\(\int \frac {x^5 (c+d x^4)}{\sqrt [4]{a+b x^4}} \, dx\) [85]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [C] (verification not implemented)
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 22, antiderivative size = 153 \[ \int \frac {x^5 \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=-\frac {2 a (3 b c-2 a d) x^2}{15 b^2 \sqrt [4]{a+b x^4}}+\frac {(3 b c-2 a d) x^2 \left (a+b x^4\right )^{3/4}}{15 b^2}+\frac {d x^6 \left (a+b x^4\right )^{3/4}}{9 b}+\frac {2 a^{3/2} (3 b c-2 a d) \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{15 b^{5/2} \sqrt [4]{a+b x^4}} \] Output:

-2/15*a*(-2*a*d+3*b*c)*x^2/b^2/(b*x^4+a)^(1/4)+1/15*(-2*a*d+3*b*c)*x^2*(b* 
x^4+a)^(3/4)/b^2+1/9*d*x^6*(b*x^4+a)^(3/4)/b+2/15*a^(3/2)*(-2*a*d+3*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 = 10.07 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.60 \[ \int \frac {x^5 \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=\frac {x^2 \left (-\left (\left (a+b x^4\right ) \left (-9 b c+6 a d-5 b d x^4\right )\right )+3 a (-3 b c+2 a d) \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^4}{a}\right )\right )}{45 b^2 \sqrt [4]{a+b x^4}} \] Input:

Integrate[(x^5*(c + d*x^4))/(a + b*x^4)^(1/4),x]
 

Output:

(x^2*(-((a + b*x^4)*(-9*b*c + 6*a*d - 5*b*d*x^4)) + 3*a*(-3*b*c + 2*a*d)*( 
1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, -((b*x^4)/a)]))/(45* 
b^2*(a + b*x^4)^(1/4))
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {959, 807, 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.

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

\(\Big \downarrow \) 959

\(\displaystyle \frac {(3 b c-2 a d) \int \frac {x^5}{\sqrt [4]{b x^4+a}}dx}{3 b}+\frac {d x^6 \left (a+b x^4\right )^{3/4}}{9 b}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {(3 b c-2 a d) \int \frac {x^4}{\sqrt [4]{b x^4+a}}dx^2}{6 b}+\frac {d x^6 \left (a+b x^4\right )^{3/4}}{9 b}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {(3 b c-2 a d) \left (\frac {2 x^2 \left (a+b x^4\right )^{3/4}}{5 b}-\frac {2 a \int \frac {1}{\sqrt [4]{b x^4+a}}dx^2}{5 b}\right )}{6 b}+\frac {d x^6 \left (a+b x^4\right )^{3/4}}{9 b}\)

\(\Big \downarrow \) 227

\(\displaystyle \frac {(3 b c-2 a d) \left (\frac {2 x^2 \left (a+b x^4\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^4}{a}+1} \int \frac {1}{\sqrt [4]{\frac {b x^4}{a}+1}}dx^2}{5 b \sqrt [4]{a+b x^4}}\right )}{6 b}+\frac {d x^6 \left (a+b x^4\right )^{3/4}}{9 b}\)

\(\Big \downarrow \) 225

\(\displaystyle \frac {(3 b c-2 a d) \left (\frac {2 x^2 \left (a+b x^4\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^4}{a}+1} \left (\frac {2 x^2}{\sqrt [4]{\frac {b x^4}{a}+1}}-\int \frac {1}{\left (\frac {b x^4}{a}+1\right )^{5/4}}dx^2\right )}{5 b \sqrt [4]{a+b x^4}}\right )}{6 b}+\frac {d x^6 \left (a+b x^4\right )^{3/4}}{9 b}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {(3 b c-2 a d) \left (\frac {2 x^2 \left (a+b x^4\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^4}{a}+1} \left (\frac {2 x^2}{\sqrt [4]{\frac {b x^4}{a}+1}}-\frac {2 \sqrt {a} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b}}\right )}{5 b \sqrt [4]{a+b x^4}}\right )}{6 b}+\frac {d x^6 \left (a+b x^4\right )^{3/4}}{9 b}\)

Input:

Int[(x^5*(c + d*x^4))/(a + b*x^4)^(1/4),x]
 

Output:

(d*x^6*(a + b*x^4)^(3/4))/(9*b) + ((3*b*c - 2*a*d)*((2*x^2*(a + b*x^4)^(3/ 
4))/(5*b) - (2*a*(1 + (b*x^4)/a)^(1/4)*((2*x^2)/(1 + (b*x^4)/a)^(1/4) - (2 
*Sqrt[a]*EllipticE[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/Sqrt[b]))/(5*b*(a 
+ b*x^4)^(1/4))))/(6*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 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]

\[\int \frac {x^{5} \left (d \,x^{4}+c \right )}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate(x^5*(d*x^4+c)/(b*x^4+a)^(1/4),x, algorithm="fricas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 4.30 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.39 \[ \int \frac {x^5 \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=\frac {c x^{6} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{6 \sqrt [4]{a}} + \frac {d x^{10} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{10 \sqrt [4]{a}} \] Input:

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

Output:

c*x**6*hyper((1/4, 3/2), (5/2,), b*x**4*exp_polar(I*pi)/a)/(6*a**(1/4)) + 
d*x**10*hyper((1/4, 5/2), (7/2,), b*x**4*exp_polar(I*pi)/a)/(10*a**(1/4))
 

Maxima [F]

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

integrate(x^5*(d*x^4+c)/(b*x^4+a)^(1/4),x, algorithm="maxima")
 

Output:

integrate((d*x^4 + c)*x^5/(b*x^4 + a)^(1/4), x)
 

Giac [F]

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

integrate(x^5*(d*x^4+c)/(b*x^4+a)^(1/4),x, algorithm="giac")
 

Output:

integrate((d*x^4 + c)*x^5/(b*x^4 + a)^(1/4), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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