3.28.20 \(\int \frac {\sqrt [4]{\frac {1+4 x^2+a x^2+6 x^4+4 a x^4+4 x^6+6 a x^6+x^8+4 a x^8+a x^{10}}{x^2}}}{x} \, dx\)

Optimal. Leaf size=249 \[ \frac {\sqrt [4]{\frac {a x^{10}+4 a x^8+6 a x^6+4 a x^4+a x^2+x^8+4 x^6+6 x^4+4 x^2+1}{x^2}} \left (\frac {\sqrt {x} \tan ^{-1}\left (\frac {\sqrt [4]{a x^2+1}}{\sqrt [4]{a} \sqrt {x}}\right )}{4 a^{3/4}}+\frac {\sqrt {x} \tanh ^{-1}\left (\frac {\sqrt [4]{a x^2+1}}{\sqrt [4]{a} \sqrt {x}}\right )}{4 a^{3/4}}+\frac {1}{2} x^2 \sqrt [4]{a x^2+1}-2 \sqrt [4]{a x^2+1}+\sqrt [4]{a} \sqrt {x} \tan ^{-1}\left (\frac {\sqrt [4]{a x^2+1}}{\sqrt [4]{a} \sqrt {x}}\right )+\sqrt [4]{a} \sqrt {x} \tanh ^{-1}\left (\frac {\sqrt [4]{a x^2+1}}{\sqrt [4]{a} \sqrt {x}}\right )\right )}{\left (x^2+1\right ) \sqrt [4]{a x^2+1}} \]

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Rubi [A]  time = 0.41, antiderivative size = 244, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 9, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6688, 6718, 453, 279, 329, 331, 298, 203, 206} \begin {gather*} -\frac {(4 a+1) \sqrt {x} \sqrt [4]{\frac {\left (x^2+1\right )^4 \left (a x^2+1\right )}{x^2}} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+1}}\right )}{4 a^{3/4} \left (x^2+1\right ) \sqrt [4]{a x^2+1}}+\frac {(4 a+1) \sqrt {x} \sqrt [4]{\frac {\left (x^2+1\right )^4 \left (a x^2+1\right )}{x^2}} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+1}}\right )}{4 a^{3/4} \left (x^2+1\right ) \sqrt [4]{a x^2+1}}+\frac {(4 a+1) x^2 \sqrt [4]{\frac {\left (x^2+1\right )^4 \left (a x^2+1\right )}{x^2}}}{2 \left (x^2+1\right )}-\frac {2 \left (a x^2+1\right ) \sqrt [4]{\frac {\left (x^2+1\right )^4 \left (a x^2+1\right )}{x^2}}}{x^2+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 + 4*x^2 + a*x^2 + 6*x^4 + 4*a*x^4 + 4*x^6 + 6*a*x^6 + x^8 + 4*a*x^8 + a*x^10)/x^2)^(1/4)/x,x]

[Out]

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

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 279

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

Rule 298

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

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 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6718

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.)*(z_)^(q_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m*w^n*z^q)^FracP
art[p])/(v^(m*FracPart[p])*w^(n*FracPart[p])*z^(q*FracPart[p])), Int[u*v^(m*p)*w^(n*p)*z^(p*q), x], x] /; Free
Q[{a, m, n, p, q}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !FreeQ[w, x] &&  !FreeQ[z, x]

Rubi steps

\begin {align*} \int \frac {\sqrt [4]{\frac {1+4 x^2+a x^2+6 x^4+4 a x^4+4 x^6+6 a x^6+x^8+4 a x^8+a x^{10}}{x^2}}}{x} \, dx &=\int \frac {\sqrt [4]{\frac {\left (1+x^2\right )^4 \left (1+a x^2\right )}{x^2}}}{x} \, dx\\ &=\frac {\left (\sqrt {x} \sqrt [4]{\frac {\left (1+x^2\right )^4 \left (1+a x^2\right )}{x^2}}\right ) \int \frac {\left (1+x^2\right ) \sqrt [4]{1+a x^2}}{x^{3/2}} \, dx}{\left (1+x^2\right ) \sqrt [4]{1+a x^2}}\\ &=-\frac {2 \left (1+a x^2\right ) \sqrt [4]{\frac {\left (1+x^2\right )^4 \left (1+a x^2\right )}{x^2}}}{1+x^2}-\frac {\left ((-1-4 a) \sqrt {x} \sqrt [4]{\frac {\left (1+x^2\right )^4 \left (1+a x^2\right )}{x^2}}\right ) \int \sqrt {x} \sqrt [4]{1+a x^2} \, dx}{\left (1+x^2\right ) \sqrt [4]{1+a x^2}}\\ &=\frac {(1+4 a) x^2 \sqrt [4]{\frac {\left (1+x^2\right )^4 \left (1+a x^2\right )}{x^2}}}{2 \left (1+x^2\right )}-\frac {2 \left (1+a x^2\right ) \sqrt [4]{\frac {\left (1+x^2\right )^4 \left (1+a x^2\right )}{x^2}}}{1+x^2}-\frac {\left ((-1-4 a) \sqrt {x} \sqrt [4]{\frac {\left (1+x^2\right )^4 \left (1+a x^2\right )}{x^2}}\right ) \int \frac {\sqrt {x}}{\left (1+a x^2\right )^{3/4}} \, dx}{4 \left (1+x^2\right ) \sqrt [4]{1+a x^2}}\\ &=\frac {(1+4 a) x^2 \sqrt [4]{\frac {\left (1+x^2\right )^4 \left (1+a x^2\right )}{x^2}}}{2 \left (1+x^2\right )}-\frac {2 \left (1+a x^2\right ) \sqrt [4]{\frac {\left (1+x^2\right )^4 \left (1+a x^2\right )}{x^2}}}{1+x^2}-\frac {\left ((-1-4 a) \sqrt {x} \sqrt [4]{\frac {\left (1+x^2\right )^4 \left (1+a x^2\right )}{x^2}}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (1+a x^4\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{2 \left (1+x^2\right ) \sqrt [4]{1+a x^2}}\\ &=\frac {(1+4 a) x^2 \sqrt [4]{\frac {\left (1+x^2\right )^4 \left (1+a x^2\right )}{x^2}}}{2 \left (1+x^2\right )}-\frac {2 \left (1+a x^2\right ) \sqrt [4]{\frac {\left (1+x^2\right )^4 \left (1+a x^2\right )}{x^2}}}{1+x^2}-\frac {\left ((-1-4 a) \sqrt {x} \sqrt [4]{\frac {\left (1+x^2\right )^4 \left (1+a x^2\right )}{x^2}}\right ) \operatorname {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+a x^2}}\right )}{2 \left (1+x^2\right ) \sqrt [4]{1+a x^2}}\\ &=\frac {(1+4 a) x^2 \sqrt [4]{\frac {\left (1+x^2\right )^4 \left (1+a x^2\right )}{x^2}}}{2 \left (1+x^2\right )}-\frac {2 \left (1+a x^2\right ) \sqrt [4]{\frac {\left (1+x^2\right )^4 \left (1+a x^2\right )}{x^2}}}{1+x^2}-\frac {\left ((-1-4 a) \sqrt {x} \sqrt [4]{\frac {\left (1+x^2\right )^4 \left (1+a x^2\right )}{x^2}}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+a x^2}}\right )}{4 \sqrt {a} \left (1+x^2\right ) \sqrt [4]{1+a x^2}}+\frac {\left ((-1-4 a) \sqrt {x} \sqrt [4]{\frac {\left (1+x^2\right )^4 \left (1+a x^2\right )}{x^2}}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+a x^2}}\right )}{4 \sqrt {a} \left (1+x^2\right ) \sqrt [4]{1+a x^2}}\\ &=\frac {(1+4 a) x^2 \sqrt [4]{\frac {\left (1+x^2\right )^4 \left (1+a x^2\right )}{x^2}}}{2 \left (1+x^2\right )}-\frac {2 \left (1+a x^2\right ) \sqrt [4]{\frac {\left (1+x^2\right )^4 \left (1+a x^2\right )}{x^2}}}{1+x^2}-\frac {(1+4 a) \sqrt {x} \sqrt [4]{\frac {\left (1+x^2\right )^4 \left (1+a x^2\right )}{x^2}} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{1+a x^2}}\right )}{4 a^{3/4} \left (1+x^2\right ) \sqrt [4]{1+a x^2}}+\frac {(1+4 a) \sqrt {x} \sqrt [4]{\frac {\left (1+x^2\right )^4 \left (1+a x^2\right )}{x^2}} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{1+a x^2}}\right )}{4 a^{3/4} \left (1+x^2\right ) \sqrt [4]{1+a x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 84, normalized size = 0.34 \begin {gather*} -\frac {2 \sqrt [4]{\frac {\left (x^2+1\right )^4 \left (a x^2+1\right )}{x^2}} \left (3 \left (a x^2+1\right )^{5/4}-(4 a+1) x^2 \, _2F_1\left (-\frac {1}{4},\frac {3}{4};\frac {7}{4};-a x^2\right )\right )}{3 \left (x^2+1\right ) \sqrt [4]{a x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 + 4*x^2 + a*x^2 + 6*x^4 + 4*a*x^4 + 4*x^6 + 6*a*x^6 + x^8 + 4*a*x^8 + a*x^10)/x^2)^(1/4)/x,x]

[Out]

(-2*(((1 + x^2)^4*(1 + a*x^2))/x^2)^(1/4)*(3*(1 + a*x^2)^(5/4) - (1 + 4*a)*x^2*Hypergeometric2F1[-1/4, 3/4, 7/
4, -(a*x^2)]))/(3*(1 + x^2)*(1 + a*x^2)^(1/4))

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IntegrateAlgebraic [A]  time = 0.29, size = 237, normalized size = 0.95 \begin {gather*} \frac {\left (-4+x^2\right ) \sqrt [4]{\frac {1+4 x^2+a x^2+6 x^4+4 a x^4+4 x^6+6 a x^6+x^8+4 a x^8+a x^{10}}{x^2}}}{2 \left (1+x^2\right )}+\frac {(-1-4 a) \tan ^{-1}\left (\frac {\sqrt [4]{a} \left (1+x^2\right )}{\sqrt [4]{\frac {1+4 x^2+a x^2+6 x^4+4 a x^4+4 x^6+6 a x^6+x^8+4 a x^8+a x^{10}}{x^2}}}\right )}{4 a^{3/4}}+\frac {(1+4 a) \tanh ^{-1}\left (\frac {\sqrt [4]{a} \left (1+x^2\right )}{\sqrt [4]{\frac {1+4 x^2+a x^2+6 x^4+4 a x^4+4 x^6+6 a x^6+x^8+4 a x^8+a x^{10}}{x^2}}}\right )}{4 a^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((1 + 4*x^2 + a*x^2 + 6*x^4 + 4*a*x^4 + 4*x^6 + 6*a*x^6 + x^8 + 4*a*x^8 + a*x^10)/x^2)^(1/4
)/x,x]

[Out]

((-4 + x^2)*((1 + 4*x^2 + a*x^2 + 6*x^4 + 4*a*x^4 + 4*x^6 + 6*a*x^6 + x^8 + 4*a*x^8 + a*x^10)/x^2)^(1/4))/(2*(
1 + x^2)) + ((-1 - 4*a)*ArcTan[(a^(1/4)*(1 + x^2))/((1 + 4*x^2 + a*x^2 + 6*x^4 + 4*a*x^4 + 4*x^6 + 6*a*x^6 + x
^8 + 4*a*x^8 + a*x^10)/x^2)^(1/4)])/(4*a^(3/4)) + ((1 + 4*a)*ArcTanh[(a^(1/4)*(1 + x^2))/((1 + 4*x^2 + a*x^2 +
6*x^4 + 4*a*x^4 + 4*x^6 + 6*a*x^6 + x^8 + 4*a*x^8 + a*x^10)/x^2)^(1/4)])/(4*a^(3/4))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x^10+4*a*x^8+x^8+6*a*x^6+4*x^6+4*a*x^4+6*x^4+a*x^2+4*x^2+1)/x^2)^(1/4)/x,x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\frac {a x^{10} + 4 \, a x^{8} + x^{8} + 6 \, a x^{6} + 4 \, x^{6} + 4 \, a x^{4} + 6 \, x^{4} + a x^{2} + 4 \, x^{2} + 1}{x^{2}}\right )^{\frac {1}{4}}}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x^10+4*a*x^8+x^8+6*a*x^6+4*x^6+4*a*x^4+6*x^4+a*x^2+4*x^2+1)/x^2)^(1/4)/x,x, algorithm="giac")

[Out]

integrate(((a*x^10 + 4*a*x^8 + x^8 + 6*a*x^6 + 4*x^6 + 4*a*x^4 + 6*x^4 + a*x^2 + 4*x^2 + 1)/x^2)^(1/4)/x, x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (\frac {a \,x^{10}+4 a \,x^{8}+x^{8}+6 a \,x^{6}+4 x^{6}+4 a \,x^{4}+6 x^{4}+a \,x^{2}+4 x^{2}+1}{x^{2}}\right )^{\frac {1}{4}}}{x}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x^10+4*a*x^8+x^8+6*a*x^6+4*x^6+4*a*x^4+6*x^4+a*x^2+4*x^2+1)/x^2)^(1/4)/x,x)

[Out]

int(((a*x^10+4*a*x^8+x^8+6*a*x^6+4*x^6+4*a*x^4+6*x^4+a*x^2+4*x^2+1)/x^2)^(1/4)/x,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\frac {a x^{10} + 4 \, a x^{8} + x^{8} + 6 \, a x^{6} + 4 \, x^{6} + 4 \, a x^{4} + 6 \, x^{4} + a x^{2} + 4 \, x^{2} + 1}{x^{2}}\right )^{\frac {1}{4}}}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x^10+4*a*x^8+x^8+6*a*x^6+4*x^6+4*a*x^4+6*x^4+a*x^2+4*x^2+1)/x^2)^(1/4)/x,x, algorithm="maxima")

[Out]

integrate(((a*x^10 + 4*a*x^8 + x^8 + 6*a*x^6 + 4*x^6 + 4*a*x^4 + 6*x^4 + a*x^2 + 4*x^2 + 1)/x^2)^(1/4)/x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (\frac {a\,x^2+4\,a\,x^4+6\,a\,x^6+4\,a\,x^8+a\,x^{10}+4\,x^2+6\,x^4+4\,x^6+x^8+1}{x^2}\right )}^{1/4}}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x^2 + 4*a*x^4 + 6*a*x^6 + 4*a*x^8 + a*x^10 + 4*x^2 + 6*x^4 + 4*x^6 + x^8 + 1)/x^2)^(1/4)/x,x)

[Out]

int(((a*x^2 + 4*a*x^4 + 6*a*x^6 + 4*a*x^8 + a*x^10 + 4*x^2 + 6*x^4 + 4*x^6 + x^8 + 1)/x^2)^(1/4)/x, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x**10+4*a*x**8+x**8+6*a*x**6+4*x**6+4*a*x**4+6*x**4+a*x**2+4*x**2+1)/x**2)**(1/4)/x,x)

[Out]

Timed out

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