3.10.2 \(\int \frac {(1+x^2) \sqrt [4]{x^2+x^6}}{x^2 (-1+x^2)} \, dx\)

Optimal. Leaf size=68 \[ \frac {2 \sqrt [4]{x^6+x^2}}{x}+\sqrt [4]{2} \tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^6+x^2}}\right )-\sqrt [4]{2} \tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^6+x^2}}\right ) \]

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Rubi [C]  time = 0.76, antiderivative size = 147, normalized size of antiderivative = 2.16, number of steps used = 20, number of rules used = 11, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {2056, 6725, 277, 329, 364, 1312, 1336, 325, 1337, 466, 510} \begin {gather*} -\frac {8 \sqrt [4]{x^6+x^2} x F_1\left (\frac {3}{8};1,\frac {3}{4};\frac {11}{8};x^4,-x^4\right )}{3 \sqrt [4]{x^4+1}}-\frac {8 \sqrt [4]{x^6+x^2} x^3 F_1\left (\frac {7}{8};1,\frac {3}{4};\frac {15}{8};x^4,-x^4\right )}{7 \sqrt [4]{x^4+1}}+\frac {4 \sqrt [4]{x^6+x^2} x \, _2F_1\left (\frac {3}{8},\frac {3}{4};\frac {11}{8};-x^4\right )}{3 \sqrt [4]{x^4+1}}+\frac {2 \sqrt [4]{x^6+x^2}}{x} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[((1 + x^2)*(x^2 + x^6)^(1/4))/(x^2*(-1 + x^2)),x]

[Out]

(2*(x^2 + x^6)^(1/4))/x - (8*x*(x^2 + x^6)^(1/4)*AppellF1[3/8, 1, 3/4, 11/8, x^4, -x^4])/(3*(1 + x^4)^(1/4)) -
(8*x^3*(x^2 + x^6)^(1/4)*AppellF1[7/8, 1, 3/4, 15/8, x^4, -x^4])/(7*(1 + x^4)^(1/4)) + (4*x*(x^2 + x^6)^(1/4)
*Hypergeometric2F1[3/8, 3/4, 11/8, -x^4])/(3*(1 + x^4)^(1/4))

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 364

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

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 1312

Int[(((f_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^4)^(p_.))/((d_.) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/(d*e), Int[(f*
x)^m*(a*e + c*d*x^2)*(a + c*x^4)^(p - 1), x], x] - Dist[(c*d^2 + a*e^2)/(d*e*f^2), Int[((f*x)^(m + 2)*(a + c*x
^4)^(p - 1))/(d + e*x^2), x], x] /; FreeQ[{a, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, 0]

Rule 1336

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(f*x)^m*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a, c, d, e, f, m, p, q}, x] && (IGtQ[p, 0] || IGtQ[q,
0] || IntegersQ[m, q])

Rule 1337

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(f*x)^m/x^m, I
nt[ExpandIntegrand[x^m*(a + c*x^4)^p, (d/(d^2 - e^2*x^4) - (e*x^2)/(d^2 - e^2*x^4))^(-q), x], x], x] /; FreeQ[
{a, c, d, e, f, m, p}, x] &&  !IntegerQ[p] && ILtQ[q, 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]

Rule 6725

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*} \int \frac {\left (1+x^2\right ) \sqrt [4]{x^2+x^6}}{x^2 \left (-1+x^2\right )} \, dx &=\frac {\sqrt [4]{x^2+x^6} \int \frac {\left (1+x^2\right ) \sqrt [4]{1+x^4}}{x^{3/2} \left (-1+x^2\right )} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}\\ &=\frac {\sqrt [4]{x^2+x^6} \int \left (\frac {\sqrt [4]{1+x^4}}{x^{3/2}}+\frac {2 \sqrt [4]{1+x^4}}{x^{3/2} \left (-1+x^2\right )}\right ) \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}\\ &=\frac {\sqrt [4]{x^2+x^6} \int \frac {\sqrt [4]{1+x^4}}{x^{3/2}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (2 \sqrt [4]{x^2+x^6}\right ) \int \frac {\sqrt [4]{1+x^4}}{x^{3/2} \left (-1+x^2\right )} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}\\ &=-\frac {2 \sqrt [4]{x^2+x^6}}{x}+\frac {\left (2 \sqrt [4]{x^2+x^6}\right ) \int \frac {x^{5/2}}{\left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}-\frac {\left (2 \sqrt [4]{x^2+x^6}\right ) \int \frac {1-x^2}{x^{3/2} \left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (4 \sqrt [4]{x^2+x^6}\right ) \int \frac {\sqrt {x}}{\left (-1+x^2\right ) \left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}\\ &=-\frac {2 \sqrt [4]{x^2+x^6}}{x}-\frac {\left (2 \sqrt [4]{x^2+x^6}\right ) \int \left (\frac {1}{x^{3/2} \left (1+x^4\right )^{3/4}}-\frac {\sqrt {x}}{\left (1+x^4\right )^{3/4}}\right ) \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (4 \sqrt [4]{x^2+x^6}\right ) \int \left (\frac {\sqrt {x}}{\left (-1+x^4\right ) \left (1+x^4\right )^{3/4}}+\frac {x^{5/2}}{\left (-1+x^4\right ) \left (1+x^4\right )^{3/4}}\right ) \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (4 \sqrt [4]{x^2+x^6}\right ) \operatorname {Subst}\left (\int \frac {x^6}{\left (1+x^8\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{1+x^4}}\\ &=-\frac {2 \sqrt [4]{x^2+x^6}}{x}+\frac {4 x^3 \sqrt [4]{x^2+x^6} \, _2F_1\left (\frac {3}{4},\frac {7}{8};\frac {15}{8};-x^4\right )}{7 \sqrt [4]{1+x^4}}-\frac {\left (2 \sqrt [4]{x^2+x^6}\right ) \int \frac {1}{x^{3/2} \left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (2 \sqrt [4]{x^2+x^6}\right ) \int \frac {\sqrt {x}}{\left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (4 \sqrt [4]{x^2+x^6}\right ) \int \frac {\sqrt {x}}{\left (-1+x^4\right ) \left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (4 \sqrt [4]{x^2+x^6}\right ) \int \frac {x^{5/2}}{\left (-1+x^4\right ) \left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}\\ &=\frac {2 \sqrt [4]{x^2+x^6}}{x}+\frac {4 x^3 \sqrt [4]{x^2+x^6} \, _2F_1\left (\frac {3}{4},\frac {7}{8};\frac {15}{8};-x^4\right )}{7 \sqrt [4]{1+x^4}}-\frac {\left (2 \sqrt [4]{x^2+x^6}\right ) \int \frac {x^{5/2}}{\left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (4 \sqrt [4]{x^2+x^6}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (1+x^8\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (8 \sqrt [4]{x^2+x^6}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (-1+x^8\right ) \left (1+x^8\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (8 \sqrt [4]{x^2+x^6}\right ) \operatorname {Subst}\left (\int \frac {x^6}{\left (-1+x^8\right ) \left (1+x^8\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{1+x^4}}\\ &=\frac {2 \sqrt [4]{x^2+x^6}}{x}-\frac {8 x \sqrt [4]{x^2+x^6} F_1\left (\frac {3}{8};1,\frac {3}{4};\frac {11}{8};x^4,-x^4\right )}{3 \sqrt [4]{1+x^4}}-\frac {8 x^3 \sqrt [4]{x^2+x^6} F_1\left (\frac {7}{8};1,\frac {3}{4};\frac {15}{8};x^4,-x^4\right )}{7 \sqrt [4]{1+x^4}}+\frac {4 x \sqrt [4]{x^2+x^6} \, _2F_1\left (\frac {3}{8},\frac {3}{4};\frac {11}{8};-x^4\right )}{3 \sqrt [4]{1+x^4}}+\frac {4 x^3 \sqrt [4]{x^2+x^6} \, _2F_1\left (\frac {3}{4},\frac {7}{8};\frac {15}{8};-x^4\right )}{7 \sqrt [4]{1+x^4}}-\frac {\left (4 \sqrt [4]{x^2+x^6}\right ) \operatorname {Subst}\left (\int \frac {x^6}{\left (1+x^8\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{1+x^4}}\\ &=\frac {2 \sqrt [4]{x^2+x^6}}{x}-\frac {8 x \sqrt [4]{x^2+x^6} F_1\left (\frac {3}{8};1,\frac {3}{4};\frac {11}{8};x^4,-x^4\right )}{3 \sqrt [4]{1+x^4}}-\frac {8 x^3 \sqrt [4]{x^2+x^6} F_1\left (\frac {7}{8};1,\frac {3}{4};\frac {15}{8};x^4,-x^4\right )}{7 \sqrt [4]{1+x^4}}+\frac {4 x \sqrt [4]{x^2+x^6} \, _2F_1\left (\frac {3}{8},\frac {3}{4};\frac {11}{8};-x^4\right )}{3 \sqrt [4]{1+x^4}}\\ \end {align*}

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Mathematica [F]  time = 0.73, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (1+x^2\right ) \sqrt [4]{x^2+x^6}}{x^2 \left (-1+x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[((1 + x^2)*(x^2 + x^6)^(1/4))/(x^2*(-1 + x^2)),x]

[Out]

Integrate[((1 + x^2)*(x^2 + x^6)^(1/4))/(x^2*(-1 + x^2)), x]

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IntegrateAlgebraic [A]  time = 0.42, size = 68, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt [4]{x^2+x^6}}{x}+\sqrt [4]{2} \tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^6}}\right )-\sqrt [4]{2} \tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^6}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((1 + x^2)*(x^2 + x^6)^(1/4))/(x^2*(-1 + x^2)),x]

[Out]

(2*(x^2 + x^6)^(1/4))/x + 2^(1/4)*ArcTan[(2^(1/4)*x)/(x^2 + x^6)^(1/4)] - 2^(1/4)*ArcTanh[(2^(1/4)*x)/(x^2 + x
^6)^(1/4)]

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fricas [B]  time = 12.53, size = 262, normalized size = 3.85 \begin {gather*} \frac {4 \cdot 2^{\frac {1}{4}} x \arctan \left (\frac {4 \cdot 2^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 2^{\frac {3}{4}} {\left (2 \cdot 2^{\frac {3}{4}} \sqrt {x^{6} + x^{2}} x + 2^{\frac {1}{4}} {\left (x^{5} + 2 \, x^{3} + x\right )}\right )} + 4 \cdot 2^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}}}{2 \, {\left (x^{5} - 2 \, x^{3} + x\right )}}\right ) - 2^{\frac {1}{4}} x \log \left (-\frac {4 \, \sqrt {2} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 2^{\frac {3}{4}} {\left (x^{5} + 2 \, x^{3} + x\right )} + 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{6} + x^{2}} x + 4 \, {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}}}{x^{5} - 2 \, x^{3} + x}\right ) + 2^{\frac {1}{4}} x \log \left (-\frac {4 \, \sqrt {2} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} - 2^{\frac {3}{4}} {\left (x^{5} + 2 \, x^{3} + x\right )} - 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{6} + x^{2}} x + 4 \, {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}}}{x^{5} - 2 \, x^{3} + x}\right ) + 8 \, {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)*(x^6+x^2)^(1/4)/x^2/(x^2-1),x, algorithm="fricas")

[Out]

1/4*(4*2^(1/4)*x*arctan(1/2*(4*2^(3/4)*(x^6 + x^2)^(1/4)*x^2 + 2^(3/4)*(2*2^(3/4)*sqrt(x^6 + x^2)*x + 2^(1/4)*
(x^5 + 2*x^3 + x)) + 4*2^(1/4)*(x^6 + x^2)^(3/4))/(x^5 - 2*x^3 + x)) - 2^(1/4)*x*log(-(4*sqrt(2)*(x^6 + x^2)^(
1/4)*x^2 + 2^(3/4)*(x^5 + 2*x^3 + x) + 4*2^(1/4)*sqrt(x^6 + x^2)*x + 4*(x^6 + x^2)^(3/4))/(x^5 - 2*x^3 + x)) +
2^(1/4)*x*log(-(4*sqrt(2)*(x^6 + x^2)^(1/4)*x^2 - 2^(3/4)*(x^5 + 2*x^3 + x) - 4*2^(1/4)*sqrt(x^6 + x^2)*x + 4
*(x^6 + x^2)^(3/4))/(x^5 - 2*x^3 + x)) + 8*(x^6 + x^2)^(1/4))/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{6} + x^{2}\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^6+x^2)^(1/4)/x^2/(x^2-1),x, algorithm="giac")

[Out]

integrate((x^6 + x^2)^(1/4)*(x^2 + 1)/((x^2 - 1)*x^2), x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (x^{2}+1\right ) \left (x^{6}+x^{2}\right )^{\frac {1}{4}}}{x^{2} \left (x^{2}-1\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+1)*(x^6+x^2)^(1/4)/x^2/(x^2-1),x)

[Out]

int((x^2+1)*(x^6+x^2)^(1/4)/x^2/(x^2-1),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{6} + x^{2}\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^6+x^2)^(1/4)/x^2/(x^2-1),x, algorithm="maxima")

[Out]

integrate((x^6 + x^2)^(1/4)*(x^2 + 1)/((x^2 - 1)*x^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (x^6+x^2\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^2 + x^6)^(1/4)*(x^2 + 1))/(x^2*(x^2 - 1)),x)

[Out]

int(((x^2 + x^6)^(1/4)*(x^2 + 1))/(x^2*(x^2 - 1)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [4]{x^{2} \left (x^{4} + 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**6+x**2)**(1/4)/x**2/(x**2-1),x)

[Out]

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

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