3.24.97 \(\int \frac {b+2 a x}{(-b+a x) (2 b+a x) \sqrt [4]{-1+b x+a x^2}} \, dx\)

Optimal. Leaf size=193 \[ \frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{a-2 b^2} \sqrt [4]{a x^2+b x-1}}{\sqrt {a-2 b^2}-\sqrt {a} \sqrt {a x^2+b x-1}}\right )}{a^{3/4} \sqrt [4]{a-2 b^2}}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\frac {\sqrt [4]{a} \sqrt {a x^2+b x-1}}{\sqrt {2} \sqrt [4]{a-2 b^2}}+\frac {\sqrt [4]{a-2 b^2}}{\sqrt {2} \sqrt [4]{a}}}{\sqrt [4]{a x^2+b x-1}}\right )}{a^{3/4} \sqrt [4]{a-2 b^2}} \]

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Rubi [A]  time = 2.28, antiderivative size = 245, normalized size of antiderivative = 1.27, number of steps used = 30, number of rules used = 13, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6742, 749, 748, 746, 399, 490, 1213, 537, 444, 63, 298, 205, 208} \begin {gather*} \frac {2 \sqrt [4]{4 a+b^2} \sqrt [4]{\frac {a \left (-a x^2-b x+1\right )}{4 a+b^2}} \tan ^{-1}\left (\frac {\sqrt [4]{4 a+b^2} \sqrt [4]{1-\frac {(2 a x+b)^2}{4 a+b^2}}}{\sqrt {2} \sqrt [4]{a-2 b^2}}\right )}{a \sqrt [4]{a-2 b^2} \sqrt [4]{a x^2+b x-1}}-\frac {2 \sqrt [4]{4 a+b^2} \sqrt [4]{\frac {a \left (-a x^2-b x+1\right )}{4 a+b^2}} \tanh ^{-1}\left (\frac {\sqrt [4]{4 a+b^2} \sqrt [4]{1-\frac {(2 a x+b)^2}{4 a+b^2}}}{\sqrt {2} \sqrt [4]{a-2 b^2}}\right )}{a \sqrt [4]{a-2 b^2} \sqrt [4]{a x^2+b x-1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

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 399

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Dist[(2*Sqrt[-((b*x^2)/a)])/x, Subst[I
nt[x^2/(Sqrt[1 - x^4/a]*(b*c - a*d + d*x^4)), x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0]

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 490

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(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)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), In
t[1/((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b,
c, d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-(f/e), -(d/c)
])

Rule 746

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(1/4)), x_Symbol] :> Dist[d, Int[1/((d^2 - e^2*x^2)*(a + c*x^
2)^(1/4)), x], x] - Dist[e, Int[x/((d^2 - e^2*x^2)*(a + c*x^2)^(1/4)), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ
[c*d^2 + a*e^2, 0]

Rule 748

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)/((d_.) + (e_.)*(x_)), x_Symbol] :> Dist[1/((-4*c)/(b^2 - 4*a*c))^
p, Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p/Simp[2*c*d - b*e + e*x, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b
, c, d, e, p}, x] && GtQ[4*a - b^2/c, 0] && IntegerQ[4*p]

Rule 749

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)/((d_.) + (e_.)*(x_)), x_Symbol] :> Dist[(a + b*x + c*x^2)^p/(-((c
*(a + b*x + c*x^2))/(b^2 - 4*a*c)))^p, Int[(-((a*c)/(b^2 - 4*a*c)) - (b*c*x)/(b^2 - 4*a*c) - (c^2*x^2)/(b^2 -
4*a*c))^p/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e, p}, x] &&  !GtQ[4*a - b^2/c, 0] && IntegerQ[4*p]

Rule 1213

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[Sqrt[-c],
Int[1/((d + e*x^2)*Sqrt[q + c*x^2]*Sqrt[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && GtQ[a, 0] && LtQ[c,
0]

Rule 6742

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

Rubi steps

\begin {align*} \int \frac {b+2 a x}{(-b+a x) (2 b+a x) \sqrt [4]{-1+b x+a x^2}} \, dx &=\int \left (\frac {1}{(-b+a x) \sqrt [4]{-1+b x+a x^2}}+\frac {1}{(2 b+a x) \sqrt [4]{-1+b x+a x^2}}\right ) \, dx\\ &=\int \frac {1}{(-b+a x) \sqrt [4]{-1+b x+a x^2}} \, dx+\int \frac {1}{(2 b+a x) \sqrt [4]{-1+b x+a x^2}} \, dx\\ &=\frac {\sqrt [4]{-\frac {a \left (-1+b x+a x^2\right )}{4 a+b^2}} \int \frac {1}{(-b+a x) \sqrt [4]{\frac {a}{4 a+b^2}-\frac {a b x}{4 a+b^2}-\frac {a^2 x^2}{4 a+b^2}}} \, dx}{\sqrt [4]{-1+b x+a x^2}}+\frac {\sqrt [4]{-\frac {a \left (-1+b x+a x^2\right )}{4 a+b^2}} \int \frac {1}{(2 b+a x) \sqrt [4]{\frac {a}{4 a+b^2}-\frac {a b x}{4 a+b^2}-\frac {a^2 x^2}{4 a+b^2}}} \, dx}{\sqrt [4]{-1+b x+a x^2}}\\ &=\frac {\left (\sqrt {2} \sqrt [4]{-\frac {a \left (-1+b x+a x^2\right )}{4 a+b^2}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (-\frac {3 a^2 b}{4 a+b^2}+a x\right ) \sqrt [4]{1-\frac {\left (4 a+b^2\right ) x^2}{a^2}}} \, dx,x,-\frac {a b}{4 a+b^2}-\frac {2 a^2 x}{4 a+b^2}\right )}{\sqrt [4]{-1+b x+a x^2}}+\frac {\left (\sqrt {2} \sqrt [4]{-\frac {a \left (-1+b x+a x^2\right )}{4 a+b^2}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\frac {3 a^2 b}{4 a+b^2}+a x\right ) \sqrt [4]{1-\frac {\left (4 a+b^2\right ) x^2}{a^2}}} \, dx,x,-\frac {a b}{4 a+b^2}-\frac {2 a^2 x}{4 a+b^2}\right )}{\sqrt [4]{-1+b x+a x^2}}\\ &=-2 \frac {\left (\sqrt {2} a \sqrt [4]{-\frac {a \left (-1+b x+a x^2\right )}{4 a+b^2}}\right ) \operatorname {Subst}\left (\int \frac {x}{\left (\frac {9 a^4 b^2}{\left (4 a+b^2\right )^2}-a^2 x^2\right ) \sqrt [4]{1-\frac {\left (4 a+b^2\right ) x^2}{a^2}}} \, dx,x,-\frac {a b}{4 a+b^2}-\frac {2 a^2 x}{4 a+b^2}\right )}{\sqrt [4]{-1+b x+a x^2}}\\ &=-2 \frac {\left (a \sqrt [4]{-\frac {a \left (-1+b x+a x^2\right )}{4 a+b^2}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\frac {9 a^4 b^2}{\left (4 a+b^2\right )^2}-a^2 x\right ) \sqrt [4]{1-\frac {\left (4 a+b^2\right ) x}{a^2}}} \, dx,x,\left (-\frac {a b}{4 a+b^2}-\frac {2 a^2 x}{4 a+b^2}\right )^2\right )}{\sqrt {2} \sqrt [4]{-1+b x+a x^2}}\\ &=2 \frac {\left (2 \sqrt {2} a^3 \sqrt [4]{-\frac {a \left (-1+b x+a x^2\right )}{4 a+b^2}}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\frac {9 a^4 b^2}{\left (4 a+b^2\right )^2}-\frac {a^4}{4 a+b^2}+\frac {a^4 x^4}{4 a+b^2}} \, dx,x,\sqrt [4]{1-\frac {(b+2 a x)^2}{4 a+b^2}}\right )}{\left (4 a+b^2\right ) \sqrt [4]{-1+b x+a x^2}}\\ &=2 \left (-\frac {\left (\sqrt {2} \sqrt {4 a+b^2} \sqrt [4]{-\frac {a \left (-1+b x+a x^2\right )}{4 a+b^2}}\right ) \operatorname {Subst}\left (\int \frac {1}{2 \sqrt {a-2 b^2}-\sqrt {4 a+b^2} x^2} \, dx,x,\sqrt [4]{1-\frac {(b+2 a x)^2}{4 a+b^2}}\right )}{a \sqrt [4]{-1+b x+a x^2}}+\frac {\left (\sqrt {2} \sqrt {4 a+b^2} \sqrt [4]{-\frac {a \left (-1+b x+a x^2\right )}{4 a+b^2}}\right ) \operatorname {Subst}\left (\int \frac {1}{2 \sqrt {a-2 b^2}+\sqrt {4 a+b^2} x^2} \, dx,x,\sqrt [4]{1-\frac {(b+2 a x)^2}{4 a+b^2}}\right )}{a \sqrt [4]{-1+b x+a x^2}}\right )\\ &=2 \left (\frac {\sqrt [4]{4 a+b^2} \sqrt [4]{\frac {a \left (1-b x-a x^2\right )}{4 a+b^2}} \tan ^{-1}\left (\frac {\sqrt [4]{4 a+b^2} \sqrt [4]{1-\frac {(b+2 a x)^2}{4 a+b^2}}}{\sqrt {2} \sqrt [4]{a-2 b^2}}\right )}{a \sqrt [4]{a-2 b^2} \sqrt [4]{-1+b x+a x^2}}-\frac {\sqrt [4]{4 a+b^2} \sqrt [4]{\frac {a \left (1-b x-a x^2\right )}{4 a+b^2}} \tanh ^{-1}\left (\frac {\sqrt [4]{4 a+b^2} \sqrt [4]{1-\frac {(b+2 a x)^2}{4 a+b^2}}}{\sqrt {2} \sqrt [4]{a-2 b^2}}\right )}{a \sqrt [4]{a-2 b^2} \sqrt [4]{-1+b x+a x^2}}\right )\\ \end {align*}

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Mathematica [F]  time = 0.40, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {b+2 a x}{(-b+a x) (2 b+a x) \sqrt [4]{-1+b x+a x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.58, size = 193, normalized size = 1.00 \begin {gather*} \frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{a-2 b^2} \sqrt [4]{-1+b x+a x^2}}{\sqrt {a-2 b^2}-\sqrt {a} \sqrt {-1+b x+a x^2}}\right )}{a^{3/4} \sqrt [4]{a-2 b^2}}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\frac {\sqrt [4]{a-2 b^2}}{\sqrt {2} \sqrt [4]{a}}+\frac {\sqrt [4]{a} \sqrt {-1+b x+a x^2}}{\sqrt {2} \sqrt [4]{a-2 b^2}}}{\sqrt [4]{-1+b x+a x^2}}\right )}{a^{3/4} \sqrt [4]{a-2 b^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.71, size = 240, normalized size = 1.24 \begin {gather*} -\frac {4 \, \arctan \left (\frac {a \sqrt {\frac {2 \, a b^{2} - a^{2}}{\sqrt {2 \, a^{3} b^{2} - a^{4}}} + \sqrt {a x^{2} + b x - 1}}}{{\left (2 \, a^{3} b^{2} - a^{4}\right )}^{\frac {1}{4}}} - \frac {{\left (a x^{2} + b x - 1\right )}^{\frac {1}{4}} a}{{\left (2 \, a^{3} b^{2} - a^{4}\right )}^{\frac {1}{4}}}\right )}{{\left (2 \, a^{3} b^{2} - a^{4}\right )}^{\frac {1}{4}}} - \frac {\log \left (\frac {2 \, a^{2} b^{2} - a^{3}}{{\left (2 \, a^{3} b^{2} - a^{4}\right )}^{\frac {3}{4}}} + {\left (a x^{2} + b x - 1\right )}^{\frac {1}{4}}\right )}{{\left (2 \, a^{3} b^{2} - a^{4}\right )}^{\frac {1}{4}}} + \frac {\log \left (-\frac {2 \, a^{2} b^{2} - a^{3}}{{\left (2 \, a^{3} b^{2} - a^{4}\right )}^{\frac {3}{4}}} + {\left (a x^{2} + b x - 1\right )}^{\frac {1}{4}}\right )}{{\left (2 \, a^{3} b^{2} - a^{4}\right )}^{\frac {1}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.15, size = 0, normalized size = 0.00 \[\int \frac {2 a x +b}{\left (a x -b \right ) \left (a x +2 b \right ) \left (a \,x^{2}+b x -1\right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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