3.23.53 \(\int \frac {(1+x^2) \sqrt [4]{x^3+x^5}}{x^2 (-1+x^2)} \, dx\)

Optimal. Leaf size=169 \[ \frac {4 \sqrt [4]{x^5+x^3}}{x}+\sqrt [4]{2} \tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^5+x^3}}\right )-\sqrt [4]{2} \tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^5+x^3}}\right )-\frac {\tan ^{-1}\left (\frac {2^{3/4} x \sqrt [4]{x^5+x^3}}{\sqrt {2} x^2-\sqrt {x^5+x^3}}\right )}{\sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\frac {x^2}{\sqrt [4]{2}}+\frac {\sqrt {x^5+x^3}}{2^{3/4}}}{x \sqrt [4]{x^5+x^3}}\right )}{\sqrt [4]{2}} \]

________________________________________________________________________________________

Rubi [C]  time = 0.18, antiderivative size = 44, normalized size of antiderivative = 0.26, number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2056, 466, 510} \begin {gather*} \frac {4 \sqrt [4]{x^5+x^3} F_1\left (-\frac {1}{8};1,-\frac {5}{4};\frac {7}{8};x^2,-x^2\right )}{x \sqrt [4]{x^2+1}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*(x^3 + x^5)^(1/4)*AppellF1[-1/8, 1, -5/4, 7/8, x^2, -x^2])/(x*(1 + x^2)^(1/4))

Rule 466

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/e^n)^p*(c + (d*x^(k*n))/e^n)^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 510

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

Rule 2056

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]

Rubi steps

\begin {align*} \int \frac {\left (1+x^2\right ) \sqrt [4]{x^3+x^5}}{x^2 \left (-1+x^2\right )} \, dx &=\frac {\sqrt [4]{x^3+x^5} \int \frac {\left (1+x^2\right )^{5/4}}{x^{5/4} \left (-1+x^2\right )} \, dx}{x^{3/4} \sqrt [4]{1+x^2}}\\ &=\frac {\left (4 \sqrt [4]{x^3+x^5}\right ) \operatorname {Subst}\left (\int \frac {\left (1+x^8\right )^{5/4}}{x^2 \left (-1+x^8\right )} \, dx,x,\sqrt [4]{x}\right )}{x^{3/4} \sqrt [4]{1+x^2}}\\ &=\frac {4 \sqrt [4]{x^3+x^5} F_1\left (-\frac {1}{8};1,-\frac {5}{4};\frac {7}{8};x^2,-x^2\right )}{x \sqrt [4]{1+x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 0.11, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (1+x^2\right ) \sqrt [4]{x^3+x^5}}{x^2 \left (-1+x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.43, size = 169, normalized size = 1.00 \begin {gather*} \frac {4 \sqrt [4]{x^3+x^5}}{x}+\sqrt [4]{2} \tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^3+x^5}}\right )-\frac {\tan ^{-1}\left (\frac {2^{3/4} x \sqrt [4]{x^3+x^5}}{\sqrt {2} x^2-\sqrt {x^3+x^5}}\right )}{\sqrt [4]{2}}-\sqrt [4]{2} \tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^3+x^5}}\right )-\frac {\tanh ^{-1}\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 )}{\sqrt [4]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 6.00, size = 1053, normalized size = 6.23

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)*(x^5+x^3)^(1/4)/x^2/(x^2-1),x, algorithm="fricas")

[Out]

1/8*(4*2^(3/4)*x*arctan(1/2*(2*x^6 + 8*x^5 + 12*x^4 + 8*x^3 + 4*2^(3/4)*(x^5 + x^3)^(3/4)*(x^2 - 6*x + 1) + 8*
sqrt(2)*sqrt(x^5 + x^3)*(x^3 + 2*x^2 + x) + 2*x^2 + sqrt(2)*(32*sqrt(2)*(x^5 + x^3)^(3/4)*x + 2^(3/4)*(x^6 - 1
6*x^5 - 2*x^4 - 16*x^3 + x^2) + 4*2^(1/4)*sqrt(x^5 + x^3)*(x^3 - 6*x^2 + x) + 8*(x^5 + x^3)^(1/4)*(x^4 + 2*x^3
+ x^2))*sqrt((4*2^(3/4)*(x^5 + x^3)^(1/4)*x^2 + sqrt(2)*(x^4 + 2*x^3 + x^2) + 8*sqrt(x^5 + x^3)*x + 4*2^(1/4)
*(x^5 + x^3)^(3/4))/(x^4 + 2*x^3 + x^2)) + 8*2^(1/4)*(x^5 + x^3)^(1/4)*(3*x^4 - 2*x^3 + 3*x^2))/(x^6 - 28*x^5
+ 6*x^4 - 28*x^3 + x^2)) - 4*2^(3/4)*x*arctan(1/2*(2*x^6 + 8*x^5 + 12*x^4 + 8*x^3 - 4*2^(3/4)*(x^5 + x^3)^(3/4
)*(x^2 - 6*x + 1) + 8*sqrt(2)*sqrt(x^5 + x^3)*(x^3 + 2*x^2 + x) + 2*x^2 + sqrt(2)*(32*sqrt(2)*(x^5 + x^3)^(3/4
)*x - 2^(3/4)*(x^6 - 16*x^5 - 2*x^4 - 16*x^3 + x^2) - 4*2^(1/4)*sqrt(x^5 + x^3)*(x^3 - 6*x^2 + x) + 8*(x^5 + x
^3)^(1/4)*(x^4 + 2*x^3 + x^2))*sqrt(-(4*2^(3/4)*(x^5 + x^3)^(1/4)*x^2 - sqrt(2)*(x^4 + 2*x^3 + x^2) - 8*sqrt(x
^5 + x^3)*x + 4*2^(1/4)*(x^5 + x^3)^(3/4))/(x^4 + 2*x^3 + x^2)) - 8*2^(1/4)*(x^5 + x^3)^(1/4)*(3*x^4 - 2*x^3 +
3*x^2))/(x^6 - 28*x^5 + 6*x^4 - 28*x^3 + x^2)) - 2^(3/4)*x*log(2*(4*2^(3/4)*(x^5 + x^3)^(1/4)*x^2 + sqrt(2)*(
x^4 + 2*x^3 + x^2) + 8*sqrt(x^5 + x^3)*x + 4*2^(1/4)*(x^5 + x^3)^(3/4))/(x^4 + 2*x^3 + x^2)) + 2^(3/4)*x*log(-
2*(4*2^(3/4)*(x^5 + x^3)^(1/4)*x^2 - sqrt(2)*(x^4 + 2*x^3 + x^2) - 8*sqrt(x^5 + x^3)*x + 4*2^(1/4)*(x^5 + x^3)
^(3/4))/(x^4 + 2*x^3 + x^2)) - 8*2^(1/4)*x*arctan(-1/2*(4*2^(3/4)*(x^5 + x^3)^(1/4)*x^2 - 2^(3/4)*(2*2^(3/4)*s
qrt(x^5 + x^3)*x + 2^(1/4)*(x^4 + 2*x^3 + x^2)) + 4*2^(1/4)*(x^5 + x^3)^(3/4))/(x^4 - 2*x^3 + x^2)) - 2*2^(1/4
)*x*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*2^(1/4)*x*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)) + 32*(x^5 + x^3)^(1/4))/x

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{5} + x^{3}\right )}^{\frac {1}{4}} {\left (x^{2} + 1\right )}}{{\left (x^{2} - 1\right )} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)*(x^5+x^3)^(1/4)/x^2/(x^2-1),x, algorithm="giac")

[Out]

integrate((x^5 + x^3)^(1/4)*(x^2 + 1)/((x^2 - 1)*x^2), x)

________________________________________________________________________________________

maple [C]  time = 28.95, size = 721, normalized size = 4.27

method	result	size
trager	\(\frac {4 \left (x^{5}+x^{3}\right )^{\frac {1}{4}}}{x}+\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{4}-2\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) x^{4}+2 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{3}+\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{2}-4 \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} \left (x^{5}+x^{3}\right )^{\frac {1}{4}} x^{2}-4 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \sqrt {x^{5}+x^{3}}\, x +4 \left (x^{5}+x^{3}\right )^{\frac {3}{4}}}{x^{2} \left (-1+x \right )^{2}}\right )}{2}-\frac {\RootOf \left (\textit {\_Z}^{4}-2\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{4}-2\right )^{3} x^{4}+2 \RootOf \left (\textit {\_Z}^{4}-2\right )^{3} x^{3}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{3} x^{2}+4 \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} \left (x^{5}+x^{3}\right )^{\frac {1}{4}} x^{2}+4 \sqrt {x^{5}+x^{3}}\, \RootOf \left (\textit {\_Z}^{4}-2\right ) x +4 \left (x^{5}+x^{3}\right )^{\frac {3}{4}}}{x^{2} \left (-1+x \right )^{2}}\right )}{2}-\frac {\ln \left (\frac {4 \RootOf \left (\textit {\_Z}^{4}-2\right )^{3} \sqrt {x^{5}+x^{3}}\, x +4 \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} \left (x^{5}+x^{3}\right )^{\frac {1}{4}} x^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right ) x^{4}+2 \RootOf \left (\textit {\_Z}^{4}-2\right ) x^{3}+4 \left (x^{5}+x^{3}\right )^{\frac {3}{4}}+\RootOf \left (\textit {\_Z}^{4}-2\right ) x^{2}}{\left (1+x \right )^{2} x^{2}}\right ) \RootOf \left (\textit {\_Z}^{4}-2\right )^{3}}{4}-\frac {\ln \left (\frac {4 \RootOf \left (\textit {\_Z}^{4}-2\right )^{3} \sqrt {x^{5}+x^{3}}\, x +4 \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} \left (x^{5}+x^{3}\right )^{\frac {1}{4}} x^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right ) x^{4}+2 \RootOf \left (\textit {\_Z}^{4}-2\right ) x^{3}+4 \left (x^{5}+x^{3}\right )^{\frac {3}{4}}+\RootOf \left (\textit {\_Z}^{4}-2\right ) x^{2}}{\left (1+x \right )^{2} x^{2}}\right ) \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right )}{4}+\frac {\RootOf \left (\textit {\_Z}^{4}-2\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (-\frac {-2 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} \sqrt {x^{5}+x^{3}}\, x +2 \RootOf \left (\textit {\_Z}^{4}-2\right )^{3} \sqrt {x^{5}+x^{3}}\, x -4 \left (x^{5}+x^{3}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-2\right ) x^{2}+\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) x^{4}+\RootOf \left (\textit {\_Z}^{4}-2\right ) x^{4}-2 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) x^{3}-2 \RootOf \left (\textit {\_Z}^{4}-2\right ) x^{3}+4 \left (x^{5}+x^{3}\right )^{\frac {3}{4}}+\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) x^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right ) x^{2}}{\left (1+x \right )^{2} x^{2}}\right )}{2}\)	\(721\)
risch	\(\text {Expression too large to display}\)	\(1752\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+1)*(x^5+x^3)^(1/4)/x^2/(x^2-1),x,method=_RETURNVERBOSE)

[Out]

4*(x^5+x^3)^(1/4)/x+1/2*RootOf(_Z^2+RootOf(_Z^4-2)^2)*ln((RootOf(_Z^4-2)^2*RootOf(_Z^2+RootOf(_Z^4-2)^2)*x^4+2
*RootOf(_Z^2+RootOf(_Z^4-2)^2)*RootOf(_Z^4-2)^2*x^3+RootOf(_Z^2+RootOf(_Z^4-2)^2)*RootOf(_Z^4-2)^2*x^2-4*RootO
f(_Z^4-2)^2*(x^5+x^3)^(1/4)*x^2-4*RootOf(_Z^2+RootOf(_Z^4-2)^2)*(x^5+x^3)^(1/2)*x+4*(x^5+x^3)^(3/4))/x^2/(-1+x
)^2)-1/2*RootOf(_Z^4-2)*ln((RootOf(_Z^4-2)^3*x^4+2*RootOf(_Z^4-2)^3*x^3+RootOf(_Z^4-2)^3*x^2+4*RootOf(_Z^4-2)^
2*(x^5+x^3)^(1/4)*x^2+4*(x^5+x^3)^(1/2)*RootOf(_Z^4-2)*x+4*(x^5+x^3)^(3/4))/x^2/(-1+x)^2)-1/4*ln((4*RootOf(_Z^
4-2)^3*(x^5+x^3)^(1/2)*x+4*RootOf(_Z^4-2)^2*(x^5+x^3)^(1/4)*x^2+RootOf(_Z^4-2)*x^4+2*RootOf(_Z^4-2)*x^3+4*(x^5
+x^3)^(3/4)+RootOf(_Z^4-2)*x^2)/(1+x)^2/x^2)*RootOf(_Z^4-2)^3-1/4*ln((4*RootOf(_Z^4-2)^3*(x^5+x^3)^(1/2)*x+4*R
ootOf(_Z^4-2)^2*(x^5+x^3)^(1/4)*x^2+RootOf(_Z^4-2)*x^4+2*RootOf(_Z^4-2)*x^3+4*(x^5+x^3)^(3/4)+RootOf(_Z^4-2)*x
^2)/(1+x)^2/x^2)*RootOf(_Z^4-2)^2*RootOf(_Z^2+RootOf(_Z^4-2)^2)+1/2*RootOf(_Z^4-2)^2*RootOf(_Z^2+RootOf(_Z^4-2
)^2)*ln(-(-2*RootOf(_Z^2+RootOf(_Z^4-2)^2)*RootOf(_Z^4-2)^2*(x^5+x^3)^(1/2)*x+2*RootOf(_Z^4-2)^3*(x^5+x^3)^(1/
2)*x-4*(x^5+x^3)^(1/4)*RootOf(_Z^2+RootOf(_Z^4-2)^2)*RootOf(_Z^4-2)*x^2+RootOf(_Z^2+RootOf(_Z^4-2)^2)*x^4+Root
Of(_Z^4-2)*x^4-2*RootOf(_Z^2+RootOf(_Z^4-2)^2)*x^3-2*RootOf(_Z^4-2)*x^3+4*(x^5+x^3)^(3/4)+RootOf(_Z^2+RootOf(_
Z^4-2)^2)*x^2+RootOf(_Z^4-2)*x^2)/(1+x)^2/x^2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{5} + x^{3}\right )}^{\frac {1}{4}} {\left (x^{2} + 1\right )}}{{\left (x^{2} - 1\right )} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^5 + x^3)^(1/4)*(x^2 + 1)/((x^2 - 1)*x^2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3 + x^5)^(1/4)*(x^2 + 1))/(x^2*(x^2 - 1)),x)

[Out]

int(((x^3 + x^5)^(1/4)*(x^2 + 1))/(x^2*(x^2 - 1)), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+1)*(x**5+x**3)**(1/4)/x**2/(x**2-1),x)

[Out]

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

________________________________________________________________________________________