\(\int \frac {(d x)^m}{(a+b x^3+c x^6)^{3/2}} \, dx\) [240]

Optimal result

Integrand size = 22, antiderivative size = 160 \[ \int \frac {(d x)^m}{\left (a+b x^3+c x^6\right )^{3/2}} \, dx=\frac {(d x)^{1+m} \sqrt {1+\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1+m}{3},\frac {3}{2},\frac {3}{2},\frac {4+m}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{a d (1+m) \sqrt {a+b x^3+c x^6}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 10.10 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.38 \[ \int \frac {(d x)^m}{\left (a+b x^3+c x^6\right )^{3/2}} \, dx=\frac {x (d x)^m \left (-b+\sqrt {b^2-4 a c}-2 c x^3\right ) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{b-\sqrt {b^2-4 a c}}} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{b+\sqrt {b^2-4 a c}}\right )^{3/2} \operatorname {AppellF1}\left (\frac {1+m}{3},\frac {3}{2},\frac {3}{2},\frac {4+m}{3},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right )}{\left (-b+\sqrt {b^2-4 a c}\right ) (1+m) \left (a+b x^3+c x^6\right )^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, 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.

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

\(\Big \downarrow \) 1721

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

\(\Big \downarrow \) 1012

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

Input:

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

Output:

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

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(sqrt(c*x^6 + b*x^3 + a)*(d*x)^m/(c^2*x^12 + 2*b*c*x^9 + (b^2 + 2* 
a*c)*x^6 + 2*a*b*x^3 + a^2), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(d x)^m}{\left (a+b x^3+c x^6\right )^{3/2}} \, dx=d^{m} \left (\int \frac {x^{m} \sqrt {c \,x^{6}+b \,x^{3}+a}}{c^{2} x^{12}+2 b c \,x^{9}+2 a c \,x^{6}+b^{2} x^{6}+2 a b \,x^{3}+a^{2}}d x \right ) \] Input:

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

Output:

d**m*int((x**m*sqrt(a + b*x**3 + c*x**6))/(a**2 + 2*a*b*x**3 + 2*a*c*x**6 
+ b**2*x**6 + 2*b*c*x**9 + c**2*x**12),x)