\(\int x^4 (a+b x^2)^{3/4} \, dx\) [847]

Optimal result

Integrand size = 15, antiderivative size = 143 \[ \int x^4 \left (a+b x^2\right )^{3/4} \, dx=\frac {8 a^3 x}{65 b^2 \sqrt [4]{a+b x^2}}-\frac {4 a^2 x \left (a+b x^2\right )^{3/4}}{65 b^2}+\frac {2 a x^3 \left (a+b x^2\right )^{3/4}}{39 b}+\frac {2}{13} x^5 \left (a+b x^2\right )^{3/4}-\frac {8 a^{7/2} \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{65 b^{5/2} \sqrt [4]{a+b x^2}} \] Output:

8/65*a^3*x/b^2/(b*x^2+a)^(1/4)-4/65*a^2*x*(b*x^2+a)^(3/4)/b^2+2/39*a*x^3*( 
b*x^2+a)^(3/4)/b+2/13*x^5*(b*x^2+a)^(3/4)-8/65*a^(7/2)*(1+b*x^2/a)^(1/4)*E 
llipticE(sin(1/2*arctan(b^(1/2)*x/a^(1/2))),2^(1/2))/b^(5/2)/(b*x^2+a)^(1/ 
4)
 

Mathematica [C] (verified)

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

Time = 7.26 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.65 \[ \int x^4 \left (a+b x^2\right )^{3/4} \, dx=\frac {2 x \left (a+b x^2\right )^{3/4} \left (\left (1+\frac {b x^2}{a}\right )^{3/4} \left (-2 a^2+a b x^2+3 b^2 x^4\right )+2 a^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{39 b^2 \left (1+\frac {b x^2}{a}\right )^{3/4}} \] Input:

Integrate[x^4*(a + b*x^2)^(3/4),x]
 

Output:

(2*x*(a + b*x^2)^(3/4)*((1 + (b*x^2)/a)^(3/4)*(-2*a^2 + a*b*x^2 + 3*b^2*x^ 
4) + 2*a^2*Hypergeometric2F1[-3/4, 1/2, 3/2, -((b*x^2)/a)]))/(39*b^2*(1 + 
(b*x^2)/a)^(3/4))
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {248, 262, 262, 227, 225, 212}

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.

\(\displaystyle \int x^4 \left (a+b x^2\right )^{3/4} \, dx\)

\(\Big \downarrow \) 248

\(\displaystyle \frac {3}{13} a \int \frac {x^4}{\sqrt [4]{b x^2+a}}dx+\frac {2}{13} x^5 \left (a+b x^2\right )^{3/4}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {3}{13} a \left (\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {2 a \int \frac {x^2}{\sqrt [4]{b x^2+a}}dx}{3 b}\right )+\frac {2}{13} x^5 \left (a+b x^2\right )^{3/4}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {3}{13} a \left (\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac {2 a \int \frac {1}{\sqrt [4]{b x^2+a}}dx}{5 b}\right )}{3 b}\right )+\frac {2}{13} x^5 \left (a+b x^2\right )^{3/4}\)

\(\Big \downarrow \) 227

\(\displaystyle \frac {3}{13} a \left (\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^2}{a}+1} \int \frac {1}{\sqrt [4]{\frac {b x^2}{a}+1}}dx}{5 b \sqrt [4]{a+b x^2}}\right )}{3 b}\right )+\frac {2}{13} x^5 \left (a+b x^2\right )^{3/4}\)

\(\Big \downarrow \) 225

\(\displaystyle \frac {3}{13} a \left (\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx\right )}{5 b \sqrt [4]{a+b x^2}}\right )}{3 b}\right )+\frac {2}{13} x^5 \left (a+b x^2\right )^{3/4}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {3}{13} a \left (\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\frac {2 \sqrt {a} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b}}\right )}{5 b \sqrt [4]{a+b x^2}}\right )}{3 b}\right )+\frac {2}{13} x^5 \left (a+b x^2\right )^{3/4}\)

Input:

Int[x^4*(a + b*x^2)^(3/4),x]
 

Output:

(2*x^5*(a + b*x^2)^(3/4))/13 + (3*a*((2*x^3*(a + b*x^2)^(3/4))/(9*b) - (2* 
a*((2*x*(a + b*x^2)^(3/4))/(5*b) - (2*a*(1 + (b*x^2)/a)^(1/4)*((2*x)/(1 + 
(b*x^2)/a)^(1/4) - (2*Sqrt[a]*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2]) 
/Sqrt[b]))/(5*b*(a + b*x^2)^(1/4))))/(3*b)))/13
 

Defintions of rubi rules used

rule 212

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) 
)*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 225

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)) 
, x] - Simp[a   Int[1/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b}, x] && GtQ[ 
a, 0] && PosQ[b/a]
 

rule 227

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(1/4)/( 
a + b*x^2)^(1/4)   Int[1/(1 + b*(x^2/a))^(1/4), 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]

\[\int x^{4} \left (b \,x^{2}+a \right )^{\frac {3}{4}}d x\]

Input:

int(x^4*(b*x^2+a)^(3/4),x)
 

Output:

int(x^4*(b*x^2+a)^(3/4),x)
 

Fricas [F]

\[ \int x^4 \left (a+b x^2\right )^{3/4} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {3}{4}} x^{4} \,d x } \] Input:

integrate(x^4*(b*x^2+a)^(3/4),x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^(3/4)*x^4, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.59 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.20 \[ \int x^4 \left (a+b x^2\right )^{3/4} \, dx=\frac {a^{\frac {3}{4}} x^{5} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} \] Input:

integrate(x**4*(b*x**2+a)**(3/4),x)
 

Output:

a**(3/4)*x**5*hyper((-3/4, 5/2), (7/2,), b*x**2*exp_polar(I*pi)/a)/5
 

Maxima [F]

\[ \int x^4 \left (a+b x^2\right )^{3/4} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {3}{4}} x^{4} \,d x } \] Input:

integrate(x^4*(b*x^2+a)^(3/4),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^(3/4)*x^4, x)
 

Giac [F]

\[ \int x^4 \left (a+b x^2\right )^{3/4} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {3}{4}} x^{4} \,d x } \] Input:

integrate(x^4*(b*x^2+a)^(3/4),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((b*x^2 + a)^(3/4)*x^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^4 \left (a+b x^2\right )^{3/4} \, dx=\int x^4\,{\left (b\,x^2+a\right )}^{3/4} \,d x \] Input:

int(x^4*(a + b*x^2)^(3/4),x)
 

Output:

int(x^4*(a + b*x^2)^(3/4), x)
 

Reduce [F]

\[ \int x^4 \left (a+b x^2\right )^{3/4} \, dx=\frac {-\frac {4 \left (b \,x^{2}+a \right )^{\frac {3}{4}} a^{2} x}{65}+\frac {2 \left (b \,x^{2}+a \right )^{\frac {3}{4}} a b \,x^{3}}{39}+\frac {2 \left (b \,x^{2}+a \right )^{\frac {3}{4}} b^{2} x^{5}}{13}+\frac {4 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}}d x \right ) a^{3}}{65}}{b^{2}} \] Input:

int(x^4*(b*x^2+a)^(3/4),x)
 

Output:

(2*( - 6*(a + b*x**2)**(3/4)*a**2*x + 5*(a + b*x**2)**(3/4)*a*b*x**3 + 15* 
(a + b*x**2)**(3/4)*b**2*x**5 + 6*int((a + b*x**2)**(3/4)/(a + b*x**2),x)* 
a**3))/(195*b**2)