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

Optimal. Leaf size=64 \[ -\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {b} \sqrt {a^2 x^3+b^2 x}}{a^2 x^2+b^2}\right )}{\sqrt {a} \sqrt {b}} \]

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Rubi [A]  time = 0.68, antiderivative size = 91, normalized size of antiderivative = 1.42, number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2056, 6733, 1699, 208} \begin {gather*} -\frac {\sqrt {2} \sqrt {x} \sqrt {a^2 x^2+b^2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {b} \sqrt {x}}{\sqrt {a^2 x^2+b^2}}\right )}{\sqrt {a} \sqrt {b} \sqrt {a^2 x^3+b^2 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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]

Rule 6733

Int[(u_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k, Subst[Int[x^(k*(m + 1) - 1)*(u /. x -> x^k
), x], x, x^(1/k)], x]] /; FractionQ[m]

Rubi steps

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

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Mathematica [C]  time = 0.40, size = 110, normalized size = 1.72 \begin {gather*} -\frac {2 i x^{3/2} \sqrt {\frac {b^2}{a^2 x^2}+1} \left (F\left (\left .i \sinh ^{-1}\left (\frac {\sqrt {\frac {i b}{a}}}{\sqrt {x}}\right )\right |-1\right )-2 \Pi \left (i;\left .i \sinh ^{-1}\left (\frac {\sqrt {\frac {i b}{a}}}{\sqrt {x}}\right )\right |-1\right )\right )}{\sqrt {\frac {i b}{a}} \sqrt {x \left (a^2 x^2+b^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*Sqrt[1 + b^2/(a^2*x^2)]*x^(3/2)*(EllipticF[I*ArcSinh[Sqrt[(I*b)/a]/Sqrt[x]], -1] - 2*EllipticPi[I, I*A
rcSinh[Sqrt[(I*b)/a]/Sqrt[x]], -1]))/(Sqrt[(I*b)/a]*Sqrt[x*(b^2 + a^2*x^2)])

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IntegrateAlgebraic [A]  time = 0.28, size = 64, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {b} \sqrt {b^2 x+a^2 x^3}}{b^2+a^2 x^2}\right )}{\sqrt {a} \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.52, size = 213, normalized size = 3.33 \begin {gather*} \left [\frac {1}{4} \, \sqrt {2} \sqrt {\frac {1}{a b}} \log \left (\frac {a^{4} x^{4} + 12 \, a^{3} b x^{3} + 6 \, a^{2} b^{2} x^{2} + 12 \, a b^{3} x + b^{4} - 4 \, \sqrt {2} {\left (a^{3} b x^{2} + 2 \, a^{2} b^{2} x + a b^{3}\right )} \sqrt {a^{2} x^{3} + b^{2} x} \sqrt {\frac {1}{a b}}}{a^{4} x^{4} - 4 \, a^{3} b x^{3} + 6 \, a^{2} b^{2} x^{2} - 4 \, a b^{3} x + b^{4}}\right ), \frac {1}{2} \, \sqrt {2} \sqrt {-\frac {1}{a b}} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {a^{2} x^{3} + b^{2} x} a b \sqrt {-\frac {1}{a b}}}{a^{2} x^{2} + 2 \, a b x + b^{2}}\right )\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*sqrt(2)*sqrt(1/(a*b))*log((a^4*x^4 + 12*a^3*b*x^3 + 6*a^2*b^2*x^2 + 12*a*b^3*x + b^4 - 4*sqrt(2)*(a^3*b*x
^2 + 2*a^2*b^2*x + a*b^3)*sqrt(a^2*x^3 + b^2*x)*sqrt(1/(a*b)))/(a^4*x^4 - 4*a^3*b*x^3 + 6*a^2*b^2*x^2 - 4*a*b^
3*x + b^4)), 1/2*sqrt(2)*sqrt(-1/(a*b))*arctan(2*sqrt(2)*sqrt(a^2*x^3 + b^2*x)*a*b*sqrt(-1/(a*b))/(a^2*x^2 + 2
*a*b*x + b^2))]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.09, size = 233, normalized size = 3.64

method	result	size
default	\(\frac {i b \sqrt {-\frac {i \left (x +\frac {i b}{a}\right ) a}{b}}\, \sqrt {2}\, \sqrt {\frac {i \left (x -\frac {i b}{a}\right ) a}{b}}\, \sqrt {\frac {i x a}{b}}\, \EllipticF \left (\sqrt {-\frac {i \left (x +\frac {i b}{a}\right ) a}{b}}, \frac {\sqrt {2}}{2}\right )}{a \sqrt {a^{2} x^{3}+b^{2} x}}+\frac {2 i b^{2} \sqrt {-\frac {i \left (x +\frac {i b}{a}\right ) a}{b}}\, \sqrt {2}\, \sqrt {\frac {i \left (x -\frac {i b}{a}\right ) a}{b}}\, \sqrt {\frac {i x a}{b}}\, \EllipticPi \left (\sqrt {-\frac {i \left (x +\frac {i b}{a}\right ) a}{b}}, -\frac {i b}{a \left (-\frac {i b}{a}-\frac {b}{a}\right )}, \frac {\sqrt {2}}{2}\right )}{a^{2} \sqrt {a^{2} x^{3}+b^{2} x}\, \left (-\frac {i b}{a}-\frac {b}{a}\right )}\)	\(233\)
elliptic	\(\frac {i b \sqrt {-\frac {i \left (x +\frac {i b}{a}\right ) a}{b}}\, \sqrt {2}\, \sqrt {\frac {i \left (x -\frac {i b}{a}\right ) a}{b}}\, \sqrt {\frac {i x a}{b}}\, \EllipticF \left (\sqrt {-\frac {i \left (x +\frac {i b}{a}\right ) a}{b}}, \frac {\sqrt {2}}{2}\right )}{a \sqrt {a^{2} x^{3}+b^{2} x}}+\frac {2 i b^{2} \sqrt {-\frac {i \left (x +\frac {i b}{a}\right ) a}{b}}\, \sqrt {2}\, \sqrt {\frac {i \left (x -\frac {i b}{a}\right ) a}{b}}\, \sqrt {\frac {i x a}{b}}\, \EllipticPi \left (\sqrt {-\frac {i \left (x +\frac {i b}{a}\right ) a}{b}}, -\frac {i b}{a \left (-\frac {i b}{a}-\frac {b}{a}\right )}, \frac {\sqrt {2}}{2}\right )}{a^{2} \sqrt {a^{2} x^{3}+b^{2} x}\, \left (-\frac {i b}{a}-\frac {b}{a}\right )}\)	\(233\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*b/a*(-I*(x+I*b/a)/b*a)^(1/2)*2^(1/2)*(I*(x-I*b/a)/b*a)^(1/2)*(I*x/b*a)^(1/2)/(a^2*x^3+b^2*x)^(1/2)*EllipticF
((-I*(x+I*b/a)/b*a)^(1/2),1/2*2^(1/2))+2*I*b^2/a^2*(-I*(x+I*b/a)/b*a)^(1/2)*2^(1/2)*(I*(x-I*b/a)/b*a)^(1/2)*(I
*x/b*a)^(1/2)/(a^2*x^3+b^2*x)^(1/2)/(-I*b/a-b/a)*EllipticPi((-I*(x+I*b/a)/b*a)^(1/2),-I*b/a/(-I*b/a-b/a),1/2*2
^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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