3.1.29 \(\int (2+3 x^2) (5+x^4)^{3/2} \, dx\) [29]

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

[Out]

1/21*x*(7*x^2+6)*(x^4+5)^(3/2)+2/7*x*(7*x^2+10)*(x^4+5)^(1/2)+20*x*(x^4+5)^(1/2)/(x^2+5^(1/2))-20*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)+10/7*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))*(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 = 0.06, antiderivative size = 197, normalized size of antiderivative = 1.00, 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 = {1191, 1212, 226, 1210} \begin {gather*} \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 \text {ArcTan}\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 \text {ArcTan}\left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{\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}} \end {gather*}

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 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1191

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 1210

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)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1212

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]

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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 4.89, size = 49, normalized size = 0.25 \begin {gather*} 5 \sqrt {5} x \left (2 \, _2F_1\left (-\frac {3}{2},\frac {1}{4};\frac {5}{4};-\frac {x^4}{5}\right )+x^2 \, _2F_1\left (-\frac {3}{2},\frac {3}{4};\frac {7}{4};-\frac {x^4}{5}\right )\right ) \end {gather*}

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

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Maple [C] Result contains complex when optimal does not.
time = 0.13, size = 192, normalized size = 0.97

method	result	size
meijerg	\(10 \sqrt {5}\, x \hypergeom \left (\left [-\frac {3}{2}, \frac {1}{4}\right ], \left [\frac {5}{4}\right ], -\frac {x^{4}}{5}\right )+5 \sqrt {5}\, x^{3} \hypergeom \left (\left [-\frac {3}{2}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -\frac {x^{4}}{5}\right )\)	\(38\)
risch	\(\frac {x \left (7 x^{6}+6 x^{4}+77 x^{2}+90\right ) \sqrt {x^{4}+5}}{21}+\frac {4 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (\EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-\EllipticE \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{\sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}+\frac {8 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{7 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\)	\(173\)
default	\(\frac {x^{7} \sqrt {x^{4}+5}}{3}+\frac {11 x^{3} \sqrt {x^{4}+5}}{3}+\frac {4 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (\EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-\EllipticE \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{\sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}+\frac {2 x^{5} \sqrt {x^{4}+5}}{7}+\frac {30 x \sqrt {x^{4}+5}}{7}+\frac {8 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{7 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\)	\(192\)
elliptic	\(\frac {x^{7} \sqrt {x^{4}+5}}{3}+\frac {11 x^{3} \sqrt {x^{4}+5}}{3}+\frac {4 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (\EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-\EllipticE \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{\sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}+\frac {2 x^{5} \sqrt {x^{4}+5}}{7}+\frac {30 x \sqrt {x^{4}+5}}{7}+\frac {8 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \EllipticF \left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{7 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\)	\(192\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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Sympy [C] Result contains complex when optimal does not.
time = 1.65, size = 158, normalized size = 0.80 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (x^4+5\right )}^{3/2}\,\left (3\,x^2+2\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4 + 5)^(3/2)*(3*x^2 + 2),x)

[Out]

int((x^4 + 5)^(3/2)*(3*x^2 + 2), x)

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