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

Optimal. Leaf size=84 \[ \frac {5 b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^2}}\right )}{2 a^{3/4}}-\frac {5 b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^2}}\right )}{2 a^{3/4}}+x \sqrt [4]{a x^4+b x^2} \]

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Rubi [A]  time = 0.22, antiderivative size = 146, normalized size of antiderivative = 1.74, number of steps used = 7, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2056, 459, 329, 331, 298, 203, 206} \begin {gather*} \frac {5 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 {5 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}}+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) + (5*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)) - (5*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))

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 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 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} \left (-b+2 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 (5 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}}\\ &=x \sqrt [4]{b x^2+a x^4}-\frac {\left (5 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}}\\ &=x \sqrt [4]{b x^2+a x^4}-\frac {\left (5 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}}\\ &=x \sqrt [4]{b x^2+a x^4}-\frac {\left (5 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 (5 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}}\\ &=x \sqrt [4]{b x^2+a x^4}+\frac {5 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 {5 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}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 123, normalized size = 1.46 \begin {gather*} \frac {x^{3/2} \left (2 a^{3/4} x^{3/2} \left (a x^2+b\right )+5 b \left (a x^2+b\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )-5 b \left (a x^2+b\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )\right )}{2 a^{3/4} \left (x^2 \left (a x^2+b\right )\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.31, size = 84, normalized size = 1.00 \begin {gather*} x \sqrt [4]{b x^2+a x^4}+\frac {5 b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{2 a^{3/4}}-\frac {5 b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{2 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) + (5*b*ArcTan[(a^(1/4)*x)/(b*x^2 + a*x^4)^(1/4)])/(2*a^(3/4)) - (5*b*ArcTanh[(a^(1/4)*
x)/(b*x^2 + a*x^4)^(1/4)])/(2*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 = 0.19, size = 209, normalized size = 2.49 \begin {gather*} \frac {8 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} b x^{2} + \frac {10 \, \sqrt {2} b^{2} \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 )}{\left (-a\right )^{\frac {3}{4}}} + \frac {10 \, \sqrt {2} b^{2} \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 )}{\left (-a\right )^{\frac {3}{4}}} + \frac {5 \, \sqrt {2} b^{2} \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 )}{\left (-a\right )^{\frac {3}{4}}} + \frac {5 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \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 )}{a}}{8 \, b} \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]

1/8*(8*(a + b/x^2)^(1/4)*b*x^2 + 10*sqrt(2)*b^2*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x^2)^(1/4))/
(-a)^(1/4))/(-a)^(3/4) + 10*sqrt(2)*b^2*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x^2)^(1/4))/(-a)^(1
/4))/(-a)^(3/4) + 5*sqrt(2)*b^2*log(sqrt(2)*(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt(a + b/x^2))/(-a)^(3
/4) + 5*sqrt(2)*(-a)^(1/4)*b^2*log(-sqrt(2)*(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt(a + b/x^2))/a)/b

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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 (b-2\,a\,x^2\right )\,{\left (a\,x^4+b\,x^2\right )}^{1/4}}{a\,x^2+b} \,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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