\(\int \frac {1}{(a+\frac {b}{x})^2 x^{7/2}} \, dx\) [1679]

Optimal result

Integrand size = 15, antiderivative size = 56 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{7/2}} \, dx=-\frac {3}{b^2 \sqrt {x}}+\frac {1}{b \sqrt {x} (b+a x)}-\frac {3 \sqrt {a} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{b^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {269, 44, 53, 65, 211} \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{7/2}} \, dx=-\frac {3 \sqrt {a} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{b^{5/2}}+\frac {1}{b \sqrt {x} (a x+b)}-\frac {3}{b^2 \sqrt {x}} \]

[In]

Int[1/((a + b/x)^2*x^(7/2)),x]

[Out]

-3/(b^2*Sqrt[x]) + 1/(b*Sqrt[x]*(b + a*x)) - (3*Sqrt[a]*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/b^(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 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 269

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{7/2}} \, dx=\frac {-2 b-3 a x}{b^2 \sqrt {x} (b+a x)}-\frac {3 \sqrt {a} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{b^{5/2}} \]

[In]

Integrate[1/((a + b/x)^2*x^(7/2)),x]

[Out]

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

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(-\frac {2 a \left (\frac {\sqrt {x}}{2 a x +2 b}+\frac {3 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{2}}-\frac {2}{\sqrt {x}\, b^{2}}\)	\(47\)
default	\(-\frac {2 a \left (\frac {\sqrt {x}}{2 a x +2 b}+\frac {3 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{2}}-\frac {2}{\sqrt {x}\, b^{2}}\)	\(47\)
risch	\(-\frac {2}{\sqrt {x}\, b^{2}}-\frac {a \sqrt {x}}{b^{2} \left (a x +b \right )}-\frac {3 a \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{b^{2} \sqrt {a b}}\)	\(48\)

[In]

int(1/(a+b/x)^2/x^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 147, normalized size of antiderivative = 2.62 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{7/2}} \, dx=\left [\frac {3 \, {\left (a x^{2} + b x\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {a x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - b}{a x + b}\right ) - 2 \, {\left (3 \, a x + 2 \, b\right )} \sqrt {x}}{2 \, {\left (a b^{2} x^{2} + b^{3} x\right )}}, \frac {3 \, {\left (a x^{2} + b x\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {\frac {a}{b}}}{a \sqrt {x}}\right ) - {\left (3 \, a x + 2 \, b\right )} \sqrt {x}}{a b^{2} x^{2} + b^{3} x}\right ] \]

[In]

integrate(1/(a+b/x)^2/x^(7/2),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 384 vs. \(2 (51) = 102\).

Time = 44.22 (sec) , antiderivative size = 384, normalized size of antiderivative = 6.86 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{7/2}} \, dx=\begin {cases} \frac {\tilde {\infty }}{\sqrt {x}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{b^{2} \sqrt {x}} & \text {for}\: a = 0 \\- \frac {2}{5 a^{2} x^{\frac {5}{2}}} & \text {for}\: b = 0 \\- \frac {3 a x^{\frac {3}{2}} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{2 a b^{2} x^{\frac {3}{2}} \sqrt {- \frac {b}{a}} + 2 b^{3} \sqrt {x} \sqrt {- \frac {b}{a}}} + \frac {3 a x^{\frac {3}{2}} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{2 a b^{2} x^{\frac {3}{2}} \sqrt {- \frac {b}{a}} + 2 b^{3} \sqrt {x} \sqrt {- \frac {b}{a}}} - \frac {6 a x \sqrt {- \frac {b}{a}}}{2 a b^{2} x^{\frac {3}{2}} \sqrt {- \frac {b}{a}} + 2 b^{3} \sqrt {x} \sqrt {- \frac {b}{a}}} - \frac {3 b \sqrt {x} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{2 a b^{2} x^{\frac {3}{2}} \sqrt {- \frac {b}{a}} + 2 b^{3} \sqrt {x} \sqrt {- \frac {b}{a}}} + \frac {3 b \sqrt {x} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{2 a b^{2} x^{\frac {3}{2}} \sqrt {- \frac {b}{a}} + 2 b^{3} \sqrt {x} \sqrt {- \frac {b}{a}}} - \frac {4 b \sqrt {- \frac {b}{a}}}{2 a b^{2} x^{\frac {3}{2}} \sqrt {- \frac {b}{a}} + 2 b^{3} \sqrt {x} \sqrt {- \frac {b}{a}}} & \text {otherwise} \end {cases} \]

[In]

integrate(1/(a+b/x)**2/x**(7/2),x)

[Out]

Piecewise((zoo/sqrt(x), Eq(a, 0) & Eq(b, 0)), (-2/(b**2*sqrt(x)), Eq(a, 0)), (-2/(5*a**2*x**(5/2)), Eq(b, 0)),
(-3*a*x**(3/2)*log(sqrt(x) - sqrt(-b/a))/(2*a*b**2*x**(3/2)*sqrt(-b/a) + 2*b**3*sqrt(x)*sqrt(-b/a)) + 3*a*x**
(3/2)*log(sqrt(x) + sqrt(-b/a))/(2*a*b**2*x**(3/2)*sqrt(-b/a) + 2*b**3*sqrt(x)*sqrt(-b/a)) - 6*a*x*sqrt(-b/a)/
(2*a*b**2*x**(3/2)*sqrt(-b/a) + 2*b**3*sqrt(x)*sqrt(-b/a)) - 3*b*sqrt(x)*log(sqrt(x) - sqrt(-b/a))/(2*a*b**2*x
**(3/2)*sqrt(-b/a) + 2*b**3*sqrt(x)*sqrt(-b/a)) + 3*b*sqrt(x)*log(sqrt(x) + sqrt(-b/a))/(2*a*b**2*x**(3/2)*sqr
t(-b/a) + 2*b**3*sqrt(x)*sqrt(-b/a)) - 4*b*sqrt(-b/a)/(2*a*b**2*x**(3/2)*sqrt(-b/a) + 2*b**3*sqrt(x)*sqrt(-b/a
)), True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{7/2}} \, dx=-\frac {a}{{\left (a b^{2} + \frac {b^{3}}{x}\right )} \sqrt {x}} + \frac {3 \, a \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{\sqrt {a b} b^{2}} - \frac {2}{b^{2} \sqrt {x}} \]

[In]

integrate(1/(a+b/x)^2/x^(7/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{7/2}} \, dx=-\frac {3 \, a \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} - \frac {3 \, a x + 2 \, b}{{\left (a x^{\frac {3}{2}} + b \sqrt {x}\right )} b^{2}} \]

[In]

integrate(1/(a+b/x)^2/x^(7/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.60 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{7/2}} \, dx=-\frac {\frac {2}{b}+\frac {3\,a\,x}{b^2}}{a\,x^{3/2}+b\,\sqrt {x}}-\frac {3\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{b^{5/2}} \]

[In]

int(1/(x^(7/2)*(a + b/x)^2),x)

[Out]

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