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

Optimal. Leaf size=322 \[ -\frac{11 \sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right ),\frac{1}{8}\right )}{75 \sqrt{x^4+3 x^2+4}}+\frac{9 \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right ),\frac{1}{8}\right )}{25 \sqrt{2} \sqrt{x^4+3 x^2+4}}+\frac{\sqrt{x^4+3 x^2+4} x}{5 \left (x^2+2\right )}+\frac{1}{5} \sqrt{\frac{11}{35}} \tan ^{-1}\left (\frac{2 \sqrt{\frac{11}{35}} x}{\sqrt{x^4+3 x^2+4}}\right )-\frac{\sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{5 \sqrt{x^4+3 x^2+4}}+\frac{187 \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} \Pi \left (-\frac{9}{280};2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{525 \sqrt{2} \sqrt{x^4+3 x^2+4}} \]

[Out]

(x*Sqrt[4 + 3*x^2 + x^4])/(5*(2 + x^2)) + (Sqrt[11/35]*ArcTan[(2*Sqrt[11/35]*x)/Sqrt[4 + 3*x^2 + x^4]])/5 - (S
qrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticE[2*ArcTan[x/Sqrt[2]], 1/8])/(5*Sqrt[4 + 3*x^2 +
x^4]) + (9*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticF[2*ArcTan[x/Sqrt[2]], 1/8])/(25*Sqrt[2]*Sqrt
[4 + 3*x^2 + x^4]) - (11*Sqrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticF[2*ArcTan[x/Sqrt[2]],
1/8])/(75*Sqrt[4 + 3*x^2 + x^4]) + (187*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticPi[-9/280, 2*Arc
Tan[x/Sqrt[2]], 1/8])/(525*Sqrt[2]*Sqrt[4 + 3*x^2 + x^4])

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Rubi [A]  time = 0.151624, antiderivative size = 322, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1208, 1197, 1103, 1195, 1216, 1706} \[ \frac{\sqrt{x^4+3 x^2+4} x}{5 \left (x^2+2\right )}+\frac{1}{5} \sqrt{\frac{11}{35}} \tan ^{-1}\left (\frac{2 \sqrt{\frac{11}{35}} x}{\sqrt{x^4+3 x^2+4}}\right )-\frac{11 \sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{75 \sqrt{x^4+3 x^2+4}}+\frac{9 \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{25 \sqrt{2} \sqrt{x^4+3 x^2+4}}-\frac{\sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{5 \sqrt{x^4+3 x^2+4}}+\frac{187 \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} \Pi \left (-\frac{9}{280};2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{525 \sqrt{2} \sqrt{x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[4 + 3*x^2 + x^4])/(5*(2 + x^2)) + (Sqrt[11/35]*ArcTan[(2*Sqrt[11/35]*x)/Sqrt[4 + 3*x^2 + x^4]])/5 - (S
qrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticE[2*ArcTan[x/Sqrt[2]], 1/8])/(5*Sqrt[4 + 3*x^2 +
x^4]) + (9*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticF[2*ArcTan[x/Sqrt[2]], 1/8])/(25*Sqrt[2]*Sqrt
[4 + 3*x^2 + x^4]) - (11*Sqrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticF[2*ArcTan[x/Sqrt[2]],
1/8])/(75*Sqrt[4 + 3*x^2 + x^4]) + (187*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticPi[-9/280, 2*Arc
Tan[x/Sqrt[2]], 1/8])/(525*Sqrt[2]*Sqrt[4 + 3*x^2 + x^4])

Rule 1208

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_)/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Dist[(e^2)^(-1), Int[(c*d -
 b*e - c*e*x^2)*(a + b*x^2 + c*x^4)^(p - 1), x], x] + Dist[(c*d^2 - b*d*e + a*e^2)/e^2, Int[(a + b*x^2 + c*x^4
)^(p - 1)/(d + e*x^2), 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] && IGtQ[p + 1/2, 0]

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]

Rule 1216

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

Rule 1706

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

Rubi steps

\begin{align*} \int \frac{\sqrt{4+3 x^2+x^4}}{7+5 x^2} \, dx &=-\left (\frac{1}{25} \int \frac{-8-5 x^2}{\sqrt{4+3 x^2+x^4}} \, dx\right )+\frac{44}{25} \int \frac{1}{\left (7+5 x^2\right ) \sqrt{4+3 x^2+x^4}} \, dx\\ &=-\left (\frac{2}{5} \int \frac{1-\frac{x^2}{2}}{\sqrt{4+3 x^2+x^4}} \, dx\right )-\frac{44}{75} \int \frac{1}{\sqrt{4+3 x^2+x^4}} \, dx+\frac{18}{25} \int \frac{1}{\sqrt{4+3 x^2+x^4}} \, dx+\frac{88}{15} \int \frac{1+\frac{x^2}{2}}{\left (7+5 x^2\right ) \sqrt{4+3 x^2+x^4}} \, dx\\ &=\frac{x \sqrt{4+3 x^2+x^4}}{5 \left (2+x^2\right )}+\frac{1}{5} \sqrt{\frac{11}{35}} \tan ^{-1}\left (\frac{2 \sqrt{\frac{11}{35}} x}{\sqrt{4+3 x^2+x^4}}\right )-\frac{\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 )}{5 \sqrt{4+3 x^2+x^4}}+\frac{9 \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 )}{25 \sqrt{2} \sqrt{4+3 x^2+x^4}}-\frac{11 \sqrt{2} \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 )}{75 \sqrt{4+3 x^2+x^4}}+\frac{187 \left (2+x^2\right ) \sqrt{\frac{4+3 x^2+x^4}{\left (2+x^2\right )^2}} \Pi \left (-\frac{9}{280};2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{525 \sqrt{2} \sqrt{4+3 x^2+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.250902, size = 283, normalized size = 0.88 \[ -\frac{\sqrt{1-\frac{2 i x^2}{\sqrt{7}-3 i}} \sqrt{1+\frac{2 i x^2}{\sqrt{7}+3 i}} \left (\left (-35 \sqrt{7}+7 i\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{-\frac{2 i}{\sqrt{7}-3 i}} x\right ),\frac{-\sqrt{7}+3 i}{\sqrt{7}+3 i}\right )+35 \left (\sqrt{7}+3 i\right ) E\left (i \sinh ^{-1}\left (\sqrt{-\frac{2 i}{-3 i+\sqrt{7}}} x\right )|\frac{3 i-\sqrt{7}}{3 i+\sqrt{7}}\right )+88 i \Pi \left (\frac{5}{14} \left (3+i \sqrt{7}\right );i \sinh ^{-1}\left (\sqrt{-\frac{2 i}{-3 i+\sqrt{7}}} x\right )|\frac{3 i-\sqrt{7}}{3 i+\sqrt{7}}\right )\right )}{350 \sqrt{2} \sqrt{-\frac{i}{\sqrt{7}-3 i}} \sqrt{x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 - ((2*I)*x^2)/(-3*I + Sqrt[7])]*Sqrt[1 + ((2*I)*x^2)/(3*I + Sqrt[7])]*(35*(3*I + Sqrt[7])*EllipticE[I
*ArcSinh[Sqrt[(-2*I)/(-3*I + Sqrt[7])]*x], (3*I - Sqrt[7])/(3*I + Sqrt[7])] + (7*I - 35*Sqrt[7])*EllipticF[I*A
rcSinh[Sqrt[(-2*I)/(-3*I + Sqrt[7])]*x], (3*I - Sqrt[7])/(3*I + Sqrt[7])] + (88*I)*EllipticPi[(5*(3 + I*Sqrt[7
]))/14, I*ArcSinh[Sqrt[(-2*I)/(-3*I + Sqrt[7])]*x], (3*I - Sqrt[7])/(3*I + Sqrt[7])]))/(350*Sqrt[2]*Sqrt[(-I)/
(-3*I + Sqrt[7])]*Sqrt[4 + 3*x^2 + x^4])

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Maple [C]  time = 0.051, size = 386, normalized size = 1.2 \begin{align*}{\frac{32}{25\,\sqrt{-6+2\,i\sqrt{7}}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}-{\frac{i}{8}}{x}^{2}\sqrt{7}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}+{\frac{i}{8}}{x}^{2}\sqrt{7}}{\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{32}{5\,\sqrt{-6+2\,i\sqrt{7}} \left ( i\sqrt{7}+3 \right ) }\sqrt{1+{\frac{3\,{x}^{2}}{8}}-{\frac{i}{8}}{x}^{2}\sqrt{7}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}+{\frac{i}{8}}{x}^{2}\sqrt{7}}{\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{32}{5\,\sqrt{-6+2\,i\sqrt{7}} \left ( i\sqrt{7}+3 \right ) }\sqrt{1+{\frac{3\,{x}^{2}}{8}}-{\frac{i}{8}}{x}^{2}\sqrt{7}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}+{\frac{i}{8}}{x}^{2}\sqrt{7}}{\it EllipticE} \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{44}{175\,\sqrt{-3/8+i/8\sqrt{7}}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}-{\frac{i}{8}}{x}^{2}\sqrt{7}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}+{\frac{i}{8}}{x}^{2}\sqrt{7}}{\it EllipticPi} \left ( \sqrt{-{\frac{3}{8}}+{\frac{i}{8}}\sqrt{7}}x,-{\frac{5}{-{\frac{21}{8}}+{\frac{7\,i}{8}}\sqrt{7}}},{\frac{\sqrt{-{\frac{3}{8}}-{\frac{i}{8}}\sqrt{7}}}{\sqrt{-{\frac{3}{8}}+{\frac{i}{8}}\sqrt{7}}}} \right ){\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+4}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/25/(-6+2*I*7^(1/2))^(1/2)*(1+3/8*x^2-1/8*I*x^2*7^(1/2))^(1/2)*(1+3/8*x^2+1/8*I*x^2*7^(1/2))^(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))-32/5/(-6+2*I*7^(1/2))^(1/2)*(1+3/
8*x^2-1/8*I*x^2*7^(1/2))^(1/2)*(1+3/8*x^2+1/8*I*x^2*7^(1/2))^(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))+32/5/(-6+2*I*7^(1/2))^(1/2)*(1+3/8*x^2-1/8*I*x^2*7^(1
/2))^(1/2)*(1+3/8*x^2+1/8*I*x^2*7^(1/2))^(1/2)/(x^4+3*x^2+4)^(1/2)/(I*7^(1/2)+3)*EllipticE(1/4*x*(-6+2*I*7^(1/
2))^(1/2),1/4*(2+6*I*7^(1/2))^(1/2))+44/175/(-3/8+1/8*I*7^(1/2))^(1/2)*(1+3/8*x^2-1/8*I*x^2*7^(1/2))^(1/2)*(1+
3/8*x^2+1/8*I*x^2*7^(1/2))^(1/2)/(x^4+3*x^2+4)^(1/2)*EllipticPi((-3/8+1/8*I*7^(1/2))^(1/2)*x,-5/7/(-3/8+1/8*I*
7^(1/2)),(-3/8-1/8*I*7^(1/2))^(1/2)/(-3/8+1/8*I*7^(1/2))^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x^4 + 3*x^2 + 4)/(5*x^2 + 7), 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}}{5 \, x^{2} + 7}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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