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

Optimal. Leaf size=170 \[ \frac{1067 \sqrt{2} \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} \text{EllipticF}\left (\tan ^{-1}(x),\frac{1}{2}\right )}{3 \sqrt{x^4+3 x^2+2}}+\frac{625}{3} \sqrt{x^4+3 x^2+2} x+\frac{637 \left (x^2+2\right ) x}{2 \sqrt{x^4+3 x^2+2}}+\frac{\left (113 x^2+145\right ) x}{2 \sqrt{x^4+3 x^2+2}}-\frac{637 \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{\sqrt{2} \sqrt{x^4+3 x^2+2}} \]

[Out]

(637*x*(2 + x^2))/(2*Sqrt[2 + 3*x^2 + x^4]) + (x*(145 + 113*x^2))/(2*Sqrt[2 + 3*x^2 + x^4]) + (625*x*Sqrt[2 +
3*x^2 + x^4])/3 - (637*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/(Sqrt[2]*Sqrt[2 + 3*x^2
+ x^4]) + (1067*Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(3*Sqrt[2 + 3*x^2 + x^4
])

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Rubi [A]  time = 0.0830864, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {1205, 1679, 1189, 1099, 1135} \[ \frac{625}{3} \sqrt{x^4+3 x^2+2} x+\frac{637 \left (x^2+2\right ) x}{2 \sqrt{x^4+3 x^2+2}}+\frac{\left (113 x^2+145\right ) x}{2 \sqrt{x^4+3 x^2+2}}+\frac{1067 \sqrt{2} \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{3 \sqrt{x^4+3 x^2+2}}-\frac{637 \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{\sqrt{2} \sqrt{x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(637*x*(2 + x^2))/(2*Sqrt[2 + 3*x^2 + x^4]) + (x*(145 + 113*x^2))/(2*Sqrt[2 + 3*x^2 + x^4]) + (625*x*Sqrt[2 +
3*x^2 + x^4])/3 - (637*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/(Sqrt[2]*Sqrt[2 + 3*x^2
+ x^4]) + (1067*Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(3*Sqrt[2 + 3*x^2 + x^4
])

Rule 1205

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{f = Coeff[Polynom
ialRemainder[(d + e*x^2)^q, a + b*x^2 + c*x^4, x], x, 0], g = Coeff[PolynomialRemainder[(d + e*x^2)^q, a + b*x
^2 + c*x^4, x], x, 2]}, Simp[(x*(a + b*x^2 + c*x^4)^(p + 1)*(a*b*g - f*(b^2 - 2*a*c) - c*(b*f - 2*a*g)*x^2))/(
2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x^2 + c*x^4)^(p + 1)*ExpandToS
um[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuotient[(d + e*x^2)^q, a + b*x^2 + c*x^4, x] + b^2*f*(2*p + 3) - 2*a*c
*f*(4*p + 5) - a*b*g + c*(4*p + 7)*(b*f - 2*a*g)*x^2, x], 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[q, 1] && LtQ[p, -1]

Rule 1679

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{q = Expon[Pq, x^2], e = Coeff[Pq, x^2,
Expon[Pq, x^2]]}, Simp[(e*x^(2*q - 3)*(a + b*x^2 + c*x^4)^(p + 1))/(c*(2*q + 4*p + 1)), x] + Dist[1/(c*(2*q +
4*p + 1)), Int[(a + b*x^2 + c*x^4)^p*ExpandToSum[c*(2*q + 4*p + 1)*Pq - a*e*(2*q - 3)*x^(2*q - 4) - b*e*(2*q
+ 2*p - 1)*x^(2*q - 2) - c*e*(2*q + 4*p + 1)*x^(2*q), x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2]
&& Expon[Pq, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] &&  !LtQ[p, -1]

Rule 1189

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

Rule 1099

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

Rule 1135

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(x*(b +
q + 2*c*x^2))/(2*c*Sqrt[a + b*x^2 + c*x^4]), x] - Simp[(Rt[(b + q)/(2*a), 2]*(2*a + (b + q)*x^2)*Sqrt[(2*a + (
b - q)*x^2)/(2*a + (b + q)*x^2)]*EllipticE[ArcTan[Rt[(b + q)/(2*a), 2]*x], (2*q)/(b + q)])/(2*c*Sqrt[a + b*x^2
+ c*x^4]), x] /; PosQ[(b + q)/a] &&  !(PosQ[(b - q)/a] && SimplerSqrtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; Fre
eQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rubi steps

\begin{align*} \int \frac{\left (7+5 x^2\right )^4}{\left (2+3 x^2+x^4\right )^{3/2}} \, dx &=\frac{x \left (145+113 x^2\right )}{2 \sqrt{2+3 x^2+x^4}}-\frac{1}{2} \int \frac{-2256-3137 x^2-1250 x^4}{\sqrt{2+3 x^2+x^4}} \, dx\\ &=\frac{x \left (145+113 x^2\right )}{2 \sqrt{2+3 x^2+x^4}}+\frac{625}{3} x \sqrt{2+3 x^2+x^4}-\frac{1}{6} \int \frac{-4268-1911 x^2}{\sqrt{2+3 x^2+x^4}} \, dx\\ &=\frac{x \left (145+113 x^2\right )}{2 \sqrt{2+3 x^2+x^4}}+\frac{625}{3} x \sqrt{2+3 x^2+x^4}+\frac{637}{2} \int \frac{x^2}{\sqrt{2+3 x^2+x^4}} \, dx+\frac{2134}{3} \int \frac{1}{\sqrt{2+3 x^2+x^4}} \, dx\\ &=\frac{637 x \left (2+x^2\right )}{2 \sqrt{2+3 x^2+x^4}}+\frac{x \left (145+113 x^2\right )}{2 \sqrt{2+3 x^2+x^4}}+\frac{625}{3} x \sqrt{2+3 x^2+x^4}-\frac{637 \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{\sqrt{2} \sqrt{2+3 x^2+x^4}}+\frac{1067 \sqrt{2} \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{3 \sqrt{2+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/(2 + 3*x^2 + x^4)^(3/2),x]

[Out]

$Aborted

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Maple [C]  time = 0.007, size = 234, normalized size = 1.4 \begin{align*} -1250\,{\frac{-9/2\,{x}^{3}-5\,x}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}+{\frac{625\,x}{3}\sqrt{{x}^{4}+3\,{x}^{2}+2}}-{{\frac{1067\,i}{3}}\sqrt{2}{\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) \sqrt{2\,{x}^{2}+4}\sqrt{{x}^{2}+1}{\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}}+{{\frac{637\,i}{4}}\sqrt{2} \left ({\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) -{\it EllipticE} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) \right ) \sqrt{2\,{x}^{2}+4}\sqrt{{x}^{2}+1}{\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}}-7000\,{\frac{5/2\,{x}^{3}+3\,x}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}-14700\,{\frac{-3/2\,{x}^{3}-2\,x}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}-13720\,{\frac{{x}^{3}+3/2\,x}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}-4802\,{\frac{-3/4\,{x}^{3}-5/4\,x}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1250*(-9/2*x^3-5*x)/(x^4+3*x^2+2)^(1/2)+625/3*x*(x^4+3*x^2+2)^(1/2)-1067/3*I*2^(1/2)*(2*x^2+4)^(1/2)*(x^2+1)^
(1/2)/(x^4+3*x^2+2)^(1/2)*EllipticF(1/2*I*x*2^(1/2),2^(1/2))+637/4*I*2^(1/2)*(2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(x^
4+3*x^2+2)^(1/2)*(EllipticF(1/2*I*x*2^(1/2),2^(1/2))-EllipticE(1/2*I*x*2^(1/2),2^(1/2)))-7000*(5/2*x^3+3*x)/(x
^4+3*x^2+2)^(1/2)-14700*(-3/2*x^3-2*x)/(x^4+3*x^2+2)^(1/2)-13720*(x^3+3/2*x)/(x^4+3*x^2+2)^(1/2)-4802*(-3/4*x^
3-5/4*x)/(x^4+3*x^2+2)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (625 \, x^{8} + 3500 \, x^{6} + 7350 \, x^{4} + 6860 \, x^{2} + 2401\right )} \sqrt{x^{4} + 3 \, x^{2} + 2}}{x^{8} + 6 \, x^{6} + 13 \, x^{4} + 12 \, x^{2} + 4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((625*x^8 + 3500*x^6 + 7350*x^4 + 6860*x^2 + 2401)*sqrt(x^4 + 3*x^2 + 2)/(x^8 + 6*x^6 + 13*x^4 + 12*x^
2 + 4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (5 x^{2} + 7\right )^{4}}{\left (\left (x^{2} + 1\right ) \left (x^{2} + 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)**4/(x**4+3*x**2+2)**(3/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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