3.18.30 \(\int \frac {x}{(-b+a x^2) \sqrt {b x+a x^3}} \, dx\)

Optimal. Leaf size=117 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {a x^3+b x}}{a x^2+b}\right )}{2 \sqrt {2} a^{3/4} b^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {a x^3+b x}}{a x^2+b}\right )}{2 \sqrt {2} a^{3/4} b^{3/4}} \]

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Rubi [A]  time = 0.46, antiderivative size = 163, normalized size of antiderivative = 1.39, number of steps used = 11, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2042, 466, 490, 1211, 220, 1699, 205, 208} \begin {gather*} \frac {\sqrt {x} \sqrt {a x^2+b} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a x^2+b}}\right )}{2 \sqrt {2} a^{3/4} b^{3/4} \sqrt {a x^3+b x}}-\frac {\sqrt {x} \sqrt {a x^2+b} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a x^2+b}}\right )}{2 \sqrt {2} a^{3/4} b^{3/4} \sqrt {a x^3+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/((-b + a*x^2)*Sqrt[b*x + a*x^3]),x]

[Out]

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

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 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

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 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 1211

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

Rule 1699

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

Rule 2042

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.))^(q_.), x_Symbol]
:> Dist[(e^IntPart[m]*(e*x)^FracPart[m]*(a*x^j + b*x^(j + n))^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a
+ b*x^n)^FracPart[p]), Int[x^(m + j*p)*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, j, m, n,
p, q}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] &&  !(EqQ[n, 1] && EqQ[j, 1])

Rubi steps

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

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Mathematica [C]  time = 0.04, size = 66, normalized size = 0.56 \begin {gather*} -\frac {2 x^2 \sqrt {\frac {a x^2+b}{b}} F_1\left (\frac {3}{4};\frac {1}{2},1;\frac {7}{4};-\frac {a x^2}{b},\frac {a x^2}{b}\right )}{3 b \sqrt {x \left (a x^2+b\right )}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[x/((-b + a*x^2)*Sqrt[b*x + a*x^3]),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.37, size = 117, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {b x+a x^3}}{b+a x^2}\right )}{2 \sqrt {2} a^{3/4} b^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {b x+a x^3}}{b+a x^2}\right )}{2 \sqrt {2} a^{3/4} b^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x/((-b + a*x^2)*Sqrt[b*x + a*x^3]),x]

[Out]

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

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fricas [B]  time = 0.86, size = 344, normalized size = 2.94 \begin {gather*} \frac {1}{2} \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \left (\frac {1}{a^{3} b^{3}}\right )^{\frac {1}{4}} \arctan \left (\frac {2 \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \sqrt {a x^{3} + b x} a b \left (\frac {1}{a^{3} b^{3}}\right )^{\frac {1}{4}}}{a x^{2} + b}\right ) - \frac {1}{8} \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \left (\frac {1}{a^{3} b^{3}}\right )^{\frac {1}{4}} \log \left (\frac {a^{2} x^{4} + 6 \, a b x^{2} + b^{2} + 4 \, {\left (4 \, \left (\frac {1}{4}\right )^{\frac {3}{4}} a^{3} b^{3} x \left (\frac {1}{a^{3} b^{3}}\right )^{\frac {3}{4}} + \left (\frac {1}{4}\right )^{\frac {1}{4}} {\left (a^{2} b x^{2} + a b^{2}\right )} \left (\frac {1}{a^{3} b^{3}}\right )^{\frac {1}{4}}\right )} \sqrt {a x^{3} + b x} + 4 \, {\left (a^{3} b^{2} x^{3} + a^{2} b^{3} x\right )} \sqrt {\frac {1}{a^{3} b^{3}}}}{a^{2} x^{4} - 2 \, a b x^{2} + b^{2}}\right ) + \frac {1}{8} \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \left (\frac {1}{a^{3} b^{3}}\right )^{\frac {1}{4}} \log \left (\frac {a^{2} x^{4} + 6 \, a b x^{2} + b^{2} - 4 \, {\left (4 \, \left (\frac {1}{4}\right )^{\frac {3}{4}} a^{3} b^{3} x \left (\frac {1}{a^{3} b^{3}}\right )^{\frac {3}{4}} + \left (\frac {1}{4}\right )^{\frac {1}{4}} {\left (a^{2} b x^{2} + a b^{2}\right )} \left (\frac {1}{a^{3} b^{3}}\right )^{\frac {1}{4}}\right )} \sqrt {a x^{3} + b x} + 4 \, {\left (a^{3} b^{2} x^{3} + a^{2} b^{3} x\right )} \sqrt {\frac {1}{a^{3} b^{3}}}}{a^{2} x^{4} - 2 \, a b x^{2} + b^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a*x^2-b)/(a*x^3+b*x)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\sqrt {a x^{3} + b x} {\left (a x^{2} - b\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a*x^2-b)/(a*x^3+b*x)^(1/2),x, algorithm="giac")

[Out]

integrate(x/(sqrt(a*x^3 + b*x)*(a*x^2 - b)), x)

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maple [C]  time = 0.11, size = 296, normalized size = 2.53

method	result	size
default	\(\frac {\sqrt {-a b}\, \sqrt {\frac {x a}{\sqrt {-a b}}+1}\, \sqrt {-\frac {2 x a}{\sqrt {-a b}}+2}\, \sqrt {-\frac {x a}{\sqrt {-a b}}}\, \EllipticPi \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, -\frac {\sqrt {-a b}}{a \left (-\frac {\sqrt {-a b}}{a}-\frac {\sqrt {a b}}{a}\right )}, \frac {\sqrt {2}}{2}\right )}{2 a^{2} \sqrt {a \,x^{3}+b x}\, \left (-\frac {\sqrt {-a b}}{a}-\frac {\sqrt {a b}}{a}\right )}+\frac {\sqrt {-a b}\, \sqrt {\frac {x a}{\sqrt {-a b}}+1}\, \sqrt {-\frac {2 x a}{\sqrt {-a b}}+2}\, \sqrt {-\frac {x a}{\sqrt {-a b}}}\, \EllipticPi \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, -\frac {\sqrt {-a b}}{a \left (-\frac {\sqrt {-a b}}{a}+\frac {\sqrt {a b}}{a}\right )}, \frac {\sqrt {2}}{2}\right )}{2 a^{2} \sqrt {a \,x^{3}+b x}\, \left (-\frac {\sqrt {-a b}}{a}+\frac {\sqrt {a b}}{a}\right )}\)	\(296\)
elliptic	\(\frac {\sqrt {-a b}\, \sqrt {\frac {x a}{\sqrt {-a b}}+1}\, \sqrt {-\frac {2 x a}{\sqrt {-a b}}+2}\, \sqrt {-\frac {x a}{\sqrt {-a b}}}\, \EllipticPi \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, -\frac {\sqrt {-a b}}{a \left (-\frac {\sqrt {-a b}}{a}-\frac {\sqrt {a b}}{a}\right )}, \frac {\sqrt {2}}{2}\right )}{2 a^{2} \sqrt {a \,x^{3}+b x}\, \left (-\frac {\sqrt {-a b}}{a}-\frac {\sqrt {a b}}{a}\right )}+\frac {\sqrt {-a b}\, \sqrt {\frac {x a}{\sqrt {-a b}}+1}\, \sqrt {-\frac {2 x a}{\sqrt {-a b}}+2}\, \sqrt {-\frac {x a}{\sqrt {-a b}}}\, \EllipticPi \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, -\frac {\sqrt {-a b}}{a \left (-\frac {\sqrt {-a b}}{a}+\frac {\sqrt {a b}}{a}\right )}, \frac {\sqrt {2}}{2}\right )}{2 a^{2} \sqrt {a \,x^{3}+b x}\, \left (-\frac {\sqrt {-a b}}{a}+\frac {\sqrt {a b}}{a}\right )}\)	\(296\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a*x^2-b)/(a*x^3+b*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/2/a^2*(-a*b)^(1/2)*(x*a/(-a*b)^(1/2)+1)^(1/2)*(-2*x*a/(-a*b)^(1/2)+2)^(1/2)*(-x*a/(-a*b)^(1/2))^(1/2)/(a*x^3
+b*x)^(1/2)/(-1/a*(-a*b)^(1/2)-1/a*(a*b)^(1/2))*EllipticPi(((x+1/a*(-a*b)^(1/2))*a/(-a*b)^(1/2))^(1/2),-1/a*(-
a*b)^(1/2)/(-1/a*(-a*b)^(1/2)-1/a*(a*b)^(1/2)),1/2*2^(1/2))+1/2/a^2*(-a*b)^(1/2)*(x*a/(-a*b)^(1/2)+1)^(1/2)*(-
2*x*a/(-a*b)^(1/2)+2)^(1/2)*(-x*a/(-a*b)^(1/2))^(1/2)/(a*x^3+b*x)^(1/2)/(-1/a*(-a*b)^(1/2)+1/a*(a*b)^(1/2))*El
lipticPi(((x+1/a*(-a*b)^(1/2))*a/(-a*b)^(1/2))^(1/2),-1/a*(-a*b)^(1/2)/(-1/a*(-a*b)^(1/2)+1/a*(a*b)^(1/2)),1/2
*2^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\sqrt {a x^{3} + b x} {\left (a x^{2} - b\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a*x^2-b)/(a*x^3+b*x)^(1/2),x, algorithm="maxima")

[Out]

integrate(x/(sqrt(a*x^3 + b*x)*(a*x^2 - b)), x)

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-x/((b*x + a*x^3)^(1/2)*(b - a*x^2)),x)

[Out]

\text{Hanged}

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a*x**2-b)/(a*x**3+b*x)**(1/2),x)

[Out]

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

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