3.643 \(\int \frac {1}{x (a+b \sqrt {c+d x})^2} \, dx\)

Optimal. Leaf size=129 \[ \frac {2 a}{\left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )}-\frac {2 \left (a^2+b^2 c\right ) \log \left (a+b \sqrt {c+d x}\right )}{\left (a^2-b^2 c\right )^2}+\frac {4 a b \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{\left (a^2-b^2 c\right )^2}+\frac {\log (x) \left (a^2+b^2 c\right )}{\left (a^2-b^2 c\right )^2} \]

[Out]

(b^2*c+a^2)*ln(x)/(-b^2*c+a^2)^2-2*(b^2*c+a^2)*ln(a+b*(d*x+c)^(1/2))/(-b^2*c+a^2)^2+4*a*b*arctanh((d*x+c)^(1/2
)/c^(1/2))*c^(1/2)/(-b^2*c+a^2)^2+2*a/(-b^2*c+a^2)/(a+b*(d*x+c)^(1/2))

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Rubi [A]  time = 0.12, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {371, 1398, 801, 635, 206, 260} \[ \frac {2 a}{\left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )}-\frac {2 \left (a^2+b^2 c\right ) \log \left (a+b \sqrt {c+d x}\right )}{\left (a^2-b^2 c\right )^2}+\frac {4 a b \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{\left (a^2-b^2 c\right )^2}+\frac {\log (x) \left (a^2+b^2 c\right )}{\left (a^2-b^2 c\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*Sqrt[c + d*x])^2),x]

[Out]

(2*a)/((a^2 - b^2*c)*(a + b*Sqrt[c + d*x])) + (4*a*b*Sqrt[c]*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(a^2 - b^2*c)^2 +
((a^2 + b^2*c)*Log[x])/(a^2 - b^2*c)^2 - (2*(a^2 + b^2*c)*Log[a + b*Sqrt[c + d*x]])/(a^2 - b^2*c)^2

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 371

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]
] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 1398

Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{g = Denominator[n]}, D
ist[g, Subst[Int[x^(g - 1)*(d + e*x^(g*n))^q*(a + c*x^(2*g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, c, d, e, p
, q}, x] && EqQ[n2, 2*n] && FractionQ[n]

Rubi steps

\begin {align*} \int \frac {1}{x \left (a+b \sqrt {c+d x}\right )^2} \, dx &=\operatorname {Subst}\left (\int \frac {1}{\left (a+b \sqrt {x}\right )^2 (-c+x)} \, dx,x,c+d x\right )\\ &=2 \operatorname {Subst}\left (\int \frac {x}{(a+b x)^2 \left (-c+x^2\right )} \, dx,x,\sqrt {c+d x}\right )\\ &=2 \operatorname {Subst}\left (\int \left (-\frac {a b}{\left (a^2-b^2 c\right ) (a+b x)^2}-\frac {b \left (a^2+b^2 c\right )}{\left (a^2-b^2 c\right )^2 (a+b x)}+\frac {2 a b c-\left (a^2+b^2 c\right ) x}{\left (a^2-b^2 c\right )^2 \left (c-x^2\right )}\right ) \, dx,x,\sqrt {c+d x}\right )\\ &=\frac {2 a}{\left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )}-\frac {2 \left (a^2+b^2 c\right ) \log \left (a+b \sqrt {c+d x}\right )}{\left (a^2-b^2 c\right )^2}+\frac {2 \operatorname {Subst}\left (\int \frac {2 a b c-\left (a^2+b^2 c\right ) x}{c-x^2} \, dx,x,\sqrt {c+d x}\right )}{\left (a^2-b^2 c\right )^2}\\ &=\frac {2 a}{\left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )}-\frac {2 \left (a^2+b^2 c\right ) \log \left (a+b \sqrt {c+d x}\right )}{\left (a^2-b^2 c\right )^2}+\frac {(4 a b c) \operatorname {Subst}\left (\int \frac {1}{c-x^2} \, dx,x,\sqrt {c+d x}\right )}{\left (a^2-b^2 c\right )^2}-\frac {\left (2 \left (a^2+b^2 c\right )\right ) \operatorname {Subst}\left (\int \frac {x}{c-x^2} \, dx,x,\sqrt {c+d x}\right )}{\left (a^2-b^2 c\right )^2}\\ &=\frac {2 a}{\left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )}+\frac {4 a b \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{\left (a^2-b^2 c\right )^2}+\frac {\left (a^2+b^2 c\right ) \log (x)}{\left (a^2-b^2 c\right )^2}-\frac {2 \left (a^2+b^2 c\right ) \log \left (a+b \sqrt {c+d x}\right )}{\left (a^2-b^2 c\right )^2}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 164, normalized size = 1.27 \[ \frac {2 a^3-2 \left (a^2+b^2 c\right ) \left (a+b \sqrt {c+d x}\right ) \log \left (a+b \sqrt {c+d x}\right )-2 a b^2 c+\left (a-b \sqrt {c}\right )^2 \log \left (\sqrt {c}-\sqrt {c+d x}\right ) \left (a+b \sqrt {c+d x}\right )+\left (a+b \sqrt {c}\right )^2 \log \left (\sqrt {c+d x}+\sqrt {c}\right ) \left (a+b \sqrt {c+d x}\right )}{\left (a^2-b^2 c\right )^2 \left (a+b \sqrt {c+d x}\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*Sqrt[c + d*x])^2),x]

[Out]

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

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fricas [A]  time = 0.55, size = 444, normalized size = 3.44 \[ \left [\frac {2 \, a^{2} b^{2} c - 2 \, a^{4} + 2 \, {\left (a b^{3} d x + a b^{3} c - a^{3} b\right )} \sqrt {c} \log \left (\frac {d x + 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left (b^{4} c^{2} - a^{4} + {\left (b^{4} c + a^{2} b^{2}\right )} d x\right )} \log \left (\sqrt {d x + c} b + a\right ) + {\left (b^{4} c^{2} - a^{4} + {\left (b^{4} c + a^{2} b^{2}\right )} d x\right )} \log \relax (x) - 2 \, {\left (a b^{3} c - a^{3} b\right )} \sqrt {d x + c}}{b^{6} c^{3} - 3 \, a^{2} b^{4} c^{2} + 3 \, a^{4} b^{2} c - a^{6} + {\left (b^{6} c^{2} - 2 \, a^{2} b^{4} c + a^{4} b^{2}\right )} d x}, \frac {2 \, a^{2} b^{2} c - 2 \, a^{4} - 4 \, {\left (a b^{3} d x + a b^{3} c - a^{3} b\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) - 2 \, {\left (b^{4} c^{2} - a^{4} + {\left (b^{4} c + a^{2} b^{2}\right )} d x\right )} \log \left (\sqrt {d x + c} b + a\right ) + {\left (b^{4} c^{2} - a^{4} + {\left (b^{4} c + a^{2} b^{2}\right )} d x\right )} \log \relax (x) - 2 \, {\left (a b^{3} c - a^{3} b\right )} \sqrt {d x + c}}{b^{6} c^{3} - 3 \, a^{2} b^{4} c^{2} + 3 \, a^{4} b^{2} c - a^{6} + {\left (b^{6} c^{2} - 2 \, a^{2} b^{4} c + a^{4} b^{2}\right )} d x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(2*a^2*b^2*c - 2*a^4 + 2*(a*b^3*d*x + a*b^3*c - a^3*b)*sqrt(c)*log((d*x + 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) -
2*(b^4*c^2 - a^4 + (b^4*c + a^2*b^2)*d*x)*log(sqrt(d*x + c)*b + a) + (b^4*c^2 - a^4 + (b^4*c + a^2*b^2)*d*x)*
log(x) - 2*(a*b^3*c - a^3*b)*sqrt(d*x + c))/(b^6*c^3 - 3*a^2*b^4*c^2 + 3*a^4*b^2*c - a^6 + (b^6*c^2 - 2*a^2*b^
4*c + a^4*b^2)*d*x), (2*a^2*b^2*c - 2*a^4 - 4*(a*b^3*d*x + a*b^3*c - a^3*b)*sqrt(-c)*arctan(sqrt(d*x + c)*sqrt
(-c)/c) - 2*(b^4*c^2 - a^4 + (b^4*c + a^2*b^2)*d*x)*log(sqrt(d*x + c)*b + a) + (b^4*c^2 - a^4 + (b^4*c + a^2*b
^2)*d*x)*log(x) - 2*(a*b^3*c - a^3*b)*sqrt(d*x + c))/(b^6*c^3 - 3*a^2*b^4*c^2 + 3*a^4*b^2*c - a^6 + (b^6*c^2 -
2*a^2*b^4*c + a^4*b^2)*d*x)]

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giac [A]  time = 0.41, size = 174, normalized size = 1.35 \[ -\frac {4 \, a b c \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{{\left (b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}\right )} \sqrt {-c}} + \frac {{\left (b^{2} c + a^{2}\right )} \log \left (-d x\right )}{b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}} - \frac {2 \, {\left (b^{3} c + a^{2} b\right )} \log \left ({\left | \sqrt {d x + c} b + a \right |}\right )}{b^{5} c^{2} - 2 \, a^{2} b^{3} c + a^{4} b} - \frac {2 \, {\left (a b^{2} c - a^{3}\right )}}{{\left (b^{2} c - a^{2}\right )}^{2} {\left (\sqrt {d x + c} b + a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*a*b*c*arctan(sqrt(d*x + c)/sqrt(-c))/((b^4*c^2 - 2*a^2*b^2*c + a^4)*sqrt(-c)) + (b^2*c + a^2)*log(-d*x)/(b^
4*c^2 - 2*a^2*b^2*c + a^4) - 2*(b^3*c + a^2*b)*log(abs(sqrt(d*x + c)*b + a))/(b^5*c^2 - 2*a^2*b^3*c + a^4*b) -
2*(a*b^2*c - a^3)/((b^2*c - a^2)^2*(sqrt(d*x + c)*b + a))

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maple [A]  time = 0.01, size = 161, normalized size = 1.25 \[ \frac {b^{2} c \ln \left (d x \right )}{\left (-b^{2} c +a^{2}\right )^{2}}-\frac {2 b^{2} c \ln \left (a +\sqrt {d x +c}\, b \right )}{\left (-b^{2} c +a^{2}\right )^{2}}+\frac {4 a b \sqrt {c}\, \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\left (-b^{2} c +a^{2}\right )^{2}}+\frac {a^{2} \ln \left (d x \right )}{\left (-b^{2} c +a^{2}\right )^{2}}-\frac {2 a^{2} \ln \left (a +\sqrt {d x +c}\, b \right )}{\left (-b^{2} c +a^{2}\right )^{2}}+\frac {2 a}{\left (-b^{2} c +a^{2}\right ) \left (a +\sqrt {d x +c}\, b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(-b^2*c+a^2)^2*ln(d*x)*b^2*c+1/(-b^2*c+a^2)^2*ln(d*x)*a^2+4*a*b*arctanh((d*x+c)^(1/2)/c^(1/2))*c^(1/2)/(-b^2
*c+a^2)^2+2*a/(-b^2*c+a^2)/(a+(d*x+c)^(1/2)*b)-2/(-b^2*c+a^2)^2*ln(a+(d*x+c)^(1/2)*b)*b^2*c-2/(-b^2*c+a^2)^2*l
n(a+(d*x+c)^(1/2)*b)*a^2

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maxima [A]  time = 1.99, size = 176, normalized size = 1.36 \[ -\frac {2 \, a b \sqrt {c} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}} + \frac {{\left (b^{2} c + a^{2}\right )} \log \left (d x\right )}{b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}} - \frac {2 \, {\left (b^{2} c + a^{2}\right )} \log \left (\sqrt {d x + c} b + a\right )}{b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}} - \frac {2 \, a}{a b^{2} c - a^{3} + {\left (b^{3} c - a^{2} b\right )} \sqrt {d x + c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.51, size = 125, normalized size = 0.97 \[ \frac {\ln \left (\sqrt {c+d\,x}-\sqrt {c}\right )}{{\left (a+b\,\sqrt {c}\right )}^2}+\ln \left (a+b\,\sqrt {c+d\,x}\right )\,\left (\frac {2}{b^2\,c-a^2}-\frac {4\,b^2\,c}{{\left (b^2\,c-a^2\right )}^2}\right )+\frac {\ln \left (\sqrt {c+d\,x}+\sqrt {c}\right )}{{\left (a-b\,\sqrt {c}\right )}^2}-\frac {2\,a}{\left (b^2\,c-a^2\right )\,\left (a+b\,\sqrt {c+d\,x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 50.51, size = 153, normalized size = 1.19 \[ - \frac {2 a b \left (\begin {cases} \frac {\sqrt {c + d x}}{a^{2}} & \text {for}\: b = 0 \\- \frac {1}{b \left (a + b \sqrt {c + d x}\right )} & \text {otherwise} \end {cases}\right )}{a^{2} - b^{2} c} - \frac {2 b \left (a^{2} + b^{2} c\right ) \left (\begin {cases} \frac {\sqrt {c + d x}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b \sqrt {c + d x} \right )}}{b} & \text {otherwise} \end {cases}\right )}{\left (a^{2} - b^{2} c\right )^{2}} - \frac {2 \left (\frac {2 a b c \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + \left (- \frac {a^{2}}{2} - \frac {b^{2} c}{2}\right ) \log {\left (- d x \right )}\right )}{\left (a^{2} - b^{2} c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*b*Piecewise((sqrt(c + d*x)/a**2, Eq(b, 0)), (-1/(b*(a + b*sqrt(c + d*x))), True))/(a**2 - b**2*c) - 2*b*(
a**2 + b**2*c)*Piecewise((sqrt(c + d*x)/a, Eq(b, 0)), (log(a + b*sqrt(c + d*x))/b, True))/(a**2 - b**2*c)**2 -
2*(2*a*b*c*atan(sqrt(c + d*x)/sqrt(-c))/sqrt(-c) + (-a**2/2 - b**2*c/2)*log(-d*x))/(a**2 - b**2*c)**2

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