\(\int \frac {x^6 (c+d x^2)}{(a+b x^2)^{3/4}} \, dx\) [382]

Optimal result

Integrand size = 22, antiderivative size = 182 \[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\frac {8 a^2 (15 b c-14 a d) x \sqrt [4]{a+b x^2}}{231 b^4}-\frac {4 a (15 b c-14 a d) x^3 \sqrt [4]{a+b x^2}}{231 b^3}+\frac {2 (15 b c-14 a d) x^5 \sqrt [4]{a+b x^2}}{165 b^2}+\frac {2 d x^7 \sqrt [4]{a+b x^2}}{15 b}-\frac {16 a^{7/2} (15 b c-14 a d) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{231 b^{9/2} \left (a+b x^2\right )^{3/4}} \] Output:

8/231*a^2*(-14*a*d+15*b*c)*x*(b*x^2+a)^(1/4)/b^4-4/231*a*(-14*a*d+15*b*c)* 
x^3*(b*x^2+a)^(1/4)/b^3+2/165*(-14*a*d+15*b*c)*x^5*(b*x^2+a)^(1/4)/b^2+2/1 
5*d*x^7*(b*x^2+a)^(1/4)/b-16/231*a^(7/2)*(-14*a*d+15*b*c)*(1+b*x^2/a)^(3/4 
)*InverseJacobiAM(1/2*arctan(b^(1/2)*x/a^(1/2)),2^(1/2))/b^(9/2)/(b*x^2+a) 
^(3/4)
 

Mathematica [C] (verified)

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

Time = 10.10 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.75 \[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\frac {2 x \left (a+b x^2\right ) \left (-280 a^3 d+20 a^2 b \left (15 c+7 d x^2\right )+7 b^3 x^4 \left (15 c+11 d x^2\right )-2 a b^2 x^2 \left (75 c+49 d x^2\right )\right )+40 a^3 (-15 b c+14 a d) x \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},-\frac {b x^2}{a}\right )}{1155 b^4 \left (a+b x^2\right )^{3/4}} \] Input:

Integrate[(x^6*(c + d*x^2))/(a + b*x^2)^(3/4),x]
 

Output:

(2*x*(a + b*x^2)*(-280*a^3*d + 20*a^2*b*(15*c + 7*d*x^2) + 7*b^3*x^4*(15*c 
 + 11*d*x^2) - 2*a*b^2*x^2*(75*c + 49*d*x^2)) + 40*a^3*(-15*b*c + 14*a*d)* 
x*(1 + (b*x^2)/a)^(3/4)*Hypergeometric2F1[1/2, 3/4, 3/2, -((b*x^2)/a)])/(1 
155*b^4*(a + b*x^2)^(3/4))
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {363, 262, 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.

\(\displaystyle \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx\)

\(\Big \downarrow \) 363

\(\displaystyle \frac {(15 b c-14 a d) \int \frac {x^6}{\left (b x^2+a\right )^{3/4}}dx}{15 b}+\frac {2 d x^7 \sqrt [4]{a+b x^2}}{15 b}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {(15 b c-14 a d) \left (\frac {2 x^5 \sqrt [4]{a+b x^2}}{11 b}-\frac {10 a \int \frac {x^4}{\left (b x^2+a\right )^{3/4}}dx}{11 b}\right )}{15 b}+\frac {2 d x^7 \sqrt [4]{a+b x^2}}{15 b}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {(15 b c-14 a d) \left (\frac {2 x^5 \sqrt [4]{a+b x^2}}{11 b}-\frac {10 a \left (\frac {2 x^3 \sqrt [4]{a+b x^2}}{7 b}-\frac {6 a \int \frac {x^2}{\left (b x^2+a\right )^{3/4}}dx}{7 b}\right )}{11 b}\right )}{15 b}+\frac {2 d x^7 \sqrt [4]{a+b x^2}}{15 b}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {(15 b c-14 a d) \left (\frac {2 x^5 \sqrt [4]{a+b x^2}}{11 b}-\frac {10 a \left (\frac {2 x^3 \sqrt [4]{a+b x^2}}{7 b}-\frac {6 a \left (\frac {2 x \sqrt [4]{a+b x^2}}{3 b}-\frac {2 a \int \frac {1}{\left (b x^2+a\right )^{3/4}}dx}{3 b}\right )}{7 b}\right )}{11 b}\right )}{15 b}+\frac {2 d x^7 \sqrt [4]{a+b x^2}}{15 b}\)

\(\Big \downarrow \) 231

\(\displaystyle \frac {(15 b c-14 a d) \left (\frac {2 x^5 \sqrt [4]{a+b x^2}}{11 b}-\frac {10 a \left (\frac {2 x^3 \sqrt [4]{a+b x^2}}{7 b}-\frac {6 a \left (\frac {2 x \sqrt [4]{a+b x^2}}{3 b}-\frac {2 a \left (\frac {b x^2}{a}+1\right )^{3/4} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{3 b \left (a+b x^2\right )^{3/4}}\right )}{7 b}\right )}{11 b}\right )}{15 b}+\frac {2 d x^7 \sqrt [4]{a+b x^2}}{15 b}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {(15 b c-14 a d) \left (\frac {2 x^5 \sqrt [4]{a+b x^2}}{11 b}-\frac {10 a \left (\frac {2 x^3 \sqrt [4]{a+b x^2}}{7 b}-\frac {6 a \left (\frac {2 x \sqrt [4]{a+b x^2}}{3 b}-\frac {4 a^{3/2} \left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{3 b^{3/2} \left (a+b x^2\right )^{3/4}}\right )}{7 b}\right )}{11 b}\right )}{15 b}+\frac {2 d x^7 \sqrt [4]{a+b x^2}}{15 b}\)

Input:

Int[(x^6*(c + d*x^2))/(a + b*x^2)^(3/4),x]
 

Output:

(2*d*x^7*(a + b*x^2)^(1/4))/(15*b) + ((15*b*c - 14*a*d)*((2*x^5*(a + b*x^2 
)^(1/4))/(11*b) - (10*a*((2*x^3*(a + b*x^2)^(1/4))/(7*b) - (6*a*((2*x*(a + 
 b*x^2)^(1/4))/(3*b) - (4*a^(3/2)*(1 + (b*x^2)/a)^(3/4)*EllipticF[ArcTan[( 
Sqrt[b]*x)/Sqrt[a]]/2, 2])/(3*b^(3/2)*(a + b*x^2)^(3/4))))/(7*b)))/(11*b)) 
)/(15*b)
 

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 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 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]

\[\int \frac {x^{6} \left (x^{2} d +c \right )}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} x^{6}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate(x^6*(d*x^2+c)/(b*x^2+a)^(3/4),x, algorithm="fricas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.23 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.33 \[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\frac {c x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{7 a^{\frac {3}{4}}} + \frac {d x^{9} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {9}{2} \\ \frac {11}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{9 a^{\frac {3}{4}}} \] Input:

integrate(x**6*(d*x**2+c)/(b*x**2+a)**(3/4),x)
 

Output:

c*x**7*hyper((3/4, 7/2), (9/2,), b*x**2*exp_polar(I*pi)/a)/(7*a**(3/4)) + 
d*x**9*hyper((3/4, 9/2), (11/2,), b*x**2*exp_polar(I*pi)/a)/(9*a**(3/4))
 

Maxima [F]

\[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} x^{6}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate(x^6*(d*x^2+c)/(b*x^2+a)^(3/4),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} x^{6}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate(x^6*(d*x^2+c)/(b*x^2+a)^(3/4),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\int \frac {x^6\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{3/4}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {x^6 \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx=\left (\int \frac {x^{8}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) d +\left (\int \frac {x^{6}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) c \] Input:

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

Output:

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