\(\int \frac {1}{(a+\frac {b}{x})^3 x^{13/2}} \, dx\) [1690]

Optimal result

Integrand size = 15, antiderivative size = 110 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{13/2}} \, dx=-\frac {63}{20 b^3 x^{5/2}}+\frac {21 a}{4 b^4 x^{3/2}}-\frac {63 a^2}{4 b^5 \sqrt {x}}+\frac {1}{2 b x^{5/2} (b+a x)^2}+\frac {9}{4 b^2 x^{5/2} (b+a x)}-\frac {63 a^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{11/2}} \]

[Out]

-63/20/b^3/x^(5/2)+21/4*a/b^4/x^(3/2)+1/2/b/x^(5/2)/(a*x+b)^2+9/4/b^2/x^(5/2)/(a*x+b)-63/4*a^(5/2)*arctan(a^(1
/2)*x^(1/2)/b^(1/2))/b^(11/2)-63/4*a^2/b^5/x^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {269, 44, 53, 65, 211} \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{13/2}} \, dx=-\frac {63 a^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{11/2}}-\frac {63 a^2}{4 b^5 \sqrt {x}}+\frac {21 a}{4 b^4 x^{3/2}}+\frac {9}{4 b^2 x^{5/2} (a x+b)}+\frac {1}{2 b x^{5/2} (a x+b)^2}-\frac {63}{20 b^3 x^{5/2}} \]

[In]

Int[1/((a + b/x)^3*x^(13/2)),x]

[Out]

-63/(20*b^3*x^(5/2)) + (21*a)/(4*b^4*x^(3/2)) - (63*a^2)/(4*b^5*Sqrt[x]) + 1/(2*b*x^(5/2)*(b + a*x)^2) + 9/(4*
b^2*x^(5/2)*(b + a*x)) - (63*a^(5/2)*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/(4*b^(11/2))

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)^3} \, dx \\ & = \frac {1}{2 b x^{5/2} (b+a x)^2}+\frac {9 \int \frac {1}{x^{7/2} (b+a x)^2} \, dx}{4 b} \\ & = \frac {1}{2 b x^{5/2} (b+a x)^2}+\frac {9}{4 b^2 x^{5/2} (b+a x)}+\frac {63 \int \frac {1}{x^{7/2} (b+a x)} \, dx}{8 b^2} \\ & = -\frac {63}{20 b^3 x^{5/2}}+\frac {1}{2 b x^{5/2} (b+a x)^2}+\frac {9}{4 b^2 x^{5/2} (b+a x)}-\frac {(63 a) \int \frac {1}{x^{5/2} (b+a x)} \, dx}{8 b^3} \\ & = -\frac {63}{20 b^3 x^{5/2}}+\frac {21 a}{4 b^4 x^{3/2}}+\frac {1}{2 b x^{5/2} (b+a x)^2}+\frac {9}{4 b^2 x^{5/2} (b+a x)}+\frac {\left (63 a^2\right ) \int \frac {1}{x^{3/2} (b+a x)} \, dx}{8 b^4} \\ & = -\frac {63}{20 b^3 x^{5/2}}+\frac {21 a}{4 b^4 x^{3/2}}-\frac {63 a^2}{4 b^5 \sqrt {x}}+\frac {1}{2 b x^{5/2} (b+a x)^2}+\frac {9}{4 b^2 x^{5/2} (b+a x)}-\frac {\left (63 a^3\right ) \int \frac {1}{\sqrt {x} (b+a x)} \, dx}{8 b^5} \\ & = -\frac {63}{20 b^3 x^{5/2}}+\frac {21 a}{4 b^4 x^{3/2}}-\frac {63 a^2}{4 b^5 \sqrt {x}}+\frac {1}{2 b x^{5/2} (b+a x)^2}+\frac {9}{4 b^2 x^{5/2} (b+a x)}-\frac {\left (63 a^3\right ) \text {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,\sqrt {x}\right )}{4 b^5} \\ & = -\frac {63}{20 b^3 x^{5/2}}+\frac {21 a}{4 b^4 x^{3/2}}-\frac {63 a^2}{4 b^5 \sqrt {x}}+\frac {1}{2 b x^{5/2} (b+a x)^2}+\frac {9}{4 b^2 x^{5/2} (b+a x)}-\frac {63 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{11/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{13/2}} \, dx=\frac {-8 b^4+24 a b^3 x-168 a^2 b^2 x^2-525 a^3 b x^3-315 a^4 x^4}{20 b^5 x^{5/2} (b+a x)^2}-\frac {63 a^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{11/2}} \]

[In]

Integrate[1/((a + b/x)^3*x^(13/2)),x]

[Out]

(-8*b^4 + 24*a*b^3*x - 168*a^2*b^2*x^2 - 525*a^3*b*x^3 - 315*a^4*x^4)/(20*b^5*x^(5/2)*(b + a*x)^2) - (63*a^(5/
2)*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/(4*b^(11/2))

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.69

method	result	size
risch	\(-\frac {2 \left (30 a^{2} x^{2}-5 a b x +b^{2}\right )}{5 b^{5} x^{\frac {5}{2}}}-\frac {a^{3} \left (\frac {\frac {15 a \,x^{\frac {3}{2}}}{4}+\frac {17 b \sqrt {x}}{4}}{\left (a x +b \right )^{2}}+\frac {63 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}\right )}{b^{5}}\)	\(76\)
derivativedivides	\(-\frac {2 a^{3} \left (\frac {\frac {15 a \,x^{\frac {3}{2}}}{8}+\frac {17 b \sqrt {x}}{8}}{\left (a x +b \right )^{2}}+\frac {63 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{5}}-\frac {2}{5 b^{3} x^{\frac {5}{2}}}-\frac {12 a^{2}}{b^{5} \sqrt {x}}+\frac {2 a}{b^{4} x^{\frac {3}{2}}}\)	\(78\)
default	\(-\frac {2 a^{3} \left (\frac {\frac {15 a \,x^{\frac {3}{2}}}{8}+\frac {17 b \sqrt {x}}{8}}{\left (a x +b \right )^{2}}+\frac {63 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{5}}-\frac {2}{5 b^{3} x^{\frac {5}{2}}}-\frac {12 a^{2}}{b^{5} \sqrt {x}}+\frac {2 a}{b^{4} x^{\frac {3}{2}}}\)	\(78\)

[In]

int(1/(a+b/x)^3/x^(13/2),x,method=_RETURNVERBOSE)

[Out]

-2/5*(30*a^2*x^2-5*a*b*x+b^2)/b^5/x^(5/2)-1/b^5*a^3*(2*(15/8*a*x^(3/2)+17/8*b*x^(1/2))/(a*x+b)^2+63/4/(a*b)^(1
/2)*arctan(a*x^(1/2)/(a*b)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.51 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{13/2}} \, dx=\left [\frac {315 \, {\left (a^{4} x^{5} + 2 \, a^{3} b x^{4} + a^{2} b^{2} x^{3}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {a x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - b}{a x + b}\right ) - 2 \, {\left (315 \, a^{4} x^{4} + 525 \, a^{3} b x^{3} + 168 \, a^{2} b^{2} x^{2} - 24 \, a b^{3} x + 8 \, b^{4}\right )} \sqrt {x}}{40 \, {\left (a^{2} b^{5} x^{5} + 2 \, a b^{6} x^{4} + b^{7} x^{3}\right )}}, \frac {315 \, {\left (a^{4} x^{5} + 2 \, a^{3} b x^{4} + a^{2} b^{2} x^{3}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {\frac {a}{b}}}{a \sqrt {x}}\right ) - {\left (315 \, a^{4} x^{4} + 525 \, a^{3} b x^{3} + 168 \, a^{2} b^{2} x^{2} - 24 \, a b^{3} x + 8 \, b^{4}\right )} \sqrt {x}}{20 \, {\left (a^{2} b^{5} x^{5} + 2 \, a b^{6} x^{4} + b^{7} x^{3}\right )}}\right ] \]

[In]

integrate(1/(a+b/x)^3/x^(13/2),x, algorithm="fricas")

[Out]

[1/40*(315*(a^4*x^5 + 2*a^3*b*x^4 + a^2*b^2*x^3)*sqrt(-a/b)*log((a*x - 2*b*sqrt(x)*sqrt(-a/b) - b)/(a*x + b))
- 2*(315*a^4*x^4 + 525*a^3*b*x^3 + 168*a^2*b^2*x^2 - 24*a*b^3*x + 8*b^4)*sqrt(x))/(a^2*b^5*x^5 + 2*a*b^6*x^4 +
b^7*x^3), 1/20*(315*(a^4*x^5 + 2*a^3*b*x^4 + a^2*b^2*x^3)*sqrt(a/b)*arctan(b*sqrt(a/b)/(a*sqrt(x))) - (315*a^
4*x^4 + 525*a^3*b*x^3 + 168*a^2*b^2*x^2 - 24*a*b^3*x + 8*b^4)*sqrt(x))/(a^2*b^5*x^5 + 2*a*b^6*x^4 + b^7*x^3)]

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{13/2}} \, dx=\text {Timed out} \]

[In]

integrate(1/(a+b/x)**3/x**(13/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{13/2}} \, dx=-\frac {\frac {15 \, a^{4}}{\sqrt {x}} + \frac {17 \, a^{3} b}{x^{\frac {3}{2}}}}{4 \, {\left (a^{2} b^{5} + \frac {2 \, a b^{6}}{x} + \frac {b^{7}}{x^{2}}\right )}} + \frac {63 \, a^{3} \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{4 \, \sqrt {a b} b^{5}} - \frac {2 \, {\left (\frac {30 \, a^{2}}{\sqrt {x}} - \frac {5 \, a b}{x^{\frac {3}{2}}} + \frac {b^{2}}{x^{\frac {5}{2}}}\right )}}{5 \, b^{5}} \]

[In]

integrate(1/(a+b/x)^3/x^(13/2),x, algorithm="maxima")

[Out]

-1/4*(15*a^4/sqrt(x) + 17*a^3*b/x^(3/2))/(a^2*b^5 + 2*a*b^6/x + b^7/x^2) + 63/4*a^3*arctan(b/(sqrt(a*b)*sqrt(x
)))/(sqrt(a*b)*b^5) - 2/5*(30*a^2/sqrt(x) - 5*a*b/x^(3/2) + b^2/x^(5/2))/b^5

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.73 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{13/2}} \, dx=-\frac {63 \, a^{3} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{5}} - \frac {15 \, a^{4} x^{\frac {3}{2}} + 17 \, a^{3} b \sqrt {x}}{4 \, {\left (a x + b\right )}^{2} b^{5}} - \frac {2 \, {\left (30 \, a^{2} x^{2} - 5 \, a b x + b^{2}\right )}}{5 \, b^{5} x^{\frac {5}{2}}} \]

[In]

integrate(1/(a+b/x)^3/x^(13/2),x, algorithm="giac")

[Out]

-63/4*a^3*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^5) - 1/4*(15*a^4*x^(3/2) + 17*a^3*b*sqrt(x))/((a*x + b)^2*b
^5) - 2/5*(30*a^2*x^2 - 5*a*b*x + b^2)/(b^5*x^(5/2))

Mupad [B] (verification not implemented)

Time = 5.96 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^{13/2}} \, dx=-\frac {\frac {2}{5\,b}+\frac {42\,a^2\,x^2}{5\,b^3}+\frac {105\,a^3\,x^3}{4\,b^4}+\frac {63\,a^4\,x^4}{4\,b^5}-\frac {6\,a\,x}{5\,b^2}}{a^2\,x^{9/2}+b^2\,x^{5/2}+2\,a\,b\,x^{7/2}}-\frac {63\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{4\,b^{11/2}} \]

[In]

int(1/(x^(13/2)*(a + b/x)^3),x)

[Out]

- (2/(5*b) + (42*a^2*x^2)/(5*b^3) + (105*a^3*x^3)/(4*b^4) + (63*a^4*x^4)/(4*b^5) - (6*a*x)/(5*b^2))/(a^2*x^(9/
2) + b^2*x^(5/2) + 2*a*b*x^(7/2)) - (63*a^(5/2)*atan((a^(1/2)*x^(1/2))/b^(1/2)))/(4*b^(11/2))