\(\int \frac {x^6 (c+d x^2)}{(a+b x^2)^{9/4}} \, dx\) [418]

Optimal result

Integrand size = 22, antiderivative size = 177 \[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=-\frac {2 a^2 (b c-a d) x}{5 b^4 \left (a+b x^2\right )^{5/4}}-\frac {2 a (36 b c-61 a d) x}{15 b^4 \sqrt [4]{a+b x^2}}+\frac {2 (3 b c-8 a d) x \left (a+b x^2\right )^{3/4}}{15 b^4}+\frac {2 d x^3 \left (a+b x^2\right )^{3/4}}{9 b^3}+\frac {16 a^{3/2} (9 b c-14 a d) \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{15 b^{9/2} \sqrt [4]{a+b x^2}} \] Output:

-2/5*a^2*(-a*d+b*c)*x/b^4/(b*x^2+a)^(5/4)-2/15*a*(-61*a*d+36*b*c)*x/b^4/(b 
*x^2+a)^(1/4)+2/15*(-8*a*d+3*b*c)*x*(b*x^2+a)^(3/4)/b^4+2/9*d*x^3*(b*x^2+a 
)^(3/4)/b^3+16/15*a^(3/2)*(-14*a*d+9*b*c)*(1+b*x^2/a)^(1/4)*EllipticE(sin( 
1/2*arctan(b^(1/2)*x/a^(1/2))),2^(1/2))/b^(9/2)/(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 = 10.10 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.73 \[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {2 x \left (-168 a^3 d+4 a^2 b \left (27 c-49 d x^2\right )-14 a b^2 x^2 \left (-9 c+d x^2\right )+b^3 x^4 \left (9 c+5 d x^2\right )+12 a (-9 b c+14 a d) \left (a+b x^2\right ) \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{45 b^4 \left (a+b x^2\right )^{5/4}} \] Input:

Integrate[(x^6*(c + d*x^2))/(a + b*x^2)^(9/4),x]
 

Output:

(2*x*(-168*a^3*d + 4*a^2*b*(27*c - 49*d*x^2) - 14*a*b^2*x^2*(-9*c + d*x^2) 
 + b^3*x^4*(9*c + 5*d*x^2) + 12*a*(-9*b*c + 14*a*d)*(a + b*x^2)*(1 + (b*x^ 
2)/a)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, -((b*x^2)/a)]))/(45*b^4*(a + 
b*x^2)^(5/4))
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {362, 250, 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.

\(\displaystyle \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx\)

\(\Big \downarrow \) 362

\(\displaystyle \frac {2 x^7 (b c-a d)}{5 a b \left (a+b x^2\right )^{5/4}}-\frac {(9 b c-14 a d) \int \frac {x^6}{\left (b x^2+a\right )^{5/4}}dx}{5 a b}\)

\(\Big \downarrow \) 250

\(\displaystyle \frac {2 x^7 (b c-a d)}{5 a b \left (a+b x^2\right )^{5/4}}-\frac {(9 b c-14 a d) \left (\frac {2 x^5}{9 b \sqrt [4]{a+b x^2}}-\frac {10 a \int \frac {x^4}{\left (b x^2+a\right )^{5/4}}dx}{9 b}\right )}{5 a b}\)

\(\Big \downarrow \) 250

\(\displaystyle \frac {2 x^7 (b c-a d)}{5 a b \left (a+b x^2\right )^{5/4}}-\frac {(9 b c-14 a d) \left (\frac {2 x^5}{9 b \sqrt [4]{a+b x^2}}-\frac {10 a \left (\frac {2 x^3}{5 b \sqrt [4]{a+b x^2}}-\frac {6 a \int \frac {x^2}{\left (b x^2+a\right )^{5/4}}dx}{5 b}\right )}{9 b}\right )}{5 a b}\)

\(\Big \downarrow \) 250

\(\displaystyle \frac {2 x^7 (b c-a d)}{5 a b \left (a+b x^2\right )^{5/4}}-\frac {(9 b c-14 a d) \left (\frac {2 x^5}{9 b \sqrt [4]{a+b x^2}}-\frac {10 a \left (\frac {2 x^3}{5 b \sqrt [4]{a+b x^2}}-\frac {6 a \left (\frac {2 x}{b \sqrt [4]{a+b x^2}}-\frac {2 a \int \frac {1}{\left (b x^2+a\right )^{5/4}}dx}{b}\right )}{5 b}\right )}{9 b}\right )}{5 a b}\)

\(\Big \downarrow \) 213

\(\displaystyle \frac {2 x^7 (b c-a d)}{5 a b \left (a+b x^2\right )^{5/4}}-\frac {(9 b c-14 a d) \left (\frac {2 x^5}{9 b \sqrt [4]{a+b x^2}}-\frac {10 a \left (\frac {2 x^3}{5 b \sqrt [4]{a+b x^2}}-\frac {6 a \left (\frac {2 x}{b \sqrt [4]{a+b x^2}}-\frac {2 \sqrt [4]{\frac {b x^2}{a}+1} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx}{b \sqrt [4]{a+b x^2}}\right )}{5 b}\right )}{9 b}\right )}{5 a b}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {2 x^7 (b c-a d)}{5 a b \left (a+b x^2\right )^{5/4}}-\frac {(9 b c-14 a d) \left (\frac {2 x^5}{9 b \sqrt [4]{a+b x^2}}-\frac {10 a \left (\frac {2 x^3}{5 b \sqrt [4]{a+b x^2}}-\frac {6 a \left (\frac {2 x}{b \sqrt [4]{a+b x^2}}-\frac {4 \sqrt {a} \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{b^{3/2} \sqrt [4]{a+b x^2}}\right )}{5 b}\right )}{9 b}\right )}{5 a b}\)

Input:

Int[(x^6*(c + d*x^2))/(a + b*x^2)^(9/4),x]
 

Output:

(2*(b*c - a*d)*x^7)/(5*a*b*(a + b*x^2)^(5/4)) - ((9*b*c - 14*a*d)*((2*x^5) 
/(9*b*(a + b*x^2)^(1/4)) - (10*a*((2*x^3)/(5*b*(a + b*x^2)^(1/4)) - (6*a*( 
(2*x)/(b*(a + b*x^2)^(1/4)) - (4*Sqrt[a]*(1 + (b*x^2)/a)^(1/4)*EllipticE[A 
rcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(b^(3/2)*(a + b*x^2)^(1/4))))/(5*b)))/(9 
*b)))/(5*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 362

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*b*e 
*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(2*a*b*(p + 1))   I 
nt[(e*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && N 
eQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) || 
  !RationalQ[m] || (ILtQ[p + 1/2, 0] && LeQ[-1, m, -2*(p + 1)]))
 

Maple [F]

\[\int \frac {x^{6} \left (x^{2} d +c \right )}{\left (b \,x^{2}+a \right )^{\frac {9}{4}}}d x\]

Input:

int(x^6*(d*x^2+c)/(b*x^2+a)^(9/4),x)
 

Output:

int(x^6*(d*x^2+c)/(b*x^2+a)^(9/4),x)
 

Fricas [F]

\[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} x^{6}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate(x^6*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="fricas")
 

Output:

integral((d*x^8 + c*x^6)*(b*x^2 + a)^(3/4)/(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2* 
b*x^2 + a^3), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 6.04 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.34 \[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {c x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{4}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{7 a^{\frac {9}{4}}} + \frac {d x^{9} {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{4}, \frac {9}{2} \\ \frac {11}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{9 a^{\frac {9}{4}}} \] Input:

integrate(x**6*(d*x**2+c)/(b*x**2+a)**(9/4),x)
 

Output:

c*x**7*hyper((9/4, 7/2), (9/2,), b*x**2*exp_polar(I*pi)/a)/(7*a**(9/4)) + 
d*x**9*hyper((9/4, 9/2), (11/2,), b*x**2*exp_polar(I*pi)/a)/(9*a**(9/4))
 

Maxima [F]

\[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} x^{6}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate(x^6*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)*x^6/(b*x^2 + a)^(9/4), x)
 

Giac [F]

\[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} x^{6}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate(x^6*(d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="giac")
 

Output:

integrate((d*x^2 + c)*x^6/(b*x^2 + a)^(9/4), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int \frac {x^6\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{9/4}} \,d x \] Input:

int((x^6*(c + d*x^2))/(a + b*x^2)^(9/4),x)
 

Output:

int((x^6*(c + d*x^2))/(a + b*x^2)^(9/4), x)
 

Reduce [F]

\[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\left (\int \frac {x^{8}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2}+2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} x^{4}}d x \right ) d +\left (\int \frac {x^{6}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2}+2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} x^{4}}d x \right ) c \] Input:

int(x^6*(d*x^2+c)/(b*x^2+a)^(9/4),x)
 

Output:

int(x**8/((a + b*x**2)**(1/4)*a**2 + 2*(a + b*x**2)**(1/4)*a*b*x**2 + (a + 
 b*x**2)**(1/4)*b**2*x**4),x)*d + int(x**6/((a + b*x**2)**(1/4)*a**2 + 2*( 
a + b*x**2)**(1/4)*a*b*x**2 + (a + b*x**2)**(1/4)*b**2*x**4),x)*c