\(\int \frac {b+a x}{(-b+a x) \sqrt {b^2 x+a^2 x^3}} \, dx\) [845]

Optimal result

Integrand size = 32, antiderivative size = 64 \[ \int \frac {b+a x}{(-b+a x) \sqrt {b^2 x+a^2 x^3}} \, dx=-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {b} \sqrt {b^2 x+a^2 x^3}}{b^2+a^2 x^2}\right )}{\sqrt {a} \sqrt {b}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.42, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2081, 6865, 1713, 214} \[ \int \frac {b+a x}{(-b+a x) \sqrt {b^2 x+a^2 x^3}} \, dx=-\frac {\sqrt {2} \sqrt {x} \sqrt {a^2 x^2+b^2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {b} \sqrt {x}}{\sqrt {a^2 x^2+b^2}}\right )}{\sqrt {a} \sqrt {b} \sqrt {a^2 x^3+b^2 x}} \]

[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 214

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

Rule 1713

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 2081

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 6865

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*} \text {integral}& = \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 ) \text {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 ) \text {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} \text {arctanh}\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*}

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.42 \[ \int \frac {b+a x}{(-b+a x) \sqrt {b^2 x+a^2 x^3}} \, dx=-\frac {\sqrt {2} \sqrt {x} \sqrt {b^2+a^2 x^2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {b} \sqrt {x}}{\sqrt {b^2+a^2 x^2}}\right )}{\sqrt {a} \sqrt {b} \sqrt {x \left (b^2+a^2 x^2\right )}} \]

[In]

Integrate[(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[x*(b^2 + a^2*x^2)]))

Maple [A] (verified)

Time = 2.55 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.62

method	result	size
default	\(-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {x \left (a^{2} x^{2}+b^{2}\right )}\, \sqrt {2}}{2 x \sqrt {a b}}\right )}{\sqrt {a b}}\)	\(40\)
pseudoelliptic	\(-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {x \left (a^{2} x^{2}+b^{2}\right )}\, \sqrt {2}}{2 x \sqrt {a b}}\right )}{\sqrt {a b}}\)	\(40\)
elliptic	\(\frac {i b \sqrt {-\frac {i \left (x +\frac {i b}{a}\right ) a}{b}}\, \sqrt {2}\, \sqrt {\frac {i \left (x -\frac {i b}{a}\right ) a}{b}}\, \sqrt {\frac {i a x}{b}}\, \operatorname {EllipticF}\left (\sqrt {-\frac {i \left (x +\frac {i b}{a}\right ) a}{b}}, \frac {\sqrt {2}}{2}\right )}{a \sqrt {a^{2} x^{3}+b^{2} x}}+\frac {2 i b^{2} \sqrt {-\frac {i \left (x +\frac {i b}{a}\right ) a}{b}}\, \sqrt {2}\, \sqrt {\frac {i \left (x -\frac {i b}{a}\right ) a}{b}}\, \sqrt {\frac {i a x}{b}}\, \operatorname {EllipticPi}\left (\sqrt {-\frac {i \left (x +\frac {i b}{a}\right ) a}{b}}, -\frac {i b}{a \left (-\frac {i b}{a}-\frac {b}{a}\right )}, \frac {\sqrt {2}}{2}\right )}{a^{2} \sqrt {a^{2} x^{3}+b^{2} x}\, \left (-\frac {i b}{a}-\frac {b}{a}\right )}\)	\(233\)

[In]

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

[Out]

-2^(1/2)/(a*b)^(1/2)*arctanh(1/2*(x*(a^2*x^2+b^2))^(1/2)/x*2^(1/2)/(a*b)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 213, normalized size of antiderivative = 3.33 \[ \int \frac {b+a x}{(-b+a x) \sqrt {b^2 x+a^2 x^3}} \, dx=\left [\frac {1}{4} \, \sqrt {2} \sqrt {\frac {1}{a b}} \log \left (\frac {a^{4} x^{4} + 12 \, a^{3} b x^{3} + 6 \, a^{2} b^{2} x^{2} + 12 \, a b^{3} x + b^{4} - 4 \, \sqrt {2} {\left (a^{3} b x^{2} + 2 \, a^{2} b^{2} x + a b^{3}\right )} \sqrt {a^{2} x^{3} + b^{2} x} \sqrt {\frac {1}{a b}}}{a^{4} x^{4} - 4 \, a^{3} b x^{3} + 6 \, a^{2} b^{2} x^{2} - 4 \, a b^{3} x + b^{4}}\right ), \frac {1}{2} \, \sqrt {2} \sqrt {-\frac {1}{a b}} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {a^{2} x^{3} + b^{2} x} a b \sqrt {-\frac {1}{a b}}}{a^{2} x^{2} + 2 \, a b x + b^{2}}\right )\right ] \]

[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))]

Sympy [F]

\[ \int \frac {b+a x}{(-b+a x) \sqrt {b^2 x+a^2 x^3}} \, dx=\int \frac {a x + b}{\sqrt {x \left (a^{2} x^{2} + b^{2}\right )} \left (a x - b\right )}\, dx \]

[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)

Maxima [F]

\[ \int \frac {b+a x}{(-b+a x) \sqrt {b^2 x+a^2 x^3}} \, dx=\int { \frac {a x + b}{\sqrt {a^{2} x^{3} + b^{2} x} {\left (a x - b\right )}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {b+a x}{(-b+a x) \sqrt {b^2 x+a^2 x^3}} \, dx=\int { \frac {a x + b}{\sqrt {a^{2} x^{3} + b^{2} x} {\left (a x - b\right )}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {b+a x}{(-b+a x) \sqrt {b^2 x+a^2 x^3}} \, dx=\text {Hanged} \]

[In]

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

[Out]

\text{Hanged}