3.23.27 \(\int \frac {(b+2 a x^2) \sqrt [4]{b x^2+a x^4}}{-b+a x^2} \, dx\)

Optimal. Leaf size=166 \[ -\frac {7 b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^2}}\right )}{2 a^{3/4}}+\frac {3 \sqrt [4]{2} b \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^2}}\right )}{a^{3/4}}+\frac {7 b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^2}}\right )}{2 a^{3/4}}-\frac {3 \sqrt [4]{2} b \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^2}}\right )}{a^{3/4}}+x \sqrt [4]{a x^4+b x^2} \]

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Rubi [A]  time = 0.42, antiderivative size = 290, normalized size of antiderivative = 1.75, number of steps used = 14, number of rules used = 10, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2056, 581, 584, 329, 331, 298, 203, 206, 466, 494} \begin {gather*} -\frac {7 b \sqrt [4]{a x^4+b x^2} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{2 a^{3/4} \sqrt {x} \sqrt [4]{a x^2+b}}+\frac {3 \sqrt [4]{2} b \sqrt [4]{a x^4+b x^2} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{a^{3/4} \sqrt {x} \sqrt [4]{a x^2+b}}+\frac {7 b \sqrt [4]{a x^4+b x^2} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{2 a^{3/4} \sqrt {x} \sqrt [4]{a x^2+b}}-\frac {3 \sqrt [4]{2} b \sqrt [4]{a x^4+b x^2} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{a^{3/4} \sqrt {x} \sqrt [4]{a x^2+b}}+x \sqrt [4]{a x^4+b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x*(b*x^2 + a*x^4)^(1/4) - (7*b*(b*x^2 + a*x^4)^(1/4)*ArcTan[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(2*a^(3/4)*S
qrt[x]*(b + a*x^2)^(1/4)) + (3*2^(1/4)*b*(b*x^2 + a*x^4)^(1/4)*ArcTan[(2^(1/4)*a^(1/4)*Sqrt[x])/(b + a*x^2)^(1
/4)])/(a^(3/4)*Sqrt[x]*(b + a*x^2)^(1/4)) + (7*b*(b*x^2 + a*x^4)^(1/4)*ArcTanh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(
1/4)])/(2*a^(3/4)*Sqrt[x]*(b + a*x^2)^(1/4)) - (3*2^(1/4)*b*(b*x^2 + a*x^4)^(1/4)*ArcTanh[(2^(1/4)*a^(1/4)*Sqr
t[x])/(b + a*x^2)^(1/4)])/(a^(3/4)*Sqrt[x]*(b + 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 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 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 494

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = Denominato
r[p]}, Dist[(k*a^(p + (m + 1)/n))/n, Subst[Int[(x^((k*(m + 1))/n - 1)*(c - (b*c - a*d)*x^k)^q)/(1 - b*x^k)^(p
+ q + (m + 1)/n + 1), x], x, x^(n/k)/(a + b*x^n)^(1/k)], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && Ration
alQ[m, p] && IntegersQ[p + (m + 1)/n, q] && LtQ[-1, p, 0]

Rule 581

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(f*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(b*g*(m + n*(p + q + 1) + 1)), x] + Dis
t[1/(b*(m + n*(p + q + 1) + 1)), Int[(g*x)^m*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*((b*e - a*f)*(m + 1) + b
*e*n*(p + q + 1)) + (d*(b*e - a*f)*(m + 1) + f*n*q*(b*c - a*d) + b*e*d*n*(p + q + 1))*x^n, x], x], x] /; FreeQ
[{a, b, c, d, e, f, g, m, p}, x] && IGtQ[n, 0] && GtQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[e + f*x^n, c + d*x^n])

Rule 584

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Sy
mbol] :> Int[ExpandIntegrand[((g*x)^m*(a + b*x^n)^p*(e + f*x^n))/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e,
f, g, m, p}, x] && IGtQ[n, 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 (b+2 a x^2\right ) \sqrt [4]{b x^2+a x^4}}{-b+a x^2} \, dx &=\frac {\sqrt [4]{b x^2+a x^4} \int \frac {\sqrt {x} \sqrt [4]{b+a x^2} \left (b+2 a x^2\right )}{-b+a x^2} \, dx}{\sqrt {x} \sqrt [4]{b+a x^2}}\\ &=x \sqrt [4]{b x^2+a x^4}+\frac {\sqrt [4]{b x^2+a x^4} \int \frac {\sqrt {x} \left (5 a b^2+7 a^2 b x^2\right )}{\left (-b+a x^2\right ) \left (b+a x^2\right )^{3/4}} \, dx}{2 a \sqrt {x} \sqrt [4]{b+a x^2}}\\ &=x \sqrt [4]{b x^2+a x^4}+\frac {\sqrt [4]{b x^2+a x^4} \int \left (\frac {7 a b \sqrt {x}}{\left (b+a x^2\right )^{3/4}}+\frac {12 a b^2 \sqrt {x}}{\left (-b+a x^2\right ) \left (b+a x^2\right )^{3/4}}\right ) \, dx}{2 a \sqrt {x} \sqrt [4]{b+a x^2}}\\ &=x \sqrt [4]{b x^2+a x^4}+\frac {\left (7 b \sqrt [4]{b x^2+a x^4}\right ) \int \frac {\sqrt {x}}{\left (b+a x^2\right )^{3/4}} \, dx}{2 \sqrt {x} \sqrt [4]{b+a x^2}}+\frac {\left (6 b^2 \sqrt [4]{b x^2+a x^4}\right ) \int \frac {\sqrt {x}}{\left (-b+a x^2\right ) \left (b+a x^2\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{b+a x^2}}\\ &=x \sqrt [4]{b x^2+a x^4}+\frac {\left (7 b \sqrt [4]{b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (b+a x^4\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{b+a x^2}}+\frac {\left (12 b^2 \sqrt [4]{b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (-b+a x^4\right ) \left (b+a x^4\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{b+a x^2}}\\ &=x \sqrt [4]{b x^2+a x^4}+\frac {\left (7 b \sqrt [4]{b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\sqrt {x} \sqrt [4]{b+a x^2}}+\frac {\left (12 b^2 \sqrt [4]{b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2}{-b+2 a b x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\sqrt {x} \sqrt [4]{b+a x^2}}\\ &=x \sqrt [4]{b x^2+a x^4}+\frac {\left (7 b \sqrt [4]{b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{2 \sqrt {a} \sqrt {x} \sqrt [4]{b+a x^2}}-\frac {\left (7 b \sqrt [4]{b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{2 \sqrt {a} \sqrt {x} \sqrt [4]{b+a x^2}}-\frac {\left (3 \sqrt {2} b \sqrt [4]{b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} \sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\sqrt {a} \sqrt {x} \sqrt [4]{b+a x^2}}+\frac {\left (3 \sqrt {2} b \sqrt [4]{b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} \sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\sqrt {a} \sqrt {x} \sqrt [4]{b+a x^2}}\\ &=x \sqrt [4]{b x^2+a x^4}-\frac {7 b \sqrt [4]{b x^2+a x^4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{2 a^{3/4} \sqrt {x} \sqrt [4]{b+a x^2}}+\frac {3 \sqrt [4]{2} b \sqrt [4]{b x^2+a x^4} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{a^{3/4} \sqrt {x} \sqrt [4]{b+a x^2}}+\frac {7 b \sqrt [4]{b x^2+a x^4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{2 a^{3/4} \sqrt {x} \sqrt [4]{b+a x^2}}-\frac {3 \sqrt [4]{2} b \sqrt [4]{b x^2+a x^4} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{a^{3/4} \sqrt {x} \sqrt [4]{b+a x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.29, size = 156, normalized size = 0.94 \begin {gather*} \frac {x^3 \left (-3 a x^2 \left (1-\frac {a^2 x^4}{b^2}\right )^{3/4} F_1\left (\frac {7}{4};\frac {3}{4},1;\frac {11}{4};-\frac {a x^2}{b},\frac {a x^2}{b}\right )-5 b \left (\frac {a x^2}{b}+1\right )^{3/4} \, _2F_1\left (\frac {3}{4},\frac {3}{4};\frac {7}{4};-\frac {2 a x^2}{b-a x^2}\right )+3 \left (a x^2+b\right ) \left (1-\frac {a x^2}{b}\right )^{3/4}\right )}{3 \left (x^2 \left (a x^2+b\right )\right )^{3/4} \left (1-\frac {a x^2}{b}\right )^{3/4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x^3*(3*(b + a*x^2)*(1 - (a*x^2)/b)^(3/4) - 3*a*x^2*(1 - (a^2*x^4)/b^2)^(3/4)*AppellF1[7/4, 3/4, 1, 11/4, -((a
*x^2)/b), (a*x^2)/b] - 5*b*(1 + (a*x^2)/b)^(3/4)*Hypergeometric2F1[3/4, 3/4, 7/4, (-2*a*x^2)/(b - a*x^2)]))/(3
*(x^2*(b + a*x^2))^(3/4)*(1 - (a*x^2)/b)^(3/4))

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IntegrateAlgebraic [A]  time = 0.60, size = 166, normalized size = 1.00 \begin {gather*} x \sqrt [4]{b x^2+a x^4}-\frac {7 b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{2 a^{3/4}}+\frac {3 \sqrt [4]{2} b \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{a^{3/4}}+\frac {7 b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{2 a^{3/4}}-\frac {3 \sqrt [4]{2} b \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{a^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x*(b*x^2 + a*x^4)^(1/4) - (7*b*ArcTan[(a^(1/4)*x)/(b*x^2 + a*x^4)^(1/4)])/(2*a^(3/4)) + (3*2^(1/4)*b*ArcTan[(2
^(1/4)*a^(1/4)*x)/(b*x^2 + a*x^4)^(1/4)])/a^(3/4) + (7*b*ArcTanh[(a^(1/4)*x)/(b*x^2 + a*x^4)^(1/4)])/(2*a^(3/4
)) - (3*2^(1/4)*b*ArcTanh[(2^(1/4)*a^(1/4)*x)/(b*x^2 + a*x^4)^(1/4)])/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((2*a*x^2+b)*(a*x^4+b*x^2)^(1/4)/(a*x^2-b),x, algorithm="fricas")

[Out]

Timed out

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giac [B]  time = 0.28, size = 394, normalized size = 2.37 \begin {gather*} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} x^{2} - \frac {3 \cdot 2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} b \arctan \left (\frac {2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{2 \, a} - \frac {3 \cdot 2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} b \arctan \left (-\frac {2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{2 \, a} + \frac {3 \cdot 2^{\frac {3}{4}} b \log \left (2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {2} \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right )}{4 \, \left (-a\right )^{\frac {3}{4}}} + \frac {3 \cdot 2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} b \log \left (-2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {2} \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right )}{4 \, a} + \frac {7 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{4 \, a} + \frac {7 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{4 \, a} + \frac {7 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right )}{8 \, a} + \frac {7 \, \sqrt {2} b \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right )}{8 \, \left (-a\right )^{\frac {3}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a + b/x^2)^(1/4)*x^2 - 3/2*2^(3/4)*(-a)^(1/4)*b*arctan(1/2*2^(1/4)*(2^(3/4)*(-a)^(1/4) + 2*(a + b/x^2)^(1/4))
/(-a)^(1/4))/a - 3/2*2^(3/4)*(-a)^(1/4)*b*arctan(-1/2*2^(1/4)*(2^(3/4)*(-a)^(1/4) - 2*(a + b/x^2)^(1/4))/(-a)^
(1/4))/a + 3/4*2^(3/4)*b*log(2^(3/4)*(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(2)*sqrt(-a) + sqrt(a + b/x^2))/(-a)^(
3/4) + 3/4*2^(3/4)*(-a)^(1/4)*b*log(-2^(3/4)*(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(2)*sqrt(-a) + sqrt(a + b/x^2)
)/a + 7/4*sqrt(2)*(-a)^(1/4)*b*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x^2)^(1/4))/(-a)^(1/4))/a + 7
/4*sqrt(2)*(-a)^(1/4)*b*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x^2)^(1/4))/(-a)^(1/4))/a + 7/8*sqr
t(2)*(-a)^(1/4)*b*log(sqrt(2)*(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt(a + b/x^2))/a + 7/8*sqrt(2)*b*log
(-sqrt(2)*(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt(a + b/x^2))/(-a)^(3/4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(-((b + 2*a*x^2)*(a*x^4 + b*x^2)^(1/4))/(b - a*x^2), 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 (a x^{2} + b\right )} \left (2 a x^{2} + b\right )}{a x^{2} - b}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*a*x**2+b)*(a*x**4+b*x**2)**(1/4)/(a*x**2-b),x)

[Out]

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

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