3.1.30 \(\int (b x+c x^2)^{4/3} \, dx\) [30]

3.1.30.1 Optimal result

Integrand size = 13, antiderivative size = 448 \[ \int \left (b x+c x^2\right )^{4/3} \, dx=\frac {3 \sqrt [3]{-\frac {c x (b+c x)}{b^2}} (b+2 c x) \left (b x+c x^2\right )^{4/3}}{55 c \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{4/3}}+\frac {3 \left (-\frac {c x (b+c x)}{b^2}\right )^{4/3} (b+2 c x) \left (b x+c x^2\right )^{4/3}}{22 c \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{4/3}}+\frac {\sqrt [3]{2} 3^{3/4} \sqrt {2-\sqrt {3}} b^2 \left (b x+c x^2\right )^{4/3} \left (1-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}\right ) \sqrt {\frac {1+2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}+2 \sqrt [3]{2} \left (-\frac {c x (b+c x)}{b^2}\right )^{2/3}}{\left (1-\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1+\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}}{1-\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}}\right ),-7+4 \sqrt {3}\right )}{55 c (b+2 c x) \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{4/3} \sqrt {-\frac {1-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}}{\left (1-\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}\right )^2}}} \]

output

3/55*(-c*x*(c*x+b)/b^2)^(1/3)*(2*c*x+b)*(c*x^2+b*x)^(4/3)/c/(-c*(c*x^2+b*x 
)/b^2)^(4/3)+3/22*(-c*x*(c*x+b)/b^2)^(4/3)*(2*c*x+b)*(c*x^2+b*x)^(4/3)/c/( 
-c*(c*x^2+b*x)/b^2)^(4/3)+1/55*2^(1/3)*3^(3/4)*b^2*(c*x^2+b*x)^(4/3)*(1-2^ 
(2/3)*(-c*x*(c*x+b)/b^2)^(1/3))*EllipticF((1-2^(2/3)*(-c*x*(c*x+b)/b^2)^(1 
/3)+3^(1/2))/(1-2^(2/3)*(-c*x*(c*x+b)/b^2)^(1/3)-3^(1/2)),2*I-I*3^(1/2))*( 
1/2*6^(1/2)-1/2*2^(1/2))*((1+2^(2/3)*(-c*x*(c*x+b)/b^2)^(1/3)+2*2^(1/3)*(- 
c*x*(c*x+b)/b^2)^(2/3))/(1-2^(2/3)*(-c*x*(c*x+b)/b^2)^(1/3)-3^(1/2))^2)^(1 
/2)/c/(2*c*x+b)/(-c*(c*x^2+b*x)/b^2)^(4/3)/((-1+2^(2/3)*(-c*x*(c*x+b)/b^2) 
^(1/3))/(1-2^(2/3)*(-c*x*(c*x+b)/b^2)^(1/3)-3^(1/2))^2)^(1/2)
 

3.1.30.2 Mathematica [C] (verified)

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

Time = 10.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.11 \[ \int \left (b x+c x^2\right )^{4/3} \, dx=\frac {3 b x^2 \sqrt [3]{x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},\frac {7}{3},\frac {10}{3},-\frac {c x}{b}\right )}{7 \sqrt [3]{1+\frac {c x}{b}}} \]

input

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

output

(3*b*x^2*(x*(b + c*x))^(1/3)*Hypergeometric2F1[-4/3, 7/3, 10/3, -((c*x)/b) 
])/(7*(1 + (c*x)/b)^(1/3))
 

3.1.30.3 Rubi [A] (warning: unable to verify)

Time = 0.35 (sec) , antiderivative size = 350, normalized size of antiderivative = 0.78, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {1087, 1087, 1093, 1090, 234, 760}

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.

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

\(\Big \downarrow \) 1087

\(\displaystyle \frac {3 (b+2 c x) \left (b x+c x^2\right )^{4/3}}{22 c}-\frac {2 b^2 \int \sqrt [3]{c x^2+b x}dx}{11 c}\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {3 (b+2 c x) \left (b x+c x^2\right )^{4/3}}{22 c}-\frac {2 b^2 \left (\frac {3 (b+2 c x) \sqrt [3]{b x+c x^2}}{10 c}-\frac {b^2 \int \frac {1}{\left (c x^2+b x\right )^{2/3}}dx}{10 c}\right )}{11 c}\)

\(\Big \downarrow \) 1093

\(\displaystyle \frac {3 (b+2 c x) \left (b x+c x^2\right )^{4/3}}{22 c}-\frac {2 b^2 \left (\frac {3 (b+2 c x) \sqrt [3]{b x+c x^2}}{10 c}-\frac {b^2 \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{2/3} \int \frac {1}{\left (-\frac {c^2 x^2}{b^2}-\frac {c x}{b}\right )^{2/3}}dx}{10 c \left (b x+c x^2\right )^{2/3}}\right )}{11 c}\)

\(\Big \downarrow \) 1090

\(\displaystyle \frac {3 (b+2 c x) \left (b x+c x^2\right )^{4/3}}{22 c}-\frac {2 b^2 \left (\frac {b^4 \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{2/3} \int \frac {1}{\left (1-\frac {b^2 \left (-\frac {2 x c^2}{b^2}-\frac {c}{b}\right )^2}{c^2}\right )^{2/3}}d\left (-\frac {2 x c^2}{b^2}-\frac {c}{b}\right )}{5\ 2^{2/3} c^3 \left (b x+c x^2\right )^{2/3}}+\frac {3 (b+2 c x) \sqrt [3]{b x+c x^2}}{10 c}\right )}{11 c}\)

\(\Big \downarrow \) 234

\(\displaystyle \frac {3 (b+2 c x) \left (b x+c x^2\right )^{4/3}}{22 c}-\frac {2 b^2 \left (\frac {3 (b+2 c x) \sqrt [3]{b x+c x^2}}{10 c}-\frac {3 b^2 \sqrt {-\frac {b^2 \left (-\frac {2 c^2 x}{b^2}-\frac {c}{b}\right )^2}{c^2}} \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{2/3} \int \frac {1}{\sqrt {-\frac {b^2 \left (-\frac {2 x c^2}{b^2}-\frac {c}{b}\right )^2}{c^2}}}d\sqrt [3]{1-\frac {b^2 \left (-\frac {2 x c^2}{b^2}-\frac {c}{b}\right )^2}{c^2}}}{10\ 2^{2/3} c \left (-\frac {2 c^2 x}{b^2}-\frac {c}{b}\right ) \left (b x+c x^2\right )^{2/3}}\right )}{11 c}\)

\(\Big \downarrow \) 760

\(\displaystyle \frac {3 (b+2 c x) \left (b x+c x^2\right )^{4/3}}{22 c}-\frac {2 b^2 \left (\frac {3^{3/4} \sqrt {2-\sqrt {3}} b^2 \left (\frac {2 c^2 x}{b^2}+\frac {c}{b}+1\right ) \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{2/3} \sqrt {\frac {\left (-\frac {2 c^2 x}{b^2}-\frac {c}{b}\right )^2+\sqrt [3]{1-\frac {b^2 \left (-\frac {2 c^2 x}{b^2}-\frac {c}{b}\right )^2}{c^2}}+1}{\left (\frac {2 c^2 x}{b^2}+\frac {c}{b}-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\frac {2 x c^2}{b^2}+\frac {c}{b}+\sqrt {3}+1}{\frac {2 x c^2}{b^2}+\frac {c}{b}-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{5\ 2^{2/3} c \left (-\frac {2 c^2 x}{b^2}-\frac {c}{b}\right ) \sqrt {-\frac {\frac {2 c^2 x}{b^2}+\frac {c}{b}+1}{\left (\frac {2 c^2 x}{b^2}+\frac {c}{b}-\sqrt {3}+1\right )^2}} \left (b x+c x^2\right )^{2/3}}+\frac {3 (b+2 c x) \sqrt [3]{b x+c x^2}}{10 c}\right )}{11 c}\)

input

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

output

(3*(b + 2*c*x)*(b*x + c*x^2)^(4/3))/(22*c) - (2*b^2*((3*(b + 2*c*x)*(b*x + 
 c*x^2)^(1/3))/(10*c) + (3^(3/4)*Sqrt[2 - Sqrt[3]]*b^2*(1 + c/b + (2*c^2*x 
)/b^2)*(-((c*(b*x + c*x^2))/b^2))^(2/3)*Sqrt[(1 + (-(c/b) - (2*c^2*x)/b^2) 
^2 + (1 - (b^2*(-(c/b) - (2*c^2*x)/b^2)^2)/c^2)^(1/3))/(1 - Sqrt[3] + c/b 
+ (2*c^2*x)/b^2)^2]*EllipticF[ArcSin[(1 + Sqrt[3] + c/b + (2*c^2*x)/b^2)/( 
1 - Sqrt[3] + c/b + (2*c^2*x)/b^2)], -7 + 4*Sqrt[3]])/(5*2^(2/3)*c*(-(c/b) 
 - (2*c^2*x)/b^2)*Sqrt[-((1 + c/b + (2*c^2*x)/b^2)/(1 - Sqrt[3] + c/b + (2 
*c^2*x)/b^2)^2)]*(b*x + c*x^2)^(2/3))))/(11*c)
 

3.1.30.3.1 Defintions of rubi rules used

rule 234

Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Simp[3*(Sqrt[b*x^2]/(2*b*x)) 
   Subst[Int[1/Sqrt[-a + x^3], x], x, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b 
}, x]
 

rule 760

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 
] && NegQ[a]
 

rule 1087

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])
 

rule 1090

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* 
(c/(b^2 - 4*a*c)))^p)   Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, 
b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]
 

rule 1093

Int[((b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b*x + c*x^2)^p/((- 
c)*((b*x + c*x^2)/b^2))^p   Int[((-c)*(x/b) - c^2*(x^2/b^2))^p, x], x] /; F 
reeQ[{b, c}, x] && (IntegerQ[4*p] || IntegerQ[3*p])
 

3.1.30.4 Maple [F]

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

input

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

output

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

3.1.30.5 Fricas [F]

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

input

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

output

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

3.1.30.6 Sympy [F]

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

input

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

output

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

3.1.30.7 Maxima [F]

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

input

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

output

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

3.1.30.8 Giac [F]

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

input

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

output

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

3.1.30.9 Mupad [B] (verification not implemented)

Time = 9.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.08 \[ \int \left (b x+c x^2\right )^{4/3} \, dx=\frac {3\,x\,{\left (c\,x^2+b\,x\right )}^{4/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {4}{3},\frac {7}{3};\ \frac {10}{3};\ -\frac {c\,x}{b}\right )}{7\,{\left (\frac {c\,x}{b}+1\right )}^{4/3}} \]

input

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

output

(3*x*(b*x + c*x^2)^(4/3)*hypergeom([-4/3, 7/3], 10/3, -(c*x)/b))/(7*((c*x) 
/b + 1)^(4/3))