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

Optimal. Leaf size=189 \[ \frac{15383 \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} \text{EllipticF}\left (\tan ^{-1}(x),\frac{1}{2}\right )}{3 \sqrt{2} \sqrt{x^4+3 x^2+2}}+625 \sqrt{x^4+3 x^2+2} x^3+\frac{5000}{3} \sqrt{x^4+3 x^2+2} x+\frac{7679 \left (x^2+2\right ) x}{2 \sqrt{x^4+3 x^2+2}}-\frac{\left (179 x^2+115\right ) x}{2 \sqrt{x^4+3 x^2+2}}-\frac{7679 \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{\sqrt{2} \sqrt{x^4+3 x^2+2}} \]

[Out]

(7679*x*(2 + x^2))/(2*Sqrt[2 + 3*x^2 + x^4]) - (x*(115 + 179*x^2))/(2*Sqrt[2 + 3*x^2 + x^4]) + (5000*x*Sqrt[2
+ 3*x^2 + x^4])/3 + 625*x^3*Sqrt[2 + 3*x^2 + x^4] - (7679*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan
[x], 1/2])/(Sqrt[2]*Sqrt[2 + 3*x^2 + x^4]) + (15383*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1
/2])/(3*Sqrt[2]*Sqrt[2 + 3*x^2 + x^4])

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Rubi [A]  time = 0.112963, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ 625 \sqrt{x^4+3 x^2+2} x^3+\frac{5000}{3} \sqrt{x^4+3 x^2+2} x+\frac{7679 \left (x^2+2\right ) x}{2 \sqrt{x^4+3 x^2+2}}-\frac{\left (179 x^2+115\right ) x}{2 \sqrt{x^4+3 x^2+2}}+\frac{15383 \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{3 \sqrt{2} \sqrt{x^4+3 x^2+2}}-\frac{7679 \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{\sqrt{2} \sqrt{x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7679*x*(2 + x^2))/(2*Sqrt[2 + 3*x^2 + x^4]) - (x*(115 + 179*x^2))/(2*Sqrt[2 + 3*x^2 + x^4]) + (5000*x*Sqrt[2
+ 3*x^2 + x^4])/3 + 625*x^3*Sqrt[2 + 3*x^2 + x^4] - (7679*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan
[x], 1/2])/(Sqrt[2]*Sqrt[2 + 3*x^2 + x^4]) + (15383*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1
/2])/(3*Sqrt[2]*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 )^5}{\left (2+3 x^2+x^4\right )^{3/2}} \, dx &=-\frac{x \left (115+179 x^2\right )}{2 \sqrt{2+3 x^2+x^4}}-\frac{1}{2} \int \frac{-16922-35179 x^2-25000 x^4-6250 x^6}{\sqrt{2+3 x^2+x^4}} \, dx\\ &=-\frac{x \left (115+179 x^2\right )}{2 \sqrt{2+3 x^2+x^4}}+625 x^3 \sqrt{2+3 x^2+x^4}-\frac{1}{10} \int \frac{-84610-138395 x^2-50000 x^4}{\sqrt{2+3 x^2+x^4}} \, dx\\ &=-\frac{x \left (115+179 x^2\right )}{2 \sqrt{2+3 x^2+x^4}}+\frac{5000}{3} x \sqrt{2+3 x^2+x^4}+625 x^3 \sqrt{2+3 x^2+x^4}-\frac{1}{30} \int \frac{-153830-115185 x^2}{\sqrt{2+3 x^2+x^4}} \, dx\\ &=-\frac{x \left (115+179 x^2\right )}{2 \sqrt{2+3 x^2+x^4}}+\frac{5000}{3} x \sqrt{2+3 x^2+x^4}+625 x^3 \sqrt{2+3 x^2+x^4}+\frac{7679}{2} \int \frac{x^2}{\sqrt{2+3 x^2+x^4}} \, dx+\frac{15383}{3} \int \frac{1}{\sqrt{2+3 x^2+x^4}} \, dx\\ &=\frac{7679 x \left (2+x^2\right )}{2 \sqrt{2+3 x^2+x^4}}-\frac{x \left (115+179 x^2\right )}{2 \sqrt{2+3 x^2+x^4}}+\frac{5000}{3} x \sqrt{2+3 x^2+x^4}+625 x^3 \sqrt{2+3 x^2+x^4}-\frac{7679 \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{15383 \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} \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)^5/(2 + 3*x^2 + x^4)^(3/2),x]

[Out]

$Aborted

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Maple [C]  time = 0.031, size = 274, normalized size = 1.5 \begin{align*} -6250\,{\frac{17/2\,{x}^{3}+9\,x}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}+625\,{x}^{3}\sqrt{{x}^{4}+3\,{x}^{2}+2}+{\frac{5000\,x}{3}\sqrt{{x}^{4}+3\,{x}^{2}+2}}+{{\frac{7679\,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}}}}-{{\frac{15383\,i}{6}}\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}}}}-43750\,{\frac{-9/2\,{x}^{3}-5\,x}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}-122500\,{\frac{5/2\,{x}^{3}+3\,x}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}-171500\,{\frac{-3/2\,{x}^{3}-2\,x}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}-120050\,{\frac{{x}^{3}+3/2\,x}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}-33614\,{\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)^5/(x^4+3*x^2+2)^(3/2),x)

[Out]

-6250*(17/2*x^3+9*x)/(x^4+3*x^2+2)^(1/2)+625*x^3*(x^4+3*x^2+2)^(1/2)+5000/3*x*(x^4+3*x^2+2)^(1/2)+7679/4*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)))-15383/6*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))-43750*(-9/2*x^3-5*x)/(x^4+3*x^2+2)^(1/2)-122500*(5/2*x^3+3*x)/(x^4+3*x^2+2)^(1/2)-171500*(-3/2*x^3-2
*x)/(x^4+3*x^2+2)^(1/2)-120050*(x^3+3/2*x)/(x^4+3*x^2+2)^(1/2)-33614*(-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 )}^{5}}{{\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)^5/(x^4+3*x^2+2)^(3/2),x, algorithm="maxima")

[Out]

integrate((5*x^2 + 7)^5/(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 (3125 \, x^{10} + 21875 \, x^{8} + 61250 \, x^{6} + 85750 \, x^{4} + 60025 \, x^{2} + 16807\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)^5/(x^4+3*x^2+2)^(3/2),x, algorithm="fricas")

[Out]

integral((3125*x^10 + 21875*x^8 + 61250*x^6 + 85750*x^4 + 60025*x^2 + 16807)*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 )^{5}}{\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)**5/(x**4+3*x**2+2)**(3/2),x)

[Out]

Integral((5*x**2 + 7)**5/((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 )}^{5}}{{\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)^5/(x^4+3*x^2+2)^(3/2),x, algorithm="giac")

[Out]

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