3.1.34 \(\int \frac {1}{(b x+c x^2)^{8/3}} \, dx\) [34]

3.1.34.1 Optimal result

Integrand size = 13, antiderivative size = 448 \[ \int \frac {1}{\left (b x+c x^2\right )^{8/3}} \, dx=\frac {3 (b+2 c x) \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{8/3}}{5 c \left (-\frac {c x (b+c x)}{b^2}\right )^{5/3} \left (b x+c x^2\right )^{8/3}}+\frac {21 (b+2 c x) \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{8/3}}{5 c \left (-\frac {c x (b+c x)}{b^2}\right )^{2/3} \left (b x+c x^2\right )^{8/3}}+\frac {14 \sqrt [3]{2} 3^{3/4} \sqrt {2-\sqrt {3}} b^2 \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{8/3} \left (1-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}\right ) \sqrt {\frac {1+2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}+2 \sqrt [3]{2} \left (-\frac {c x (b+c x)}{b^2}\right )^{2/3}}{\left (1-\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1+\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}}{1-\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}}\right ),-7+4 \sqrt {3}\right )}{5 c (b+2 c x) \left (b x+c x^2\right )^{8/3} \sqrt {-\frac {1-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}}{\left (1-\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}\right )^2}}} \]

output

3/5*(2*c*x+b)*(-c*(c*x^2+b*x)/b^2)^(8/3)/c/(-c*x*(c*x+b)/b^2)^(5/3)/(c*x^2 
+b*x)^(8/3)+21/5*(2*c*x+b)*(-c*(c*x^2+b*x)/b^2)^(8/3)/c/(-c*x*(c*x+b)/b^2) 
^(2/3)/(c*x^2+b*x)^(8/3)+14/5*2^(1/3)*3^(3/4)*b^2*(-c*(c*x^2+b*x)/b^2)^(8/ 
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*x^2+b*x)^(8/3)/((-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)
 

3.1.34.2 Mathematica [C] (verified)

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

Time = 10.02 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.11 \[ \int \frac {1}{\left (b x+c x^2\right )^{8/3}} \, dx=-\frac {3 \left (1+\frac {c x}{b}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {5}{3},\frac {8}{3},-\frac {2}{3},-\frac {c x}{b}\right )}{5 b^2 x (x (b+c x))^{2/3}} \]

input

Integrate[(b*x + c*x^2)^(-8/3),x]
 

output

(-3*(1 + (c*x)/b)^(2/3)*Hypergeometric2F1[-5/3, 8/3, -2/3, -((c*x)/b)])/(5 
*b^2*x*(x*(b + c*x))^(2/3))
 

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

Time = 0.35 (sec) , antiderivative size = 343, normalized size of antiderivative = 0.77, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {1089, 1089, 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.

\(\displaystyle \int \frac {1}{\left (b x+c x^2\right )^{8/3}} \, dx\)

\(\Big \downarrow \) 1089

\(\displaystyle -\frac {14 c \int \frac {1}{\left (c x^2+b x\right )^{5/3}}dx}{5 b^2}-\frac {3 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/3}}\)

\(\Big \downarrow \) 1089

\(\displaystyle -\frac {14 c \left (-\frac {c \int \frac {1}{\left (c x^2+b x\right )^{2/3}}dx}{b^2}-\frac {3 (b+2 c x)}{2 b^2 \left (b x+c x^2\right )^{2/3}}\right )}{5 b^2}-\frac {3 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/3}}\)

\(\Big \downarrow \) 1093

\(\displaystyle -\frac {14 c \left (-\frac {c \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{2/3} \int \frac {1}{\left (-\frac {c^2 x^2}{b^2}-\frac {c x}{b}\right )^{2/3}}dx}{b^2 \left (b x+c x^2\right )^{2/3}}-\frac {3 (b+2 c x)}{2 b^2 \left (b x+c x^2\right )^{2/3}}\right )}{5 b^2}-\frac {3 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/3}}\)

\(\Big \downarrow \) 1090

\(\displaystyle -\frac {14 c \left (\frac {\sqrt [3]{2} \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{2/3} \int \frac {1}{\left (1-\frac {b^2 \left (-\frac {2 x c^2}{b^2}-\frac {c}{b}\right )^2}{c^2}\right )^{2/3}}d\left (-\frac {2 x c^2}{b^2}-\frac {c}{b}\right )}{c \left (b x+c x^2\right )^{2/3}}-\frac {3 (b+2 c x)}{2 b^2 \left (b x+c x^2\right )^{2/3}}\right )}{5 b^2}-\frac {3 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/3}}\)

\(\Big \downarrow \) 234

\(\displaystyle -\frac {14 c \left (-\frac {3 c \sqrt {-\frac {b^2 \left (-\frac {2 c^2 x}{b^2}-\frac {c}{b}\right )^2}{c^2}} \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{2/3} \int \frac {1}{\sqrt {-\frac {b^2 \left (-\frac {2 x c^2}{b^2}-\frac {c}{b}\right )^2}{c^2}}}d\sqrt [3]{1-\frac {b^2 \left (-\frac {2 x c^2}{b^2}-\frac {c}{b}\right )^2}{c^2}}}{2^{2/3} b^2 \left (-\frac {2 c^2 x}{b^2}-\frac {c}{b}\right ) \left (b x+c x^2\right )^{2/3}}-\frac {3 (b+2 c x)}{2 b^2 \left (b x+c x^2\right )^{2/3}}\right )}{5 b^2}-\frac {3 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/3}}\)

\(\Big \downarrow \) 760

\(\displaystyle -\frac {14 c \left (\frac {\sqrt [3]{2} 3^{3/4} \sqrt {2-\sqrt {3}} c \left (\frac {2 c^2 x}{b^2}+\frac {c}{b}+1\right ) \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{2/3} \sqrt {\frac {\left (-\frac {2 c^2 x}{b^2}-\frac {c}{b}\right )^2+\sqrt [3]{1-\frac {b^2 \left (-\frac {2 c^2 x}{b^2}-\frac {c}{b}\right )^2}{c^2}}+1}{\left (\frac {2 c^2 x}{b^2}+\frac {c}{b}-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\frac {2 x c^2}{b^2}+\frac {c}{b}+\sqrt {3}+1}{\frac {2 x c^2}{b^2}+\frac {c}{b}-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{b^2 \left (-\frac {2 c^2 x}{b^2}-\frac {c}{b}\right ) \sqrt {-\frac {\frac {2 c^2 x}{b^2}+\frac {c}{b}+1}{\left (\frac {2 c^2 x}{b^2}+\frac {c}{b}-\sqrt {3}+1\right )^2}} \left (b x+c x^2\right )^{2/3}}-\frac {3 (b+2 c x)}{2 b^2 \left (b x+c x^2\right )^{2/3}}\right )}{5 b^2}-\frac {3 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/3}}\)

input

Int[(b*x + c*x^2)^(-8/3),x]
 

output

(-3*(b + 2*c*x))/(5*b^2*(b*x + c*x^2)^(5/3)) - (14*c*((-3*(b + 2*c*x))/(2* 
b^2*(b*x + c*x^2)^(2/3)) + (2^(1/3)*3^(3/4)*Sqrt[2 - Sqrt[3]]*c*(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]])/(b^2*(-(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))))/(5*b^2)
 

3.1.34.3.1 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 1089

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

3.1.34.4 Maple [F]

\[\int \frac {1}{\left (c \,x^{2}+b x \right )^{\frac {8}{3}}}d x\]

input

int(1/(c*x^2+b*x)^(8/3),x)
 

output

int(1/(c*x^2+b*x)^(8/3),x)
 

3.1.34.5 Fricas [F]

\[ \int \frac {1}{\left (b x+c x^2\right )^{8/3}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x\right )}^{\frac {8}{3}}} \,d x } \]

input

integrate(1/(c*x^2+b*x)^(8/3),x, algorithm="fricas")
 

output

integral((c*x^2 + b*x)^(1/3)/(c^3*x^6 + 3*b*c^2*x^5 + 3*b^2*c*x^4 + b^3*x^ 
3), x)
 

3.1.34.6 Sympy [F]

\[ \int \frac {1}{\left (b x+c x^2\right )^{8/3}} \, dx=\int \frac {1}{\left (b x + c x^{2}\right )^{\frac {8}{3}}}\, dx \]

input

integrate(1/(c*x**2+b*x)**(8/3),x)
 

output

Integral((b*x + c*x**2)**(-8/3), x)
 

3.1.34.7 Maxima [F]

\[ \int \frac {1}{\left (b x+c x^2\right )^{8/3}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x\right )}^{\frac {8}{3}}} \,d x } \]

input

integrate(1/(c*x^2+b*x)^(8/3),x, algorithm="maxima")
 

output

integrate((c*x^2 + b*x)^(-8/3), x)
 

3.1.34.8 Giac [F]

\[ \int \frac {1}{\left (b x+c x^2\right )^{8/3}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x\right )}^{\frac {8}{3}}} \,d x } \]

input

integrate(1/(c*x^2+b*x)^(8/3),x, algorithm="giac")
 

output

integrate((c*x^2 + b*x)^(-8/3), x)
 

3.1.34.9 Mupad [B] (verification not implemented)

Time = 9.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.08 \[ \int \frac {1}{\left (b x+c x^2\right )^{8/3}} \, dx=-\frac {3\,x\,{\left (\frac {c\,x}{b}+1\right )}^{8/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{3},\frac {8}{3};\ -\frac {2}{3};\ -\frac {c\,x}{b}\right )}{5\,{\left (c\,x^2+b\,x\right )}^{8/3}} \]

input

int(1/(b*x + c*x^2)^(8/3),x)
 

output

-(3*x*((c*x)/b + 1)^(8/3)*hypergeom([-5/3, 8/3], -2/3, -(c*x)/b))/(5*(b*x 
+ c*x^2)^(8/3))