\(\int \frac {A+B x}{x (a+b x)^{3/2}} \, dx\) [439]

Optimal result

Integrand size = 18, antiderivative size = 50 \[ \int \frac {A+B x}{x (a+b x)^{3/2}} \, dx=\frac {2 (A b-a B)}{a b \sqrt {a+b x}}-\frac {2 A \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 50, 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} \[ \int \frac {A+B x}{x (a+b x)^{3/2}} \, dx=\frac {2 (A b-a B)}{a b \sqrt {a+b x}}-\frac {2 A \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}} \]

[In]

Int[(A + B*x)/(x*(a + b*x)^(3/2)),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x}{x (a+b x)^{3/2}} \, dx=-\frac {2 (-A b+a B)}{a b \sqrt {a+b x}}-\frac {2 A \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}} \]

[In]

Integrate[(A + B*x)/(x*(a + b*x)^(3/2)),x]

[Out]

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

Maple [A] (verified)

Time = 1.40 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.92

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-A b +B a \right )}{a \sqrt {b x +a}}-\frac {2 A b \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}}{b}\)	\(46\)
default	\(\frac {-\frac {2 \left (-A b +B a \right )}{a \sqrt {b x +a}}-\frac {2 A b \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}}{b}\)	\(46\)
pseudoelliptic	\(-\frac {2 \left (A \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b \sqrt {b x +a}-A b \sqrt {a}+B \,a^{\frac {3}{2}}\right )}{a^{\frac {3}{2}} b \sqrt {b x +a}}\)	\(51\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 151, normalized size of antiderivative = 3.02 \[ \int \frac {A+B x}{x (a+b x)^{3/2}} \, dx=\left [\frac {{\left (A b^{2} x + A a b\right )} \sqrt {a} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (B a^{2} - A a b\right )} \sqrt {b x + a}}{a^{2} b^{2} x + a^{3} b}, \frac {2 \, {\left ({\left (A b^{2} x + A a b\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - {\left (B a^{2} - A a b\right )} \sqrt {b x + a}\right )}}{a^{2} b^{2} x + a^{3} b}\right ] \]

[In]

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

[Out]

[((A*b^2*x + A*a*b)*sqrt(a)*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(B*a^2 - A*a*b)*sqrt(b*x + a))/(a
^2*b^2*x + a^3*b), 2*((A*b^2*x + A*a*b)*sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) - (B*a^2 - A*a*b)*sqrt(b*x +
a))/(a^2*b^2*x + a^3*b)]

Sympy [A] (verification not implemented)

Time = 1.68 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.30 \[ \int \frac {A+B x}{x (a+b x)^{3/2}} \, dx=\begin {cases} \frac {2 A \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{a \sqrt {- a}} - \frac {2 \left (- A b + B a\right )}{a b \sqrt {a + b x}} & \text {for}\: b \neq 0 \\\frac {A \log {\left (B x \right )} + B x}{a^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate((B*x+A)/x/(b*x+a)**(3/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.14 \[ \int \frac {A+B x}{x (a+b x)^{3/2}} \, dx=\frac {A \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} - \frac {2 \, {\left (B a - A b\right )}}{\sqrt {b x + a} a b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x}{x (a+b x)^{3/2}} \, dx=\frac {2 \, A \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} - \frac {2 \, {\left (B a - A b\right )}}{\sqrt {b x + a} a b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.84 \[ \int \frac {A+B x}{x (a+b x)^{3/2}} \, dx=\frac {2\,\left (A\,b-B\,a\right )}{a\,b\,\sqrt {a+b\,x}}-\frac {2\,A\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )}{a^{3/2}} \]

[In]

int((A + B*x)/(x*(a + b*x)^(3/2)),x)

[Out]

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