\(\int \frac {(a+b x^3+c x^6)^{3/2}}{x^2} \, dx\) [202]

Optimal result

Integrand size = 20, antiderivative size = 139 \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^2} \, dx=-\frac {a \sqrt {a+b x^3+c x^6} \operatorname {AppellF1}\left (-\frac {1}{3},-\frac {3}{2},-\frac {3}{2},\frac {2}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{x \sqrt {1+\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}}} \] Output:

-a*(c*x^6+b*x^3+a)^(1/2)*AppellF1(-1/3,-3/2,-3/2,2/3,-2*c*x^3/(b-(-4*a*c+b 
^2)^(1/2)),-2*c*x^3/(b+(-4*a*c+b^2)^(1/2)))/x/(1+2*c*x^3/(b-(-4*a*c+b^2)^( 
1/2)))^(1/2)/(1+2*c*x^3/(b+(-4*a*c+b^2)^(1/2)))^(1/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(379\) vs. \(2(139)=278\).

Time = 10.59 (sec) , antiderivative size = 379, normalized size of antiderivative = 2.73 \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^2} \, dx=\frac {10 \left (-80 a^2-61 a b x^3+19 b^2 x^6-70 a c x^6+29 b c x^9+10 c^2 x^{12}\right )+810 a b x^3 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},\frac {1}{2},\frac {5}{3},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right )+27 \left (b^2+20 a c\right ) x^6 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},\frac {1}{2},\frac {8}{3},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right )}{800 x \sqrt {a+b x^3+c x^6}} \] Input:

Integrate[(a + b*x^3 + c*x^6)^(3/2)/x^2,x]
 

Output:

(10*(-80*a^2 - 61*a*b*x^3 + 19*b^2*x^6 - 70*a*c*x^6 + 29*b*c*x^9 + 10*c^2* 
x^12) + 810*a*b*x^3*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b - Sqrt[b^2 - 
 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b + Sqrt[b^2 - 4*a*c])]* 
AppellF1[2/3, 1/2, 1/2, 5/3, (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^3) 
/(-b + Sqrt[b^2 - 4*a*c])] + 27*(b^2 + 20*a*c)*x^6*Sqrt[(b - Sqrt[b^2 - 4* 
a*c] + 2*c*x^3)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c 
*x^3)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[5/3, 1/2, 1/2, 8/3, (-2*c*x^3)/(b 
+ Sqrt[b^2 - 4*a*c]), (2*c*x^3)/(-b + Sqrt[b^2 - 4*a*c])])/(800*x*Sqrt[a + 
 b*x^3 + c*x^6])
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1721, 1012}

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.

\(\displaystyle \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^2} \, dx\)

\(\Big \downarrow \) 1721

\(\displaystyle \frac {a \sqrt {a+b x^3+c x^6} \int \frac {\left (\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1\right )^{3/2} \left (\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}+1\right )^{3/2}}{x^2}dx}{\sqrt {\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1}}\)

\(\Big \downarrow \) 1012

\(\displaystyle -\frac {a \sqrt {a+b x^3+c x^6} \operatorname {AppellF1}\left (-\frac {1}{3},-\frac {3}{2},-\frac {3}{2},\frac {2}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{x \sqrt {\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1}}\)

Input:

Int[(a + b*x^3 + c*x^6)^(3/2)/x^2,x]
 

Output:

-((a*Sqrt[a + b*x^3 + c*x^6]*AppellF1[-1/3, -3/2, -3/2, 2/3, (-2*c*x^3)/(b 
 - Sqrt[b^2 - 4*a*c]), (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c])])/(x*Sqrt[1 + (2 
*c*x^3)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^3)/(b + Sqrt[b^2 - 4*a*c] 
)]))
 

Defintions of rubi rules used

rule 1012

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m 
+ 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, 
 b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n 
 - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 1721

Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x 
_Symbol] :> Simp[a^IntPart[p]*((a + b*x^n + c*x^(2*n))^FracPart[p]/((1 + 2* 
c*(x^n/(b + Rt[b^2 - 4*a*c, 2])))^FracPart[p]*(1 + 2*c*(x^n/(b - Rt[b^2 - 4 
*a*c, 2])))^FracPart[p]))   Int[(d*x)^m*(1 + 2*c*(x^n/(b + Sqrt[b^2 - 4*a*c 
])))^p*(1 + 2*c*(x^n/(b - Sqrt[b^2 - 4*a*c])))^p, x], x] /; FreeQ[{a, b, c, 
 d, m, n, p}, x] && EqQ[n2, 2*n]
 

Maple [F]

\[\int \frac {\left (c \,x^{6}+b \,x^{3}+a \right )^{\frac {3}{2}}}{x^{2}}d x\]

Input:

int((c*x^6+b*x^3+a)^(3/2)/x^2,x)
 

Output:

int((c*x^6+b*x^3+a)^(3/2)/x^2,x)
 

Fricas [F]

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

integrate((c*x^6+b*x^3+a)^(3/2)/x^2,x, algorithm="fricas")
 

Output:

integral((c*x^6 + b*x^3 + a)^(3/2)/x^2, x)
 

Sympy [F]

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

integrate((c*x**6+b*x**3+a)**(3/2)/x**2,x)
 

Output:

Integral((a + b*x**3 + c*x**6)**(3/2)/x**2, x)
 

Maxima [F]

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

integrate((c*x^6+b*x^3+a)^(3/2)/x^2,x, algorithm="maxima")
 

Output:

integrate((c*x^6 + b*x^3 + a)^(3/2)/x^2, x)
 

Giac [F]

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

integrate((c*x^6+b*x^3+a)^(3/2)/x^2,x, algorithm="giac")
 

Output:

integrate((c*x^6 + b*x^3 + a)^(3/2)/x^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^2} \, dx=\int \frac {{\left (c\,x^6+b\,x^3+a\right )}^{3/2}}{x^2} \,d x \] Input:

int((a + b*x^3 + c*x^6)^(3/2)/x^2,x)
 

Output:

int((a + b*x^3 + c*x^6)^(3/2)/x^2, x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^3+c x^6\right )^{3/2}}{x^2} \, dx=\frac {488 \sqrt {c \,x^{6}+b \,x^{3}+a}\, a +38 \sqrt {c \,x^{6}+b \,x^{3}+a}\, b \,x^{3}+20 \sqrt {c \,x^{6}+b \,x^{3}+a}\, c \,x^{6}+648 \left (\int \frac {\sqrt {c \,x^{6}+b \,x^{3}+a}}{c \,x^{8}+b \,x^{5}+a \,x^{2}}d x \right ) a^{2} x -756 \left (\int \frac {\sqrt {c \,x^{6}+b \,x^{3}+a}\, x^{4}}{c \,x^{6}+b \,x^{3}+a}d x \right ) a c x +27 \left (\int \frac {\sqrt {c \,x^{6}+b \,x^{3}+a}\, x^{4}}{c \,x^{6}+b \,x^{3}+a}d x \right ) b^{2} x}{160 x} \] Input:

int((c*x^6+b*x^3+a)^(3/2)/x^2,x)
 

Output:

(488*sqrt(a + b*x**3 + c*x**6)*a + 38*sqrt(a + b*x**3 + c*x**6)*b*x**3 + 2 
0*sqrt(a + b*x**3 + c*x**6)*c*x**6 + 648*int(sqrt(a + b*x**3 + c*x**6)/(a* 
x**2 + b*x**5 + c*x**8),x)*a**2*x - 756*int((sqrt(a + b*x**3 + c*x**6)*x** 
4)/(a + b*x**3 + c*x**6),x)*a*c*x + 27*int((sqrt(a + b*x**3 + c*x**6)*x**4 
)/(a + b*x**3 + c*x**6),x)*b**2*x)/(160*x)