\(\int \frac {1}{(a+\frac {b}{x})^2 x^{11/2}} \, dx\) [1681]

Optimal result

Integrand size = 15, antiderivative size = 84 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{11/2}} \, dx=-\frac {7}{5 b^2 x^{5/2}}+\frac {7 a}{3 b^3 x^{3/2}}-\frac {7 a^2}{b^4 \sqrt {x}}+\frac {1}{b x^{5/2} (b+a x)}-\frac {7 a^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{b^{9/2}} \]

[Out]

-7/5/b^2/x^(5/2)+7/3*a/b^3/x^(3/2)+1/b/x^(5/2)/(a*x+b)-7*a^(5/2)*arctan(a^(1/2)*x^(1/2)/b^(1/2))/b^(9/2)-7*a^2
/b^4/x^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {269, 44, 53, 65, 211} \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{11/2}} \, dx=-\frac {7 a^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{b^{9/2}}-\frac {7 a^2}{b^4 \sqrt {x}}+\frac {7 a}{3 b^3 x^{3/2}}+\frac {1}{b x^{5/2} (a x+b)}-\frac {7}{5 b^2 x^{5/2}} \]

[In]

Int[1/((a + b/x)^2*x^(11/2)),x]

[Out]

-7/(5*b^2*x^(5/2)) + (7*a)/(3*b^3*x^(3/2)) - (7*a^2)/(b^4*Sqrt[x]) + 1/(b*x^(5/2)*(b + a*x)) - (7*a^(5/2)*ArcT
an[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/b^(9/2)

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^{7/2} (b+a x)^2} \, dx \\ & = \frac {1}{b x^{5/2} (b+a x)}+\frac {7 \int \frac {1}{x^{7/2} (b+a x)} \, dx}{2 b} \\ & = -\frac {7}{5 b^2 x^{5/2}}+\frac {1}{b x^{5/2} (b+a x)}-\frac {(7 a) \int \frac {1}{x^{5/2} (b+a x)} \, dx}{2 b^2} \\ & = -\frac {7}{5 b^2 x^{5/2}}+\frac {7 a}{3 b^3 x^{3/2}}+\frac {1}{b x^{5/2} (b+a x)}+\frac {\left (7 a^2\right ) \int \frac {1}{x^{3/2} (b+a x)} \, dx}{2 b^3} \\ & = -\frac {7}{5 b^2 x^{5/2}}+\frac {7 a}{3 b^3 x^{3/2}}-\frac {7 a^2}{b^4 \sqrt {x}}+\frac {1}{b x^{5/2} (b+a x)}-\frac {\left (7 a^3\right ) \int \frac {1}{\sqrt {x} (b+a x)} \, dx}{2 b^4} \\ & = -\frac {7}{5 b^2 x^{5/2}}+\frac {7 a}{3 b^3 x^{3/2}}-\frac {7 a^2}{b^4 \sqrt {x}}+\frac {1}{b x^{5/2} (b+a x)}-\frac {\left (7 a^3\right ) \text {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,\sqrt {x}\right )}{b^4} \\ & = -\frac {7}{5 b^2 x^{5/2}}+\frac {7 a}{3 b^3 x^{3/2}}-\frac {7 a^2}{b^4 \sqrt {x}}+\frac {1}{b x^{5/2} (b+a x)}-\frac {7 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{b^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.94 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{11/2}} \, dx=\frac {-6 b^3+14 a b^2 x-70 a^2 b x^2-105 a^3 x^3}{15 b^4 x^{5/2} (b+a x)}-\frac {7 a^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{b^{9/2}} \]

[In]

Integrate[1/((a + b/x)^2*x^(11/2)),x]

[Out]

(-6*b^3 + 14*a*b^2*x - 70*a^2*b*x^2 - 105*a^3*x^3)/(15*b^4*x^(5/2)*(b + a*x)) - (7*a^(5/2)*ArcTan[(Sqrt[a]*Sqr
t[x])/Sqrt[b]])/b^(9/2)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(-\frac {2}{5 b^{2} x^{\frac {5}{2}}}+\frac {4 a}{3 b^{3} x^{\frac {3}{2}}}-\frac {6 a^{2}}{b^{4} \sqrt {x}}-\frac {2 a^{3} \left (\frac {\sqrt {x}}{2 a x +2 b}+\frac {7 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{4}}\)	\(69\)
default	\(-\frac {2}{5 b^{2} x^{\frac {5}{2}}}+\frac {4 a}{3 b^{3} x^{\frac {3}{2}}}-\frac {6 a^{2}}{b^{4} \sqrt {x}}-\frac {2 a^{3} \left (\frac {\sqrt {x}}{2 a x +2 b}+\frac {7 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{4}}\)	\(69\)
risch	\(-\frac {2 \left (45 a^{2} x^{2}-10 a b x +3 b^{2}\right )}{15 b^{4} x^{\frac {5}{2}}}-\frac {a^{3} \sqrt {x}}{b^{4} \left (a x +b \right )}-\frac {7 a^{3} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{b^{4} \sqrt {a b}}\)	\(71\)

[In]

int(1/(a+b/x)^2/x^(11/2),x,method=_RETURNVERBOSE)

[Out]

-2/5/b^2/x^(5/2)+4/3*a/b^3/x^(3/2)-6*a^2/b^4/x^(1/2)-2*a^3/b^4*(1/2*x^(1/2)/(a*x+b)+7/2/(a*b)^(1/2)*arctan(a*x
^(1/2)/(a*b)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.50 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{11/2}} \, dx=\left [\frac {105 \, {\left (a^{3} x^{4} + a^{2} b x^{3}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {a x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - b}{a x + b}\right ) - 2 \, {\left (105 \, a^{3} x^{3} + 70 \, a^{2} b x^{2} - 14 \, a b^{2} x + 6 \, b^{3}\right )} \sqrt {x}}{30 \, {\left (a b^{4} x^{4} + b^{5} x^{3}\right )}}, \frac {105 \, {\left (a^{3} x^{4} + a^{2} b x^{3}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {\frac {a}{b}}}{a \sqrt {x}}\right ) - {\left (105 \, a^{3} x^{3} + 70 \, a^{2} b x^{2} - 14 \, a b^{2} x + 6 \, b^{3}\right )} \sqrt {x}}{15 \, {\left (a b^{4} x^{4} + b^{5} x^{3}\right )}}\right ] \]

[In]

integrate(1/(a+b/x)^2/x^(11/2),x, algorithm="fricas")

[Out]

[1/30*(105*(a^3*x^4 + a^2*b*x^3)*sqrt(-a/b)*log((a*x - 2*b*sqrt(x)*sqrt(-a/b) - b)/(a*x + b)) - 2*(105*a^3*x^3
+ 70*a^2*b*x^2 - 14*a*b^2*x + 6*b^3)*sqrt(x))/(a*b^4*x^4 + b^5*x^3), 1/15*(105*(a^3*x^4 + a^2*b*x^3)*sqrt(a/b
)*arctan(b*sqrt(a/b)/(a*sqrt(x))) - (105*a^3*x^3 + 70*a^2*b*x^2 - 14*a*b^2*x + 6*b^3)*sqrt(x))/(a*b^4*x^4 + b^
5*x^3)]

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{11/2}} \, dx=\text {Timed out} \]

[In]

integrate(1/(a+b/x)**2/x**(11/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{11/2}} \, dx=-\frac {a^{3}}{{\left (a b^{4} + \frac {b^{5}}{x}\right )} \sqrt {x}} + \frac {7 \, a^{3} \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{\sqrt {a b} b^{4}} - \frac {2 \, {\left (\frac {45 \, a^{2}}{\sqrt {x}} - \frac {10 \, a b}{x^{\frac {3}{2}}} + \frac {3 \, b^{2}}{x^{\frac {5}{2}}}\right )}}{15 \, b^{4}} \]

[In]

integrate(1/(a+b/x)^2/x^(11/2),x, algorithm="maxima")

[Out]

-a^3/((a*b^4 + b^5/x)*sqrt(x)) + 7*a^3*arctan(b/(sqrt(a*b)*sqrt(x)))/(sqrt(a*b)*b^4) - 2/15*(45*a^2/sqrt(x) -
10*a*b/x^(3/2) + 3*b^2/x^(5/2))/b^4

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{11/2}} \, dx=-\frac {7 \, a^{3} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{4}} - \frac {a^{3} \sqrt {x}}{{\left (a x + b\right )} b^{4}} - \frac {2 \, {\left (45 \, a^{2} x^{2} - 10 \, a b x + 3 \, b^{2}\right )}}{15 \, b^{4} x^{\frac {5}{2}}} \]

[In]

integrate(1/(a+b/x)^2/x^(11/2),x, algorithm="giac")

[Out]

-7*a^3*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4) - a^3*sqrt(x)/((a*x + b)*b^4) - 2/15*(45*a^2*x^2 - 10*a*b*x
+ 3*b^2)/(b^4*x^(5/2))

Mupad [B] (verification not implemented)

Time = 5.74 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^{11/2}} \, dx=-\frac {\frac {2}{5\,b}+\frac {14\,a^2\,x^2}{3\,b^3}+\frac {7\,a^3\,x^3}{b^4}-\frac {14\,a\,x}{15\,b^2}}{a\,x^{7/2}+b\,x^{5/2}}-\frac {7\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{b^{9/2}} \]

[In]

int(1/(x^(11/2)*(a + b/x)^2),x)

[Out]

- (2/(5*b) + (14*a^2*x^2)/(3*b^3) + (7*a^3*x^3)/b^4 - (14*a*x)/(15*b^2))/(a*x^(7/2) + b*x^(5/2)) - (7*a^(5/2)*
atan((a^(1/2)*x^(1/2))/b^(1/2)))/b^(9/2)