\(\int x^4 (a-b x^2)^{5/4} \, dx\) [865]

Optimal result

Integrand size = 16, antiderivative size = 148 \[ \int x^4 \left (a-b x^2\right )^{5/4} \, dx=-\frac {4 a^3 x \sqrt [4]{a-b x^2}}{231 b^2}-\frac {2 a^2 x^3 \sqrt [4]{a-b x^2}}{231 b}+\frac {2}{33} a x^5 \sqrt [4]{a-b x^2}+\frac {2}{15} x^5 \left (a-b x^2\right )^{5/4}+\frac {8 a^{9/2} \left (1-\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{231 b^{5/2} \left (a-b x^2\right )^{3/4}} \] Output:

-4/231*a^3*x*(-b*x^2+a)^(1/4)/b^2-2/231*a^2*x^3*(-b*x^2+a)^(1/4)/b+2/33*a* 
x^5*(-b*x^2+a)^(1/4)+2/15*x^5*(-b*x^2+a)^(5/4)+8/231*a^(9/2)*(1-b*x^2/a)^( 
3/4)*InverseJacobiAM(1/2*arcsin(b^(1/2)*x/a^(1/2)),2^(1/2))/b^(5/2)/(-b*x^ 
2+a)^(3/4)
 

Mathematica [C] (verified)

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

Time = 7.80 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.55 \[ \int x^4 \left (a-b x^2\right )^{5/4} \, dx=\frac {2 x \sqrt [4]{a-b x^2} \left (-\left (a-b x^2\right )^2 \left (6 a+11 b x^2\right )+\frac {6 a^3 \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{2},\frac {3}{2},\frac {b x^2}{a}\right )}{\sqrt [4]{1-\frac {b x^2}{a}}}\right )}{165 b^2} \] Input:

Integrate[x^4*(a - b*x^2)^(5/4),x]
 

Output:

(2*x*(a - b*x^2)^(1/4)*(-((a - b*x^2)^2*(6*a + 11*b*x^2)) + (6*a^3*Hyperge 
ometric2F1[-5/4, 1/2, 3/2, (b*x^2)/a])/(1 - (b*x^2)/a)^(1/4)))/(165*b^2)
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {248, 248, 262, 262, 231, 230}

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.

\(\displaystyle \int x^4 \left (a-b x^2\right )^{5/4} \, dx\)

\(\Big \downarrow \) 248

\(\displaystyle \frac {1}{3} a \int x^4 \sqrt [4]{a-b x^2}dx+\frac {2}{15} x^5 \left (a-b x^2\right )^{5/4}\)

\(\Big \downarrow \) 248

\(\displaystyle \frac {1}{3} a \left (\frac {1}{11} a \int \frac {x^4}{\left (a-b x^2\right )^{3/4}}dx+\frac {2}{11} x^5 \sqrt [4]{a-b x^2}\right )+\frac {2}{15} x^5 \left (a-b x^2\right )^{5/4}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {1}{3} a \left (\frac {1}{11} a \left (\frac {6 a \int \frac {x^2}{\left (a-b x^2\right )^{3/4}}dx}{7 b}-\frac {2 x^3 \sqrt [4]{a-b x^2}}{7 b}\right )+\frac {2}{11} x^5 \sqrt [4]{a-b x^2}\right )+\frac {2}{15} x^5 \left (a-b x^2\right )^{5/4}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {1}{3} a \left (\frac {1}{11} a \left (\frac {6 a \left (\frac {2 a \int \frac {1}{\left (a-b x^2\right )^{3/4}}dx}{3 b}-\frac {2 x \sqrt [4]{a-b x^2}}{3 b}\right )}{7 b}-\frac {2 x^3 \sqrt [4]{a-b x^2}}{7 b}\right )+\frac {2}{11} x^5 \sqrt [4]{a-b x^2}\right )+\frac {2}{15} x^5 \left (a-b x^2\right )^{5/4}\)

\(\Big \downarrow \) 231

\(\displaystyle \frac {1}{3} a \left (\frac {1}{11} a \left (\frac {6 a \left (\frac {2 a \left (1-\frac {b x^2}{a}\right )^{3/4} \int \frac {1}{\left (1-\frac {b x^2}{a}\right )^{3/4}}dx}{3 b \left (a-b x^2\right )^{3/4}}-\frac {2 x \sqrt [4]{a-b x^2}}{3 b}\right )}{7 b}-\frac {2 x^3 \sqrt [4]{a-b x^2}}{7 b}\right )+\frac {2}{11} x^5 \sqrt [4]{a-b x^2}\right )+\frac {2}{15} x^5 \left (a-b x^2\right )^{5/4}\)

\(\Big \downarrow \) 230

\(\displaystyle \frac {1}{3} a \left (\frac {1}{11} a \left (\frac {6 a \left (\frac {4 a^{3/2} \left (1-\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{3 b^{3/2} \left (a-b x^2\right )^{3/4}}-\frac {2 x \sqrt [4]{a-b x^2}}{3 b}\right )}{7 b}-\frac {2 x^3 \sqrt [4]{a-b x^2}}{7 b}\right )+\frac {2}{11} x^5 \sqrt [4]{a-b x^2}\right )+\frac {2}{15} x^5 \left (a-b x^2\right )^{5/4}\)

Input:

Int[x^4*(a - b*x^2)^(5/4),x]
 

Output:

(2*x^5*(a - b*x^2)^(5/4))/15 + (a*((2*x^5*(a - b*x^2)^(1/4))/11 + (a*((-2* 
x^3*(a - b*x^2)^(1/4))/(7*b) + (6*a*((-2*x*(a - b*x^2)^(1/4))/(3*b) + (4*a 
^(3/2)*(1 - (b*x^2)/a)^(3/4)*EllipticF[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/ 
(3*b^(3/2)*(a - b*x^2)^(3/4))))/(7*b)))/11))/3
 

Defintions of rubi rules used

rule 230

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[-b/a, 2] 
))*EllipticF[(1/2)*ArcSin[Rt[-b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ 
[a, 0] && NegQ[b/a]
 

rule 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(3/4)/( 
a + b*x^2)^(3/4)   Int[1/(1 + b*(x^2/a))^(3/4), x], x] /; FreeQ[{a, b}, x] 
&& PosQ[a]
 

rule 248

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

Maple [F]

\[\int x^{4} \left (-b \,x^{2}+a \right )^{\frac {5}{4}}d x\]

Input:

int(x^4*(-b*x^2+a)^(5/4),x)
 

Output:

int(x^4*(-b*x^2+a)^(5/4),x)
 

Fricas [F]

\[ \int x^4 \left (a-b x^2\right )^{5/4} \, dx=\int { {\left (-b x^{2} + a\right )}^{\frac {5}{4}} x^{4} \,d x } \] Input:

integrate(x^4*(-b*x^2+a)^(5/4),x, algorithm="fricas")
 

Output:

integral(-(b*x^6 - a*x^4)*(-b*x^2 + a)^(1/4), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.70 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.21 \[ \int x^4 \left (a-b x^2\right )^{5/4} \, dx=\frac {a^{\frac {5}{4}} x^{5} {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{5} \] Input:

integrate(x**4*(-b*x**2+a)**(5/4),x)
 

Output:

a**(5/4)*x**5*hyper((-5/4, 5/2), (7/2,), b*x**2*exp_polar(2*I*pi)/a)/5
 

Maxima [F]

\[ \int x^4 \left (a-b x^2\right )^{5/4} \, dx=\int { {\left (-b x^{2} + a\right )}^{\frac {5}{4}} x^{4} \,d x } \] Input:

integrate(x^4*(-b*x^2+a)^(5/4),x, algorithm="maxima")
 

Output:

integrate((-b*x^2 + a)^(5/4)*x^4, x)
 

Giac [F]

\[ \int x^4 \left (a-b x^2\right )^{5/4} \, dx=\int { {\left (-b x^{2} + a\right )}^{\frac {5}{4}} x^{4} \,d x } \] Input:

integrate(x^4*(-b*x^2+a)^(5/4),x, algorithm="giac")
 

Output:

integrate((-b*x^2 + a)^(5/4)*x^4, x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int x^4 \left (a-b x^2\right )^{5/4} \, dx=\int x^4\,{\left (a-b\,x^2\right )}^{5/4} \,d x \] Input:

int(x^4*(a - b*x^2)^(5/4),x)
 

Output:

int(x^4*(a - b*x^2)^(5/4), x)
 

Reduce [F]

\[ \int x^4 \left (a-b x^2\right )^{5/4} \, dx=\frac {-\frac {4 \left (-b \,x^{2}+a \right )^{\frac {1}{4}} a^{3} x}{231}-\frac {2 \left (-b \,x^{2}+a \right )^{\frac {1}{4}} a^{2} b \,x^{3}}{231}+\frac {32 \left (-b \,x^{2}+a \right )^{\frac {1}{4}} a \,b^{2} x^{5}}{165}-\frac {2 \left (-b \,x^{2}+a \right )^{\frac {1}{4}} b^{3} x^{7}}{15}+\frac {4 \left (\int \frac {1}{\left (-b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) a^{4}}{231}}{b^{2}} \] Input:

int(x^4*(-b*x^2+a)^(5/4),x)
 

Output:

(2*( - 10*(a - b*x**2)**(1/4)*a**3*x - 5*(a - b*x**2)**(1/4)*a**2*b*x**3 + 
 112*(a - b*x**2)**(1/4)*a*b**2*x**5 - 77*(a - b*x**2)**(1/4)*b**3*x**7 + 
10*int((a - b*x**2)**(1/4)/(a - b*x**2),x)*a**4))/(1155*b**2)