Integrand size = 15, antiderivative size = 126 \[ \int \frac {1}{x^6 \left (a+b x^2\right )^{5/4}} \, dx=-\frac {1}{5 a x^5 \sqrt [4]{a+b x^2}}+\frac {11 b}{30 a^2 x^3 \sqrt [4]{a+b x^2}}-\frac {77 b^2}{60 a^3 x \sqrt [4]{a+b x^2}}-\frac {77 b^{5/2} \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{20 a^{7/2} \sqrt [4]{a+b x^2}} \] Output:
-1/5/a/x^5/(b*x^2+a)^(1/4)+11/30*b/a^2/x^3/(b*x^2+a)^(1/4)-77/60*b^2/a^3/x /(b*x^2+a)^(1/4)-77/20*b^(5/2)*(1+b*x^2/a)^(1/4)*EllipticE(sin(1/2*arctan( b^(1/2)*x/a^(1/2))),2^(1/2))/a^(7/2)/(b*x^2+a)^(1/4)
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 = 0.43 \[ \int \frac {1}{x^6 \left (a+b x^2\right )^{5/4}} \, dx=-\frac {\sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {5}{4},-\frac {3}{2},-\frac {b x^2}{a}\right )}{5 a x^5 \sqrt [4]{a+b x^2}} \] Input:
Integrate[1/(x^6*(a + b*x^2)^(5/4)),x]
Output:
-1/5*((1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[-5/2, 5/4, -3/2, -((b*x^2)/a )])/(a*x^5*(a + b*x^2)^(1/4))
Time = 0.21 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {251, 251, 251, 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.
\(\displaystyle \int \frac {1}{x^6 \left (a+b x^2\right )^{5/4}} \, dx\) |
\(\Big \downarrow \) 251 |
\(\displaystyle -\frac {11 b \int \frac {1}{x^4 \left (b x^2+a\right )^{5/4}}dx}{10 a}-\frac {1}{5 a x^5 \sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 251 |
\(\displaystyle -\frac {11 b \left (-\frac {7 b \int \frac {1}{x^2 \left (b x^2+a\right )^{5/4}}dx}{6 a}-\frac {1}{3 a x^3 \sqrt [4]{a+b x^2}}\right )}{10 a}-\frac {1}{5 a x^5 \sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 251 |
\(\displaystyle -\frac {11 b \left (-\frac {7 b \left (-\frac {3 b \int \frac {1}{\left (b x^2+a\right )^{5/4}}dx}{2 a}-\frac {1}{a x \sqrt [4]{a+b x^2}}\right )}{6 a}-\frac {1}{3 a x^3 \sqrt [4]{a+b x^2}}\right )}{10 a}-\frac {1}{5 a x^5 \sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 213 |
\(\displaystyle -\frac {11 b \left (-\frac {7 b \left (-\frac {3 b \sqrt [4]{\frac {b x^2}{a}+1} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx}{2 a^2 \sqrt [4]{a+b x^2}}-\frac {1}{a x \sqrt [4]{a+b x^2}}\right )}{6 a}-\frac {1}{3 a x^3 \sqrt [4]{a+b x^2}}\right )}{10 a}-\frac {1}{5 a x^5 \sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle -\frac {11 b \left (-\frac {7 b \left (-\frac {3 \sqrt {b} \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{a^{3/2} \sqrt [4]{a+b x^2}}-\frac {1}{a x \sqrt [4]{a+b x^2}}\right )}{6 a}-\frac {1}{3 a x^3 \sqrt [4]{a+b x^2}}\right )}{10 a}-\frac {1}{5 a x^5 \sqrt [4]{a+b x^2}}\) |
Input:
Int[1/(x^6*(a + b*x^2)^(5/4)),x]
Output:
-1/5*1/(a*x^5*(a + b*x^2)^(1/4)) - (11*b*(-1/3*1/(a*x^3*(a + b*x^2)^(1/4)) - (7*b*(-(1/(a*x*(a + b*x^2)^(1/4))) - (3*Sqrt[b]*(1 + (b*x^2)/a)^(1/4)*E llipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(a^(3/2)*(a + b*x^2)^(1/4))))/ (6*a)))/(10*a)
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]
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]
Int[((c_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[(c*x)^ (m + 1)/(a*c*(m + 1)*(a + b*x^2)^(1/4)), x] - Simp[b*((2*m + 1)/(2*a*c^2*(m + 1))) Int[(c*x)^(m + 2)/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b, c}, x ] && PosQ[b/a] && IntegerQ[2*m] && LtQ[m, -1]
\[\int \frac {1}{x^{6} \left (b \,x^{2}+a \right )^{\frac {5}{4}}}d x\]
Input:
int(1/x^6/(b*x^2+a)^(5/4),x)
Output:
int(1/x^6/(b*x^2+a)^(5/4),x)
\[ \int \frac {1}{x^6 \left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} x^{6}} \,d x } \] Input:
integrate(1/x^6/(b*x^2+a)^(5/4),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(3/4)/(b^2*x^10 + 2*a*b*x^8 + a^2*x^6), x)
Result contains complex when optimal does not.
Time = 0.68 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.25 \[ \int \frac {1}{x^6 \left (a+b x^2\right )^{5/4}} \, dx=- \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, \frac {5}{4} \\ - \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5 a^{\frac {5}{4}} x^{5}} \] Input:
integrate(1/x**6/(b*x**2+a)**(5/4),x)
Output:
-hyper((-5/2, 5/4), (-3/2,), b*x**2*exp_polar(I*pi)/a)/(5*a**(5/4)*x**5)
\[ \int \frac {1}{x^6 \left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} x^{6}} \,d x } \] Input:
integrate(1/x^6/(b*x^2+a)^(5/4),x, algorithm="maxima")
Output:
integrate(1/((b*x^2 + a)^(5/4)*x^6), x)
\[ \int \frac {1}{x^6 \left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} x^{6}} \,d x } \] Input:
integrate(1/x^6/(b*x^2+a)^(5/4),x, algorithm="giac")
Output:
integrate(1/((b*x^2 + a)^(5/4)*x^6), x)
Timed out. \[ \int \frac {1}{x^6 \left (a+b x^2\right )^{5/4}} \, dx=\int \frac {1}{x^6\,{\left (b\,x^2+a\right )}^{5/4}} \,d x \] Input:
int(1/(x^6*(a + b*x^2)^(5/4)),x)
Output:
int(1/(x^6*(a + b*x^2)^(5/4)), x)
\[ \int \frac {1}{x^6 \left (a+b x^2\right )^{5/4}} \, dx=\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a \,x^{6}+\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{8}}d x \] Input:
int(1/x^6/(b*x^2+a)^(5/4),x)
Output:
int(1/((a + b*x**2)**(1/4)*a*x**6 + (a + b*x**2)**(1/4)*b*x**8),x)