\(\int \frac {(c x)^{7/2}}{(3 a-2 a x^2)^{3/2}} \, dx\) [653]

Optimal result

Integrand size = 22, antiderivative size = 126 \[ \int \frac {(c x)^{7/2}}{\left (3 a-2 a x^2\right )^{3/2}} \, dx=\frac {c (c x)^{5/2}}{2 a \sqrt {3 a-2 a x^2}}+\frac {5 c^3 \sqrt {c x} \sqrt {3 a-2 a x^2}}{12 a^2}-\frac {5 c^{7/2} \sqrt {3-2 x^2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{\frac {2}{3}} \sqrt {c x}}{\sqrt {c}}\right ),-1\right )}{4 \sqrt [4]{6} a \sqrt {3 a-2 a x^2}} \] Output:

1/2*c*(c*x)^(5/2)/a/(-2*a*x^2+3*a)^(1/2)+5/12*c^3*(c*x)^(1/2)*(-2*a*x^2+3* 
a)^(1/2)/a^2-5/24*c^(7/2)*(-2*x^2+3)^(1/2)*EllipticF(1/3*2^(1/4)*3^(3/4)*( 
c*x)^(1/2)/c^(1/2),I)*6^(3/4)/a/(-2*a*x^2+3*a)^(1/2)
 

Mathematica [C] (verified)

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

Time = 10.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.53 \[ \int \frac {(c x)^{7/2}}{\left (3 a-2 a x^2\right )^{3/2}} \, dx=-\frac {c^3 \sqrt {c x} \left (-15+4 x^2+5 \sqrt {9-6 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {2 x^2}{3}\right )\right )}{12 a \sqrt {a \left (3-2 x^2\right )}} \] Input:

Integrate[(c*x)^(7/2)/(3*a - 2*a*x^2)^(3/2),x]
 

Output:

-1/12*(c^3*Sqrt[c*x]*(-15 + 4*x^2 + 5*Sqrt[9 - 6*x^2]*Hypergeometric2F1[1/ 
4, 1/2, 5/4, (2*x^2)/3]))/(a*Sqrt[a*(3 - 2*x^2)])
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {252, 262, 266, 765, 762}

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.

\(\displaystyle \int \frac {(c x)^{7/2}}{\left (3 a-2 a x^2\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {c (c x)^{5/2}}{2 a \sqrt {3 a-2 a x^2}}-\frac {5 c^2 \int \frac {(c x)^{3/2}}{\sqrt {3 a-2 a x^2}}dx}{4 a}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {c (c x)^{5/2}}{2 a \sqrt {3 a-2 a x^2}}-\frac {5 c^2 \left (\frac {1}{2} c^2 \int \frac {1}{\sqrt {c x} \sqrt {3 a-2 a x^2}}dx-\frac {c \sqrt {3 a-2 a x^2} \sqrt {c x}}{3 a}\right )}{4 a}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {c (c x)^{5/2}}{2 a \sqrt {3 a-2 a x^2}}-\frac {5 c^2 \left (c \int \frac {1}{\sqrt {3 a-2 a x^2}}d\sqrt {c x}-\frac {c \sqrt {3 a-2 a x^2} \sqrt {c x}}{3 a}\right )}{4 a}\)

\(\Big \downarrow \) 765

\(\displaystyle \frac {c (c x)^{5/2}}{2 a \sqrt {3 a-2 a x^2}}-\frac {5 c^2 \left (\frac {c \sqrt {3-2 x^2} \int \frac {1}{\sqrt {1-\frac {2 x^2}{3}}}d\sqrt {c x}}{\sqrt {3} \sqrt {3 a-2 a x^2}}-\frac {c \sqrt {3 a-2 a x^2} \sqrt {c x}}{3 a}\right )}{4 a}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {c (c x)^{5/2}}{2 a \sqrt {3 a-2 a x^2}}-\frac {5 c^2 \left (\frac {c^{3/2} \sqrt {3-2 x^2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{\frac {2}{3}} \sqrt {c x}}{\sqrt {c}}\right ),-1\right )}{\sqrt [4]{6} \sqrt {3 a-2 a x^2}}-\frac {c \sqrt {3 a-2 a x^2} \sqrt {c x}}{3 a}\right )}{4 a}\)

Input:

Int[(c*x)^(7/2)/(3*a - 2*a*x^2)^(3/2),x]
 

Output:

(c*(c*x)^(5/2))/(2*a*Sqrt[3*a - 2*a*x^2]) - (5*c^2*(-1/3*(c*Sqrt[c*x]*Sqrt 
[3*a - 2*a*x^2])/a + (c^(3/2)*Sqrt[3 - 2*x^2]*EllipticF[ArcSin[((2/3)^(1/4 
)*Sqrt[c*x])/Sqrt[c]], -1])/(6^(1/4)*Sqrt[3*a - 2*a*x^2])))/(4*a)
 

Defintions of rubi rules used

rule 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

rule 262

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

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 762

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) 
)*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] 
 && GtQ[a, 0]
 

rule 765

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[Sqrt[1 + b*(x^4/a)]/Sqrt 
[a + b*x^4]   Int[1/Sqrt[1 + b*(x^4/a)], x], x] /; FreeQ[{a, b}, x] && NegQ 
[b/a] &&  !GtQ[a, 0]
 

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.06

method	result	size
default	\(\frac {c^{3} \sqrt {c x}\, \sqrt {-a \left (2 x^{2}-3\right )}\, \left (5 \sqrt {\left (2 x +\sqrt {3}\, \sqrt {2}\right ) \sqrt {3}\, \sqrt {2}}\, \sqrt {\left (-2 x +\sqrt {3}\, \sqrt {2}\right ) \sqrt {3}\, \sqrt {2}}\, \sqrt {-\sqrt {3}\, \sqrt {2}\, x}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\left (2 x +\sqrt {3}\, \sqrt {2}\right ) \sqrt {3}\, \sqrt {2}}}{6}, \frac {\sqrt {2}}{2}\right )+16 x^{3}-60 x \right )}{48 x \,a^{2} \left (2 x^{2}-3\right )}\)	\(134\)
elliptic	\(\frac {\sqrt {c x}\, \sqrt {-c x a \left (2 x^{2}-3\right )}\, \left (\frac {3 c^{4} x}{4 a \sqrt {-2 \left (x^{2}-\frac {3}{2}\right ) a c x}}+\frac {c^{3} \sqrt {-2 a c \,x^{3}+3 a c x}}{6 a^{2}}-\frac {5 c^{4} \sqrt {6}\, \sqrt {3}\, \sqrt {\left (x +\frac {\sqrt {6}}{2}\right ) \sqrt {6}}\, \sqrt {-6 \left (x -\frac {\sqrt {6}}{2}\right ) \sqrt {6}}\, \sqrt {-3 \sqrt {6}\, x}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {\left (x +\frac {\sqrt {6}}{2}\right ) \sqrt {6}}}{3}, \frac {\sqrt {2}}{2}\right )}{432 a \sqrt {-2 a c \,x^{3}+3 a c x}}\right )}{c x \sqrt {-a \left (2 x^{2}-3\right )}}\)	\(172\)

Input:

int((c*x)^(7/2)/(-2*a*x^2+3*a)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/48*c^3*(c*x)^(1/2)*(-a*(2*x^2-3))^(1/2)*(5*((2*x+3^(1/2)*2^(1/2))*3^(1/2 
)*2^(1/2))^(1/2)*((-2*x+3^(1/2)*2^(1/2))*3^(1/2)*2^(1/2))^(1/2)*(-3^(1/2)* 
2^(1/2)*x)^(1/2)*EllipticF(1/6*3^(1/2)*2^(1/2)*((2*x+3^(1/2)*2^(1/2))*3^(1 
/2)*2^(1/2))^(1/2),1/2*2^(1/2))+16*x^3-60*x)/x/a^2/(2*x^2-3)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.64 \[ \int \frac {(c x)^{7/2}}{\left (3 a-2 a x^2\right )^{3/2}} \, dx=\frac {15 \, \sqrt {2} {\left (2 \, c^{3} x^{2} - 3 \, c^{3}\right )} \sqrt {-a c} {\rm weierstrassPInverse}\left (6, 0, x\right ) + 2 \, {\left (4 \, c^{3} x^{2} - 15 \, c^{3}\right )} \sqrt {-2 \, a x^{2} + 3 \, a} \sqrt {c x}}{24 \, {\left (2 \, a^{2} x^{2} - 3 \, a^{2}\right )}} \] Input:

integrate((c*x)^(7/2)/(-2*a*x^2+3*a)^(3/2),x, algorithm="fricas")
 

Output:

1/24*(15*sqrt(2)*(2*c^3*x^2 - 3*c^3)*sqrt(-a*c)*weierstrassPInverse(6, 0, 
x) + 2*(4*c^3*x^2 - 15*c^3)*sqrt(-2*a*x^2 + 3*a)*sqrt(c*x))/(2*a^2*x^2 - 3 
*a^2)
 

Sympy [A] (verification not implemented)

Time = 12.11 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.40 \[ \int \frac {(c x)^{7/2}}{\left (3 a-2 a x^2\right )^{3/2}} \, dx=\frac {\sqrt {3} c^{\frac {7}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {2 x^{2} e^{2 i \pi }}{3}} \right )}}{18 a^{\frac {3}{2}} \Gamma \left (\frac {13}{4}\right )} \] Input:

integrate((c*x)**(7/2)/(-2*a*x**2+3*a)**(3/2),x)
 

Output:

sqrt(3)*c**(7/2)*x**(9/2)*gamma(9/4)*hyper((3/2, 9/4), (13/4,), 2*x**2*exp 
_polar(2*I*pi)/3)/(18*a**(3/2)*gamma(13/4))
 

Maxima [F]

\[ \int \frac {(c x)^{7/2}}{\left (3 a-2 a x^2\right )^{3/2}} \, dx=\int { \frac {\left (c x\right )^{\frac {7}{2}}}{{\left (-2 \, a x^{2} + 3 \, a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((c*x)^(7/2)/(-2*a*x^2+3*a)^(3/2),x, algorithm="maxima")
 

Output:

integrate((c*x)^(7/2)/(-2*a*x^2 + 3*a)^(3/2), x)
 

Giac [F]

\[ \int \frac {(c x)^{7/2}}{\left (3 a-2 a x^2\right )^{3/2}} \, dx=\int { \frac {\left (c x\right )^{\frac {7}{2}}}{{\left (-2 \, a x^{2} + 3 \, a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((c*x)^(7/2)/(-2*a*x^2+3*a)^(3/2),x, algorithm="giac")
 

Output:

integrate((c*x)^(7/2)/(-2*a*x^2 + 3*a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c x)^{7/2}}{\left (3 a-2 a x^2\right )^{3/2}} \, dx=\int \frac {{\left (c\,x\right )}^{7/2}}{{\left (3\,a-2\,a\,x^2\right )}^{3/2}} \,d x \] Input:

int((c*x)^(7/2)/(3*a - 2*a*x^2)^(3/2),x)
 

Output:

int((c*x)^(7/2)/(3*a - 2*a*x^2)^(3/2), x)
 

Reduce [F]

\[ \int \frac {(c x)^{7/2}}{\left (3 a-2 a x^2\right )^{3/2}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, c^{3} \left (4 \sqrt {x}\, \sqrt {-2 x^{2}+3}\, x^{2}-30 \sqrt {x}\, \sqrt {-2 x^{2}+3}-90 \left (\int \frac {\sqrt {x}\, \sqrt {-2 x^{2}+3}}{4 x^{5}-12 x^{3}+9 x}d x \right ) x^{2}+135 \left (\int \frac {\sqrt {x}\, \sqrt {-2 x^{2}+3}}{4 x^{5}-12 x^{3}+9 x}d x \right )\right )}{12 a^{2} \left (2 x^{2}-3\right )} \] Input:

int((c*x)^(7/2)/(-2*a*x^2+3*a)^(3/2),x)
 

Output:

(sqrt(c)*sqrt(a)*c**3*(4*sqrt(x)*sqrt( - 2*x**2 + 3)*x**2 - 30*sqrt(x)*sqr 
t( - 2*x**2 + 3) - 90*int((sqrt(x)*sqrt( - 2*x**2 + 3))/(4*x**5 - 12*x**3 
+ 9*x),x)*x**2 + 135*int((sqrt(x)*sqrt( - 2*x**2 + 3))/(4*x**5 - 12*x**3 + 
 9*x),x)))/(12*a**2*(2*x**2 - 3))