\(\int \frac {1+x^4}{(1-x^4) \sqrt [4]{x^3+x^5}} \, dx\) [2341]

Optimal result

Integrand size = 26, antiderivative size = 185 \[ \int \frac {1+x^4}{\left (1-x^4\right ) \sqrt [4]{x^3+x^5}} \, dx=\frac {2 \left (x^3+x^5\right )^{3/4}}{x^2 \left (1+x^2\right )}+\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^3+x^5}}\right )}{2 \sqrt [4]{2}}-\frac {\arctan \left (\frac {2^{3/4} x \sqrt [4]{x^3+x^5}}{\sqrt {2} x^2-\sqrt {x^3+x^5}}\right )}{2\ 2^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^3+x^5}}\right )}{2 \sqrt [4]{2}}+\frac {\text {arctanh}\left (\frac {\frac {x^2}{\sqrt [4]{2}}+\frac {\sqrt {x^3+x^5}}{2^{3/4}}}{x \sqrt [4]{x^3+x^5}}\right )}{2\ 2^{3/4}} \]

[Out]

2*(x^5+x^3)^(3/4)/x^2/(x^2+1)+1/4*arctan(2^(1/4)*x/(x^5+x^3)^(1/4))*2^(3/4)-1/4*arctan(2^(3/4)*x*(x^5+x^3)^(1/
4)/(2^(1/2)*x^2-(x^5+x^3)^(1/2)))*2^(1/4)+1/4*arctanh(2^(1/4)*x/(x^5+x^3)^(1/4))*2^(3/4)+1/4*arctanh((1/2*x^2*
2^(3/4)+1/2*(x^5+x^3)^(1/2)*2^(1/4))/x/(x^5+x^3)^(1/4))*2^(1/4)

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 0.37 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.44, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2081, 6847, 6857, 251, 1418, 440} \[ \int \frac {1+x^4}{\left (1-x^4\right ) \sqrt [4]{x^3+x^5}} \, dx=\frac {8 x \sqrt [4]{x^2+1} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {5}{4},\frac {9}{8},x^2,-x^2\right )}{\sqrt [4]{x^5+x^3}}-\frac {4 x \sqrt [4]{x^2+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-x^2\right )}{\sqrt [4]{x^5+x^3}} \]

[In]

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

[Out]

(8*x*(1 + x^2)^(1/4)*AppellF1[1/8, 1, 5/4, 9/8, x^2, -x^2])/(x^3 + x^5)^(1/4) - (4*x*(1 + x^2)^(1/4)*Hypergeom
etric2F1[1/8, 1/4, 9/8, -x^2])/(x^3 + x^5)^(1/4)

Rule 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 1418

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((a_) + (c_.)*(x_)^(n2_))^(p_.), x_Symbol] :> Int[(d + e*x^n)^(p + q)*(a/d
+ (c/e)*x^n)^p, x] /; FreeQ[{a, c, d, e, n, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && IntegerQ[p]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rule 6847

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (x^{3/4} \sqrt [4]{1+x^2}\right ) \int \frac {1+x^4}{x^{3/4} \sqrt [4]{1+x^2} \left (1-x^4\right )} \, dx}{\sqrt [4]{x^3+x^5}} \\ & = \frac {\left (4 x^{3/4} \sqrt [4]{1+x^2}\right ) \text {Subst}\left (\int \frac {1+x^{16}}{\sqrt [4]{1+x^8} \left (1-x^{16}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{x^3+x^5}} \\ & = \frac {\left (4 x^{3/4} \sqrt [4]{1+x^2}\right ) \text {Subst}\left (\int \left (-\frac {1}{\sqrt [4]{1+x^8}}+\frac {2}{\sqrt [4]{1+x^8} \left (1-x^{16}\right )}\right ) \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{x^3+x^5}} \\ & = -\frac {\left (4 x^{3/4} \sqrt [4]{1+x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1+x^8}} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{x^3+x^5}}+\frac {\left (8 x^{3/4} \sqrt [4]{1+x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1+x^8} \left (1-x^{16}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{x^3+x^5}} \\ & = -\frac {4 x \sqrt [4]{1+x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-x^2\right )}{\sqrt [4]{x^3+x^5}}+\frac {\left (8 x^{3/4} \sqrt [4]{1+x^2}\right ) \text {Subst}\left (\int \frac {1}{\left (1-x^8\right ) \left (1+x^8\right )^{5/4}} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{x^3+x^5}} \\ & = \frac {8 x \sqrt [4]{1+x^2} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {5}{4},\frac {9}{8},x^2,-x^2\right )}{\sqrt [4]{x^3+x^5}}-\frac {4 x \sqrt [4]{1+x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-x^2\right )}{\sqrt [4]{x^3+x^5}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.27 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.21 \[ \int \frac {1+x^4}{\left (1-x^4\right ) \sqrt [4]{x^3+x^5}} \, dx=\frac {x^{3/4} \left (8 \sqrt [4]{x}+2^{3/4} \sqrt [4]{1+x^2} \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{x}}{\sqrt [4]{1+x^2}}\right )-\sqrt [4]{2} \sqrt [4]{1+x^2} \arctan \left (\frac {2^{3/4} \sqrt [4]{x} \sqrt [4]{1+x^2}}{\sqrt {2} \sqrt {x}-\sqrt {1+x^2}}\right )+2^{3/4} \sqrt [4]{1+x^2} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{x}}{\sqrt [4]{1+x^2}}\right )+\sqrt [4]{2} \sqrt [4]{1+x^2} \text {arctanh}\left (\frac {2 \sqrt [4]{2} \sqrt [4]{x} \sqrt [4]{1+x^2}}{2 \sqrt {x}+\sqrt {2} \sqrt {1+x^2}}\right )\right )}{4 \sqrt [4]{x^3+x^5}} \]

[In]

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

[Out]

(x^(3/4)*(8*x^(1/4) + 2^(3/4)*(1 + x^2)^(1/4)*ArcTan[(2^(1/4)*x^(1/4))/(1 + x^2)^(1/4)] - 2^(1/4)*(1 + x^2)^(1
/4)*ArcTan[(2^(3/4)*x^(1/4)*(1 + x^2)^(1/4))/(Sqrt[2]*Sqrt[x] - Sqrt[1 + x^2])] + 2^(3/4)*(1 + x^2)^(1/4)*ArcT
anh[(2^(1/4)*x^(1/4))/(1 + x^2)^(1/4)] + 2^(1/4)*(1 + x^2)^(1/4)*ArcTanh[(2*2^(1/4)*x^(1/4)*(1 + x^2)^(1/4))/(
2*Sqrt[x] + Sqrt[2]*Sqrt[1 + x^2])]))/(4*(x^3 + x^5)^(1/4))

Maple [A] (verified)

Time = 39.50 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.52

method	result	size
pseudoelliptic	\(\frac {\ln \left (\frac {-2^{\frac {1}{4}} x -\left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{2^{\frac {1}{4}} x -\left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}\right ) 2^{\frac {3}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-2 \arctan \left (\frac {2^{\frac {3}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{2 x}\right ) 2^{\frac {3}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-\ln \left (\frac {-2^{\frac {3}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{3} \left (x^{2}+1\right )}}{2^{\frac {3}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{3} \left (x^{2}+1\right )}}\right ) 2^{\frac {1}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-2 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}+x}{x}\right ) 2^{\frac {1}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-2 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-x}{x}\right ) 2^{\frac {1}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}+16 x}{8 \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}\)	\(281\)
risch	\(\text {Expression too large to display}\)	\(733\)
trager	\(\text {Expression too large to display}\)	\(744\)

[In]

int((x^4+1)/(-x^4+1)/(x^5+x^3)^(1/4),x,method=_RETURNVERBOSE)

[Out]

1/8*(ln((-2^(1/4)*x-(x^3*(x^2+1))^(1/4))/(2^(1/4)*x-(x^3*(x^2+1))^(1/4)))*2^(3/4)*(x^3*(x^2+1))^(1/4)-2*arctan
(1/2*2^(3/4)/x*(x^3*(x^2+1))^(1/4))*2^(3/4)*(x^3*(x^2+1))^(1/4)-ln((-2^(3/4)*(x^3*(x^2+1))^(1/4)*x+2^(1/2)*x^2
+(x^3*(x^2+1))^(1/2))/(2^(3/4)*(x^3*(x^2+1))^(1/4)*x+2^(1/2)*x^2+(x^3*(x^2+1))^(1/2)))*2^(1/4)*(x^3*(x^2+1))^(
1/4)-2*arctan((2^(1/4)*(x^3*(x^2+1))^(1/4)+x)/x)*2^(1/4)*(x^3*(x^2+1))^(1/4)-2*arctan((2^(1/4)*(x^3*(x^2+1))^(
1/4)-x)/x)*2^(1/4)*(x^3*(x^2+1))^(1/4)+16*x)/(x^3*(x^2+1))^(1/4)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 8.25 (sec) , antiderivative size = 783, normalized size of antiderivative = 4.23 \[ \int \frac {1+x^4}{\left (1-x^4\right ) \sqrt [4]{x^3+x^5}} \, dx=\text {Too large to display} \]

[In]

integrate((x^4+1)/(-x^4+1)/(x^5+x^3)^(1/4),x, algorithm="fricas")

[Out]

1/16*(2^(3/4)*(x^4 + x^2)*log((4*sqrt(2)*(x^5 + x^3)^(1/4)*x^2 + 2^(3/4)*(x^4 + 2*x^3 + x^2) + 4*2^(1/4)*sqrt(
x^5 + x^3)*x + 4*(x^5 + x^3)^(3/4))/(x^4 - 2*x^3 + x^2)) - 2^(3/4)*(x^4 + x^2)*log((4*sqrt(2)*(x^5 + x^3)^(1/4
)*x^2 - 2^(3/4)*(x^4 + 2*x^3 + x^2) - 4*2^(1/4)*sqrt(x^5 + x^3)*x + 4*(x^5 + x^3)^(3/4))/(x^4 - 2*x^3 + x^2))
- 2^(3/4)*(-I*x^4 - I*x^2)*log(-(4*sqrt(2)*(x^5 + x^3)^(1/4)*x^2 - 2^(3/4)*(I*x^4 + 2*I*x^3 + I*x^2) + 4*I*2^(
1/4)*sqrt(x^5 + x^3)*x - 4*(x^5 + x^3)^(3/4))/(x^4 - 2*x^3 + x^2)) - 2^(3/4)*(I*x^4 + I*x^2)*log(-(4*sqrt(2)*(
x^5 + x^3)^(1/4)*x^2 - 2^(3/4)*(-I*x^4 - 2*I*x^3 - I*x^2) - 4*I*2^(1/4)*sqrt(x^5 + x^3)*x - 4*(x^5 + x^3)^(3/4
))/(x^4 - 2*x^3 + x^2)) + 2^(1/4)*(-(I - 1)*x^4 - (I - 1)*x^2)*log(-2*(4*I*sqrt(2)*(x^5 + x^3)^(1/4)*x^2 + (2*
I + 2)*2^(3/4)*sqrt(x^5 + x^3)*x - 2^(1/4)*((I - 1)*x^4 - (2*I - 2)*x^3 + (I - 1)*x^2) + 4*(x^5 + x^3)^(3/4))/
(x^4 + 2*x^3 + x^2)) + 2^(1/4)*((I - 1)*x^4 + (I - 1)*x^2)*log(-2*(4*I*sqrt(2)*(x^5 + x^3)^(1/4)*x^2 - (2*I +
2)*2^(3/4)*sqrt(x^5 + x^3)*x - 2^(1/4)*(-(I - 1)*x^4 + (2*I - 2)*x^3 - (I - 1)*x^2) + 4*(x^5 + x^3)^(3/4))/(x^
4 + 2*x^3 + x^2)) + 2^(1/4)*((I + 1)*x^4 + (I + 1)*x^2)*log(-2*(-4*I*sqrt(2)*(x^5 + x^3)^(1/4)*x^2 - (2*I - 2)
*2^(3/4)*sqrt(x^5 + x^3)*x - 2^(1/4)*(-(I + 1)*x^4 + (2*I + 2)*x^3 - (I + 1)*x^2) + 4*(x^5 + x^3)^(3/4))/(x^4
+ 2*x^3 + x^2)) + 2^(1/4)*(-(I + 1)*x^4 - (I + 1)*x^2)*log(-2*(-4*I*sqrt(2)*(x^5 + x^3)^(1/4)*x^2 + (2*I - 2)*
2^(3/4)*sqrt(x^5 + x^3)*x - 2^(1/4)*((I + 1)*x^4 - (2*I + 2)*x^3 + (I + 1)*x^2) + 4*(x^5 + x^3)^(3/4))/(x^4 +
2*x^3 + x^2)) + 32*(x^5 + x^3)^(3/4))/(x^4 + x^2)

Sympy [F]

\[ \int \frac {1+x^4}{\left (1-x^4\right ) \sqrt [4]{x^3+x^5}} \, dx=- \int \frac {x^{4}}{x^{4} \sqrt [4]{x^{5} + x^{3}} - \sqrt [4]{x^{5} + x^{3}}}\, dx - \int \frac {1}{x^{4} \sqrt [4]{x^{5} + x^{3}} - \sqrt [4]{x^{5} + x^{3}}}\, dx \]

[In]

integrate((x**4+1)/(-x**4+1)/(x**5+x**3)**(1/4),x)

[Out]

-Integral(x**4/(x**4*(x**5 + x**3)**(1/4) - (x**5 + x**3)**(1/4)), x) - Integral(1/(x**4*(x**5 + x**3)**(1/4)
- (x**5 + x**3)**(1/4)), x)

Maxima [F]

\[ \int \frac {1+x^4}{\left (1-x^4\right ) \sqrt [4]{x^3+x^5}} \, dx=\int { -\frac {x^{4} + 1}{{\left (x^{5} + x^{3}\right )}^{\frac {1}{4}} {\left (x^{4} - 1\right )}} \,d x } \]

[In]

integrate((x^4+1)/(-x^4+1)/(x^5+x^3)^(1/4),x, algorithm="maxima")

[Out]

-integrate((x^4 + 1)/((x^5 + x^3)^(1/4)*(x^4 - 1)), x)

Giac [F]

\[ \int \frac {1+x^4}{\left (1-x^4\right ) \sqrt [4]{x^3+x^5}} \, dx=\int { -\frac {x^{4} + 1}{{\left (x^{5} + x^{3}\right )}^{\frac {1}{4}} {\left (x^{4} - 1\right )}} \,d x } \]

[In]

integrate((x^4+1)/(-x^4+1)/(x^5+x^3)^(1/4),x, algorithm="giac")

[Out]

integrate(-(x^4 + 1)/((x^5 + x^3)^(1/4)*(x^4 - 1)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1+x^4}{\left (1-x^4\right ) \sqrt [4]{x^3+x^5}} \, dx=\int -\frac {x^4+1}{{\left (x^5+x^3\right )}^{1/4}\,\left (x^4-1\right )} \,d x \]

[In]

int(-(x^4 + 1)/((x^3 + x^5)^(1/4)*(x^4 - 1)),x)

[Out]

int(-(x^4 + 1)/((x^3 + x^5)^(1/4)*(x^4 - 1)), x)