\(\int \frac {1}{x^2 \sqrt {(a+b x)^3}} \, dx\) [149]

Optimal result

Integrand size = 15, antiderivative size = 72 \[ \int \frac {1}{x^2 \sqrt {(a+b x)^3}} \, dx=\frac {-\frac {3 b}{a^2}-\frac {1}{a x}}{\sqrt {a+b x}}-\frac {3 b \log \left (\frac {-\sqrt {a}+\sqrt {a+b x}}{\sqrt {a}+\sqrt {a+b x}}\right )}{2 a^{5/2}} \]

[Out]

(-1/a/x-3*b/a^2)/(b*x+a)^(1/2)-3/2*b/a^(5/2)*ln(((b*x+a)^(1/2)-a^(1/2))/((b*x+a)^(1/2)+a^(1/2)))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.24, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1973, 44, 53, 65, 214} \[ \int \frac {1}{x^2 \sqrt {(a+b x)^3}} \, dx=-\frac {3 b (a+b x)}{a^2 \sqrt {(a+b x)^3}}+\frac {3 b \left (\frac {b x}{a}+1\right )^{3/2} \text {arctanh}\left (\sqrt {\frac {b x}{a}+1}\right )}{a \sqrt {(a+b x)^3}}-\frac {a+b x}{a x \sqrt {(a+b x)^3}} \]

[In]

Int[1/(x^2*Sqrt[(a + b*x)^3]),x]

[Out]

(-3*b*(a + b*x))/(a^2*Sqrt[(a + b*x)^3]) - (a + b*x)/(a*x*Sqrt[(a + b*x)^3]) + (3*b*(1 + (b*x)/a)^(3/2)*ArcTan
h[Sqrt[1 + (b*x)/a]])/(a*Sqrt[(a + b*x)^3])

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 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]

Rule 1973

Int[(u_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Dist[Simp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/
a))^(p*q)], Int[u*(1 + b*(x^n/a))^(p*q), x], x] /; FreeQ[{a, b, c, n, p, q}, x] &&  !GeQ[a, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (1+\frac {b x}{a}\right )^{3/2} \int \frac {1}{x^2 \left (1+\frac {b x}{a}\right )^{3/2}} \, dx}{\sqrt {(a+b x)^3}} \\ & = -\frac {a+b x}{a x \sqrt {(a+b x)^3}}-\frac {\left (3 b \left (1+\frac {b x}{a}\right )^{3/2}\right ) \int \frac {1}{x \left (1+\frac {b x}{a}\right )^{3/2}} \, dx}{2 a \sqrt {(a+b x)^3}} \\ & = -\frac {3 b (a+b x)}{a^2 \sqrt {(a+b x)^3}}-\frac {a+b x}{a x \sqrt {(a+b x)^3}}-\frac {\left (3 b \left (1+\frac {b x}{a}\right )^{3/2}\right ) \int \frac {1}{x \sqrt {1+\frac {b x}{a}}} \, dx}{2 a \sqrt {(a+b x)^3}} \\ & = -\frac {3 b (a+b x)}{a^2 \sqrt {(a+b x)^3}}-\frac {a+b x}{a x \sqrt {(a+b x)^3}}-\frac {\left (3 \left (1+\frac {b x}{a}\right )^{3/2}\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {a x^2}{b}} \, dx,x,\sqrt {1+\frac {b x}{a}}\right )}{\sqrt {(a+b x)^3}} \\ & = -\frac {3 b (a+b x)}{a^2 \sqrt {(a+b x)^3}}-\frac {a+b x}{a x \sqrt {(a+b x)^3}}+\frac {3 b \left (1+\frac {b x}{a}\right )^{3/2} \text {arctanh}\left (\sqrt {1+\frac {b x}{a}}\right )}{a \sqrt {(a+b x)^3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^2 \sqrt {(a+b x)^3}} \, dx=-\frac {(a+b x) \left (\sqrt {a} (a+3 b x)-3 b x \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{a^{5/2} x \sqrt {(a+b x)^3}} \]

[In]

Integrate[1/(x^2*Sqrt[(a + b*x)^3]),x]

[Out]

-(((a + b*x)*(Sqrt[a]*(a + 3*b*x) - 3*b*x*Sqrt[a + b*x]*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]))/(a^(5/2)*x*Sqrt[(a +
b*x)^3]))

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.81

method	result	size
default	\(\frac {\left (b x +a \right ) \left (3 \sqrt {b x +a}\, \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b x -3 \sqrt {a}\, b x -a^{\frac {3}{2}}\right )}{\sqrt {\left (b x +a \right )^{3}}\, a^{\frac {5}{2}} x}\)	\(58\)
risch	\(-\frac {\left (b x +a \right )^{2}}{a^{2} x \sqrt {\left (b x +a \right )^{3}}}-\frac {b \left (\frac {4}{\sqrt {b x +a}}-\frac {6 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}\right ) \left (b x +a \right )^{\frac {3}{2}}}{2 a^{2} \sqrt {\left (b x +a \right )^{3}}}\)	\(75\)

[In]

int(1/x^2/((b*x+a)^3)^(1/2),x,method=_RETURNVERBOSE)

[Out]

(b*x+a)*(3*(b*x+a)^(1/2)*arctanh((b*x+a)^(1/2)/a^(1/2))*b*x-3*a^(1/2)*b*x-a^(3/2))/((b*x+a)^3)^(1/2)/a^(5/2)/x

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 145 vs. \(2 (58) = 116\).

Time = 0.26 (sec) , antiderivative size = 309, normalized size of antiderivative = 4.29 \[ \int \frac {1}{x^2 \sqrt {(a+b x)^3}} \, dx=\left [\frac {3 \, {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \sqrt {a} \log \left (\frac {b^{2} x^{2} + 3 \, a b x + 2 \, a^{2} + 2 \, \sqrt {b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}} \sqrt {a}}{b x^{2} + a x}\right ) - 2 \, \sqrt {b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}} {\left (3 \, a b x + a^{2}\right )}}{2 \, {\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}}, -\frac {3 \, {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}} \sqrt {-a}}{a b x + a^{2}}\right ) + \sqrt {b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}} {\left (3 \, a b x + a^{2}\right )}}{a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x}\right ] \]

[In]

integrate(1/x^2/((b*x+a)^3)^(1/2),x, algorithm="fricas")

[Out]

[1/2*(3*(b^3*x^3 + 2*a*b^2*x^2 + a^2*b*x)*sqrt(a)*log((b^2*x^2 + 3*a*b*x + 2*a^2 + 2*sqrt(b^3*x^3 + 3*a*b^2*x^
2 + 3*a^2*b*x + a^3)*sqrt(a))/(b*x^2 + a*x)) - 2*sqrt(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*(3*a*b*x + a^2)
)/(a^3*b^2*x^3 + 2*a^4*b*x^2 + a^5*x), -(3*(b^3*x^3 + 2*a*b^2*x^2 + a^2*b*x)*sqrt(-a)*arctan(sqrt(b^3*x^3 + 3*
a*b^2*x^2 + 3*a^2*b*x + a^3)*sqrt(-a)/(a*b*x + a^2)) + sqrt(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*(3*a*b*x
+ a^2))/(a^3*b^2*x^3 + 2*a^4*b*x^2 + a^5*x)]

Sympy [F]

\[ \int \frac {1}{x^2 \sqrt {(a+b x)^3}} \, dx=\int \frac {1}{x^{2} \sqrt {\left (a + b x\right )^{3}}}\, dx \]

[In]

integrate(1/x**2/((b*x+a)**3)**(1/2),x)

[Out]

Integral(1/(x**2*sqrt((a + b*x)**3)), x)

Maxima [F]

\[ \int \frac {1}{x^2 \sqrt {(a+b x)^3}} \, dx=\int { \frac {1}{\sqrt {{\left (b x + a\right )}^{3}} x^{2}} \,d x } \]

[In]

integrate(1/x^2/((b*x+a)^3)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt((b*x + a)^3)*x^2), x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^2 \sqrt {(a+b x)^3}} \, dx=-\frac {3 \, b \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} - \frac {3 \, {\left (b x + a\right )} b - 2 \, a b}{{\left ({\left (b x + a\right )}^{\frac {3}{2}} - \sqrt {b x + a} a\right )} a^{2}} \]

[In]

integrate(1/x^2/((b*x+a)^3)^(1/2),x, algorithm="giac")

[Out]

-3*b*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^2) - (3*(b*x + a)*b - 2*a*b)/(((b*x + a)^(3/2) - sqrt(b*x + a)
*a)*a^2)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^2 \sqrt {(a+b x)^3}} \, dx=\int \frac {1}{x^2\,\sqrt {{\left (a+b\,x\right )}^3}} \,d x \]

[In]

int(1/(x^2*((a + b*x)^3)^(1/2)),x)

[Out]

int(1/(x^2*((a + b*x)^3)^(1/2)), x)