\(\int \frac {1}{x^6 (a+b x^2)^{7/4}} \, dx\) [919]

Optimal result

Integrand size = 15, antiderivative size = 147 \[ \int \frac {1}{x^6 \left (a+b x^2\right )^{7/4}} \, dx=\frac {2}{3 a x^5 \left (a+b x^2\right )^{3/4}}-\frac {13 \sqrt [4]{a+b x^2}}{15 a^2 x^5}+\frac {13 b \sqrt [4]{a+b x^2}}{10 a^3 x^3}-\frac {13 b^2 \sqrt [4]{a+b x^2}}{4 a^4 x}-\frac {13 b^{5/2} \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{4 a^{7/2} \left (a+b x^2\right )^{3/4}} \] Output:

2/3/a/x^5/(b*x^2+a)^(3/4)-13/15*(b*x^2+a)^(1/4)/a^2/x^5+13/10*b*(b*x^2+a)^ 
(1/4)/a^3/x^3-13/4*b^2*(b*x^2+a)^(1/4)/a^4/x-13/4*b^(5/2)*(1+b*x^2/a)^(3/4 
)*InverseJacobiAM(1/2*arctan(b^(1/2)*x/a^(1/2)),2^(1/2))/a^(7/2)/(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 = 10.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.37 \[ \int \frac {1}{x^6 \left (a+b x^2\right )^{7/4}} \, dx=-\frac {\left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {7}{4},-\frac {3}{2},-\frac {b x^2}{a}\right )}{5 a x^5 \left (a+b x^2\right )^{3/4}} \] Input:

Integrate[1/(x^6*(a + b*x^2)^(7/4)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {253, 264, 264, 264, 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.

\(\displaystyle \int \frac {1}{x^6 \left (a+b x^2\right )^{7/4}} \, dx\)

\(\Big \downarrow \) 253

\(\displaystyle \frac {13 \int \frac {1}{x^6 \left (b x^2+a\right )^{3/4}}dx}{3 a}+\frac {2}{3 a x^5 \left (a+b x^2\right )^{3/4}}\)

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 231

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

\(\Big \downarrow \) 229

\(\displaystyle \frac {13 \left (-\frac {9 b \left (-\frac {5 b \left (-\frac {\sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{\sqrt {a} \left (a+b x^2\right )^{3/4}}-\frac {\sqrt [4]{a+b x^2}}{a x}\right )}{6 a}-\frac {\sqrt [4]{a+b x^2}}{3 a x^3}\right )}{10 a}-\frac {\sqrt [4]{a+b x^2}}{5 a x^5}\right )}{3 a}+\frac {2}{3 a x^5 \left (a+b x^2\right )^{3/4}}\)

Input:

Int[1/(x^6*(a + b*x^2)^(7/4)),x]
 

Output:

2/(3*a*x^5*(a + b*x^2)^(3/4)) + (13*(-1/5*(a + b*x^2)^(1/4)/(a*x^5) - (9*b 
*(-1/3*(a + b*x^2)^(1/4)/(a*x^3) - (5*b*(-((a + b*x^2)^(1/4)/(a*x)) - (Sqr 
t[b]*(1 + (b*x^2)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(S 
qrt[a]*(a + b*x^2)^(3/4))))/(6*a)))/(10*a)))/(3*a)
 

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 253

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x 
)^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 
2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m 
}, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

Maple [F]

\[\int \frac {1}{x^{6} \left (b \,x^{2}+a \right )^{\frac {7}{4}}}d x\]

Input:

int(1/x^6/(b*x^2+a)^(7/4),x)
 

Output:

int(1/x^6/(b*x^2+a)^(7/4),x)
 

Fricas [F]

\[ \int \frac {1}{x^6 \left (a+b x^2\right )^{7/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {7}{4}} x^{6}} \,d x } \] Input:

integrate(1/x^6/(b*x^2+a)^(7/4),x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^(1/4)/(b^2*x^10 + 2*a*b*x^8 + a^2*x^6), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.81 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.22 \[ \int \frac {1}{x^6 \left (a+b x^2\right )^{7/4}} \, dx=- \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, \frac {7}{4} \\ - \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5 a^{\frac {7}{4}} x^{5}} \] Input:

integrate(1/x**6/(b*x**2+a)**(7/4),x)
 

Output:

-hyper((-5/2, 7/4), (-3/2,), b*x**2*exp_polar(I*pi)/a)/(5*a**(7/4)*x**5)
 

Maxima [F]

\[ \int \frac {1}{x^6 \left (a+b x^2\right )^{7/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {7}{4}} x^{6}} \,d x } \] Input:

integrate(1/x^6/(b*x^2+a)^(7/4),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^2 + a)^(7/4)*x^6), x)
 

Giac [F]

\[ \int \frac {1}{x^6 \left (a+b x^2\right )^{7/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {7}{4}} x^{6}} \,d x } \] Input:

integrate(1/x^6/(b*x^2+a)^(7/4),x, algorithm="giac")
 

Output:

integrate(1/((b*x^2 + a)^(7/4)*x^6), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^6 \left (a+b x^2\right )^{7/4}} \, dx=\int \frac {1}{x^6\,{\left (b\,x^2+a\right )}^{7/4}} \,d x \] Input:

int(1/(x^6*(a + b*x^2)^(7/4)),x)
 

Output:

int(1/(x^6*(a + b*x^2)^(7/4)), x)
 

Reduce [F]

\[ \int \frac {1}{x^6 \left (a+b x^2\right )^{7/4}} \, dx=\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} a \,x^{6}+\left (b \,x^{2}+a \right )^{\frac {3}{4}} b \,x^{8}}d x \] Input:

int(1/x^6/(b*x^2+a)^(7/4),x)
 

Output:

int(1/((a + b*x**2)**(3/4)*a*x**6 + (a + b*x**2)**(3/4)*b*x**8),x)