\(\int \frac {1}{(a+\frac {b}{x}) x^{7/2}} \, dx\) [1671]

Optimal result

Integrand size = 15, antiderivative size = 53 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^{7/2}} \, dx=-\frac {2}{3 b x^{3/2}}+\frac {2 a}{b^2 \sqrt {x}}+\frac {2 a^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{b^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {269, 53, 65, 211} \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^{7/2}} \, dx=\frac {2 a^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{b^{5/2}}+\frac {2 a}{b^2 \sqrt {x}}-\frac {2}{3 b x^{3/2}} \]

[In]

Int[1/((a + b/x)*x^(7/2)),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^{7/2}} \, dx=-\frac {2 (b-3 a x)}{3 b^2 x^{3/2}}+\frac {2 a^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{b^{5/2}} \]

[In]

Integrate[1/((a + b/x)*x^(7/2)),x]

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.79

method	result	size
risch	\(\frac {2 a x -\frac {2 b}{3}}{b^{2} x^{\frac {3}{2}}}+\frac {2 a^{2} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{b^{2} \sqrt {a b}}\)	\(42\)
derivativedivides	\(-\frac {2}{3 b \,x^{\frac {3}{2}}}+\frac {2 a}{\sqrt {x}\, b^{2}}+\frac {2 a^{2} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{b^{2} \sqrt {a b}}\)	\(43\)
default	\(-\frac {2}{3 b \,x^{\frac {3}{2}}}+\frac {2 a}{\sqrt {x}\, b^{2}}+\frac {2 a^{2} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{b^{2} \sqrt {a b}}\)	\(43\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 118, normalized size of antiderivative = 2.23 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^{7/2}} \, dx=\left [\frac {3 \, a x^{2} \sqrt {-\frac {a}{b}} \log \left (\frac {a x + 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - b}{a x + b}\right ) + 2 \, {\left (3 \, a x - b\right )} \sqrt {x}}{3 \, b^{2} x^{2}}, -\frac {2 \, {\left (3 \, a x^{2} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {\frac {a}{b}}}{a \sqrt {x}}\right ) - {\left (3 \, a x - b\right )} \sqrt {x}\right )}}{3 \, b^{2} x^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (49) = 98\).

Time = 5.24 (sec) , antiderivative size = 107, normalized size of antiderivative = 2.02 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^{7/2}} \, dx=\begin {cases} \frac {\tilde {\infty }}{x^{\frac {3}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{3 b x^{\frac {3}{2}}} & \text {for}\: a = 0 \\- \frac {2}{5 a x^{\frac {5}{2}}} & \text {for}\: b = 0 \\\frac {a \log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{b^{2} \sqrt {- \frac {b}{a}}} - \frac {a \log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{b^{2} \sqrt {- \frac {b}{a}}} + \frac {2 a}{b^{2} \sqrt {x}} - \frac {2}{3 b x^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^{7/2}} \, dx=-\frac {2 \, a^{2} \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{\sqrt {a b} b^{2}} + \frac {2 \, {\left (\frac {3 \, a}{\sqrt {x}} - \frac {b}{x^{\frac {3}{2}}}\right )}}{3 \, b^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.77 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^{7/2}} \, dx=\frac {2 \, a^{2} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {2 \, {\left (3 \, a x - b\right )}}{3 \, b^{2} x^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.72 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^{7/2}} \, dx=\frac {2\,a^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{b^{5/2}}-\frac {\frac {2}{3\,b}-\frac {2\,a\,x}{b^2}}{x^{3/2}} \]

[In]

int(1/(x^(7/2)*(a + b/x)),x)

[Out]

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