3.4.49 \(\int \frac {1}{x^2 (a+b x)^{3/2}} \, dx\) [349]

Optimal. Leaf size=57 \[ -\frac {3 b}{a^2 \sqrt {a+b x}}-\frac {1}{a x \sqrt {a+b x}}+\frac {3 b \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {44, 53, 65, 214} \begin {gather*} \frac {3 b \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {3 b}{a^2 \sqrt {a+b x}}-\frac {1}{a x \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x)^(3/2)),x]

[Out]

(-3*b)/(a^2*Sqrt[a + b*x]) - 1/(a*x*Sqrt[a + b*x]) + (3*b*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/a^(5/2)

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]

Rubi steps

\begin {align*} \int \frac {1}{x^2 (a+b x)^{3/2}} \, dx &=\frac {2}{a x \sqrt {a+b x}}+\frac {3 \int \frac {1}{x^2 \sqrt {a+b x}} \, dx}{a}\\ &=\frac {2}{a x \sqrt {a+b x}}-\frac {3 \sqrt {a+b x}}{a^2 x}-\frac {(3 b) \int \frac {1}{x \sqrt {a+b x}} \, dx}{2 a^2}\\ &=\frac {2}{a x \sqrt {a+b x}}-\frac {3 \sqrt {a+b x}}{a^2 x}-\frac {3 \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{a^2}\\ &=\frac {2}{a x \sqrt {a+b x}}-\frac {3 \sqrt {a+b x}}{a^2 x}+\frac {3 b \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 49, normalized size = 0.86 \begin {gather*} \frac {-a-3 b x}{a^2 x \sqrt {a+b x}}+\frac {3 b \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + b*x)^(3/2)),x]

[Out]

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

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Maple [A]
time = 0.10, size = 54, normalized size = 0.95

method	result	size
risch	\(-\frac {\sqrt {b x +a}}{a^{2} x}-\frac {b \left (\frac {4}{\sqrt {b x +a}}-\frac {6 \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}\right )}{2 a^{2}}\)	\(50\)
derivativedivides	\(2 b \left (-\frac {1}{a^{2} \sqrt {b x +a}}+\frac {-\frac {\sqrt {b x +a}}{2 b x}+\frac {3 \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}}{a^{2}}\right )\)	\(54\)
default	\(2 b \left (-\frac {1}{a^{2} \sqrt {b x +a}}+\frac {-\frac {\sqrt {b x +a}}{2 b x}+\frac {3 \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}}{a^{2}}\right )\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.47, size = 76, normalized size = 1.33 \begin {gather*} -\frac {3 \, {\left (b x + a\right )} b - 2 \, a b}{{\left (b x + a\right )}^{\frac {3}{2}} a^{2} - \sqrt {b x + a} a^{3}} - \frac {3 \, b \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{2 \, a^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*(b*x + a)*b - 2*a*b)/((b*x + a)^(3/2)*a^2 - sqrt(b*x + a)*a^3) - 3/2*b*log((sqrt(b*x + a) - sqrt(a))/(sqrt
(b*x + a) + sqrt(a)))/a^(5/2)

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Fricas [A]
time = 0.46, size = 151, normalized size = 2.65 \begin {gather*} \left [\frac {3 \, {\left (b^{2} x^{2} + a b x\right )} \sqrt {a} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (3 \, a b x + a^{2}\right )} \sqrt {b x + a}}{2 \, {\left (a^{3} b x^{2} + a^{4} x\right )}}, -\frac {3 \, {\left (b^{2} x^{2} + a b x\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (3 \, a b x + a^{2}\right )} \sqrt {b x + a}}{a^{3} b x^{2} + a^{4} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 1.77, size = 73, normalized size = 1.28 \begin {gather*} - \frac {1}{a \sqrt {b} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {3 \sqrt {b}}{a^{2} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {3 b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{a^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(a*sqrt(b)*x**(3/2)*sqrt(a/(b*x) + 1)) - 3*sqrt(b)/(a**2*sqrt(x)*sqrt(a/(b*x) + 1)) + 3*b*asinh(sqrt(a)/(sq
rt(b)*sqrt(x)))/a**(5/2)

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Giac [A]
time = 0.74, size = 64, normalized size = 1.12 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Mupad [B]
time = 0.12, size = 60, normalized size = 1.05 \begin {gather*} \frac {3\,b\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {\frac {2\,b}{a}-\frac {3\,b\,\left (a+b\,x\right )}{a^2}}{a\,\sqrt {a+b\,x}-{\left (a+b\,x\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a + b*x)^(3/2)),x)

[Out]

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

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