3.1060 \(\int \frac{x^2}{(a-b x^2)^{3/4} (2 a-b x^2)} \, dx\)

Optimal. Leaf size=119 \[ \frac{\tan ^{-1}\left (\frac{a^{3/4} \left (1-\frac{\sqrt{a-b x^2}}{\sqrt{a}}\right )}{\sqrt{b} x \sqrt [4]{a-b x^2}}\right )}{\sqrt [4]{a} b^{3/2}}-\frac{\tanh ^{-1}\left (\frac{a^{3/4} \left (\frac{\sqrt{a-b x^2}}{\sqrt{a}}+1\right )}{\sqrt{b} x \sqrt [4]{a-b x^2}}\right )}{\sqrt [4]{a} b^{3/2}} \]

[Out]

ArcTan[(a^(3/4)*(1 - Sqrt[a - b*x^2]/Sqrt[a]))/(Sqrt[b]*x*(a - b*x^2)^(1/4))]/(a^(1/4)*b^(3/2)) - ArcTanh[(a^(
3/4)*(1 + Sqrt[a - b*x^2]/Sqrt[a]))/(Sqrt[b]*x*(a - b*x^2)^(1/4))]/(a^(1/4)*b^(3/2))

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Rubi [A]  time = 0.0371274, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.036, Rules used = {441} \[ \frac{\tan ^{-1}\left (\frac{a^{3/4} \left (1-\frac{\sqrt{a-b x^2}}{\sqrt{a}}\right )}{\sqrt{b} x \sqrt [4]{a-b x^2}}\right )}{\sqrt [4]{a} b^{3/2}}-\frac{\tanh ^{-1}\left (\frac{a^{3/4} \left (\frac{\sqrt{a-b x^2}}{\sqrt{a}}+1\right )}{\sqrt{b} x \sqrt [4]{a-b x^2}}\right )}{\sqrt [4]{a} b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/((a - b*x^2)^(3/4)*(2*a - b*x^2)),x]

[Out]

ArcTan[(a^(3/4)*(1 - Sqrt[a - b*x^2]/Sqrt[a]))/(Sqrt[b]*x*(a - b*x^2)^(1/4))]/(a^(1/4)*b^(3/2)) - ArcTanh[(a^(
3/4)*(1 + Sqrt[a - b*x^2]/Sqrt[a]))/(Sqrt[b]*x*(a - b*x^2)^(1/4))]/(a^(1/4)*b^(3/2))

Rule 441

Int[(x_)^2/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> -Simp[(b*ArcTan[(b + Rt[b^2/a, 4]
^2*Sqrt[a + b*x^2])/(Rt[b^2/a, 4]^3*x*(a + b*x^2)^(1/4))])/(a*d*Rt[b^2/a, 4]^3), x] + Simp[(b*ArcTanh[(b - Rt[
b^2/a, 4]^2*Sqrt[a + b*x^2])/(Rt[b^2/a, 4]^3*x*(a + b*x^2)^(1/4))])/(a*d*Rt[b^2/a, 4]^3), x] /; FreeQ[{a, b, c
, d}, x] && EqQ[b*c - 2*a*d, 0] && PosQ[b^2/a]

Rubi steps

\begin{align*} \int \frac{x^2}{\left (a-b x^2\right )^{3/4} \left (2 a-b x^2\right )} \, dx &=\frac{\tan ^{-1}\left (\frac{a^{3/4} \left (1-\frac{\sqrt{a-b x^2}}{\sqrt{a}}\right )}{\sqrt{b} x \sqrt [4]{a-b x^2}}\right )}{\sqrt [4]{a} b^{3/2}}-\frac{\tanh ^{-1}\left (\frac{a^{3/4} \left (1+\frac{\sqrt{a-b x^2}}{\sqrt{a}}\right )}{\sqrt{b} x \sqrt [4]{a-b x^2}}\right )}{\sqrt [4]{a} b^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0603899, size = 68, normalized size = 0.57 \[ \frac{x^3 \left (\frac{a-b x^2}{a}\right )^{3/4} F_1\left (\frac{3}{2};\frac{3}{4},1;\frac{5}{2};\frac{b x^2}{a},\frac{b x^2}{2 a}\right )}{6 a \left (a-b x^2\right )^{3/4}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^2/((a - b*x^2)^(3/4)*(2*a - b*x^2)),x]

[Out]

(x^3*((a - b*x^2)/a)^(3/4)*AppellF1[3/2, 3/4, 1, 5/2, (b*x^2)/a, (b*x^2)/(2*a)])/(6*a*(a - b*x^2)^(3/4))

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Maple [F]  time = 0.056, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{2}}{-b{x}^{2}+2\,a} \left ( -b{x}^{2}+a \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{x^{2}}{{\left (b x^{2} - 2 \, a\right )}{\left (-b x^{2} + a\right )}^{\frac{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.63471, size = 571, normalized size = 4.8 \begin{align*} -2 \, \left (\frac{1}{4}\right )^{\frac{1}{4}} \left (-\frac{1}{a b^{6}}\right )^{\frac{1}{4}} \arctan \left (\frac{4 \,{\left (\sqrt{\frac{1}{2}} \left (\frac{1}{4}\right )^{\frac{3}{4}} a b^{4} x \sqrt{\frac{b^{4} x^{2} \sqrt{-\frac{1}{a b^{6}}} + 2 \, \sqrt{-b x^{2} + a}}{x^{2}}} \left (-\frac{1}{a b^{6}}\right )^{\frac{3}{4}} - \left (\frac{1}{4}\right )^{\frac{3}{4}}{\left (-b x^{2} + a\right )}^{\frac{1}{4}} a b^{4} \left (-\frac{1}{a b^{6}}\right )^{\frac{3}{4}}\right )}}{x}\right ) - \frac{1}{2} \, \left (\frac{1}{4}\right )^{\frac{1}{4}} \left (-\frac{1}{a b^{6}}\right )^{\frac{1}{4}} \log \left (\frac{\left (\frac{1}{4}\right )^{\frac{1}{4}} b^{2} x \left (-\frac{1}{a b^{6}}\right )^{\frac{1}{4}} +{\left (-b x^{2} + a\right )}^{\frac{1}{4}}}{x}\right ) + \frac{1}{2} \, \left (\frac{1}{4}\right )^{\frac{1}{4}} \left (-\frac{1}{a b^{6}}\right )^{\frac{1}{4}} \log \left (-\frac{\left (\frac{1}{4}\right )^{\frac{1}{4}} b^{2} x \left (-\frac{1}{a b^{6}}\right )^{\frac{1}{4}} -{\left (-b x^{2} + a\right )}^{\frac{1}{4}}}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(1/4)^(1/4)*(-1/(a*b^6))^(1/4)*arctan(4*(sqrt(1/2)*(1/4)^(3/4)*a*b^4*x*sqrt((b^4*x^2*sqrt(-1/(a*b^6)) + 2*s
qrt(-b*x^2 + a))/x^2)*(-1/(a*b^6))^(3/4) - (1/4)^(3/4)*(-b*x^2 + a)^(1/4)*a*b^4*(-1/(a*b^6))^(3/4))/x) - 1/2*(
1/4)^(1/4)*(-1/(a*b^6))^(1/4)*log(((1/4)^(1/4)*b^2*x*(-1/(a*b^6))^(1/4) + (-b*x^2 + a)^(1/4))/x) + 1/2*(1/4)^(
1/4)*(-1/(a*b^6))^(1/4)*log(-((1/4)^(1/4)*b^2*x*(-1/(a*b^6))^(1/4) - (-b*x^2 + a)^(1/4))/x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x^{2}}{- 2 a \left (a - b x^{2}\right )^{\frac{3}{4}} + b x^{2} \left (a - b x^{2}\right )^{\frac{3}{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(x**2/(-2*a*(a - b*x**2)**(3/4) + b*x**2*(a - b*x**2)**(3/4)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{x^{2}}{{\left (b x^{2} - 2 \, a\right )}{\left (-b x^{2} + a\right )}^{\frac{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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