3.11 \(\int \frac{A+B x^2}{(a+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=262 \[ -\frac{\left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (\sqrt{a} B-A \sqrt{c}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{4 a^{5/4} c^{3/4} \sqrt{a+c x^4}}+\frac{B \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 a^{3/4} c^{3/4} \sqrt{a+c x^4}}+\frac{x \left (A+B x^2\right )}{2 a \sqrt{a+c x^4}}-\frac{B x \sqrt{a+c x^4}}{2 a \sqrt{c} \left (\sqrt{a}+\sqrt{c} x^2\right )} \]

[Out]

(x*(A + B*x^2))/(2*a*Sqrt[a + c*x^4]) - (B*x*Sqrt[a + c*x^4])/(2*a*Sqrt[c]*(Sqrt[a] + Sqrt[c]*x^2)) + (B*(Sqrt
[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/
(2*a^(3/4)*c^(3/4)*Sqrt[a + c*x^4]) - ((Sqrt[a]*B - A*Sqrt[c])*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[
a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(4*a^(5/4)*c^(3/4)*Sqrt[a + c*x^4])

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Rubi [A]  time = 0.0964995, antiderivative size = 262, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {1179, 1198, 220, 1196} \[ -\frac{\left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (\sqrt{a} B-A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 a^{5/4} c^{3/4} \sqrt{a+c x^4}}+\frac{B \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 a^{3/4} c^{3/4} \sqrt{a+c x^4}}+\frac{x \left (A+B x^2\right )}{2 a \sqrt{a+c x^4}}-\frac{B x \sqrt{a+c x^4}}{2 a \sqrt{c} \left (\sqrt{a}+\sqrt{c} x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(a + c*x^4)^(3/2),x]

[Out]

(x*(A + B*x^2))/(2*a*Sqrt[a + c*x^4]) - (B*x*Sqrt[a + c*x^4])/(2*a*Sqrt[c]*(Sqrt[a] + Sqrt[c]*x^2)) + (B*(Sqrt
[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/
(2*a^(3/4)*c^(3/4)*Sqrt[a + c*x^4]) - ((Sqrt[a]*B - A*Sqrt[c])*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[
a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(4*a^(5/4)*c^(3/4)*Sqrt[a + c*x^4])

Rule 1179

Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)*(a + c*x^4)^(p + 1))/(
4*a*(p + 1)), x] + Dist[1/(4*a*(p + 1)), Int[Simp[d*(4*p + 5) + e*(4*p + 7)*x^2, x]*(a + c*x^4)^(p + 1), x], x
] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntegerQ[2*p]

Rule 1198

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(e + d*q)/q, Int
[1/Sqrt[a + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a
, c, d, e}, x] && PosQ[c/a]

Rule 220

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

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{A+B x^2}{\left (a+c x^4\right )^{3/2}} \, dx &=\frac{x \left (A+B x^2\right )}{2 a \sqrt{a+c x^4}}-\frac{\int \frac{-A+B x^2}{\sqrt{a+c x^4}} \, dx}{2 a}\\ &=\frac{x \left (A+B x^2\right )}{2 a \sqrt{a+c x^4}}+\frac{\left (A-\frac{\sqrt{a} B}{\sqrt{c}}\right ) \int \frac{1}{\sqrt{a+c x^4}} \, dx}{2 a}+\frac{B \int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{a+c x^4}} \, dx}{2 \sqrt{a} \sqrt{c}}\\ &=\frac{x \left (A+B x^2\right )}{2 a \sqrt{a+c x^4}}-\frac{B x \sqrt{a+c x^4}}{2 a \sqrt{c} \left (\sqrt{a}+\sqrt{c} x^2\right )}+\frac{B \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 a^{3/4} c^{3/4} \sqrt{a+c x^4}}+\frac{\left (A-\frac{\sqrt{a} B}{\sqrt{c}}\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 a^{5/4} \sqrt [4]{c} \sqrt{a+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0417099, size = 99, normalized size = 0.38 \[ \frac{3 A x \sqrt{\frac{c x^4}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{c x^4}{a}\right )+2 B x^3 \sqrt{\frac{c x^4}{a}+1} \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\frac{c x^4}{a}\right )+3 A x}{6 a \sqrt{a+c x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(a + c*x^4)^(3/2),x]

[Out]

(3*A*x + 3*A*x*Sqrt[1 + (c*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^4)/a)] + 2*B*x^3*Sqrt[1 + (c*x^4)/a
]*Hypergeometric2F1[3/4, 3/2, 7/4, -((c*x^4)/a)])/(6*a*Sqrt[a + c*x^4])

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Maple [C]  time = 0.004, size = 217, normalized size = 0.8 \begin{align*} B \left ({\frac{{x}^{3}}{2\,a}{\frac{1}{\sqrt{ \left ({x}^{4}+{\frac{a}{c}} \right ) c}}}}-{{\frac{i}{2}}\sqrt{1-{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}} \left ({\it EllipticF} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ) -{\it EllipticE} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ) \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{c{x}^{4}+a}}}{\frac{1}{\sqrt{c}}}} \right ) +A \left ({\frac{x}{2\,a}{\frac{1}{\sqrt{ \left ({x}^{4}+{\frac{a}{c}} \right ) c}}}}+{\frac{1}{2\,a}\sqrt{1-{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{c{x}^{4}+a}}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/(c*x^4+a)^(3/2),x)

[Out]

B*(1/2/a*x^3/((x^4+a/c)*c)^(1/2)-1/2*I/a^(1/2)/(I/a^(1/2)*c^(1/2))^(1/2)*(1-I/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1+I/
a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)/c^(1/2)*(EllipticF(x*(I/a^(1/2)*c^(1/2))^(1/2),I)-EllipticE(x*(I/a^
(1/2)*c^(1/2))^(1/2),I)))+A*(1/2/a*x/((x^4+a/c)*c)^(1/2)+1/2/a/(I/a^(1/2)*c^(1/2))^(1/2)*(1-I/a^(1/2)*c^(1/2)*
x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)*EllipticF(x*(I/a^(1/2)*c^(1/2))^(1/2),I))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(c*x^4+a)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)/(c*x^4 + a)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{4} + a}{\left (B x^{2} + A\right )}}{c^{2} x^{8} + 2 \, a c x^{4} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(c*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^4 + a)*(B*x^2 + A)/(c^2*x^8 + 2*a*c*x^4 + a^2), x)

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Sympy [C]  time = 8.42081, size = 78, normalized size = 0.3 \begin{align*} \frac{A x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{3}{2} \\ \frac{5}{4} \end{matrix}\middle |{\frac{c x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac{3}{2}} \Gamma \left (\frac{5}{4}\right )} + \frac{B x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{4}, \frac{3}{2} \\ \frac{7}{4} \end{matrix}\middle |{\frac{c x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac{3}{2}} \Gamma \left (\frac{7}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/(c*x**4+a)**(3/2),x)

[Out]

A*x*gamma(1/4)*hyper((1/4, 3/2), (5/4,), c*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*gamma(5/4)) + B*x**3*gamma(3/4)
*hyper((3/4, 3/2), (7/4,), c*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*gamma(7/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(c*x^4+a)^(3/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)/(c*x^4 + a)^(3/2), x)