right )^{\frac {10}{3}}}\, dx \] Input:

integrate((c*x**2+b*x+a)**(4/3)/(2*c*d*x+b*d)**(10/3),x)
 

Output:

Integral((a + b*x + c*x**2)**(4/3)/(d*(b + 2*c*x))**(10/3), x)
 

Maxima [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {10}{3}}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(10/3),x, algorithm="maxima")
 

Output:

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(10/3), x)
 

Giac [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {10}{3}}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(10/3),x, algorithm="giac")
 

Output:

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(10/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{4/3}}{{\left (b\,d+2\,c\,d\,x\right )}^{10/3}} \,d x \] Input:

int((a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(10/3),x)
 

Output:

int((a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(10/3), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{10/3}} \, dx=\frac {\left (\int \frac {\left (c \,x^{2}+b x +a \right )^{\frac {1}{3}}}{\left (2 c x +b \right )^{\frac {1}{3}} b^{3}+6 \left (2 c x +b \right )^{\frac {1}{3}} b^{2} c x +12 \left (2 c x +b \right )^{\frac {1}{3}} b \,c^{2} x^{2}+8 \left (2 c x +b \right )^{\frac {1}{3}} c^{3} x^{3}}d x \right ) a +\left (\int \frac {\left (c \,x^{2}+b x +a \right )^{\frac {1}{3}} x^{2}}{\left (2 c x +b \right )^{\frac {1}{3}} b^{3}+6 \left (2 c x +b \right )^{\frac {1}{3}} b^{2} c x +12 \left (2 c x +b \right )^{\frac {1}{3}} b \,c^{2} x^{2}+8 \left (2 c x +b \right )^{\frac {1}{3}} c^{3} x^{3}}d x \right ) c +\left (\int \frac {\left (c \,x^{2}+b x +a \right )^{\frac {1}{3}} x}{\left (2 c x +b \right )^{\frac {1}{3}} b^{3}+6 \left (2 c x +b \right )^{\frac {1}{3}} b^{2} c x +12 \left (2 c x +b \right )^{\frac {1}{3}} b \,c^{2} x^{2}+8 \left (2 c x +b \right )^{\frac {1}{3}} c^{3} x^{3}}d x \right ) b}{d^{\frac {10}{3}}} \] Input:

int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(10/3),x)
 

Output:

(int((a + b*x + c*x**2)**(1/3)/((b + 2*c*x)**(1/3)*b**3 + 6*(b + 2*c*x)**( 
1/3)*b**2*c*x + 12*(b + 2*c*x)**(1/3)*b*c**2*x**2 + 8*(b + 2*c*x)**(1/3)*c 
**3*x**3),x)*a + int(((a + b*x + c*x**2)**(1/3)*x**2)/((b + 2*c*x)**(1/3)* 
b**3 + 6*(b + 2*c*x)**(1/3)*b**2*c*x + 12*(b + 2*c*x)**(1/3)*b*c**2*x**2 + 
 8*(b + 2*c*x)**(1/3)*c**3*x**3),x)*c + int(((a + b*x + c*x**2)**(1/3)*x)/ 
((b + 2*c*x)**(1/3)*b**3 + 6*(b + 2*c*x)**(1/3)*b**2*c*x + 12*(b + 2*c*x)* 
*(1/3)*b*c**2*x**2 + 8*(b + 2*c*x)**(1/3)*c**3*x**3),x)*b)/(d**(1/3)*d**3)