3.21.96 \(\int \frac {1}{(2 b+a x) \sqrt [4]{b x^2+a x^3}} \, dx\)

Optimal. Leaf size=152 \[ \frac {\tanh ^{-1}\left (\frac {2 \sqrt {a} \sqrt [4]{b} x \sqrt [4]{a x^3+b x^2}}{2 \sqrt {b} \sqrt {a x^3+b x^2}+a x^2}\right )}{2 \sqrt {a} b^{3/4}}-\frac {\tan ^{-1}\left (\frac {\frac {\sqrt [4]{b} \sqrt {a x^3+b x^2}}{\sqrt {a}}-\frac {\sqrt {a} x^2}{2 \sqrt [4]{b}}}{x \sqrt [4]{a x^3+b x^2}}\right )}{2 \sqrt {a} b^{3/4}} \]

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Rubi [A]  time = 0.37, antiderivative size = 172, normalized size of antiderivative = 1.13, number of steps used = 12, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {2056, 107, 106, 490, 1211, 221, 1699, 203, 206} \begin {gather*} \frac {\sqrt {-\frac {a x}{b}} \sqrt [4]{a x+b} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a x+b}}{\sqrt [4]{-b} \sqrt {-\frac {a x}{b}}}\right )}{\sqrt {2} a \sqrt [4]{-b} \sqrt [4]{a x^3+b x^2}}-\frac {\sqrt {-\frac {a x}{b}} \sqrt [4]{a x+b} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a x+b}}{\sqrt [4]{-b} \sqrt {-\frac {a x}{b}}}\right )}{\sqrt {2} a \sqrt [4]{-b} \sqrt [4]{a x^3+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((2*b + a*x)*(b*x^2 + a*x^3)^(1/4)),x]

[Out]

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

Rule 106

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

Rule 107

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(1/4)), x_Symbol] :> Dist[Sqrt[-((f*
(c + d*x))/(d*e - c*f))]/Sqrt[c + d*x], Int[1/((a + b*x)*Sqrt[-((c*f)/(d*e - c*f)) - (d*f*x)/(d*e - c*f)]*(e +
f*x)^(1/4)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] &&  !GtQ[-(f/(d*e - c*f)), 0]

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 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 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 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 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 {1}{(2 b+a x) \sqrt [4]{b x^2+a x^3}} \, dx &=\frac {\left (\sqrt {x} \sqrt [4]{b+a x}\right ) \int \frac {1}{\sqrt {x} \sqrt [4]{b+a x} (2 b+a x)} \, dx}{\sqrt [4]{b x^2+a x^3}}\\ &=\frac {\left (\sqrt {-\frac {a x}{b}} \sqrt [4]{b+a x}\right ) \int \frac {1}{\sqrt {-\frac {a x}{b}} \sqrt [4]{b+a x} (2 b+a x)} \, dx}{\sqrt [4]{b x^2+a x^3}}\\ &=-\frac {\left (4 \sqrt {-\frac {a x}{b}} \sqrt [4]{b+a x}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (-a b-a x^4\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x}\right )}{\sqrt [4]{b x^2+a x^3}}\\ &=-\frac {\left (2 \sqrt {-\frac {a x}{b}} \sqrt [4]{b+a x}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {-b}-x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x}\right )}{a \sqrt [4]{b x^2+a x^3}}+\frac {\left (2 \sqrt {-\frac {a x}{b}} \sqrt [4]{b+a x}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {-b}+x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x}\right )}{a \sqrt [4]{b x^2+a x^3}}\\ &=\frac {\left (\sqrt {-\frac {a x}{b}} \sqrt [4]{b+a x}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {-b}-x^2}{\left (\sqrt {-b}+x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x}\right )}{a \sqrt {-b} \sqrt [4]{b x^2+a x^3}}-\frac {\left (\sqrt {-\frac {a x}{b}} \sqrt [4]{b+a x}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {-b}+x^2}{\left (\sqrt {-b}-x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x}\right )}{a \sqrt {-b} \sqrt [4]{b x^2+a x^3}}\\ &=-\frac {\left (\sqrt {-\frac {a x}{b}} \sqrt [4]{b+a x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-b}-2 x^2} \, dx,x,\frac {\sqrt [4]{b+a x}}{\sqrt {-\frac {a x}{b}}}\right )}{a \sqrt [4]{b x^2+a x^3}}+\frac {\left (\sqrt {-\frac {a x}{b}} \sqrt [4]{b+a x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-b}+2 x^2} \, dx,x,\frac {\sqrt [4]{b+a x}}{\sqrt {-\frac {a x}{b}}}\right )}{a \sqrt [4]{b x^2+a x^3}}\\ &=\frac {\sqrt {-\frac {a x}{b}} \sqrt [4]{b+a x} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b+a x}}{\sqrt [4]{-b} \sqrt {-\frac {a x}{b}}}\right )}{\sqrt {2} a \sqrt [4]{-b} \sqrt [4]{b x^2+a x^3}}-\frac {\sqrt {-\frac {a x}{b}} \sqrt [4]{b+a x} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b+a x}}{\sqrt [4]{-b} \sqrt {-\frac {a x}{b}}}\right )}{\sqrt {2} a \sqrt [4]{-b} \sqrt [4]{b x^2+a x^3}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 58, normalized size = 0.38 \begin {gather*} \frac {x \sqrt [4]{\frac {a x+b}{b}} F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};-\frac {a x}{b},-\frac {a x}{2 b}\right )}{b \sqrt [4]{x^2 (a x+b)}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((2*b + a*x)*(b*x^2 + a*x^3)^(1/4)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.46, size = 152, normalized size = 1.00 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {-\frac {\sqrt {a} x^2}{2 \sqrt [4]{b}}+\frac {\sqrt [4]{b} \sqrt {b x^2+a x^3}}{\sqrt {a}}}{x \sqrt [4]{b x^2+a x^3}}\right )}{2 \sqrt {a} b^{3/4}}+\frac {\tanh ^{-1}\left (\frac {2 \sqrt {a} \sqrt [4]{b} x \sqrt [4]{b x^2+a x^3}}{a x^2+2 \sqrt {b} \sqrt {b x^2+a x^3}}\right )}{2 \sqrt {a} b^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((2*b + a*x)*(b*x^2 + a*x^3)^(1/4)),x]

[Out]

-1/2*ArcTan[(-1/2*(Sqrt[a]*x^2)/b^(1/4) + (b^(1/4)*Sqrt[b*x^2 + a*x^3])/Sqrt[a])/(x*(b*x^2 + a*x^3)^(1/4))]/(S
qrt[a]*b^(3/4)) + ArcTanh[(2*Sqrt[a]*b^(1/4)*x*(b*x^2 + a*x^3)^(1/4))/(a*x^2 + 2*Sqrt[b]*Sqrt[b*x^2 + a*x^3])]
/(2*Sqrt[a]*b^(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(1/(a*x+2*b)/(a*x^3+b*x^2)^(1/4),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 {1}{{\left (a x^{3} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x + 2 \, b\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+2*b)/(a*x^3+b*x^2)^(1/4),x, algorithm="giac")

[Out]

integrate(1/((a*x^3 + b*x^2)^(1/4)*(a*x + 2*b)), x)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a x +2 b \right ) \left (a \,x^{3}+b \,x^{2}\right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+2*b)/(a*x^3+b*x^2)^(1/4),x)

[Out]

int(1/(a*x+2*b)/(a*x^3+b*x^2)^(1/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (a x^{3} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x + 2 \, b\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+2*b)/(a*x^3+b*x^2)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((a*x^3 + b*x^2)^(1/4)*(a*x + 2*b)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\left (2\,b+a\,x\right )\,{\left (a\,x^3+b\,x^2\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((2*b + a*x)*(a*x^3 + b*x^2)^(1/4)),x)

[Out]

int(1/((2*b + a*x)*(a*x^3 + b*x^2)^(1/4)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+2*b)/(a*x**3+b*x**2)**(1/4),x)

[Out]

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

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