\(\int \frac {x^9 (c+d x^4)}{(a+b x^4)^{9/4}} \, dx\) [160]

Optimal result

Integrand size = 22, antiderivative size = 151 \[ \int \frac {x^9 \left (c+d x^4\right )}{\left (a+b x^4\right )^{9/4}} \, dx=\frac {a (b c-a d) x^2}{5 b^3 \left (a+b x^4\right )^{5/4}}+\frac {(5 b c-12 a d) x^2}{5 b^3 \sqrt [4]{a+b x^4}}+\frac {d x^2 \left (a+b x^4\right )^{3/4}}{5 b^3}-\frac {12 \sqrt {a} (b c-2 a d) \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{5 b^{7/2} \sqrt [4]{a+b x^4}} \] Output:

1/5*a*(-a*d+b*c)*x^2/b^3/(b*x^4+a)^(5/4)+1/5*(-12*a*d+5*b*c)*x^2/b^3/(b*x^ 
4+a)^(1/4)+1/5*d*x^2*(b*x^4+a)^(3/4)/b^3-12/5*a^(1/2)*(-2*a*d+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^(7/2) 
/(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 = 10.09 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.71 \[ \int \frac {x^9 \left (c+d x^4\right )}{\left (a+b x^4\right )^{9/4}} \, dx=\frac {x^2 \left (12 a^2 d+b^2 x^4 \left (-7 c+d x^4\right )+a b \left (-6 c+14 d x^4\right )+6 (b c-2 a d) \left (a+b x^4\right ) \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^4}{a}\right )\right )}{5 b^3 \left (a+b x^4\right )^{5/4}} \] Input:

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

Output:

(x^2*(12*a^2*d + b^2*x^4*(-7*c + d*x^4) + a*b*(-6*c + 14*d*x^4) + 6*(b*c - 
 2*a*d)*(a + b*x^4)*(1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, 
 -((b*x^4)/a)]))/(5*b^3*(a + b*x^4)^(5/4))
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {957, 807, 250, 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.

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

\(\Big \downarrow \) 957

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

\(\Big \downarrow \) 807

\(\displaystyle \frac {x^{10} (b c-a d)}{5 a b \left (a+b x^4\right )^{5/4}}-\frac {(b c-2 a d) \int \frac {x^8}{\left (b x^4+a\right )^{5/4}}dx^2}{2 a b}\)

\(\Big \downarrow \) 250

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

\(\Big \downarrow \) 250

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

\(\Big \downarrow \) 213

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

\(\Big \downarrow \) 212

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

Input:

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

Output:

((b*c - a*d)*x^10)/(5*a*b*(a + b*x^4)^(5/4)) - ((b*c - 2*a*d)*((2*x^6)/(5* 
b*(a + b*x^4)^(1/4)) - (6*a*((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))))/(5*b)))/(2*a*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 957

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a 
*b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* 
(p + 1))   Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, 
 m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N 
eQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 
, m, (-n)*(p + 1)]))
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((d*x^13 + c*x^9)*(b*x^4 + a)^(3/4)/(b^3*x^12 + 3*a*b^2*x^8 + 3*a^ 
2*b*x^4 + a^3), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 41.38 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.40 \[ \int \frac {x^9 \left (c+d x^4\right )}{\left (a+b x^4\right )^{9/4}} \, dx=\frac {c x^{10} {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{10 a^{\frac {9}{4}}} + \frac {d x^{14} {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{4}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{14 a^{\frac {9}{4}}} \] Input:

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

Output:

c*x**10*hyper((9/4, 5/2), (7/2,), b*x**4*exp_polar(I*pi)/a)/(10*a**(9/4)) 
+ d*x**14*hyper((9/4, 7/2), (9/2,), b*x**4*exp_polar(I*pi)/a)/(14*a**(9/4) 
)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^9 \left (c+d x^4\right )}{\left (a+b x^4\right )^{9/4}} \, dx=\left (\int \frac {x^{13}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2}+2 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a b \,x^{4}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{2} x^{8}}d x \right ) d +\left (\int \frac {x^{9}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2}+2 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a b \,x^{4}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{2} x^{8}}d x \right ) c \] Input:

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

Output:

int(x**13/((a + b*x**4)**(1/4)*a**2 + 2*(a + b*x**4)**(1/4)*a*b*x**4 + (a 
+ b*x**4)**(1/4)*b**2*x**8),x)*d + int(x**9/((a + b*x**4)**(1/4)*a**2 + 2* 
(a + b*x**4)**(1/4)*a*b*x**4 + (a + b*x**4)**(1/4)*b**2*x**8),x)*c