\(\int (d x)^m (a+b x^3+c x^6)^{3/2} \, dx\) [237]

Optimal result

Integrand size = 22, antiderivative size = 158 \[ \int (d x)^m \left (a+b x^3+c x^6\right )^{3/2} \, dx=\frac {a (d x)^{1+m} \sqrt {a+b x^3+c x^6} \operatorname {AppellF1}\left (\frac {1+m}{3},-\frac {3}{2},-\frac {3}{2},\frac {4+m}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{d (1+m) \sqrt {1+\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}}} \] Output:

a*(d*x)^(1+m)*(c*x^6+b*x^3+a)^(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)))/d/(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)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(357\) vs. \(2(158)=316\).

Time = 2.35 (sec) , antiderivative size = 357, normalized size of antiderivative = 2.26 \[ \int (d x)^m \left (a+b x^3+c x^6\right )^{3/2} \, dx=\frac {x (d x)^m \sqrt {a+b x^3+c x^6} \left (a \left (28+11 m+m^2\right ) \operatorname {AppellF1}\left (\frac {1+m}{3},-\frac {1}{2},-\frac {1}{2},\frac {4+m}{3},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right )+(1+m) x^3 \left (b (7+m) \operatorname {AppellF1}\left (\frac {4+m}{3},-\frac {1}{2},-\frac {1}{2},\frac {7+m}{3},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right )+c (4+m) x^3 \operatorname {AppellF1}\left (\frac {7+m}{3},-\frac {1}{2},-\frac {1}{2},\frac {10+m}{3},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right )\right )\right )}{(1+m) (4+m) (7+m) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{b+\sqrt {b^2-4 a c}}}} \] Input:

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

Output:

(x*(d*x)^m*Sqrt[a + b*x^3 + c*x^6]*(a*(28 + 11*m + m^2)*AppellF1[(1 + m)/3 
, -1/2, -1/2, (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])] + (1 + m)*x^3*(b*(7 + m)*AppellF1[(4 + m)/3, -1/2, 
-1/2, (7 + m)/3, (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^3)/(-b + Sqrt[ 
b^2 - 4*a*c])] + c*(4 + m)*x^3*AppellF1[(7 + m)/3, -1/2, -1/2, (10 + m)/3, 
 (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^3)/(-b + Sqrt[b^2 - 4*a*c])])) 
)/((1 + m)*(4 + m)*(7 + m)*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b - Sqr 
t[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b + Sqrt[b^2 - 4* 
a*c])])
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 158, 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.

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

\(\Big \downarrow \) 1721

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

\(\Big \downarrow \) 1012

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

Input:

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

Output:

(a*(d*x)^(1 + m)*Sqrt[a + b*x^3 + c*x^6]*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])])/(d*(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])])
 

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]

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

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]

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

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

Output:

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

Sympy [F]

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

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

(d**m*(8*x**m*sqrt(a + b*x**3 + c*x**6)*a*c*m**2*x + 160*x**m*sqrt(a + b*x 
**3 + c*x**6)*a*c*m*x + 728*x**m*sqrt(a + b*x**3 + c*x**6)*a*c*x + 54*x**m 
*sqrt(a + b*x**3 + c*x**6)*b**2*x + 8*x**m*sqrt(a + b*x**3 + c*x**6)*b*c*m 
**2*x**4 + 124*x**m*sqrt(a + b*x**3 + c*x**6)*b*c*m*x**4 + 368*x**m*sqrt(a 
 + b*x**3 + c*x**6)*b*c*x**4 + 8*x**m*sqrt(a + b*x**3 + c*x**6)*c**2*m**2* 
x**7 + 88*x**m*sqrt(a + b*x**3 + c*x**6)*c**2*m*x**7 + 224*x**m*sqrt(a + b 
*x**3 + c*x**6)*c**2*x**7 + 216*int((x**m*sqrt(a + b*x**3 + c*x**6)*x**3)/ 
(a*m**3 + 21*a*m**2 + 138*a*m + 280*a + b*m**3*x**3 + 21*b*m**2*x**3 + 138 
*b*m*x**3 + 280*b*x**3 + c*m**3*x**6 + 21*c*m**2*x**6 + 138*c*m*x**6 + 280 
*c*x**6),x)*a*b*c*m**4 + 5724*int((x**m*sqrt(a + b*x**3 + c*x**6)*x**3)/(a 
*m**3 + 21*a*m**2 + 138*a*m + 280*a + b*m**3*x**3 + 21*b*m**2*x**3 + 138*b 
*m*x**3 + 280*b*x**3 + c*m**3*x**6 + 21*c*m**2*x**6 + 138*c*m*x**6 + 280*c 
*x**6),x)*a*b*c*m**3 + 54756*int((x**m*sqrt(a + b*x**3 + c*x**6)*x**3)/(a* 
m**3 + 21*a*m**2 + 138*a*m + 280*a + b*m**3*x**3 + 21*b*m**2*x**3 + 138*b* 
m*x**3 + 280*b*x**3 + c*m**3*x**6 + 21*c*m**2*x**6 + 138*c*m*x**6 + 280*c* 
x**6),x)*a*b*c*m**2 + 224424*int((x**m*sqrt(a + b*x**3 + c*x**6)*x**3)/(a* 
m**3 + 21*a*m**2 + 138*a*m + 280*a + b*m**3*x**3 + 21*b*m**2*x**3 + 138*b* 
m*x**3 + 280*b*x**3 + c*m**3*x**6 + 21*c*m**2*x**6 + 138*c*m*x**6 + 280*c* 
x**6),x)*a*b*c*m + 332640*int((x**m*sqrt(a + b*x**3 + c*x**6)*x**3)/(a*m** 
3 + 21*a*m**2 + 138*a*m + 280*a + b*m**3*x**3 + 21*b*m**2*x**3 + 138*b*...