3.317 \(\int \frac {(1+x^4)^{3/4}}{(2+x^4)^2} \, dx\)

Optimal. Leaf size=74 \[ \frac {\left (x^4+1\right )^{3/4} x}{8 \left (x^4+2\right )}+\frac {3 \tan ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )}{16\ 2^{3/4}}+\frac {3 \tanh ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )}{16\ 2^{3/4}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {378, 377, 212, 206, 203} \[ \frac {\left (x^4+1\right )^{3/4} x}{8 \left (x^4+2\right )}+\frac {3 \tan ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )}{16\ 2^{3/4}}+\frac {3 \tanh ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )}{16\ 2^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + x^4)^(3/4)/(2 + x^4)^2,x]

[Out]

(x*(1 + x^4)^(3/4))/(8*(2 + x^4)) + (3*ArcTan[x/(2^(1/4)*(1 + x^4)^(1/4))])/(16*2^(3/4)) + (3*ArcTanh[x/(2^(1/
4)*(1 + x^4)^(1/4))])/(16*2^(3/4))

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 378

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^q)/(a*n*(p + 1)), x] - Dist[(c*q)/(a*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0] && GtQ[q, 0] && NeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {\left (1+x^4\right )^{3/4}}{\left (2+x^4\right )^2} \, dx &=\frac {x \left (1+x^4\right )^{3/4}}{8 \left (2+x^4\right )}+\frac {3}{8} \int \frac {1}{\sqrt [4]{1+x^4} \left (2+x^4\right )} \, dx\\ &=\frac {x \left (1+x^4\right )^{3/4}}{8 \left (2+x^4\right )}+\frac {3}{8} \operatorname {Subst}\left (\int \frac {1}{2-x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )\\ &=\frac {x \left (1+x^4\right )^{3/4}}{8 \left (2+x^4\right )}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {2}-x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{16 \sqrt {2}}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {2}+x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{16 \sqrt {2}}\\ &=\frac {x \left (1+x^4\right )^{3/4}}{8 \left (2+x^4\right )}+\frac {3 \tan ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{1+x^4}}\right )}{16\ 2^{3/4}}+\frac {3 \tanh ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{1+x^4}}\right )}{16\ 2^{3/4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.01, size = 41, normalized size = 0.55 \[ \frac {x \, _2F_1\left (-\frac {3}{4},\frac {1}{4};\frac {5}{4};-\frac {x^4}{x^4+2}\right )}{2\ 2^{3/4} \sqrt [4]{x^4+2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(1 + x^4)^(3/4)/(2 + x^4)^2,x]

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.22, size = 74, normalized size = 1.00 \[ \frac {\left (x^4+1\right )^{3/4} x}{8 \left (x^4+2\right )}+\frac {3 \tan ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )}{16\ 2^{3/4}}+\frac {3 \tanh ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )}{16\ 2^{3/4}} \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + x^4)^(3/4)/(2 + x^4)^2,x]

[Out]

(x*(1 + x^4)^(3/4))/(8*(2 + x^4)) + (3*ArcTan[x/(2^(1/4)*(1 + x^4)^(1/4))])/(16*2^(3/4)) + (3*ArcTanh[x/(2^(1/
4)*(1 + x^4)^(1/4))])/(16*2^(3/4))

________________________________________________________________________________________

fricas [B]  time = 11.89, size = 242, normalized size = 3.27 \[ -\frac {12 \cdot 8^{\frac {3}{4}} {\left (x^{4} + 2\right )} \arctan \left (-\frac {8^{\frac {3}{4}} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 4 \cdot 8^{\frac {1}{4}} {\left (x^{4} + 1\right )}^{\frac {3}{4}} x - 2^{\frac {1}{4}} {\left (8^{\frac {3}{4}} \sqrt {x^{4} + 1} x^{2} + 8^{\frac {1}{4}} {\left (3 \, x^{4} + 2\right )}\right )}}{2 \, {\left (x^{4} + 2\right )}}\right ) - 3 \cdot 8^{\frac {3}{4}} {\left (x^{4} + 2\right )} \log \left (\frac {8 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 8 \cdot 8^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} + 8^{\frac {3}{4}} {\left (3 \, x^{4} + 2\right )} + 16 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} + 2}\right ) + 3 \cdot 8^{\frac {3}{4}} {\left (x^{4} + 2\right )} \log \left (\frac {8 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} - 8 \cdot 8^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} - 8^{\frac {3}{4}} {\left (3 \, x^{4} + 2\right )} + 16 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} + 2}\right ) - 64 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{512 \, {\left (x^{4} + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/512*(12*8^(3/4)*(x^4 + 2)*arctan(-1/2*(8^(3/4)*(x^4 + 1)^(1/4)*x^3 + 4*8^(1/4)*(x^4 + 1)^(3/4)*x - 2^(1/4)*
(8^(3/4)*sqrt(x^4 + 1)*x^2 + 8^(1/4)*(3*x^4 + 2)))/(x^4 + 2)) - 3*8^(3/4)*(x^4 + 2)*log((8*sqrt(2)*(x^4 + 1)^(
1/4)*x^3 + 8*8^(1/4)*sqrt(x^4 + 1)*x^2 + 8^(3/4)*(3*x^4 + 2) + 16*(x^4 + 1)^(3/4)*x)/(x^4 + 2)) + 3*8^(3/4)*(x
^4 + 2)*log((8*sqrt(2)*(x^4 + 1)^(1/4)*x^3 - 8*8^(1/4)*sqrt(x^4 + 1)*x^2 - 8^(3/4)*(3*x^4 + 2) + 16*(x^4 + 1)^
(3/4)*x)/(x^4 + 2)) - 64*(x^4 + 1)^(3/4)*x)/(x^4 + 2)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (x^{4} + 1\right )}^{\frac {3}{4}}}{{\left (x^{4} + 2\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 2.83, size = 228, normalized size = 3.08

method	result	size
risch	\(\frac {\left (x^{4}+1\right )^{\frac {3}{4}} x}{8 x^{4}+16}+\frac {3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (\frac {2 \sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) x^{2}+2 \left (x^{4}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{3}-3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) x^{4}-4 \left (x^{4}+1\right )^{\frac {3}{4}} x -2 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right )}{x^{4}+2}\right )}{64}+\frac {3 \RootOf \left (\textit {\_Z}^{4}-2\right ) \ln \left (-\frac {2 \sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{4}-2\right )^{3} x^{2}+2 \left (x^{4}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{3}+3 \RootOf \left (\textit {\_Z}^{4}-2\right ) x^{4}+4 \left (x^{4}+1\right )^{\frac {3}{4}} x +2 \RootOf \left (\textit {\_Z}^{4}-2\right )}{x^{4}+2}\right )}{64}\)	\(228\)
trager	\(\frac {\left (x^{4}+1\right )^{\frac {3}{4}} x}{8 x^{4}+16}-\frac {3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (-\frac {2 \sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) x^{2}-2 \left (x^{4}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{3}-3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) x^{4}+4 \left (x^{4}+1\right )^{\frac {3}{4}} x -2 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right )}{x^{4}+2}\right )}{64}-\frac {3 \RootOf \left (\textit {\_Z}^{4}-2\right ) \ln \left (-\frac {-2 \sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{4}-2\right )^{3} x^{2}+2 \left (x^{4}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{3}-3 \RootOf \left (\textit {\_Z}^{4}-2\right ) x^{4}+4 \left (x^{4}+1\right )^{\frac {3}{4}} x -2 \RootOf \left (\textit {\_Z}^{4}-2\right )}{x^{4}+2}\right )}{64}\)	\(229\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x*(x^4+1)^(3/4)/(x^4+2)+3/64*RootOf(_Z^2+RootOf(_Z^4-2)^2)*ln((2*(x^4+1)^(1/2)*RootOf(_Z^4-2)^2*RootOf(_Z^
2+RootOf(_Z^4-2)^2)*x^2+2*(x^4+1)^(1/4)*RootOf(_Z^4-2)^2*x^3-3*RootOf(_Z^2+RootOf(_Z^4-2)^2)*x^4-4*(x^4+1)^(3/
4)*x-2*RootOf(_Z^2+RootOf(_Z^4-2)^2))/(x^4+2))+3/64*RootOf(_Z^4-2)*ln(-(2*(x^4+1)^(1/2)*RootOf(_Z^4-2)^3*x^2+2
*(x^4+1)^(1/4)*RootOf(_Z^4-2)^2*x^3+3*RootOf(_Z^4-2)*x^4+4*(x^4+1)^(3/4)*x+2*RootOf(_Z^4-2))/(x^4+2))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (x^{4} + 1\right )}^{\frac {3}{4}}}{{\left (x^{4} + 2\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (x^4+1\right )}^{3/4}}{{\left (x^4+2\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4 + 1)^(3/4)/(x^4 + 2)^2,x)

[Out]

int((x^4 + 1)^(3/4)/(x^4 + 2)^2, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x^{4} + 1\right )^{\frac {3}{4}}}{\left (x^{4} + 2\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________