\(\int \frac {A+B x}{x^2 \sqrt {a+b x}} \, dx\) [429]

Optimal result

Integrand size = 18, antiderivative size = 49 \[ \int \frac {A+B x}{x^2 \sqrt {a+b x}} \, dx=-\frac {A \sqrt {a+b x}}{a x}+\frac {(A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {79, 65, 214} \[ \int \frac {A+B x}{x^2 \sqrt {a+b x}} \, dx=\frac {(A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {A \sqrt {a+b x}}{a x} \]

[In]

Int[(A + B*x)/(x^2*Sqrt[a + b*x]),x]

[Out]

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

Rule 65

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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {A \sqrt {a+b x}}{a x}+\frac {\left (-\frac {A b}{2}+a B\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx}{a} \\ & = -\frac {A \sqrt {a+b x}}{a x}+\frac {\left (2 \left (-\frac {A b}{2}+a B\right )\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{a b} \\ & = -\frac {A \sqrt {a+b x}}{a x}+\frac {(A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x}{x^2 \sqrt {a+b x}} \, dx=-\frac {A \sqrt {a+b x}}{a x}+\frac {(A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}} \]

[In]

Integrate[(A + B*x)/(x^2*Sqrt[a + b*x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {\left (A b -2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}-\frac {A \sqrt {b x +a}}{a x}\)	\(42\)
default	\(\frac {\left (A b -2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}-\frac {A \sqrt {b x +a}}{a x}\)	\(42\)
risch	\(\frac {\left (A b -2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}-\frac {A \sqrt {b x +a}}{a x}\)	\(42\)
pseudoelliptic	\(\frac {\left (A b -2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}-\frac {A \sqrt {b x +a}}{a x}\)	\(42\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 111, normalized size of antiderivative = 2.27 \[ \int \frac {A+B x}{x^2 \sqrt {a+b x}} \, dx=\left [-\frac {{\left (2 \, B a - A b\right )} \sqrt {a} x \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, \sqrt {b x + a} A a}{2 \, a^{2} x}, \frac {{\left (2 \, B a - A b\right )} \sqrt {-a} x \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - \sqrt {b x + a} A a}{a^{2} x}\right ] \]

[In]

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

[Out]

[-1/2*((2*B*a - A*b)*sqrt(a)*x*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*sqrt(b*x + a)*A*a)/(a^2*x), ((
2*B*a - A*b)*sqrt(-a)*x*arctan(sqrt(b*x + a)*sqrt(-a)/a) - sqrt(b*x + a)*A*a)/(a^2*x)]

Sympy [A] (verification not implemented)

Time = 5.50 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.69 \[ \int \frac {A+B x}{x^2 \sqrt {a+b x}} \, dx=- \frac {A \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{a \sqrt {x}} + \frac {A b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{a^{\frac {3}{2}}} + B \left (\begin {cases} \frac {2 \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} & \text {for}\: b \neq 0 \\\frac {\log {\left (x \right )}}{\sqrt {a}} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate((B*x+A)/x**2/(b*x+a)**(1/2),x)

[Out]

-A*sqrt(b)*sqrt(a/(b*x) + 1)/(a*sqrt(x)) + A*b*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/a**(3/2) + B*Piecewise((2*atan
(sqrt(a + b*x)/sqrt(-a))/sqrt(-a), Ne(b, 0)), (log(x)/sqrt(a), True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.51 \[ \int \frac {A+B x}{x^2 \sqrt {a+b x}} \, dx=-\frac {1}{2} \, b {\left (\frac {2 \, \sqrt {b x + a} A}{{\left (b x + a\right )} a - a^{2}} - \frac {{\left (2 \, B a - A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}} b}\right )} \]

[In]

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

[Out]

-1/2*b*(2*sqrt(b*x + a)*A/((b*x + a)*a - a^2) - (2*B*a - A*b)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + s
qrt(a)))/(a^(3/2)*b))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.18 \[ \int \frac {A+B x}{x^2 \sqrt {a+b x}} \, dx=-\frac {\frac {\sqrt {b x + a} A b}{a x} - \frac {{\left (2 \, B a b - A b^{2}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a}}{b} \]

[In]

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

[Out]

-(sqrt(b*x + a)*A*b/(a*x) - (2*B*a*b - A*b^2)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a))/b

Mupad [B] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.84 \[ \int \frac {A+B x}{x^2 \sqrt {a+b x}} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (A\,b-2\,B\,a\right )}{a^{3/2}}-\frac {A\,\sqrt {a+b\,x}}{a\,x} \]

[In]

int((A + B*x)/(x^2*(a + b*x)^(1/2)),x)

[Out]

(atanh((a + b*x)^(1/2)/a^(1/2))*(A*b - 2*B*a))/a^(3/2) - (A*(a + b*x)^(1/2))/(a*x)