\(\int \frac {x^5 (c+d x^4)}{(a+b x^4)^{5/4}} \, dx\) [121]

Optimal result

Integrand size = 22, antiderivative size = 122 \[ \int \frac {x^5 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {(5 b c-6 a d) x^2}{5 b^2 \sqrt [4]{a+b x^4}}+\frac {d x^6}{5 b \sqrt [4]{a+b x^4}}-\frac {2 \sqrt {a} (5 b c-6 a d) \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{5 b^{5/2} \sqrt [4]{a+b x^4}} \] Output:

1/5*(-6*a*d+5*b*c)*x^2/b^2/(b*x^4+a)^(1/4)+1/5*d*x^6/b/(b*x^4+a)^(1/4)-2/5 
*a^(1/2)*(-6*a*d+5*b*c)*(1+b*x^4/a)^(1/4)*EllipticE(sin(1/2*arctan(b^(1/2) 
*x^2/a^(1/2))),2^(1/2))/b^(5/2)/(b*x^4+a)^(1/4)
 

Mathematica [C] (verified)

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

Time = 10.06 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.65 \[ \int \frac {x^5 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {x^2 \left (-5 b c+6 a d+b d x^4+(5 b c-6 a d) \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^4}{a}\right )\right )}{5 b^2 \sqrt [4]{a+b x^4}} \] Input:

Integrate[(x^5*(c + d*x^4))/(a + b*x^4)^(5/4),x]
 

Output:

(x^2*(-5*b*c + 6*a*d + b*d*x^4 + (5*b*c - 6*a*d)*(1 + (b*x^4)/a)^(1/4)*Hyp 
ergeometric2F1[1/4, 1/2, 3/2, -((b*x^4)/a)]))/(5*b^2*(a + b*x^4)^(1/4))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {959, 807, 250, 213, 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.

\(\displaystyle \int \frac {x^5 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx\)

\(\Big \downarrow \) 959

\(\displaystyle \frac {(5 b c-6 a d) \int \frac {x^5}{\left (b x^4+a\right )^{5/4}}dx}{5 b}+\frac {d x^6}{5 b \sqrt [4]{a+b x^4}}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {(5 b c-6 a d) \int \frac {x^4}{\left (b x^4+a\right )^{5/4}}dx^2}{10 b}+\frac {d x^6}{5 b \sqrt [4]{a+b x^4}}\)

\(\Big \downarrow \) 250

\(\displaystyle \frac {(5 b c-6 a d) \left (\frac {2 x^2}{b \sqrt [4]{a+b x^4}}-\frac {2 a \int \frac {1}{\left (b x^4+a\right )^{5/4}}dx^2}{b}\right )}{10 b}+\frac {d x^6}{5 b \sqrt [4]{a+b x^4}}\)

\(\Big \downarrow \) 213

\(\displaystyle \frac {(5 b c-6 a d) \left (\frac {2 x^2}{b \sqrt [4]{a+b x^4}}-\frac {2 \sqrt [4]{\frac {b x^4}{a}+1} \int \frac {1}{\left (\frac {b x^4}{a}+1\right )^{5/4}}dx^2}{b \sqrt [4]{a+b x^4}}\right )}{10 b}+\frac {d x^6}{5 b \sqrt [4]{a+b x^4}}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {(5 b c-6 a d) \left (\frac {2 x^2}{b \sqrt [4]{a+b x^4}}-\frac {4 \sqrt {a} \sqrt [4]{\frac {b x^4}{a}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{b^{3/2} \sqrt [4]{a+b x^4}}\right )}{10 b}+\frac {d x^6}{5 b \sqrt [4]{a+b x^4}}\)

Input:

Int[(x^5*(c + d*x^4))/(a + b*x^4)^(5/4),x]
 

Output:

(d*x^6)/(5*b*(a + b*x^4)^(1/4)) + ((5*b*c - 6*a*d)*((2*x^2)/(b*(a + b*x^4) 
^(1/4)) - (4*Sqrt[a]*(1 + (b*x^4)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x^2)/ 
Sqrt[a]]/2, 2])/(b^(3/2)*(a + b*x^4)^(1/4))))/(10*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 213

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

rule 250

Int[((c_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[2*c*(( 
c*x)^(m - 1)/(b*(2*m - 3)*(a + b*x^2)^(1/4))), x] - Simp[2*a*c^2*((m - 1)/( 
b*(2*m - 3)))   Int[(c*x)^(m - 2)/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b, 
 c}, x] && PosQ[b/a] && IntegerQ[2*m] && GtQ[m, 3/2]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 959

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

Maple [F]

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

Input:

int(x^5*(d*x^4+c)/(b*x^4+a)^(5/4),x)
 

Output:

int(x^5*(d*x^4+c)/(b*x^4+a)^(5/4),x)
 

Fricas [F]

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

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

Output:

integral((d*x^9 + c*x^5)*(b*x^4 + a)^(3/4)/(b^2*x^8 + 2*a*b*x^4 + a^2), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 15.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.49 \[ \int \frac {x^5 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {c x^{6} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{6 a^{\frac {5}{4}}} + \frac {d x^{10} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{10 a^{\frac {5}{4}}} \] Input:

integrate(x**5*(d*x**4+c)/(b*x**4+a)**(5/4),x)
 

Output:

c*x**6*hyper((5/4, 3/2), (5/2,), b*x**4*exp_polar(I*pi)/a)/(6*a**(5/4)) + 
d*x**10*hyper((5/4, 5/2), (7/2,), b*x**4*exp_polar(I*pi)/a)/(10*a**(5/4))
 

Maxima [F]

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

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

Output:

integrate((d*x^4 + c)*x^5/(b*x^4 + a)^(5/4), x)
 

Giac [F]

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

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

Output:

integrate((d*x^4 + c)*x^5/(b*x^4 + a)^(5/4), x)
 

Mupad [F(-1)]

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

int((x^5*(c + d*x^4))/(a + b*x^4)^(5/4),x)
 

Output:

int((x^5*(c + d*x^4))/(a + b*x^4)^(5/4), x)
 

Reduce [F]

\[ \int \frac {x^5 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/4}} \, dx=\left (\int \frac {x^{9}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a +\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{4}}d x \right ) d +\left (\int \frac {x^{5}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a +\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{4}}d x \right ) c \] Input:

int(x^5*(d*x^4+c)/(b*x^4+a)^(5/4),x)
 

Output:

int(x**9/((a + b*x**4)**(1/4)*a + (a + b*x**4)**(1/4)*b*x**4),x)*d + int(x 
**5/((a + b*x**4)**(1/4)*a + (a + b*x**4)**(1/4)*b*x**4),x)*c