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

Optimal. Leaf size=395 \[ \frac{\left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right ) \left (-\sqrt{a} \sqrt{c} (A e+B d)-a B e+A c d\right )}{4 a^{5/4} c^{5/4} \sqrt{a+c x^4}}+\frac{B e \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{2 \sqrt [4]{a} c^{5/4} \sqrt{a+c x^4}}+\frac{\left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} (A e+B d) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 a^{3/4} c^{3/4} \sqrt{a+c x^4}}-\frac{x \sqrt{a+c x^4} (A e+B d)}{2 a \sqrt{c} \left (\sqrt{a}+\sqrt{c} x^2\right )}+\frac{x \left (-a B e+c x^2 (A e+B d)+A c d\right )}{2 a c \sqrt{a+c x^4}} \]

[Out]

(x*(A*c*d - a*B*e + c*(B*d + A*e)*x^2))/(2*a*c*Sqrt[a + c*x^4]) - ((B*d + A*e)*x*Sqrt[a + c*x^4])/(2*a*Sqrt[c]
*(Sqrt[a] + Sqrt[c]*x^2)) + ((B*d + A*e)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*E
llipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(2*a^(3/4)*c^(3/4)*Sqrt[a + c*x^4]) + (B*e*(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])/(2*a^(1/4)*c^(5/
4)*Sqrt[a + c*x^4]) + ((A*c*d - a*B*e - Sqrt[a]*Sqrt[c]*(B*d + A*e))*(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^(5/4)*Sqrt[a + c*x^4])

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Rubi [A]  time = 0.327119, antiderivative size = 395, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {1721, 1179, 1198, 220, 1196} \[ \frac{\left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right ) \left (-\sqrt{a} \sqrt{c} (A e+B d)-a B e+A c d\right )}{4 a^{5/4} c^{5/4} \sqrt{a+c x^4}}+\frac{\left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} (A e+B d) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 a^{3/4} c^{3/4} \sqrt{a+c x^4}}-\frac{x \sqrt{a+c x^4} (A e+B d)}{2 a \sqrt{c} \left (\sqrt{a}+\sqrt{c} x^2\right )}+\frac{x \left (-a B e+c x^2 (A e+B d)+A c d\right )}{2 a c \sqrt{a+c x^4}}+\frac{B e \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 \sqrt [4]{a} c^{5/4} \sqrt{a+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(A*c*d - a*B*e + c*(B*d + A*e)*x^2))/(2*a*c*Sqrt[a + c*x^4]) - ((B*d + A*e)*x*Sqrt[a + c*x^4])/(2*a*Sqrt[c]
*(Sqrt[a] + Sqrt[c]*x^2)) + ((B*d + A*e)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*E
llipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(2*a^(3/4)*c^(3/4)*Sqrt[a + c*x^4]) + (B*e*(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])/(2*a^(1/4)*c^(5/
4)*Sqrt[a + c*x^4]) + ((A*c*d - a*B*e - Sqrt[a]*Sqrt[c]*(B*d + A*e))*(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^(5/4)*Sqrt[a + c*x^4])

Rule 1721

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int[ExpandIntegrand[1/Sqrt[a +
c*x^4], Px*(d + e*x^2)^q*(a + c*x^4)^(p + 1/2), x], x] /; FreeQ[{a, c, d, e}, x] && PolyQ[Px, x^2] && NeQ[c*d^
2 + a*e^2, 0] && IntegerQ[p + 1/2] && IntegerQ[q]

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

Mathematica [C]  time = 0.113368, size = 126, normalized size = 0.32 \[ \frac{2 c x^3 \sqrt{\frac{c x^4}{a}+1} \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\frac{c x^4}{a}\right ) (A e+B d)+3 x \sqrt{\frac{c x^4}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{c x^4}{a}\right ) (a B e+A c d)+3 x (A c d-a B e)}{6 a c \sqrt{a+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.006, size = 320, normalized size = 0.8 \begin{align*} Be \left ( -{\frac{x}{2\,c}{\frac{1}{\sqrt{ \left ({x}^{4}+{\frac{a}{c}} \right ) c}}}}+{\frac{1}{2\,c}\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 ) + \left ( Ae+Bd \right ) \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 ) +Ad \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)*(e*x^2+d)/(c*x^4+a)^(3/2),x)

[Out]

B*e*(-1/2/c*x/((x^4+a/c)*c)^(1/2)+1/2/c/(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))+(A*e+B*d)*(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*d*(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{{\left (B x^{2} + A\right )}{\left (e x^{2} + d\right )}}{{\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)*(e*x^2+d)/(c*x^4+a)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*(e*x^2 + d)/(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{{\left (B e x^{4} +{\left (B d + A e\right )} x^{2} + A d\right )} \sqrt{c x^{4} + a}}{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)*(e*x^2+d)/(c*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 15.8711, size = 167, normalized size = 0.42 \begin{align*} \frac{A d 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{A e 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 )} + \frac{B d 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 )} + \frac{B e x^{5} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{5}{4}, \frac{3}{2} \\ \frac{9}{4} \end{matrix}\middle |{\frac{c x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac{3}{2}} \Gamma \left (\frac{9}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*d*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)) + A*e*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)) + B*d*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)) + B*e*x**5*gamma(5/4)*hyper((5/4, 3/2), (
9/4,), c*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*gamma(9/4))

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

[Out]

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