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

Optimal. Leaf size=226 \[ \frac{4628 \sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right ),\frac{1}{8}\right )}{33 \sqrt{x^4+3 x^2+4}}+\frac{25}{11} x \left (x^4+3 x^2+4\right )^{5/2}+\frac{1}{693} x \left (2240 x^2+6831\right ) \left (x^4+3 x^2+4\right )^{3/2}+\frac{x \left (18253 x^2+64533\right ) \sqrt{x^4+3 x^2+4}}{1155}+\frac{175346 x \sqrt{x^4+3 x^2+4}}{1155 \left (x^2+2\right )}-\frac{175346 \sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{1155 \sqrt{x^4+3 x^2+4}} \]

[Out]

(175346*x*Sqrt[4 + 3*x^2 + x^4])/(1155*(2 + x^2)) + (x*(64533 + 18253*x^2)*Sqrt[4 + 3*x^2 + x^4])/1155 + (x*(6
831 + 2240*x^2)*(4 + 3*x^2 + x^4)^(3/2))/693 + (25*x*(4 + 3*x^2 + x^4)^(5/2))/11 - (175346*Sqrt[2]*(2 + x^2)*S
qrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticE[2*ArcTan[x/Sqrt[2]], 1/8])/(1155*Sqrt[4 + 3*x^2 + x^4]) + (4628*S
qrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticF[2*ArcTan[x/Sqrt[2]], 1/8])/(33*Sqrt[4 + 3*x^2 +
x^4])

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Rubi [A]  time = 0.0979694, antiderivative size = 226, 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 = {1206, 1176, 1197, 1103, 1195} \[ \frac{25}{11} x \left (x^4+3 x^2+4\right )^{5/2}+\frac{1}{693} x \left (2240 x^2+6831\right ) \left (x^4+3 x^2+4\right )^{3/2}+\frac{x \left (18253 x^2+64533\right ) \sqrt{x^4+3 x^2+4}}{1155}+\frac{175346 x \sqrt{x^4+3 x^2+4}}{1155 \left (x^2+2\right )}+\frac{4628 \sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{33 \sqrt{x^4+3 x^2+4}}-\frac{175346 \sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{1155 \sqrt{x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(175346*x*Sqrt[4 + 3*x^2 + x^4])/(1155*(2 + x^2)) + (x*(64533 + 18253*x^2)*Sqrt[4 + 3*x^2 + x^4])/1155 + (x*(6
831 + 2240*x^2)*(4 + 3*x^2 + x^4)^(3/2))/693 + (25*x*(4 + 3*x^2 + x^4)^(5/2))/11 - (175346*Sqrt[2]*(2 + x^2)*S
qrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticE[2*ArcTan[x/Sqrt[2]], 1/8])/(1155*Sqrt[4 + 3*x^2 + x^4]) + (4628*S
qrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticF[2*ArcTan[x/Sqrt[2]], 1/8])/(33*Sqrt[4 + 3*x^2 +
x^4])

Rule 1206

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

Rule 1176

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(2*b*e*p + c*d*(4*p
+ 3) + c*e*(4*p + 1)*x^2)*(a + b*x^2 + c*x^4)^p)/(c*(4*p + 1)*(4*p + 3)), x] + Dist[(2*p)/(c*(4*p + 1)*(4*p +
3)), Int[Simp[2*a*c*d*(4*p + 3) - a*b*e + (2*a*c*e*(4*p + 1) + b*c*d*(4*p + 3) - b^2*e*(2*p + 1))*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] && GtQ[p, 0] && FractionQ[p] && 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 \left (7+5 x^2\right )^2 \left (4+3 x^2+x^4\right )^{3/2} \, dx &=\frac{25}{11} x \left (4+3 x^2+x^4\right )^{5/2}+\frac{1}{11} \int \left (439+320 x^2\right ) \left (4+3 x^2+x^4\right )^{3/2} \, dx\\ &=\frac{1}{693} x \left (6831+2240 x^2\right ) \left (4+3 x^2+x^4\right )^{3/2}+\frac{25}{11} x \left (4+3 x^2+x^4\right )^{5/2}+\frac{1}{231} \int \left (27768+18253 x^2\right ) \sqrt{4+3 x^2+x^4} \, dx\\ &=\frac{x \left (64533+18253 x^2\right ) \sqrt{4+3 x^2+x^4}}{1155}+\frac{1}{693} x \left (6831+2240 x^2\right ) \left (4+3 x^2+x^4\right )^{3/2}+\frac{25}{11} x \left (4+3 x^2+x^4\right )^{5/2}+\frac{\int \frac{891684+526038 x^2}{\sqrt{4+3 x^2+x^4}} \, dx}{3465}\\ &=\frac{x \left (64533+18253 x^2\right ) \sqrt{4+3 x^2+x^4}}{1155}+\frac{1}{693} x \left (6831+2240 x^2\right ) \left (4+3 x^2+x^4\right )^{3/2}+\frac{25}{11} x \left (4+3 x^2+x^4\right )^{5/2}-\frac{350692 \int \frac{1-\frac{x^2}{2}}{\sqrt{4+3 x^2+x^4}} \, dx}{1155}+\frac{18512}{33} \int \frac{1}{\sqrt{4+3 x^2+x^4}} \, dx\\ &=\frac{175346 x \sqrt{4+3 x^2+x^4}}{1155 \left (2+x^2\right )}+\frac{x \left (64533+18253 x^2\right ) \sqrt{4+3 x^2+x^4}}{1155}+\frac{1}{693} x \left (6831+2240 x^2\right ) \left (4+3 x^2+x^4\right )^{3/2}+\frac{25}{11} x \left (4+3 x^2+x^4\right )^{5/2}-\frac{175346 \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 )}{1155 \sqrt{4+3 x^2+x^4}}+\frac{4628 \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 )}{33 \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)^2*(4 + 3*x^2 + x^4)^(3/2),x]

[Out]

$Aborted

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Maple [C]  time = 0.007, size = 292, normalized size = 1.3 \begin{align*}{\frac{25\,{x}^{9}}{11}\sqrt{{x}^{4}+3\,{x}^{2}+4}}+{\frac{1670\,{x}^{7}}{99}\sqrt{{x}^{4}+3\,{x}^{2}+4}}+{\frac{1222\,{x}^{5}}{21}\sqrt{{x}^{4}+3\,{x}^{2}+4}}+{\frac{391024\,{x}^{3}}{3465}\sqrt{{x}^{4}+3\,{x}^{2}+4}}+{\frac{50691\,x}{385}\sqrt{{x}^{4}+3\,{x}^{2}+4}}+{\frac{396304}{385\,\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{5611072}{1155\,\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}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/11*x^9*(x^4+3*x^2+4)^(1/2)+1670/99*x^7*(x^4+3*x^2+4)^(1/2)+1222/21*x^5*(x^4+3*x^2+4)^(1/2)+391024/3465*x^3*
(x^4+3*x^2+4)^(1/2)+50691/385*x*(x^4+3*x^2+4)^(1/2)+396304/385/(-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))-5611072/1155/(-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)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (25 \, x^{8} + 145 \, x^{6} + 359 \, x^{4} + 427 \, x^{2} + 196\right )} \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)^2*(x^4+3*x^2+4)^(3/2),x, algorithm="fricas")

[Out]

integral((25*x^8 + 145*x^6 + 359*x^4 + 427*x^2 + 196)*sqrt(x^4 + 3*x^2 + 4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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