3.829 \(\int \frac {x^6}{(a+b x^2)^{3/4}} \, dx\)

Optimal. Leaf size=124 \[ -\frac {80 a^{7/2} \left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{77 b^{7/2} \left (a+b x^2\right )^{3/4}}+\frac {40 a^2 x \sqrt [4]{a+b x^2}}{77 b^3}-\frac {20 a x^3 \sqrt [4]{a+b x^2}}{77 b^2}+\frac {2 x^5 \sqrt [4]{a+b x^2}}{11 b} \]

[Out]

40/77*a^2*x*(b*x^2+a)^(1/4)/b^3-20/77*a*x^3*(b*x^2+a)^(1/4)/b^2+2/11*x^5*(b*x^2+a)^(1/4)/b-80/77*a^(7/2)*(1+b*
x^2/a)^(3/4)*(cos(1/2*arctan(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arctan(x*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2
*arctan(x*b^(1/2)/a^(1/2))),2^(1/2))/b^(7/2)/(b*x^2+a)^(3/4)

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {321, 233, 231} \[ \frac {40 a^2 x \sqrt [4]{a+b x^2}}{77 b^3}-\frac {80 a^{7/2} \left (\frac {b x^2}{a}+1\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{77 b^{7/2} \left (a+b x^2\right )^{3/4}}-\frac {20 a x^3 \sqrt [4]{a+b x^2}}{77 b^2}+\frac {2 x^5 \sqrt [4]{a+b x^2}}{11 b} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(a + b*x^2)^(3/4),x]

[Out]

(40*a^2*x*(a + b*x^2)^(1/4))/(77*b^3) - (20*a*x^3*(a + b*x^2)^(1/4))/(77*b^2) + (2*x^5*(a + b*x^2)^(1/4))/(11*
b) - (80*a^(7/2)*(1 + (b*x^2)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(77*b^(7/2)*(a + b*x^2)^(3
/4))

Rule 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(3/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 233

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Dist[(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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {x^6}{\left (a+b x^2\right )^{3/4}} \, dx &=\frac {2 x^5 \sqrt [4]{a+b x^2}}{11 b}-\frac {(10 a) \int \frac {x^4}{\left (a+b x^2\right )^{3/4}} \, dx}{11 b}\\ &=-\frac {20 a x^3 \sqrt [4]{a+b x^2}}{77 b^2}+\frac {2 x^5 \sqrt [4]{a+b x^2}}{11 b}+\frac {\left (60 a^2\right ) \int \frac {x^2}{\left (a+b x^2\right )^{3/4}} \, dx}{77 b^2}\\ &=\frac {40 a^2 x \sqrt [4]{a+b x^2}}{77 b^3}-\frac {20 a x^3 \sqrt [4]{a+b x^2}}{77 b^2}+\frac {2 x^5 \sqrt [4]{a+b x^2}}{11 b}-\frac {\left (40 a^3\right ) \int \frac {1}{\left (a+b x^2\right )^{3/4}} \, dx}{77 b^3}\\ &=\frac {40 a^2 x \sqrt [4]{a+b x^2}}{77 b^3}-\frac {20 a x^3 \sqrt [4]{a+b x^2}}{77 b^2}+\frac {2 x^5 \sqrt [4]{a+b x^2}}{11 b}-\frac {\left (40 a^3 \left (1+\frac {b x^2}{a}\right )^{3/4}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{3/4}} \, dx}{77 b^3 \left (a+b x^2\right )^{3/4}}\\ &=\frac {40 a^2 x \sqrt [4]{a+b x^2}}{77 b^3}-\frac {20 a x^3 \sqrt [4]{a+b x^2}}{77 b^2}+\frac {2 x^5 \sqrt [4]{a+b x^2}}{11 b}-\frac {80 a^{7/2} \left (1+\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{77 b^{7/2} \left (a+b x^2\right )^{3/4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.04, size = 90, normalized size = 0.73 \[ \frac {2 \left (-20 a^3 x \left (\frac {b x^2}{a}+1\right )^{3/4} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};-\frac {b x^2}{a}\right )+20 a^3 x+10 a^2 b x^3-3 a b^2 x^5+7 b^3 x^7\right )}{77 b^3 \left (a+b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(a + b*x^2)^(3/4),x]

[Out]

(2*(20*a^3*x + 10*a^2*b*x^3 - 3*a*b^2*x^5 + 7*b^3*x^7 - 20*a^3*x*(1 + (b*x^2)/a)^(3/4)*Hypergeometric2F1[1/2,
3/4, 3/2, -((b*x^2)/a)]))/(77*b^3*(a + b*x^2)^(3/4))

________________________________________________________________________________________

fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{6}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^6/(b*x^2 + a)^(3/4), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{6}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^6/(b*x^2 + a)^(3/4), x)

________________________________________________________________________________________

maple [F]  time = 0.30, size = 0, normalized size = 0.00 \[ \int \frac {x^{6}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{6}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^6/(b*x^2 + a)^(3/4), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^6}{{\left (b\,x^2+a\right )}^{3/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(a + b*x^2)^(3/4),x)

[Out]

int(x^6/(a + b*x^2)^(3/4), x)

________________________________________________________________________________________

sympy [C]  time = 0.96, size = 27, normalized size = 0.22 \[ \frac {x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{7 a^{\frac {3}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________