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

Optimal. Leaf size=181 \[ -\frac{3 \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right ),\frac{1}{8}\right )}{4 \sqrt{2} \sqrt{x^4+3 x^2+4}}-\frac{19 \sqrt{x^4+3 x^2+4} x}{28 \left (x^2+2\right )}+\frac{\left (19 x^2+53\right ) x}{28 \sqrt{x^4+3 x^2+4}}+\frac{19 \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{14 \sqrt{2} \sqrt{x^4+3 x^2+4}} \]

[Out]

(x*(53 + 19*x^2))/(28*Sqrt[4 + 3*x^2 + x^4]) - (19*x*Sqrt[4 + 3*x^2 + x^4])/(28*(2 + x^2)) + (19*(2 + x^2)*Sqr
t[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticE[2*ArcTan[x/Sqrt[2]], 1/8])/(14*Sqrt[2]*Sqrt[4 + 3*x^2 + x^4]) - (3*
(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticF[2*ArcTan[x/Sqrt[2]], 1/8])/(4*Sqrt[2]*Sqrt[4 + 3*x^2 +
x^4])

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Rubi [A]  time = 0.0523354, antiderivative size = 181, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {1178, 1197, 1103, 1195} \[ -\frac{19 \sqrt{x^4+3 x^2+4} x}{28 \left (x^2+2\right )}+\frac{\left (19 x^2+53\right ) x}{28 \sqrt{x^4+3 x^2+4}}-\frac{3 \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{4 \sqrt{2} \sqrt{x^4+3 x^2+4}}+\frac{19 \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{14 \sqrt{2} \sqrt{x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(53 + 19*x^2))/(28*Sqrt[4 + 3*x^2 + x^4]) - (19*x*Sqrt[4 + 3*x^2 + x^4])/(28*(2 + x^2)) + (19*(2 + x^2)*Sqr
t[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticE[2*ArcTan[x/Sqrt[2]], 1/8])/(14*Sqrt[2]*Sqrt[4 + 3*x^2 + x^4]) - (3*
(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticF[2*ArcTan[x/Sqrt[2]], 1/8])/(4*Sqrt[2]*Sqrt[4 + 3*x^2 +
x^4])

Rule 1178

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(a*b*e - d*(b^2 - 2*
a*c) - c*(b*d - 2*a*e)*x^2)*(a + b*x^2 + c*x^4)^(p + 1))/(2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)
*(b^2 - 4*a*c)), Int[Simp[(2*p + 3)*d*b^2 - a*b*e - 2*a*c*d*(4*p + 5) + (4*p + 7)*(d*b - 2*a*e)*c*x^2, x]*(a +
b*x^2 + c*x^4)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e
^2, 0] && LtQ[p, -1] && IntegerQ[2*p]

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}{\left (4+3 x^2+x^4\right )^{3/2}} \, dx &=\frac{x \left (53+19 x^2\right )}{28 \sqrt{4+3 x^2+x^4}}+\frac{1}{28} \int \frac{-4-19 x^2}{\sqrt{4+3 x^2+x^4}} \, dx\\ &=\frac{x \left (53+19 x^2\right )}{28 \sqrt{4+3 x^2+x^4}}+\frac{19}{14} \int \frac{1-\frac{x^2}{2}}{\sqrt{4+3 x^2+x^4}} \, dx-\frac{3}{2} \int \frac{1}{\sqrt{4+3 x^2+x^4}} \, dx\\ &=\frac{x \left (53+19 x^2\right )}{28 \sqrt{4+3 x^2+x^4}}-\frac{19 x \sqrt{4+3 x^2+x^4}}{28 \left (2+x^2\right )}+\frac{19 \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 )}{14 \sqrt{2} \sqrt{4+3 x^2+x^4}}-\frac{3 \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 )}{4 \sqrt{2} \sqrt{4+3 x^2+x^4}}\\ \end{align*}

Mathematica [F]  time = 0, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

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Maple [C]  time = 0.004, size = 255, normalized size = 1.4 \begin{align*} -10\,{\frac{-1/7\,{x}^{3}-3/14\,x}{\sqrt{{x}^{4}+3\,{x}^{2}+4}}}-{\frac{4}{7\,\sqrt{-6+2\,i\sqrt{7}}}\sqrt{1- \left ( -{\frac{3}{8}}+{\frac{i}{8}}\sqrt{7} \right ){x}^{2}}\sqrt{1- \left ( -{\frac{3}{8}}-{\frac{i}{8}}\sqrt{7} \right ){x}^{2}}{\it EllipticF} \left ({\frac{x\sqrt{-6+2\,i\sqrt{7}}}{4}},{\frac{\sqrt{2+6\,i\sqrt{7}}}{4}} \right ){\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+4}}}}+{\frac{152}{7\,\sqrt{-6+2\,i\sqrt{7}} \left ( i\sqrt{7}+3 \right ) }\sqrt{1- \left ( -{\frac{3}{8}}+{\frac{i}{8}}\sqrt{7} \right ){x}^{2}}\sqrt{1- \left ( -{\frac{3}{8}}-{\frac{i}{8}}\sqrt{7} \right ){x}^{2}} \left ({\it EllipticF} \left ({\frac{x\sqrt{-6+2\,i\sqrt{7}}}{4}},{\frac{\sqrt{2+6\,i\sqrt{7}}}{4}} \right ) -{\it EllipticE} \left ({\frac{x\sqrt{-6+2\,i\sqrt{7}}}{4}},{\frac{\sqrt{2+6\,i\sqrt{7}}}{4}} \right ) \right ){\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+4}}}}-14\,{\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+4}} \left ({\frac{x}{56}}+{\frac{3\,{x}^{3}}{56}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-10*(-1/7*x^3-3/14*x)/(x^4+3*x^2+4)^(1/2)-4/7/(-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))+152/7/(-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)))-14*(1/56*x+3/56*x^3)/(x^4+3*x^2+4)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{5 \, x^{2} + 7}{{\left (x^{4} + 3 \, x^{2} + 4\right )}^{\frac{3}{2}}}\,{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)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{4} + 3 \, x^{2} + 4}{\left (5 \, x^{2} + 7\right )}}{x^{8} + 6 \, x^{6} + 17 \, x^{4} + 24 \, x^{2} + 16}, 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)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(x^4 + 3*x^2 + 4)*(5*x^2 + 7)/(x^8 + 6*x^6 + 17*x^4 + 24*x^2 + 16), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{5 x^{2} + 7}{\left (\left (x^{2} - x + 2\right ) \left (x^{2} + x + 2\right )\right )^{\frac{3}{2}}}\, 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)**(3/2),x)

[Out]

Integral((5*x**2 + 7)/((x**2 - x + 2)*(x**2 + x + 2))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{5 \, x^{2} + 7}{{\left (x^{4} + 3 \, x^{2} + 4\right )}^{\frac{3}{2}}}\,{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)^(3/2),x, algorithm="giac")

[Out]

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