\(\int x^9 (a+b x^4)^{5/4} \, dx\) [510]

Optimal result

Integrand size = 15, antiderivative size = 146 \[ \int x^9 \left (a+b x^4\right )^{5/4} \, dx=-\frac {2 a^3 x^2 \sqrt [4]{a+b x^4}}{231 b^2}+\frac {a^2 x^6 \sqrt [4]{a+b x^4}}{231 b}+\frac {1}{33} a x^{10} \sqrt [4]{a+b x^4}+\frac {1}{15} x^{10} \left (a+b x^4\right )^{5/4}+\frac {4 a^{9/2} \left (1+\frac {b x^4}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{231 b^{5/2} \left (a+b x^4\right )^{3/4}} \] Output:

-2/231*a^3*x^2*(b*x^4+a)^(1/4)/b^2+1/231*a^2*x^6*(b*x^4+a)^(1/4)/b+1/33*a* 
x^10*(b*x^4+a)^(1/4)+1/15*x^10*(b*x^4+a)^(5/4)+4/231*a^(9/2)*(1+b*x^4/a)^( 
3/4)*InverseJacobiAM(1/2*arctan(b^(1/2)*x^2/a^(1/2)),2^(1/2))/b^(5/2)/(b*x 
^4+a)^(3/4)
 

Mathematica [C] (verified)

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

Time = 7.85 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.55 \[ \int x^9 \left (a+b x^4\right )^{5/4} \, dx=\frac {x^2 \sqrt [4]{a+b x^4} \left (-\left (\left (6 a-11 b x^4\right ) \left (a+b x^4\right )^2\right )+\frac {6 a^3 \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^4}{a}\right )}{\sqrt [4]{1+\frac {b x^4}{a}}}\right )}{165 b^2} \] Input:

Integrate[x^9*(a + b*x^4)^(5/4),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {807, 248, 248, 262, 262, 231, 229}

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.

\(\displaystyle \int x^9 \left (a+b x^4\right )^{5/4} \, dx\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {1}{2} \int x^8 \left (b x^4+a\right )^{5/4}dx^2\)

\(\Big \downarrow \) 248

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} a \int x^8 \sqrt [4]{b x^4+a}dx^2+\frac {2}{15} x^{10} \left (a+b x^4\right )^{5/4}\right )\)

\(\Big \downarrow \) 248

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} a \left (\frac {1}{11} a \int \frac {x^8}{\left (b x^4+a\right )^{3/4}}dx^2+\frac {2}{11} x^{10} \sqrt [4]{a+b x^4}\right )+\frac {2}{15} x^{10} \left (a+b x^4\right )^{5/4}\right )\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} a \left (\frac {1}{11} a \left (\frac {2 x^6 \sqrt [4]{a+b x^4}}{7 b}-\frac {6 a \int \frac {x^4}{\left (b x^4+a\right )^{3/4}}dx^2}{7 b}\right )+\frac {2}{11} x^{10} \sqrt [4]{a+b x^4}\right )+\frac {2}{15} x^{10} \left (a+b x^4\right )^{5/4}\right )\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} a \left (\frac {1}{11} a \left (\frac {2 x^6 \sqrt [4]{a+b x^4}}{7 b}-\frac {6 a \left (\frac {2 x^2 \sqrt [4]{a+b x^4}}{3 b}-\frac {2 a \int \frac {1}{\left (b x^4+a\right )^{3/4}}dx^2}{3 b}\right )}{7 b}\right )+\frac {2}{11} x^{10} \sqrt [4]{a+b x^4}\right )+\frac {2}{15} x^{10} \left (a+b x^4\right )^{5/4}\right )\)

\(\Big \downarrow \) 231

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} a \left (\frac {1}{11} a \left (\frac {2 x^6 \sqrt [4]{a+b x^4}}{7 b}-\frac {6 a \left (\frac {2 x^2 \sqrt [4]{a+b x^4}}{3 b}-\frac {2 a \left (\frac {b x^4}{a}+1\right )^{3/4} \int \frac {1}{\left (\frac {b x^4}{a}+1\right )^{3/4}}dx^2}{3 b \left (a+b x^4\right )^{3/4}}\right )}{7 b}\right )+\frac {2}{11} x^{10} \sqrt [4]{a+b x^4}\right )+\frac {2}{15} x^{10} \left (a+b x^4\right )^{5/4}\right )\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} a \left (\frac {1}{11} a \left (\frac {2 x^6 \sqrt [4]{a+b x^4}}{7 b}-\frac {6 a \left (\frac {2 x^2 \sqrt [4]{a+b x^4}}{3 b}-\frac {4 a^{3/2} \left (\frac {b x^4}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{3 b^{3/2} \left (a+b x^4\right )^{3/4}}\right )}{7 b}\right )+\frac {2}{11} x^{10} \sqrt [4]{a+b x^4}\right )+\frac {2}{15} x^{10} \left (a+b x^4\right )^{5/4}\right )\)

Input:

Int[x^9*(a + b*x^4)^(5/4),x]
 

Output:

((2*x^10*(a + b*x^4)^(5/4))/15 + (a*((2*x^10*(a + b*x^4)^(1/4))/11 + (a*(( 
2*x^6*(a + b*x^4)^(1/4))/(7*b) - (6*a*((2*x^2*(a + b*x^4)^(1/4))/(3*b) - ( 
4*a^(3/2)*(1 + (b*x^4)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 
 2])/(3*b^(3/2)*(a + b*x^4)^(3/4))))/(7*b)))/11))/3)/2
 

Defintions of rubi rules used

rule 229

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

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]
 

Maple [F]

\[\int x^{9} \left (b \,x^{4}+a \right )^{\frac {5}{4}}d x\]

Input:

int(x^9*(b*x^4+a)^(5/4),x)
 

Output:

int(x^9*(b*x^4+a)^(5/4),x)
 

Fricas [F]

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

integrate(x^9*(b*x^4+a)^(5/4),x, algorithm="fricas")
 

Output:

integral((b*x^13 + a*x^9)*(b*x^4 + a)^(1/4), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.00 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.20 \[ \int x^9 \left (a+b x^4\right )^{5/4} \, dx=\frac {a^{\frac {5}{4}} x^{10} {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{10} \] Input:

integrate(x**9*(b*x**4+a)**(5/4),x)
 

Output:

a**(5/4)*x**10*hyper((-5/4, 5/2), (7/2,), b*x**4*exp_polar(I*pi)/a)/10
 

Maxima [F]

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

integrate(x^9*(b*x^4+a)^(5/4),x, algorithm="maxima")
 

Output:

integrate((b*x^4 + a)^(5/4)*x^9, x)
 

Giac [F]

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

integrate(x^9*(b*x^4+a)^(5/4),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((b*x^4 + a)^(5/4)*x^9, x)
 

Mupad [F(-1)]

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

int(x^9*(a + b*x^4)^(5/4),x)
 

Output:

int(x^9*(a + b*x^4)^(5/4), x)
 

Reduce [F]

\[ \int x^9 \left (a+b x^4\right )^{5/4} \, dx=\frac {-10 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3} x^{2}+5 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b \,x^{6}+112 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{2} x^{10}+77 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{3} x^{14}+20 \left (\int \frac {x}{\left (b \,x^{4}+a \right )^{\frac {3}{4}}}d x \right ) a^{4}}{1155 b^{2}} \] Input:

int(x^9*(b*x^4+a)^(5/4),x)
 

Output:

( - 10*(a + b*x**4)**(1/4)*a**3*x**2 + 5*(a + b*x**4)**(1/4)*a**2*b*x**6 + 
 112*(a + b*x**4)**(1/4)*a*b**2*x**10 + 77*(a + b*x**4)**(1/4)*b**3*x**14 
+ 20*int(((a + b*x**4)**(1/4)*x)/(a + b*x**4),x)*a**4)/(1155*b**2)