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

Optimal. Leaf size=130 \[ -\frac {a-b \sqrt {c+d x}}{x \left (a^2-b^2 c\right )}+\frac {a b^2 d \log (x)}{\left (a^2-b^2 c\right )^2}-\frac {2 a b^2 d \log \left (a+b \sqrt {c+d x}\right )}{\left (a^2-b^2 c\right )^2}+\frac {b d \left (a^2+b^2 c\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{\sqrt {c} \left (a^2-b^2 c\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {371, 1398, 823, 801, 635, 206, 260} \[ -\frac {a-b \sqrt {c+d x}}{x \left (a^2-b^2 c\right )}+\frac {a b^2 d \log (x)}{\left (a^2-b^2 c\right )^2}-\frac {2 a b^2 d \log \left (a+b \sqrt {c+d x}\right )}{\left (a^2-b^2 c\right )^2}+\frac {b d \left (a^2+b^2 c\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{\sqrt {c} \left (a^2-b^2 c\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*Sqrt[c + d*x])),x]

[Out]

-((a - b*Sqrt[c + d*x])/((a^2 - b^2*c)*x)) + (b*(a^2 + b^2*c)*d*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(Sqrt[c]*(a^2
- b^2*c)^2) + (a*b^2*d*Log[x])/(a^2 - b^2*c)^2 - (2*a*b^2*d*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 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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^2 \left (a+b \sqrt {c+d x}\right )} \, dx &=d \operatorname {Subst}\left (\int \frac {1}{\left (a+b \sqrt {x}\right ) (-c+x)^2} \, dx,x,c+d x\right )\\ &=(2 d) \operatorname {Subst}\left (\int \frac {x}{(a+b x) \left (-c+x^2\right )^2} \, dx,x,\sqrt {c+d x}\right )\\ &=-\frac {a-b \sqrt {c+d x}}{\left (a^2-b^2 c\right ) x}+\frac {d \operatorname {Subst}\left (\int \frac {-a b c+b^2 c x}{(a+b x) \left (-c+x^2\right )} \, dx,x,\sqrt {c+d x}\right )}{c \left (a^2-b^2 c\right )}\\ &=-\frac {a-b \sqrt {c+d x}}{\left (a^2-b^2 c\right ) x}+\frac {d \operatorname {Subst}\left (\int \left (-\frac {2 a b^3 c}{\left (a^2-b^2 c\right ) (a+b x)}-\frac {b c \left (a^2+b^2 c-2 a b x\right )}{\left (-a^2+b^2 c\right ) \left (c-x^2\right )}\right ) \, dx,x,\sqrt {c+d x}\right )}{c \left (a^2-b^2 c\right )}\\ &=-\frac {a-b \sqrt {c+d x}}{\left (a^2-b^2 c\right ) x}-\frac {2 a b^2 d \log \left (a+b \sqrt {c+d x}\right )}{\left (a^2-b^2 c\right )^2}+\frac {(b d) \operatorname {Subst}\left (\int \frac {a^2+b^2 c-2 a b x}{c-x^2} \, dx,x,\sqrt {c+d x}\right )}{\left (a^2-b^2 c\right )^2}\\ &=-\frac {a-b \sqrt {c+d x}}{\left (a^2-b^2 c\right ) x}-\frac {2 a b^2 d \log \left (a+b \sqrt {c+d x}\right )}{\left (a^2-b^2 c\right )^2}-\frac {\left (2 a b^2 d\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 {\left (b \left (a^2+b^2 c\right ) d\right ) \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 {a-b \sqrt {c+d x}}{\left (a^2-b^2 c\right ) x}+\frac {b \left (a^2+b^2 c\right ) d \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{\sqrt {c} \left (a^2-b^2 c\right )^2}+\frac {a b^2 d \log (x)}{\left (a^2-b^2 c\right )^2}-\frac {2 a b^2 d \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 = 0.19, size = 144, normalized size = 1.11 \[ \frac {\sqrt {c} \left (-\left (a^2-b^2 c\right ) \left (a-b \sqrt {c+d x}\right )-a b^2 d x \log \left (a^2-b^2 (c+d x)\right )+a b^2 d x \log (x)\right )+b d x \left (a^2+b^2 c\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )-2 a b^2 \sqrt {c} d x \tanh ^{-1}\left (\frac {b \sqrt {c+d x}}{a}\right )}{\sqrt {c} x \left (a^2-b^2 c\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.61, size = 283, normalized size = 2.18 \[ \left [-\frac {4 \, a b^{2} c d x \log \left (\sqrt {d x + c} b + a\right ) - 2 \, a b^{2} c d x \log \relax (x) - 2 \, a b^{2} c^{2} - {\left (b^{3} c + a^{2} b\right )} \sqrt {c} d x \log \left (\frac {d x + 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, a^{3} c + 2 \, {\left (b^{3} c^{2} - a^{2} b c\right )} \sqrt {d x + c}}{2 \, {\left (b^{4} c^{3} - 2 \, a^{2} b^{2} c^{2} + a^{4} c\right )} x}, -\frac {2 \, a b^{2} c d x \log \left (\sqrt {d x + c} b + a\right ) - a b^{2} c d x \log \relax (x) - a b^{2} c^{2} + {\left (b^{3} c + a^{2} b\right )} \sqrt {-c} d x \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) + a^{3} c + {\left (b^{3} c^{2} - a^{2} b c\right )} \sqrt {d x + c}}{{\left (b^{4} c^{3} - 2 \, a^{2} b^{2} c^{2} + a^{4} c\right )} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(4*a*b^2*c*d*x*log(sqrt(d*x + c)*b + a) - 2*a*b^2*c*d*x*log(x) - 2*a*b^2*c^2 - (b^3*c + a^2*b)*sqrt(c)*d
*x*log((d*x + 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 2*a^3*c + 2*(b^3*c^2 - a^2*b*c)*sqrt(d*x + c))/((b^4*c^3 - 2
*a^2*b^2*c^2 + a^4*c)*x), -(2*a*b^2*c*d*x*log(sqrt(d*x + c)*b + a) - a*b^2*c*d*x*log(x) - a*b^2*c^2 + (b^3*c +
a^2*b)*sqrt(-c)*d*x*arctan(sqrt(d*x + c)*sqrt(-c)/c) + a^3*c + (b^3*c^2 - a^2*b*c)*sqrt(d*x + c))/((b^4*c^3 -
2*a^2*b^2*c^2 + a^4*c)*x)]

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giac [A]  time = 0.46, size = 191, normalized size = 1.47 \[ -\frac {2 \, a b^{3} d \log \left ({\left | \sqrt {d x + c} b + a \right |}\right )}{b^{5} c^{2} - 2 \, a^{2} b^{3} c + a^{4} b} + \frac {a b^{2} d \log \left (-d x\right )}{b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}} - \frac {{\left (b^{3} c d + a^{2} b d\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{{\left (b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}\right )} \sqrt {-c}} + \frac {a b^{2} c d - a^{3} d - {\left (b^{3} c d - a^{2} b d\right )} \sqrt {d x + c}}{{\left (b^{2} c - a^{2}\right )}^{2} d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 216, normalized size = 1.66 \[ \frac {b^{3} \sqrt {c}\, d \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\left (-b^{2} c +a^{2}\right )^{2}}+\frac {a \,b^{2} d \ln \left (d x \right )}{\left (-b^{2} c +a^{2}\right )^{2}}-\frac {2 a \,b^{2} d \ln \left (a +\sqrt {d x +c}\, b \right )}{\left (-b^{2} c +a^{2}\right )^{2}}+\frac {a^{2} b d \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\left (-b^{2} c +a^{2}\right )^{2} \sqrt {c}}+\frac {a \,b^{2} c}{\left (-b^{2} c +a^{2}\right )^{2} x}-\frac {\sqrt {d x +c}\, b^{3} c}{\left (-b^{2} c +a^{2}\right )^{2} x}-\frac {a^{3}}{\left (-b^{2} c +a^{2}\right )^{2} x}+\frac {\sqrt {d x +c}\, a^{2} b}{\left (-b^{2} c +a^{2}\right )^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.04, size = 191, normalized size = 1.47 \[ \frac {1}{2} \, {\left (\frac {2 \, a b^{2} \log \left (d x\right )}{b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}} - \frac {4 \, a b^{2} \log \left (\sqrt {d x + c} b + a\right )}{b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}} - \frac {{\left (b^{3} c + a^{2} b\right )} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{{\left (b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}\right )} \sqrt {c}} + \frac {2 \, {\left (\sqrt {d x + c} b - a\right )}}{b^{2} c^{2} - a^{2} c - {\left (b^{2} c - a^{2}\right )} {\left (d x + c\right )}}\right )} d \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.59, size = 220, normalized size = 1.69 \[ \frac {\ln \left (\sqrt {c+d\,x}-\sqrt {c}\right )\,\left (4\,a\,b^2\,c\,d-b\,\sqrt {c}\,d\,\left (2\,a^2+2\,c\,b^2\right )\right )}{4\,a^4\,c-8\,a^2\,b^2\,c^2+4\,b^4\,c^3}+\frac {\ln \left (\sqrt {c+d\,x}+\sqrt {c}\right )\,\left (4\,a\,b^2\,c\,d+b\,\sqrt {c}\,d\,\left (2\,a^2+2\,c\,b^2\right )\right )}{4\,a^4\,c-8\,a^2\,b^2\,c^2+4\,b^4\,c^3}+\frac {\frac {a\,d}{b^2\,c-a^2}-\frac {b\,d\,\sqrt {c+d\,x}}{b^2\,c-a^2}}{d\,x}-\frac {2\,a\,b^2\,d\,\ln \left (a+b\,\sqrt {c+d\,x}\right )}{{\left (b^2\,c-a^2\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log((c + d*x)^(1/2) - c^(1/2))*(4*a*b^2*c*d - b*c^(1/2)*d*(2*b^2*c + 2*a^2)))/(4*a^4*c + 4*b^4*c^3 - 8*a^2*b^
2*c^2) + (log((c + d*x)^(1/2) + c^(1/2))*(4*a*b^2*c*d + b*c^(1/2)*d*(2*b^2*c + 2*a^2)))/(4*a^4*c + 4*b^4*c^3 -
8*a^2*b^2*c^2) + ((a*d)/(b^2*c - a^2) - (b*d*(c + d*x)^(1/2))/(b^2*c - a^2))/(d*x) - (2*a*b^2*d*log(a + b*(c
+ d*x)^(1/2)))/(b^2*c - a^2)^2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{2} \left (a + b \sqrt {c + d x}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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