\(\int \frac {\sqrt {a+b x}}{x^2 \sqrt {c+d x}} \, dx\) [581]

Optimal result

Integrand size = 22, antiderivative size = 77 \[ \int \frac {\sqrt {a+b x}}{x^2 \sqrt {c+d x}} \, dx=-\frac {\sqrt {a+b x} \sqrt {c+d x}}{c x}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a} c^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {96, 95, 214} \[ \int \frac {\sqrt {a+b x}}{x^2 \sqrt {c+d x}} \, dx=-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a} c^{3/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x}}{c x} \]

[In]

Int[Sqrt[a + b*x]/(x^2*Sqrt[c + d*x]),x]

[Out]

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

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 96

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

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

Mathematica [A] (verified)

Time = 10.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {a+b x}}{x^2 \sqrt {c+d x}} \, dx=-\frac {\sqrt {a+b x} \sqrt {c+d x}}{c x}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a} c^{3/2}} \]

[In]

Integrate[Sqrt[a + b*x]/(x^2*Sqrt[c + d*x]),x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(146\) vs. \(2(61)=122\).

Time = 0.54 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.91

method	result	size
default	\(\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (\ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a d x -\ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) b c x -2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\right )}{2 c \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, x \sqrt {a c}}\)	\(147\)

[In]

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

[Out]

1/2*(b*x+a)^(1/2)*(d*x+c)^(1/2)/c*(ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*a*d*x-ln((a
*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*b*c*x-2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/((b*x+
a)*(d*x+c))^(1/2)/x/(a*c)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 252, normalized size of antiderivative = 3.27 \[ \int \frac {\sqrt {a+b x}}{x^2 \sqrt {c+d x}} \, dx=\left [-\frac {\sqrt {a c} {\left (b c - a d\right )} x \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} + 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \, \sqrt {b x + a} \sqrt {d x + c} a c}{4 \, a c^{2} x}, \frac {\sqrt {-a c} {\left (b c - a d\right )} x \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) - 2 \, \sqrt {b x + a} \sqrt {d x + c} a c}{2 \, a c^{2} x}\right ] \]

[In]

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

[Out]

[-1/4*(sqrt(a*c)*(b*c - a*d)*x*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 + 4*(2*a*c + (b*c + a*d)*x
)*sqrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + 4*sqrt(b*x + a)*sqrt(d*x + c)*a*c)/(
a*c^2*x), 1/2*(sqrt(-a*c)*(b*c - a*d)*x*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x +
c)/(a*b*c*d*x^2 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x)) - 2*sqrt(b*x + a)*sqrt(d*x + c)*a*c)/(a*c^2*x)]

Sympy [F]

\[ \int \frac {\sqrt {a+b x}}{x^2 \sqrt {c+d x}} \, dx=\int \frac {\sqrt {a + b x}}{x^{2} \sqrt {c + d x}}\, dx \]

[In]

integrate((b*x+a)**(1/2)/x**2/(d*x+c)**(1/2),x)

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+b x}}{x^2 \sqrt {c+d x}} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 410 vs. \(2 (61) = 122\).

Time = 0.54 (sec) , antiderivative size = 410, normalized size of antiderivative = 5.32 \[ \int \frac {\sqrt {a+b x}}{x^2 \sqrt {c+d x}} \, dx=-\frac {b {\left (\frac {{\left (\sqrt {b d} b^{2} c - \sqrt {b d} a b d\right )} \arctan \left (-\frac {b^{2} c + a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} b c} + \frac {2 \, {\left (\sqrt {b d} b^{4} c^{2} - 2 \, \sqrt {b d} a b^{3} c d + \sqrt {b d} a^{2} b^{2} d^{2} - \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{2} c - \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b d\right )}}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b d + {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4}\right )} c}\right )}}{{\left | b \right |}} \]

[In]

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

[Out]

-b*((sqrt(b*d)*b^2*c - sqrt(b*d)*a*b*d)*arctan(-1/2*(b^2*c + a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (
b*x + a)*b*d - a*b*d))^2)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*b*c) + 2*(sqrt(b*d)*b^4*c^2 - 2*sqrt(b*d)*a*b^3*
c*d + sqrt(b*d)*a^2*b^2*d^2 - sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^2*
c - sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b*d)/((b^4*c^2 - 2*a*b^3*c*d
+ a^2*b^2*d^2 - 2*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^2*c - 2*(sqrt(b*d)*sqrt
(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b*d + (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b
*d - a*b*d))^4)*c))/abs(b)

Mupad [B] (verification not implemented)

Time = 5.90 (sec) , antiderivative size = 439, normalized size of antiderivative = 5.70 \[ \int \frac {\sqrt {a+b x}}{x^2 \sqrt {c+d x}} \, dx=\frac {\frac {\left (\frac {c\,b^2}{4}+\frac {a\,d\,b}{4}\right )\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {a}\,c^{3/2}\,d\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}-\frac {b^2}{4\,c\,d}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2\,\left (\frac {a^2\,d^2}{4}-\frac {3\,a\,b\,c\,d}{4}+\frac {b^2\,c^2}{4}\right )}{a\,c^2\,d\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}}{\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}+\frac {b\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{d\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}-\frac {\left (a\,d+b\,c\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{\sqrt {a}\,\sqrt {c}\,d\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}}-\frac {d\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{4\,c\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}+\frac {\ln \left (\frac {\left (\sqrt {c}\,\sqrt {a+b\,x}-\sqrt {a}\,\sqrt {c+d\,x}\right )\,\left (b\,\sqrt {c}-\frac {\sqrt {a}\,d\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {c+d\,x}-\sqrt {c}}\right )}{\sqrt {c+d\,x}-\sqrt {c}}\right )\,\left (\sqrt {a}\,b\,c^{3/2}-a^{3/2}\,\sqrt {c}\,d\right )}{2\,a\,c^2}-\frac {\ln \left (\frac {\sqrt {a+b\,x}-\sqrt {a}}{\sqrt {c+d\,x}-\sqrt {c}}\right )\,\left (\sqrt {a}\,b\,c^{3/2}-a^{3/2}\,\sqrt {c}\,d\right )}{2\,a\,c^2} \]

[In]

int((a + b*x)^(1/2)/(x^2*(c + d*x)^(1/2)),x)

[Out]

((((b^2*c)/4 + (a*b*d)/4)*((a + b*x)^(1/2) - a^(1/2)))/(a^(1/2)*c^(3/2)*d*((c + d*x)^(1/2) - c^(1/2))) - b^2/(
4*c*d) + (((a + b*x)^(1/2) - a^(1/2))^2*((a^2*d^2)/4 + (b^2*c^2)/4 - (3*a*b*c*d)/4))/(a*c^2*d*((c + d*x)^(1/2)
- c^(1/2))^2))/(((a + b*x)^(1/2) - a^(1/2))^3/((c + d*x)^(1/2) - c^(1/2))^3 + (b*((a + b*x)^(1/2) - a^(1/2)))
/(d*((c + d*x)^(1/2) - c^(1/2))) - ((a*d + b*c)*((a + b*x)^(1/2) - a^(1/2))^2)/(a^(1/2)*c^(1/2)*d*((c + d*x)^(
1/2) - c^(1/2))^2)) - (d*((a + b*x)^(1/2) - a^(1/2)))/(4*c*((c + d*x)^(1/2) - c^(1/2))) + (log(((c^(1/2)*(a +
b*x)^(1/2) - a^(1/2)*(c + d*x)^(1/2))*(b*c^(1/2) - (a^(1/2)*d*((a + b*x)^(1/2) - a^(1/2)))/((c + d*x)^(1/2) -
c^(1/2))))/((c + d*x)^(1/2) - c^(1/2)))*(a^(1/2)*b*c^(3/2) - a^(3/2)*c^(1/2)*d))/(2*a*c^2) - (log(((a + b*x)^(
1/2) - a^(1/2))/((c + d*x)^(1/2) - c^(1/2)))*(a^(1/2)*b*c^(3/2) - a^(3/2)*c^(1/2)*d))/(2*a*c^2)