\(\int \frac {x^4 (c+d x^2)}{\sqrt [4]{a+b x^2}} \, dx\) [371]

Optimal result

Integrand size = 22, antiderivative size = 180 \[ \int \frac {x^4 \left (c+d x^2\right )}{\sqrt [4]{a+b x^2}} \, dx=\frac {8 a^2 (13 b c-10 a d) x}{195 b^3 \sqrt [4]{a+b x^2}}-\frac {4 a (13 b c-10 a d) x \left (a+b x^2\right )^{3/4}}{195 b^3}+\frac {2 (13 b c-10 a d) x^3 \left (a+b x^2\right )^{3/4}}{117 b^2}+\frac {2 d x^5 \left (a+b x^2\right )^{3/4}}{13 b}-\frac {8 a^{5/2} (13 b c-10 a d) \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{195 b^{7/2} \sqrt [4]{a+b x^2}} \] Output:

8/195*a^2*(-10*a*d+13*b*c)*x/b^3/(b*x^2+a)^(1/4)-4/195*a*(-10*a*d+13*b*c)* 
x*(b*x^2+a)^(3/4)/b^3+2/117*(-10*a*d+13*b*c)*x^3*(b*x^2+a)^(3/4)/b^2+2/13* 
d*x^5*(b*x^2+a)^(3/4)/b-8/195*a^(5/2)*(-10*a*d+13*b*c)*(1+b*x^2/a)^(1/4)*E 
llipticE(sin(1/2*arctan(b^(1/2)*x/a^(1/2))),2^(1/2))/b^(7/2)/(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 = 10.07 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.63 \[ \int \frac {x^4 \left (c+d x^2\right )}{\sqrt [4]{a+b x^2}} \, dx=\frac {2 x \left (\left (a+b x^2\right ) \left (60 a^2 d+5 b^2 x^2 \left (13 c+9 d x^2\right )-2 a b \left (39 c+25 d x^2\right )\right )+6 a^2 (13 b c-10 a d) \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{585 b^3 \sqrt [4]{a+b x^2}} \] Input:

Integrate[(x^4*(c + d*x^2))/(a + b*x^2)^(1/4),x]
 

Output:

(2*x*((a + b*x^2)*(60*a^2*d + 5*b^2*x^2*(13*c + 9*d*x^2) - 2*a*b*(39*c + 2 
5*d*x^2)) + 6*a^2*(13*b*c - 10*a*d)*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F 
1[1/4, 1/2, 3/2, -((b*x^2)/a)]))/(585*b^3*(a + b*x^2)^(1/4))
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {363, 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.

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

\(\Big \downarrow \) 363

\(\displaystyle \frac {(13 b c-10 a d) \int \frac {x^4}{\sqrt [4]{b x^2+a}}dx}{13 b}+\frac {2 d x^5 \left (a+b x^2\right )^{3/4}}{13 b}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {(13 b c-10 a d) \left (\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {2 a \int \frac {x^2}{\sqrt [4]{b x^2+a}}dx}{3 b}\right )}{13 b}+\frac {2 d x^5 \left (a+b x^2\right )^{3/4}}{13 b}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {(13 b c-10 a d) \left (\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac {2 a \int \frac {1}{\sqrt [4]{b x^2+a}}dx}{5 b}\right )}{3 b}\right )}{13 b}+\frac {2 d x^5 \left (a+b x^2\right )^{3/4}}{13 b}\)

\(\Big \downarrow \) 227

\(\displaystyle \frac {(13 b c-10 a d) \left (\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^2}{a}+1} \int \frac {1}{\sqrt [4]{\frac {b x^2}{a}+1}}dx}{5 b \sqrt [4]{a+b x^2}}\right )}{3 b}\right )}{13 b}+\frac {2 d x^5 \left (a+b x^2\right )^{3/4}}{13 b}\)

\(\Big \downarrow \) 225

\(\displaystyle \frac {(13 b c-10 a d) \left (\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx\right )}{5 b \sqrt [4]{a+b x^2}}\right )}{3 b}\right )}{13 b}+\frac {2 d x^5 \left (a+b x^2\right )^{3/4}}{13 b}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {(13 b c-10 a d) \left (\frac {2 x^3 \left (a+b x^2\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\frac {2 \sqrt {a} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b}}\right )}{5 b \sqrt [4]{a+b x^2}}\right )}{3 b}\right )}{13 b}+\frac {2 d x^5 \left (a+b x^2\right )^{3/4}}{13 b}\)

Input:

Int[(x^4*(c + d*x^2))/(a + b*x^2)^(1/4),x]
 

Output:

(2*d*x^5*(a + b*x^2)^(3/4))/(13*b) + ((13*b*c - 10*a*d)*((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*b)
 

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 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 363

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), 
 x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3))   Int[(e*x)^ 
m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d 
, 0] && NeQ[m + 2*p + 3, 0]
 

Maple [F]

\[\int \frac {x^{4} \left (x^{2} d +c \right )}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}}d x\]

Input:

int(x^4*(d*x^2+c)/(b*x^2+a)^(1/4),x)
 

Output:

int(x^4*(d*x^2+c)/(b*x^2+a)^(1/4),x)
 

Fricas [F]

\[ \int \frac {x^4 \left (c+d x^2\right )}{\sqrt [4]{a+b x^2}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} x^{4}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate(x^4*(d*x^2+c)/(b*x^2+a)^(1/4),x, algorithm="fricas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.15 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.33 \[ \int \frac {x^4 \left (c+d x^2\right )}{\sqrt [4]{a+b x^2}} \, dx=\frac {c x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5 \sqrt [4]{a}} + \frac {d x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{7 \sqrt [4]{a}} \] Input:

integrate(x**4*(d*x**2+c)/(b*x**2+a)**(1/4),x)
 

Output:

c*x**5*hyper((1/4, 5/2), (7/2,), b*x**2*exp_polar(I*pi)/a)/(5*a**(1/4)) + 
d*x**7*hyper((1/4, 7/2), (9/2,), b*x**2*exp_polar(I*pi)/a)/(7*a**(1/4))
 

Maxima [F]

\[ \int \frac {x^4 \left (c+d x^2\right )}{\sqrt [4]{a+b x^2}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} x^{4}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate(x^4*(d*x^2+c)/(b*x^2+a)^(1/4),x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)*x^4/(b*x^2 + a)^(1/4), x)
 

Giac [F]

\[ \int \frac {x^4 \left (c+d x^2\right )}{\sqrt [4]{a+b x^2}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} x^{4}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate(x^4*(d*x^2+c)/(b*x^2+a)^(1/4),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((d*x^2 + c)*x^4/(b*x^2 + a)^(1/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 \left (c+d x^2\right )}{\sqrt [4]{a+b x^2}} \, dx=\int \frac {x^4\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{1/4}} \,d x \] Input:

int((x^4*(c + d*x^2))/(a + b*x^2)^(1/4),x)
 

Output:

int((x^4*(c + d*x^2))/(a + b*x^2)^(1/4), x)
 

Reduce [F]

\[ \int \frac {x^4 \left (c+d x^2\right )}{\sqrt [4]{a+b x^2}} \, dx=\left (\int \frac {x^{6}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}}d x \right ) d +\left (\int \frac {x^{4}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}}d x \right ) c \] Input:

int(x^4*(d*x^2+c)/(b*x^2+a)^(1/4),x)
 

Output:

int(x**6/(a + b*x**2)**(1/4),x)*d + int(x**4/(a + b*x**2)**(1/4),x)*c