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

Optimal. Leaf size=77 \[ -\frac {(b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a} c^{3/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x}}{c x} \]

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {94, 93, 208} \begin {gather*} -\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}}-\frac {\sqrt {a+b x} \sqrt {c+d x}}{c x} \end {gather*}

Antiderivative was successfully verified.

[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 93

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 94

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[p, 1] &&  !SumSimplerQ[m, 1])

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]

Rubi steps

\begin {align*} \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) \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) \operatorname {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]  time = 0.06, size = 77, normalized size = 1.00 \begin {gather*} -\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}}-\frac {\sqrt {a+b x} \sqrt {c+d x}}{c x} \end {gather*}

Antiderivative was successfully verified.

[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))

________________________________________________________________________________________

IntegrateAlgebraic [B]  time = 0.60, size = 161, normalized size = 2.09 \begin {gather*} -\frac {\sqrt {\frac {b}{d}} \left (b c \sqrt {d}-a d^{3/2}\right ) \tanh ^{-1}\left (\frac {d \sqrt {\frac {b}{d}} \sqrt {c+d x} \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}-b (c+d x)+b c}{\sqrt {a} \sqrt {b} \sqrt {c} \sqrt {d}}\right )}{\sqrt {a} \sqrt {b} c^{3/2}}-\frac {\sqrt {c+d x} \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}{c x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 1.79, size = 252, normalized size = 3.27 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

________________________________________________________________________________________

giac [B]  time = 3.31, size = 410, normalized size = 5.32 \begin {gather*} -\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 |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [B]  time = 0.02, size = 147, normalized size = 1.91 \begin {gather*} \frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (a d x \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-b c x \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\right )}{2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, c x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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 details)Is a*d-b*c zero or nonzero?

________________________________________________________________________________________

mupad [B]  time = 5.18, size = 439, normalized size = 5.70 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________