\(\int \frac {x^6 (c+d x^2)}{(a+b x^2)^{5/4}} \, dx\) [394]

Optimal result

Integrand size = 22, antiderivative size = 182 \[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {8 a^2 (13 b c-14 a d) x}{39 b^4 \sqrt [4]{a+b x^2}}-\frac {4 a (13 b c-14 a d) x^3}{117 b^3 \sqrt [4]{a+b x^2}}+\frac {2 (13 b c-14 a d) x^5}{117 b^2 \sqrt [4]{a+b x^2}}+\frac {2 d x^7}{13 b \sqrt [4]{a+b x^2}}-\frac {16 a^{5/2} (13 b c-14 a d) \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{39 b^{9/2} \sqrt [4]{a+b x^2}} \] Output:

8/39*a^2*(-14*a*d+13*b*c)*x/b^4/(b*x^2+a)^(1/4)-4/117*a*(-14*a*d+13*b*c)*x 
^3/b^3/(b*x^2+a)^(1/4)+2/117*(-14*a*d+13*b*c)*x^5/b^2/(b*x^2+a)^(1/4)+2/13 
*d*x^7/b/(b*x^2+a)^(1/4)-16/39*a^(5/2)*(-14*a*d+13*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 = 10.08 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.69 \[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {2 x \left (168 a^3 d-4 a^2 b \left (39 c-7 d x^2\right )-2 a b^2 x^2 \left (13 c+7 d x^2\right )+b^3 x^4 \left (13 c+9 d x^2\right )+12 a^2 (13 b c-14 a d) \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{117 b^4 \sqrt [4]{a+b x^2}} \] Input:

Integrate[(x^6*(c + d*x^2))/(a + b*x^2)^(5/4),x]
 

Output:

(2*x*(168*a^3*d - 4*a^2*b*(39*c - 7*d*x^2) - 2*a*b^2*x^2*(13*c + 7*d*x^2) 
+ b^3*x^4*(13*c + 9*d*x^2) + 12*a^2*(13*b*c - 14*a*d)*(1 + (b*x^2)/a)^(1/4 
)*Hypergeometric2F1[1/4, 1/2, 3/2, -((b*x^2)/a)]))/(117*b^4*(a + b*x^2)^(1 
/4))
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {363, 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.

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

\(\Big \downarrow \) 363

\(\displaystyle \frac {(13 b c-14 a d) \int \frac {x^6}{\left (b x^2+a\right )^{5/4}}dx}{13 b}+\frac {2 d x^7}{13 b \sqrt [4]{a+b x^2}}\)

\(\Big \downarrow \) 250

\(\displaystyle \frac {(13 b c-14 a d) \left (\frac {2 x^5}{9 b \sqrt [4]{a+b x^2}}-\frac {10 a \int \frac {x^4}{\left (b x^2+a\right )^{5/4}}dx}{9 b}\right )}{13 b}+\frac {2 d x^7}{13 b \sqrt [4]{a+b x^2}}\)

\(\Big \downarrow \) 250

\(\displaystyle \frac {(13 b c-14 a d) \left (\frac {2 x^5}{9 b \sqrt [4]{a+b x^2}}-\frac {10 a \left (\frac {2 x^3}{5 b \sqrt [4]{a+b x^2}}-\frac {6 a \int \frac {x^2}{\left (b x^2+a\right )^{5/4}}dx}{5 b}\right )}{9 b}\right )}{13 b}+\frac {2 d x^7}{13 b \sqrt [4]{a+b x^2}}\)

\(\Big \downarrow \) 250

\(\displaystyle \frac {(13 b c-14 a d) \left (\frac {2 x^5}{9 b \sqrt [4]{a+b x^2}}-\frac {10 a \left (\frac {2 x^3}{5 b \sqrt [4]{a+b x^2}}-\frac {6 a \left (\frac {2 x}{b \sqrt [4]{a+b x^2}}-\frac {2 a \int \frac {1}{\left (b x^2+a\right )^{5/4}}dx}{b}\right )}{5 b}\right )}{9 b}\right )}{13 b}+\frac {2 d x^7}{13 b \sqrt [4]{a+b x^2}}\)

\(\Big \downarrow \) 213

\(\displaystyle \frac {(13 b c-14 a d) \left (\frac {2 x^5}{9 b \sqrt [4]{a+b x^2}}-\frac {10 a \left (\frac {2 x^3}{5 b \sqrt [4]{a+b x^2}}-\frac {6 a \left (\frac {2 x}{b \sqrt [4]{a+b x^2}}-\frac {2 \sqrt [4]{\frac {b x^2}{a}+1} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx}{b \sqrt [4]{a+b x^2}}\right )}{5 b}\right )}{9 b}\right )}{13 b}+\frac {2 d x^7}{13 b \sqrt [4]{a+b x^2}}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {(13 b c-14 a d) \left (\frac {2 x^5}{9 b \sqrt [4]{a+b x^2}}-\frac {10 a \left (\frac {2 x^3}{5 b \sqrt [4]{a+b x^2}}-\frac {6 a \left (\frac {2 x}{b \sqrt [4]{a+b x^2}}-\frac {4 \sqrt {a} \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{b^{3/2} \sqrt [4]{a+b x^2}}\right )}{5 b}\right )}{9 b}\right )}{13 b}+\frac {2 d x^7}{13 b \sqrt [4]{a+b x^2}}\)

Input:

Int[(x^6*(c + d*x^2))/(a + b*x^2)^(5/4),x]
 

Output:

(2*d*x^7)/(13*b*(a + b*x^2)^(1/4)) + ((13*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[ArcTan[(Sqr 
t[b]*x)/Sqrt[a]]/2, 2])/(b^(3/2)*(a + b*x^2)^(1/4))))/(5*b)))/(9*b)))/(13* 
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 363

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), 
 x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3))   Int[(e*x)^ 
m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d 
, 0] && NeQ[m + 2*p + 3, 0]
 

Maple [F]

\[\int \frac {x^{6} \left (x^{2} d +c \right )}{\left (b \,x^{2}+a \right )^{\frac {5}{4}}}d x\]

Input:

int(x^6*(d*x^2+c)/(b*x^2+a)^(5/4),x)
 

Output:

int(x^6*(d*x^2+c)/(b*x^2+a)^(5/4),x)
 

Fricas [F]

\[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} x^{6}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(x^6*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="fricas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.56 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.33 \[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {c x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{7 a^{\frac {5}{4}}} + \frac {d x^{9} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {9}{2} \\ \frac {11}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{9 a^{\frac {5}{4}}} \] Input:

integrate(x**6*(d*x**2+c)/(b*x**2+a)**(5/4),x)
 

Output:

c*x**7*hyper((5/4, 7/2), (9/2,), b*x**2*exp_polar(I*pi)/a)/(7*a**(5/4)) + 
d*x**9*hyper((5/4, 9/2), (11/2,), b*x**2*exp_polar(I*pi)/a)/(9*a**(5/4))
 

Maxima [F]

\[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} x^{6}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(x^6*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)*x^6/(b*x^2 + a)^(5/4), x)
 

Giac [F]

\[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} x^{6}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(x^6*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="giac")
 

Output:

integrate((d*x^2 + c)*x^6/(b*x^2 + a)^(5/4), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int \frac {x^6\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{5/4}} \,d x \] Input:

int((x^6*(c + d*x^2))/(a + b*x^2)^(5/4),x)
 

Output:

int((x^6*(c + d*x^2))/(a + b*x^2)^(5/4), x)
 

Reduce [F]

\[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\left (\int \frac {x^{8}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) d +\left (\int \frac {x^{6}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) c \] Input:

int(x^6*(d*x^2+c)/(b*x^2+a)^(5/4),x)
 

Output:

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