Integrand size = 15, antiderivative size = 121 \[ \int \frac {x^6}{\left (a+b x^2\right )^{7/4}} \, dx=-\frac {2 x^5}{3 b \left (a+b x^2\right )^{3/4}}-\frac {40 a x \sqrt [4]{a+b x^2}}{21 b^3}+\frac {20 x^3 \sqrt [4]{a+b x^2}}{21 b^2}+\frac {80 a^{5/2} \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{21 b^{7/2} \left (a+b x^2\right )^{3/4}} \] Output:
-2/3*x^5/b/(b*x^2+a)^(3/4)-40/21*a*x*(b*x^2+a)^(1/4)/b^3+20/21*x^3*(b*x^2+ a)^(1/4)/b^2+80/21*a^(5/2)*(1+b*x^2/a)^(3/4)*InverseJacobiAM(1/2*arctan(b^ (1/2)*x/a^(1/2)),2^(1/2))/b^(7/2)/(b*x^2+a)^(3/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 7.65 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.65 \[ \int \frac {x^6}{\left (a+b x^2\right )^{7/4}} \, dx=\frac {-40 a^2 x-20 a b x^3+6 b^2 x^5+40 a^2 x \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},-\frac {b x^2}{a}\right )}{21 b^3 \left (a+b x^2\right )^{3/4}} \] Input:
Integrate[x^6/(a + b*x^2)^(7/4),x]
Output:
(-40*a^2*x - 20*a*b*x^3 + 6*b^2*x^5 + 40*a^2*x*(1 + (b*x^2)/a)^(3/4)*Hyper geometric2F1[1/2, 3/4, 3/2, -((b*x^2)/a)])/(21*b^3*(a + b*x^2)^(3/4))
Time = 0.21 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {252, 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.
\(\displaystyle \int \frac {x^6}{\left (a+b x^2\right )^{7/4}} \, dx\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {10 \int \frac {x^4}{\left (b x^2+a\right )^{3/4}}dx}{3 b}-\frac {2 x^5}{3 b \left (a+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {10 \left (\frac {2 x^3 \sqrt [4]{a+b x^2}}{7 b}-\frac {6 a \int \frac {x^2}{\left (b x^2+a\right )^{3/4}}dx}{7 b}\right )}{3 b}-\frac {2 x^5}{3 b \left (a+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {10 \left (\frac {2 x^3 \sqrt [4]{a+b x^2}}{7 b}-\frac {6 a \left (\frac {2 x \sqrt [4]{a+b x^2}}{3 b}-\frac {2 a \int \frac {1}{\left (b x^2+a\right )^{3/4}}dx}{3 b}\right )}{7 b}\right )}{3 b}-\frac {2 x^5}{3 b \left (a+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 231 |
\(\displaystyle \frac {10 \left (\frac {2 x^3 \sqrt [4]{a+b x^2}}{7 b}-\frac {6 a \left (\frac {2 x \sqrt [4]{a+b x^2}}{3 b}-\frac {2 a \left (\frac {b x^2}{a}+1\right )^{3/4} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{3 b \left (a+b x^2\right )^{3/4}}\right )}{7 b}\right )}{3 b}-\frac {2 x^5}{3 b \left (a+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 229 |
\(\displaystyle \frac {10 \left (\frac {2 x^3 \sqrt [4]{a+b x^2}}{7 b}-\frac {6 a \left (\frac {2 x \sqrt [4]{a+b x^2}}{3 b}-\frac {4 a^{3/2} \left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{3 b^{3/2} \left (a+b x^2\right )^{3/4}}\right )}{7 b}\right )}{3 b}-\frac {2 x^5}{3 b \left (a+b x^2\right )^{3/4}}\) |
Input:
Int[x^6/(a + b*x^2)^(7/4),x]
Output:
(-2*x^5)/(3*b*(a + b*x^2)^(3/4)) + (10*((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)*Ell ipticF[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(3*b^(3/2)*(a + b*x^2)^(3/4))))/ (7*b)))/(3*b)
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]
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]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
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]
\[\int \frac {x^{6}}{\left (b \,x^{2}+a \right )^{\frac {7}{4}}}d x\]
Input:
int(x^6/(b*x^2+a)^(7/4),x)
Output:
int(x^6/(b*x^2+a)^(7/4),x)
\[ \int \frac {x^6}{\left (a+b x^2\right )^{7/4}} \, dx=\int { \frac {x^{6}}{{\left (b x^{2} + a\right )}^{\frac {7}{4}}} \,d x } \] Input:
integrate(x^6/(b*x^2+a)^(7/4),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(1/4)*x^6/(b^2*x^4 + 2*a*b*x^2 + a^2), x)
Result contains complex when optimal does not.
Time = 0.54 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.22 \[ \int \frac {x^6}{\left (a+b x^2\right )^{7/4}} \, dx=\frac {x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{7 a^{\frac {7}{4}}} \] Input:
integrate(x**6/(b*x**2+a)**(7/4),x)
Output:
x**7*hyper((7/4, 7/2), (9/2,), b*x**2*exp_polar(I*pi)/a)/(7*a**(7/4))
\[ \int \frac {x^6}{\left (a+b x^2\right )^{7/4}} \, dx=\int { \frac {x^{6}}{{\left (b x^{2} + a\right )}^{\frac {7}{4}}} \,d x } \] Input:
integrate(x^6/(b*x^2+a)^(7/4),x, algorithm="maxima")
Output:
integrate(x^6/(b*x^2 + a)^(7/4), x)
\[ \int \frac {x^6}{\left (a+b x^2\right )^{7/4}} \, dx=\int { \frac {x^{6}}{{\left (b x^{2} + a\right )}^{\frac {7}{4}}} \,d x } \] Input:
integrate(x^6/(b*x^2+a)^(7/4),x, algorithm="giac")
Output:
integrate(x^6/(b*x^2 + a)^(7/4), x)
Timed out. \[ \int \frac {x^6}{\left (a+b x^2\right )^{7/4}} \, dx=\int \frac {x^6}{{\left (b\,x^2+a\right )}^{7/4}} \,d x \] Input:
int(x^6/(a + b*x^2)^(7/4),x)
Output:
int(x^6/(a + b*x^2)^(7/4), x)
\[ \int \frac {x^6}{\left (a+b x^2\right )^{7/4}} \, dx=\int \frac {x^{6}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} a +\left (b \,x^{2}+a \right )^{\frac {3}{4}} b \,x^{2}}d x \] Input:
int(x^6/(b*x^2+a)^(7/4),x)
Output:
int(x**6/((a + b*x**2)**(3/4)*a + (a + b*x**2)**(3/4)*b*x**2),x)