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

Optimal. Leaf size=197 \[ \frac{10 \sqrt [4]{5} \left (7+2 \sqrt{5}\right ) \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right ),\frac{1}{2}\right )}{7 \sqrt{x^4+5}}+\frac{1}{21} x \left (7 x^2+6\right ) \left (x^4+5\right )^{3/2}+\frac{2}{7} x \left (7 x^2+10\right ) \sqrt{x^4+5}+\frac{20 x \sqrt{x^4+5}}{x^2+\sqrt{5}}-\frac{20 \sqrt [4]{5} \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{\sqrt{x^4+5}} \]

[Out]

(20*x*Sqrt[5 + x^4])/(Sqrt[5] + x^2) + (2*x*(10 + 7*x^2)*Sqrt[5 + x^4])/7 + (x*(6 + 7*x^2)*(5 + x^4)^(3/2))/21
- (20*5^(1/4)*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticE[2*ArcTan[x/5^(1/4)], 1/2])/Sqrt[5 +
x^4] + (10*5^(1/4)*(7 + 2*Sqrt[5])*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticF[2*ArcTan[x/5^(
1/4)], 1/2])/(7*Sqrt[5 + x^4])

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Rubi [A]  time = 0.0817294, antiderivative size = 197, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {1177, 1198, 220, 1196} \[ \frac{1}{21} x \left (7 x^2+6\right ) \left (x^4+5\right )^{3/2}+\frac{2}{7} x \left (7 x^2+10\right ) \sqrt{x^4+5}+\frac{20 x \sqrt{x^4+5}}{x^2+\sqrt{5}}+\frac{10 \sqrt [4]{5} \left (7+2 \sqrt{5}\right ) \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{7 \sqrt{x^4+5}}-\frac{20 \sqrt [4]{5} \left (x^2+\sqrt{5}\right ) \sqrt{\frac{x^4+5}{\left (x^2+\sqrt{5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{\sqrt{x^4+5}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x^2)*(5 + x^4)^(3/2),x]

[Out]

(20*x*Sqrt[5 + x^4])/(Sqrt[5] + x^2) + (2*x*(10 + 7*x^2)*Sqrt[5 + x^4])/7 + (x*(6 + 7*x^2)*(5 + x^4)^(3/2))/21
- (20*5^(1/4)*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticE[2*ArcTan[x/5^(1/4)], 1/2])/Sqrt[5 +
x^4] + (10*5^(1/4)*(7 + 2*Sqrt[5])*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticF[2*ArcTan[x/5^(
1/4)], 1/2])/(7*Sqrt[5 + x^4])

Rule 1177

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

Rule 1198

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

Rule 220

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

Rule 1196

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

Rubi steps

\begin{align*} \int \left (2+3 x^2\right ) \left (5+x^4\right )^{3/2} \, dx &=\frac{1}{21} x \left (6+7 x^2\right ) \left (5+x^4\right )^{3/2}+\frac{1}{21} \int \left (180+210 x^2\right ) \sqrt{5+x^4} \, dx\\ &=\frac{2}{7} x \left (10+7 x^2\right ) \sqrt{5+x^4}+\frac{1}{21} x \left (6+7 x^2\right ) \left (5+x^4\right )^{3/2}+\frac{1}{315} \int \frac{9000+6300 x^2}{\sqrt{5+x^4}} \, dx\\ &=\frac{2}{7} x \left (10+7 x^2\right ) \sqrt{5+x^4}+\frac{1}{21} x \left (6+7 x^2\right ) \left (5+x^4\right )^{3/2}-\left (20 \sqrt{5}\right ) \int \frac{1-\frac{x^2}{\sqrt{5}}}{\sqrt{5+x^4}} \, dx+\frac{1}{7} \left (20 \left (10+7 \sqrt{5}\right )\right ) \int \frac{1}{\sqrt{5+x^4}} \, dx\\ &=\frac{20 x \sqrt{5+x^4}}{\sqrt{5}+x^2}+\frac{2}{7} x \left (10+7 x^2\right ) \sqrt{5+x^4}+\frac{1}{21} x \left (6+7 x^2\right ) \left (5+x^4\right )^{3/2}-\frac{20 \sqrt [4]{5} \left (\sqrt{5}+x^2\right ) \sqrt{\frac{5+x^4}{\left (\sqrt{5}+x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{\sqrt{5+x^4}}+\frac{10 \sqrt [4]{5} \left (7+2 \sqrt{5}\right ) \left (\sqrt{5}+x^2\right ) \sqrt{\frac{5+x^4}{\left (\sqrt{5}+x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt [4]{5}}\right )|\frac{1}{2}\right )}{7 \sqrt{5+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0129397, size = 49, normalized size = 0.25 \[ 5 \sqrt{5} x \left (x^2 \, _2F_1\left (-\frac{3}{2},\frac{3}{4};\frac{7}{4};-\frac{x^4}{5}\right )+2 \, _2F_1\left (-\frac{3}{2},\frac{1}{4};\frac{5}{4};-\frac{x^4}{5}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

5*Sqrt[5]*x*(2*Hypergeometric2F1[-3/2, 1/4, 5/4, -x^4/5] + x^2*Hypergeometric2F1[-3/2, 3/4, 7/4, -x^4/5])

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Maple [C]  time = 0.013, size = 192, normalized size = 1. \begin{align*}{\frac{{x}^{7}}{3}\sqrt{{x}^{4}+5}}+{\frac{11\,{x}^{3}}{3}\sqrt{{x}^{4}+5}}+{\frac{4\,i}{\sqrt{i\sqrt{5}}}\sqrt{25-5\,i\sqrt{5}{x}^{2}}\sqrt{25+5\,i\sqrt{5}{x}^{2}} \left ({\it EllipticF} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ) -{\it EllipticE} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ) \right ){\frac{1}{\sqrt{{x}^{4}+5}}}}+{\frac{2\,{x}^{5}}{7}\sqrt{{x}^{4}+5}}+{\frac{30\,x}{7}\sqrt{{x}^{4}+5}}+{\frac{8\,\sqrt{5}}{7\,\sqrt{i\sqrt{5}}}\sqrt{25-5\,i\sqrt{5}{x}^{2}}\sqrt{25+5\,i\sqrt{5}{x}^{2}}{\it EllipticF} \left ({\frac{x\sqrt{5}\sqrt{i\sqrt{5}}}{5}},i \right ){\frac{1}{\sqrt{{x}^{4}+5}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^7*(x^4+5)^(1/2)+11/3*x^3*(x^4+5)^(1/2)+4*I/(I*5^(1/2))^(1/2)*(25-5*I*5^(1/2)*x^2)^(1/2)*(25+5*I*5^(1/2)*
x^2)^(1/2)/(x^4+5)^(1/2)*(EllipticF(1/5*x*5^(1/2)*(I*5^(1/2))^(1/2),I)-EllipticE(1/5*x*5^(1/2)*(I*5^(1/2))^(1/
2),I))+2/7*x^5*(x^4+5)^(1/2)+30/7*x*(x^4+5)^(1/2)+8/7*5^(1/2)/(I*5^(1/2))^(1/2)*(25-5*I*5^(1/2)*x^2)^(1/2)*(25
+5*I*5^(1/2)*x^2)^(1/2)/(x^4+5)^(1/2)*EllipticF(1/5*x*5^(1/2)*(I*5^(1/2))^(1/2),I)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (3 \, x^{6} + 2 \, x^{4} + 15 \, x^{2} + 10\right )} \sqrt{x^{4} + 5}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((3*x^6 + 2*x^4 + 15*x^2 + 10)*sqrt(x^4 + 5), x)

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Sympy [C]  time = 2.92852, size = 158, normalized size = 0.8 \begin{align*} \frac{3 \sqrt{5} x^{7} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{4 \Gamma \left (\frac{11}{4}\right )} + \frac{\sqrt{5} x^{5} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{2 \Gamma \left (\frac{9}{4}\right )} + \frac{15 \sqrt{5} x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{4 \Gamma \left (\frac{7}{4}\right )} + \frac{5 \sqrt{5} x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{x^{4} e^{i \pi }}{5}} \right )}}{2 \Gamma \left (\frac{5}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*sqrt(5)*x**7*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), x**4*exp_polar(I*pi)/5)/(4*gamma(11/4)) + sqrt(5)*x**5*g
amma(5/4)*hyper((-1/2, 5/4), (9/4,), x**4*exp_polar(I*pi)/5)/(2*gamma(9/4)) + 15*sqrt(5)*x**3*gamma(3/4)*hyper
((-1/2, 3/4), (7/4,), x**4*exp_polar(I*pi)/5)/(4*gamma(7/4)) + 5*sqrt(5)*x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,
), x**4*exp_polar(I*pi)/5)/(2*gamma(5/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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