3.9.3 \(\int (1+x^4)^{3/2} \, dx\) [803]

Optimal. Leaf size=72 \[ \frac {2}{7} x \sqrt {1+x^4}+\frac {1}{7} x \left (1+x^4\right )^{3/2}+\frac {2 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{7 \sqrt {1+x^4}} \]

[Out]

1/7*x*(x^4+1)^(3/2)+2/7*x*(x^4+1)^(1/2)+2/7*(x^2+1)*(cos(2*arctan(x))^2)^(1/2)/cos(2*arctan(x))*EllipticF(sin(
2*arctan(x)),1/2*2^(1/2))*((x^4+1)/(x^2+1)^2)^(1/2)/(x^4+1)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {201, 226} \begin {gather*} \frac {2 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \text {ArcTan}(x)\left |\frac {1}{2}\right .\right )}{7 \sqrt {x^4+1}}+\frac {1}{7} x \left (x^4+1\right )^{3/2}+\frac {2}{7} x \sqrt {x^4+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x^4)^(3/2),x]

[Out]

(2*x*Sqrt[1 + x^4])/7 + (x*(1 + x^4)^(3/2))/7 + (2*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x]
, 1/2])/(7*Sqrt[1 + x^4])

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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]

Rubi steps

\begin {align*} \int \left (1+x^4\right )^{3/2} \, dx &=\frac {1}{7} x \left (1+x^4\right )^{3/2}+\frac {6}{7} \int \sqrt {1+x^4} \, dx\\ &=\frac {2}{7} x \sqrt {1+x^4}+\frac {1}{7} x \left (1+x^4\right )^{3/2}+\frac {4}{7} \int \frac {1}{\sqrt {1+x^4}} \, dx\\ &=\frac {2}{7} x \sqrt {1+x^4}+\frac {1}{7} x \left (1+x^4\right )^{3/2}+\frac {2 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{7 \sqrt {1+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 = 3.48, size = 17, normalized size = 0.24 \begin {gather*} x \, _2F_1\left (-\frac {3}{2},\frac {1}{4};\frac {5}{4};-x^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^4)^(3/2),x]

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.18, size = 84, normalized size = 1.17

method	result	size
meijerg	\(x \hypergeom \left (\left [-\frac {3}{2}, \frac {1}{4}\right ], \left [\frac {5}{4}\right ], -x^{4}\right )\)	\(14\)
risch	\(\frac {x \left (x^{4}+3\right ) \sqrt {x^{4}+1}}{7}+\frac {4 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{7 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(77\)
default	\(\frac {x^{5} \sqrt {x^{4}+1}}{7}+\frac {3 x \sqrt {x^{4}+1}}{7}+\frac {4 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{7 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(84\)
elliptic	\(\frac {x^{5} \sqrt {x^{4}+1}}{7}+\frac {3 x \sqrt {x^{4}+1}}{7}+\frac {4 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{7 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4+1)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/7*x^5*(x^4+1)^(1/2)+3/7*x*(x^4+1)^(1/2)+4/7/(1/2*2^(1/2)+1/2*I*2^(1/2))*(1-I*x^2)^(1/2)*(1+I*x^2)^(1/2)/(x^4
+1)^(1/2)*EllipticF(x*(1/2*2^(1/2)+1/2*I*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((x^4+1)^(3/2),x, algorithm="maxima")

[Out]

integrate((x^4 + 1)^(3/2), x)

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Fricas [C] Result contains complex when optimal does not.
time = 0.08, size = 32, normalized size = 0.44 \begin {gather*} \frac {1}{7} \, {\left (x^{5} + 3 \, x\right )} \sqrt {x^{4} + 1} + \frac {4}{7} i \, \sqrt {i} F(\arcsin \left (\frac {\sqrt {i}}{x}\right )\,|\,-1) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+1)^(3/2),x, algorithm="fricas")

[Out]

1/7*(x^5 + 3*x)*sqrt(x^4 + 1) + 4/7*I*sqrt(I)*elliptic_f(arcsin(sqrt(I)/x), -1)

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Sympy [C] Result contains complex when optimal does not.
time = 0.33, size = 29, normalized size = 0.40 \begin {gather*} \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4+1)**(3/2),x)

[Out]

x*gamma(1/4)*hyper((-3/2, 1/4), (5/4,), x**4*exp_polar(I*pi))/(4*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((x^4+1)^(3/2),x, algorithm="giac")

[Out]

integrate((x^4 + 1)^(3/2), x)

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Mupad [B]
time = 1.02, size = 12, normalized size = 0.17 \begin {gather*} x\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{4};\ \frac {5}{4};\ -x^4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4 + 1)^(3/2),x)

[Out]

x*hypergeom([-3/2, 1/4], 5/4, -x^4)

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