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

Optimal. Leaf size=201 \[ -\frac {2 \left (27-2 x^2\right ) \sqrt {x^4+5}}{3 x}+\frac {36 x \sqrt {x^4+5}}{x^2+\sqrt {5}}+\frac {2 \sqrt [4]{5} \left (27+2 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{3 \sqrt {x^4+5}}-\frac {36 \sqrt [4]{5} \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{\sqrt {x^4+5}}-\frac {\left (10-9 x^2\right ) \left (x^4+5\right )^{3/2}}{15 x^3} \]

[Out]

-1/15*(-9*x^2+10)*(x^4+5)^(3/2)/x^3-2/3*(-2*x^2+27)*(x^4+5)^(1/2)/x+36*x*(x^4+5)^(1/2)/(x^2+5^(1/2))-36*5^(1/4
)*(cos(2*arctan(1/5*x*5^(3/4)))^2)^(1/2)/cos(2*arctan(1/5*x*5^(3/4)))*EllipticE(sin(2*arctan(1/5*x*5^(3/4))),1
/2*2^(1/2))*(x^2+5^(1/2))*((x^4+5)/(x^2+5^(1/2))^2)^(1/2)/(x^4+5)^(1/2)+2/3*5^(1/4)*(cos(2*arctan(1/5*x*5^(3/4
)))^2)^(1/2)/cos(2*arctan(1/5*x*5^(3/4)))*EllipticF(sin(2*arctan(1/5*x*5^(3/4))),1/2*2^(1/2))*(x^2+5^(1/2))*(2
7+2*5^(1/2))*((x^4+5)/(x^2+5^(1/2))^2)^(1/2)/(x^4+5)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1272, 1198, 220, 1196} \[ -\frac {\left (10-9 x^2\right ) \left (x^4+5\right )^{3/2}}{15 x^3}-\frac {2 \left (27-2 x^2\right ) \sqrt {x^4+5}}{3 x}+\frac {36 x \sqrt {x^4+5}}{x^2+\sqrt {5}}+\frac {2 \sqrt [4]{5} \left (27+2 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{3 \sqrt {x^4+5}}-\frac {36 \sqrt [4]{5} \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{\sqrt {x^4+5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(27 - 2*x^2)*Sqrt[5 + x^4])/(3*x) + (36*x*Sqrt[5 + x^4])/(Sqrt[5] + x^2) - ((10 - 9*x^2)*(5 + x^4)^(3/2))/
(15*x^3) - (36*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] + (2*5^(1/4)*(27 + 2*Sqrt[5])*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticF[2*ArcT
an[x/5^(1/4)], 1/2])/(3*Sqrt[5 + x^4])

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]

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 1272

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

Rubi steps

\begin {align*} \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^4} \, dx &=-\frac {\left (10-9 x^2\right ) \left (5+x^4\right )^{3/2}}{15 x^3}-\frac {2}{5} \int \frac {\left (-45-10 x^2\right ) \sqrt {5+x^4}}{x^2} \, dx\\ &=-\frac {2 \left (27-2 x^2\right ) \sqrt {5+x^4}}{3 x}-\frac {\left (10-9 x^2\right ) \left (5+x^4\right )^{3/2}}{15 x^3}+\frac {4}{15} \int \frac {50+135 x^2}{\sqrt {5+x^4}} \, dx\\ &=-\frac {2 \left (27-2 x^2\right ) \sqrt {5+x^4}}{3 x}-\frac {\left (10-9 x^2\right ) \left (5+x^4\right )^{3/2}}{15 x^3}-\left (36 \sqrt {5}\right ) \int \frac {1-\frac {x^2}{\sqrt {5}}}{\sqrt {5+x^4}} \, dx+\frac {1}{3} \left (4 \left (10+27 \sqrt {5}\right )\right ) \int \frac {1}{\sqrt {5+x^4}} \, dx\\ &=-\frac {2 \left (27-2 x^2\right ) \sqrt {5+x^4}}{3 x}+\frac {36 x \sqrt {5+x^4}}{\sqrt {5}+x^2}-\frac {\left (10-9 x^2\right ) \left (5+x^4\right )^{3/2}}{15 x^3}-\frac {36 \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 {2 \sqrt [4]{5} \left (27+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 )}{3 \sqrt {5+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 54, normalized size = 0.27 \[ -\frac {5 \sqrt {5} \left (2 \, _2F_1\left (-\frac {3}{2},-\frac {3}{4};\frac {1}{4};-\frac {x^4}{5}\right )+9 x^2 \, _2F_1\left (-\frac {3}{2},-\frac {1}{4};\frac {3}{4};-\frac {x^4}{5}\right )\right )}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*Sqrt[5]*(2*Hypergeometric2F1[-3/2, -3/4, 1/4, -1/5*x^4] + 9*x^2*Hypergeometric2F1[-3/2, -1/4, 3/4, -1/5*x^
4]))/(3*x^3)

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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (3 \, x^{6} + 2 \, x^{4} + 15 \, x^{2} + 10\right )} \sqrt {x^{4} + 5}}{x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (x^{4} + 5\right )}^{\frac {3}{2}} {\left (3 \, x^{2} + 2\right )}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.02, size = 192, normalized size = 0.96 \[ \frac {3 \sqrt {x^{4}+5}\, x^{3}}{5}+\frac {2 \sqrt {x^{4}+5}\, x}{3}+\frac {8 \sqrt {5}\, \sqrt {-5 i \sqrt {5}\, x^{2}+25}\, \sqrt {5 i \sqrt {5}\, x^{2}+25}\, \EllipticF \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )}{15 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}-\frac {15 \sqrt {x^{4}+5}}{x}-\frac {10 \sqrt {x^{4}+5}}{3 x^{3}}+\frac {36 i \sqrt {-5 i \sqrt {5}\, x^{2}+25}\, \sqrt {5 i \sqrt {5}\, x^{2}+25}\, \left (-\EllipticE \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )+\EllipticF \left (\frac {\sqrt {5}\, \sqrt {i \sqrt {5}}\, x}{5}, i\right )\right )}{5 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15*(x^4+5)^(1/2)/x+3/5*(x^4+5)^(1/2)*x^3+36/5*I/(I*5^(1/2))^(1/2)*(-5*I*5^(1/2)*x^2+25)^(1/2)*(5*I*5^(1/2)*x^
2+25)^(1/2)/(x^4+5)^(1/2)*(EllipticF(1/5*5^(1/2)*(I*5^(1/2))^(1/2)*x,I)-EllipticE(1/5*5^(1/2)*(I*5^(1/2))^(1/2
)*x,I))-10/3*(x^4+5)^(1/2)/x^3+2/3*(x^4+5)^(1/2)*x+8/15*5^(1/2)/(I*5^(1/2))^(1/2)*(-5*I*5^(1/2)*x^2+25)^(1/2)*
(5*I*5^(1/2)*x^2+25)^(1/2)/(x^4+5)^(1/2)*EllipticF(1/5*5^(1/2)*(I*5^(1/2))^(1/2)*x,I)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (x^{4} + 5\right )}^{\frac {3}{2}} {\left (3 \, x^{2} + 2\right )}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (x^4+5\right )}^{3/2}\,\left (3\,x^2+2\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 4.12, size = 163, normalized size = 0.81 \[ \frac {3 \sqrt {5} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {\sqrt {5} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{2 \Gamma \left (\frac {5}{4}\right )} + \frac {15 \sqrt {5} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} + \frac {5 \sqrt {5} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{2 x^{3} \Gamma \left (\frac {1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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