\(\int \frac {x^{3/2}}{(a+\frac {b}{x})^2} \, dx\) [1674]

Optimal result

Integrand size = 15, antiderivative size = 85 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^2} \, dx=\frac {7 b^2 \sqrt {x}}{a^4}-\frac {7 b x^{3/2}}{3 a^3}+\frac {7 x^{5/2}}{5 a^2}-\frac {x^{7/2}}{a (b+a x)}-\frac {7 b^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{9/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 85, 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, 43, 52, 65, 211} \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^2} \, dx=-\frac {7 b^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{9/2}}+\frac {7 b^2 \sqrt {x}}{a^4}-\frac {7 b x^{3/2}}{3 a^3}+\frac {7 x^{5/2}}{5 a^2}-\frac {x^{7/2}}{a (a x+b)} \]

[In]

Int[x^(3/2)/(a + b/x)^2,x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 {x^{7/2}}{(b+a x)^2} \, dx \\ & = -\frac {x^{7/2}}{a (b+a x)}+\frac {7 \int \frac {x^{5/2}}{b+a x} \, dx}{2 a} \\ & = \frac {7 x^{5/2}}{5 a^2}-\frac {x^{7/2}}{a (b+a x)}-\frac {(7 b) \int \frac {x^{3/2}}{b+a x} \, dx}{2 a^2} \\ & = -\frac {7 b x^{3/2}}{3 a^3}+\frac {7 x^{5/2}}{5 a^2}-\frac {x^{7/2}}{a (b+a x)}+\frac {\left (7 b^2\right ) \int \frac {\sqrt {x}}{b+a x} \, dx}{2 a^3} \\ & = \frac {7 b^2 \sqrt {x}}{a^4}-\frac {7 b x^{3/2}}{3 a^3}+\frac {7 x^{5/2}}{5 a^2}-\frac {x^{7/2}}{a (b+a x)}-\frac {\left (7 b^3\right ) \int \frac {1}{\sqrt {x} (b+a x)} \, dx}{2 a^4} \\ & = \frac {7 b^2 \sqrt {x}}{a^4}-\frac {7 b x^{3/2}}{3 a^3}+\frac {7 x^{5/2}}{5 a^2}-\frac {x^{7/2}}{a (b+a x)}-\frac {\left (7 b^3\right ) \text {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,\sqrt {x}\right )}{a^4} \\ & = \frac {7 b^2 \sqrt {x}}{a^4}-\frac {7 b x^{3/2}}{3 a^3}+\frac {7 x^{5/2}}{5 a^2}-\frac {x^{7/2}}{a (b+a x)}-\frac {7 b^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.93 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^2} \, dx=\frac {\sqrt {x} \left (105 b^3+70 a b^2 x-14 a^2 b x^2+6 a^3 x^3\right )}{15 a^4 (b+a x)}-\frac {7 b^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{9/2}} \]

[In]

Integrate[x^(3/2)/(a + b/x)^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(\frac {\frac {2 a^{2} x^{\frac {5}{2}}}{5}-\frac {4 a b \,x^{\frac {3}{2}}}{3}+6 b^{2} \sqrt {x}}{a^{4}}-\frac {2 b^{3} \left (-\frac {\sqrt {x}}{2 \left (a x +b \right )}+\frac {7 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{4}}\)	\(70\)
default	\(\frac {\frac {2 a^{2} x^{\frac {5}{2}}}{5}-\frac {4 a b \,x^{\frac {3}{2}}}{3}+6 b^{2} \sqrt {x}}{a^{4}}-\frac {2 b^{3} \left (-\frac {\sqrt {x}}{2 \left (a x +b \right )}+\frac {7 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{4}}\)	\(70\)
risch	\(\frac {2 \left (3 a^{2} x^{2}-10 a b x +45 b^{2}\right ) \sqrt {x}}{15 a^{4}}+\frac {b^{3} \sqrt {x}}{a^{4} \left (a x +b \right )}-\frac {7 b^{3} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{a^{4} \sqrt {a b}}\)	\(70\)

[In]

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

[Out]

2/a^4*(1/5*a^2*x^(5/2)-2/3*a*b*x^(3/2)+3*b^2*x^(1/2))-2*b^3/a^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 = 0.25 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.21 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^2} \, dx=\left [\frac {105 \, {\left (a b^{2} x + b^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {a x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - b}{a x + b}\right ) + 2 \, {\left (6 \, a^{3} x^{3} - 14 \, a^{2} b x^{2} + 70 \, a b^{2} x + 105 \, b^{3}\right )} \sqrt {x}}{30 \, {\left (a^{5} x + a^{4} b\right )}}, -\frac {105 \, {\left (a b^{2} x + b^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {x} \sqrt {\frac {b}{a}}}{b}\right ) - {\left (6 \, a^{3} x^{3} - 14 \, a^{2} b x^{2} + 70 \, a b^{2} x + 105 \, b^{3}\right )} \sqrt {x}}{15 \, {\left (a^{5} x + a^{4} b\right )}}\right ] \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 442 vs. \(2 (78) = 156\).

Time = 7.74 (sec) , antiderivative size = 442, normalized size of antiderivative = 5.20 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^2} \, dx=\begin {cases} \tilde {\infty } x^{\frac {9}{2}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {9}{2}}}{9 b^{2}} & \text {for}\: a = 0 \\\frac {2 x^{\frac {5}{2}}}{5 a^{2}} & \text {for}\: b = 0 \\\frac {12 a^{4} x^{\frac {7}{2}} \sqrt {- \frac {b}{a}}}{30 a^{6} x \sqrt {- \frac {b}{a}} + 30 a^{5} b \sqrt {- \frac {b}{a}}} - \frac {28 a^{3} b x^{\frac {5}{2}} \sqrt {- \frac {b}{a}}}{30 a^{6} x \sqrt {- \frac {b}{a}} + 30 a^{5} b \sqrt {- \frac {b}{a}}} + \frac {140 a^{2} b^{2} x^{\frac {3}{2}} \sqrt {- \frac {b}{a}}}{30 a^{6} x \sqrt {- \frac {b}{a}} + 30 a^{5} b \sqrt {- \frac {b}{a}}} + \frac {210 a b^{3} \sqrt {x} \sqrt {- \frac {b}{a}}}{30 a^{6} x \sqrt {- \frac {b}{a}} + 30 a^{5} b \sqrt {- \frac {b}{a}}} - \frac {105 a b^{3} x \log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{30 a^{6} x \sqrt {- \frac {b}{a}} + 30 a^{5} b \sqrt {- \frac {b}{a}}} + \frac {105 a b^{3} x \log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{30 a^{6} x \sqrt {- \frac {b}{a}} + 30 a^{5} b \sqrt {- \frac {b}{a}}} - \frac {105 b^{4} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{30 a^{6} x \sqrt {- \frac {b}{a}} + 30 a^{5} b \sqrt {- \frac {b}{a}}} + \frac {105 b^{4} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{30 a^{6} x \sqrt {- \frac {b}{a}} + 30 a^{5} b \sqrt {- \frac {b}{a}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((zoo*x**(9/2), Eq(a, 0) & Eq(b, 0)), (2*x**(9/2)/(9*b**2), Eq(a, 0)), (2*x**(5/2)/(5*a**2), Eq(b, 0)
), (12*a**4*x**(7/2)*sqrt(-b/a)/(30*a**6*x*sqrt(-b/a) + 30*a**5*b*sqrt(-b/a)) - 28*a**3*b*x**(5/2)*sqrt(-b/a)/
(30*a**6*x*sqrt(-b/a) + 30*a**5*b*sqrt(-b/a)) + 140*a**2*b**2*x**(3/2)*sqrt(-b/a)/(30*a**6*x*sqrt(-b/a) + 30*a
**5*b*sqrt(-b/a)) + 210*a*b**3*sqrt(x)*sqrt(-b/a)/(30*a**6*x*sqrt(-b/a) + 30*a**5*b*sqrt(-b/a)) - 105*a*b**3*x
*log(sqrt(x) - sqrt(-b/a))/(30*a**6*x*sqrt(-b/a) + 30*a**5*b*sqrt(-b/a)) + 105*a*b**3*x*log(sqrt(x) + sqrt(-b/
a))/(30*a**6*x*sqrt(-b/a) + 30*a**5*b*sqrt(-b/a)) - 105*b**4*log(sqrt(x) - sqrt(-b/a))/(30*a**6*x*sqrt(-b/a) +
30*a**5*b*sqrt(-b/a)) + 105*b**4*log(sqrt(x) + sqrt(-b/a))/(30*a**6*x*sqrt(-b/a) + 30*a**5*b*sqrt(-b/a)), Tru
e))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.91 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^2} \, dx=\frac {6 \, a^{3} - \frac {14 \, a^{2} b}{x} + \frac {70 \, a b^{2}}{x^{2}} + \frac {105 \, b^{3}}{x^{3}}}{15 \, {\left (\frac {a^{5}}{x^{\frac {5}{2}}} + \frac {a^{4} b}{x^{\frac {7}{2}}}\right )}} + \frac {7 \, b^{3} \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{\sqrt {a b} a^{4}} \]

[In]

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

[Out]

1/15*(6*a^3 - 14*a^2*b/x + 70*a*b^2/x^2 + 105*b^3/x^3)/(a^5/x^(5/2) + a^4*b/x^(7/2)) + 7*b^3*arctan(b/(sqrt(a*
b)*sqrt(x)))/(sqrt(a*b)*a^4)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.89 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^2} \, dx=-\frac {7 \, b^{3} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{4}} + \frac {b^{3} \sqrt {x}}{{\left (a x + b\right )} a^{4}} + \frac {2 \, {\left (3 \, a^{8} x^{\frac {5}{2}} - 10 \, a^{7} b x^{\frac {3}{2}} + 45 \, a^{6} b^{2} \sqrt {x}\right )}}{15 \, a^{10}} \]

[In]

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

[Out]

-7*b^3*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4) + b^3*sqrt(x)/((a*x + b)*a^4) + 2/15*(3*a^8*x^(5/2) - 10*a^
7*b*x^(3/2) + 45*a^6*b^2*sqrt(x))/a^10

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.80 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^2} \, dx=\frac {2\,x^{5/2}}{5\,a^2}-\frac {4\,b\,x^{3/2}}{3\,a^3}+\frac {6\,b^2\,\sqrt {x}}{a^4}+\frac {b^3\,\sqrt {x}}{x\,a^5+b\,a^4}-\frac {7\,b^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{a^{9/2}} \]

[In]

int(x^(3/2)/(a + b/x)^2,x)

[Out]

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