3.366 \(\int \frac{7+5 x^2}{\sqrt{4+3 x^2+x^4}} \, dx\)

Optimal. Leaf size=151 \[ \frac{17 \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right ),\frac{1}{8}\right )}{2 \sqrt{2} \sqrt{x^4+3 x^2+4}}+\frac{5 \sqrt{x^4+3 x^2+4} x}{x^2+2}-\frac{5 \sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{\sqrt{x^4+3 x^2+4}} \]

[Out]

(5*x*Sqrt[4 + 3*x^2 + x^4])/(2 + x^2) - (5*Sqrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticE[2*A
rcTan[x/Sqrt[2]], 1/8])/Sqrt[4 + 3*x^2 + x^4] + (17*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticF[2*
ArcTan[x/Sqrt[2]], 1/8])/(2*Sqrt[2]*Sqrt[4 + 3*x^2 + x^4])

________________________________________________________________________________________

Rubi [A]  time = 0.0361226, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {1197, 1103, 1195} \[ \frac{5 \sqrt{x^4+3 x^2+4} x}{x^2+2}+\frac{17 \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{2 \sqrt{2} \sqrt{x^4+3 x^2+4}}-\frac{5 \sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{\sqrt{x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

Int[(7 + 5*x^2)/Sqrt[4 + 3*x^2 + x^4],x]

[Out]

(5*x*Sqrt[4 + 3*x^2 + x^4])/(2 + x^2) - (5*Sqrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticE[2*A
rcTan[x/Sqrt[2]], 1/8])/Sqrt[4 + 3*x^2 + x^4] + (17*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticF[2*
ArcTan[x/Sqrt[2]], 1/8])/(2*Sqrt[2]*Sqrt[4 + 3*x^2 + x^4])

Rule 1197

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

Rule 1103

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1195

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

Rubi steps

\begin{align*} \int \frac{7+5 x^2}{\sqrt{4+3 x^2+x^4}} \, dx &=-\left (10 \int \frac{1-\frac{x^2}{2}}{\sqrt{4+3 x^2+x^4}} \, dx\right )+17 \int \frac{1}{\sqrt{4+3 x^2+x^4}} \, dx\\ &=\frac{5 x \sqrt{4+3 x^2+x^4}}{2+x^2}-\frac{5 \sqrt{2} \left (2+x^2\right ) \sqrt{\frac{4+3 x^2+x^4}{\left (2+x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{\sqrt{4+3 x^2+x^4}}+\frac{17 \left (2+x^2\right ) \sqrt{\frac{4+3 x^2+x^4}{\left (2+x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{2 \sqrt{2} \sqrt{4+3 x^2+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.177229, size = 214, normalized size = 1.42 \[ \frac{\sqrt{1-\frac{2 i x^2}{\sqrt{7}-3 i}} \sqrt{1+\frac{2 i x^2}{\sqrt{7}+3 i}} \left (\left (5 \sqrt{7}+i\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{-\frac{2 i}{\sqrt{7}-3 i}} x\right ),\frac{-\sqrt{7}+3 i}{\sqrt{7}+3 i}\right )-5 \left (\sqrt{7}+3 i\right ) E\left (i \sinh ^{-1}\left (\sqrt{-\frac{2 i}{-3 i+\sqrt{7}}} x\right )|\frac{3 i-\sqrt{7}}{3 i+\sqrt{7}}\right )\right )}{2 \sqrt{2} \sqrt{-\frac{i}{\sqrt{7}-3 i}} \sqrt{x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(7 + 5*x^2)/Sqrt[4 + 3*x^2 + x^4],x]

[Out]

(Sqrt[1 - ((2*I)*x^2)/(-3*I + Sqrt[7])]*Sqrt[1 + ((2*I)*x^2)/(3*I + Sqrt[7])]*(-5*(3*I + Sqrt[7])*EllipticE[I*
ArcSinh[Sqrt[(-2*I)/(-3*I + Sqrt[7])]*x], (3*I - Sqrt[7])/(3*I + Sqrt[7])] + (I + 5*Sqrt[7])*EllipticF[I*ArcSi
nh[Sqrt[(-2*I)/(-3*I + Sqrt[7])]*x], (3*I - Sqrt[7])/(3*I + Sqrt[7])]))/(2*Sqrt[2]*Sqrt[(-I)/(-3*I + Sqrt[7])]
*Sqrt[4 + 3*x^2 + x^4])

________________________________________________________________________________________

Maple [C]  time = 0.007, size = 209, normalized size = 1.4 \begin{align*} -160\,{\frac{\sqrt{1- \left ( -3/8+i/8\sqrt{7} \right ){x}^{2}}\sqrt{1- \left ( -3/8-i/8\sqrt{7} \right ){x}^{2}} \left ({\it EllipticF} \left ( 1/4\,x\sqrt{-6+2\,i\sqrt{7}},1/4\,\sqrt{2+6\,i\sqrt{7}} \right ) -{\it EllipticE} \left ( 1/4\,x\sqrt{-6+2\,i\sqrt{7}},1/4\,\sqrt{2+6\,i\sqrt{7}} \right ) \right ) }{\sqrt{-6+2\,i\sqrt{7}}\sqrt{{x}^{4}+3\,{x}^{2}+4} \left ( i\sqrt{7}+3 \right ) }}+28\,{\frac{\sqrt{1- \left ( -3/8+i/8\sqrt{7} \right ){x}^{2}}\sqrt{1- \left ( -3/8-i/8\sqrt{7} \right ){x}^{2}}{\it EllipticF} \left ( 1/4\,x\sqrt{-6+2\,i\sqrt{7}},1/4\,\sqrt{2+6\,i\sqrt{7}} \right ) }{\sqrt{-6+2\,i\sqrt{7}}\sqrt{{x}^{4}+3\,{x}^{2}+4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^2+7)/(x^4+3*x^2+4)^(1/2),x)

[Out]

-160/(-6+2*I*7^(1/2))^(1/2)*(1-(-3/8+1/8*I*7^(1/2))*x^2)^(1/2)*(1-(-3/8-1/8*I*7^(1/2))*x^2)^(1/2)/(x^4+3*x^2+4
)^(1/2)/(I*7^(1/2)+3)*(EllipticF(1/4*x*(-6+2*I*7^(1/2))^(1/2),1/4*(2+6*I*7^(1/2))^(1/2))-EllipticE(1/4*x*(-6+2
*I*7^(1/2))^(1/2),1/4*(2+6*I*7^(1/2))^(1/2)))+28/(-6+2*I*7^(1/2))^(1/2)*(1-(-3/8+1/8*I*7^(1/2))*x^2)^(1/2)*(1-
(-3/8-1/8*I*7^(1/2))*x^2)^(1/2)/(x^4+3*x^2+4)^(1/2)*EllipticF(1/4*x*(-6+2*I*7^(1/2))^(1/2),1/4*(2+6*I*7^(1/2))
^(1/2))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{5 \, x^{2} + 7}{\sqrt{x^{4} + 3 \, x^{2} + 4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+7)/(x^4+3*x^2+4)^(1/2),x, algorithm="maxima")

[Out]

integrate((5*x^2 + 7)/sqrt(x^4 + 3*x^2 + 4), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{5 \, x^{2} + 7}{\sqrt{x^{4} + 3 \, x^{2} + 4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+7)/(x^4+3*x^2+4)^(1/2),x, algorithm="fricas")

[Out]

integral((5*x^2 + 7)/sqrt(x^4 + 3*x^2 + 4), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{5 x^{2} + 7}{\sqrt{\left (x^{2} - x + 2\right ) \left (x^{2} + x + 2\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x**2+7)/(x**4+3*x**2+4)**(1/2),x)

[Out]

Integral((5*x**2 + 7)/sqrt((x**2 - x + 2)*(x**2 + x + 2)), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{5 \, x^{2} + 7}{\sqrt{x^{4} + 3 \, x^{2} + 4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+7)/(x^4+3*x^2+4)^(1/2),x, algorithm="giac")

[Out]

integrate((5*x^2 + 7)/sqrt(x^4 + 3*x^2 + 4), x)