3.1673 \(\int \frac {x^{5/2}}{(a+\frac {b}{x})^2} \, dx\)

Optimal. Leaf size=98 \[ \frac {9 b^{7/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{11/2}}-\frac {9 b^3 \sqrt {x}}{a^5}+\frac {3 b^2 x^{3/2}}{a^4}-\frac {9 b x^{5/2}}{5 a^3}+\frac {9 x^{7/2}}{7 a^2}-\frac {x^{9/2}}{a (a x+b)} \]

[Out]

3*b^2*x^(3/2)/a^4-9/5*b*x^(5/2)/a^3+9/7*x^(7/2)/a^2-x^(9/2)/a/(a*x+b)+9*b^(7/2)*arctan(a^(1/2)*x^(1/2)/b^(1/2)
)/a^(11/2)-9*b^3*x^(1/2)/a^5

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Rubi [A]  time = 0.04, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {263, 47, 50, 63, 205} \[ \frac {3 b^2 x^{3/2}}{a^4}-\frac {9 b^3 \sqrt {x}}{a^5}+\frac {9 b^{7/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{11/2}}-\frac {9 b x^{5/2}}{5 a^3}+\frac {9 x^{7/2}}{7 a^2}-\frac {x^{9/2}}{a (a x+b)} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(a + b/x)^2,x]

[Out]

(-9*b^3*Sqrt[x])/a^5 + (3*b^2*x^(3/2))/a^4 - (9*b*x^(5/2))/(5*a^3) + (9*x^(7/2))/(7*a^2) - x^(9/2)/(a*(b + a*x
)) + (9*b^(7/2)*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/a^(11/2)

Rule 47

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},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

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 63

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 205

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

Rule 263

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

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Mathematica [C]  time = 0.01, size = 27, normalized size = 0.28 \[ \frac {2 x^{11/2} \, _2F_1\left (2,\frac {11}{2};\frac {13}{2};-\frac {a x}{b}\right )}{11 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(a + b/x)^2,x]

[Out]

(2*x^(11/2)*Hypergeometric2F1[2, 11/2, 13/2, -((a*x)/b)])/(11*b^2)

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fricas [A]  time = 1.01, size = 209, normalized size = 2.13 \[ \left [\frac {315 \, {\left (a b^{3} x + b^{4}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {a x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - b}{a x + b}\right ) + 2 \, {\left (10 \, a^{4} x^{4} - 18 \, a^{3} b x^{3} + 42 \, a^{2} b^{2} x^{2} - 210 \, a b^{3} x - 315 \, b^{4}\right )} \sqrt {x}}{70 \, {\left (a^{6} x + a^{5} b\right )}}, \frac {315 \, {\left (a b^{3} x + b^{4}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {x} \sqrt {\frac {b}{a}}}{b}\right ) + {\left (10 \, a^{4} x^{4} - 18 \, a^{3} b x^{3} + 42 \, a^{2} b^{2} x^{2} - 210 \, a b^{3} x - 315 \, b^{4}\right )} \sqrt {x}}{35 \, {\left (a^{6} x + a^{5} b\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(a+b/x)^2,x, algorithm="fricas")

[Out]

[1/70*(315*(a*b^3*x + b^4)*sqrt(-b/a)*log((a*x + 2*a*sqrt(x)*sqrt(-b/a) - b)/(a*x + b)) + 2*(10*a^4*x^4 - 18*a
^3*b*x^3 + 42*a^2*b^2*x^2 - 210*a*b^3*x - 315*b^4)*sqrt(x))/(a^6*x + a^5*b), 1/35*(315*(a*b^3*x + b^4)*sqrt(b/
a)*arctan(a*sqrt(x)*sqrt(b/a)/b) + (10*a^4*x^4 - 18*a^3*b*x^3 + 42*a^2*b^2*x^2 - 210*a*b^3*x - 315*b^4)*sqrt(x
))/(a^6*x + a^5*b)]

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giac [A]  time = 0.15, size = 88, normalized size = 0.90 \[ \frac {9 \, b^{4} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{5}} - \frac {b^{4} \sqrt {x}}{{\left (a x + b\right )} a^{5}} + \frac {2 \, {\left (5 \, a^{12} x^{\frac {7}{2}} - 14 \, a^{11} b x^{\frac {5}{2}} + 35 \, a^{10} b^{2} x^{\frac {3}{2}} - 140 \, a^{9} b^{3} \sqrt {x}\right )}}{35 \, a^{14}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(a+b/x)^2,x, algorithm="giac")

[Out]

9*b^4*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^5) - b^4*sqrt(x)/((a*x + b)*a^5) + 2/35*(5*a^12*x^(7/2) - 14*a^
11*b*x^(5/2) + 35*a^10*b^2*x^(3/2) - 140*a^9*b^3*sqrt(x))/a^14

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maple [A]  time = 0.01, size = 83, normalized size = 0.85 \[ \frac {2 x^{\frac {7}{2}}}{7 a^{2}}-\frac {4 b \,x^{\frac {5}{2}}}{5 a^{3}}+\frac {9 b^{4} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a^{5}}+\frac {2 b^{2} x^{\frac {3}{2}}}{a^{4}}-\frac {b^{4} \sqrt {x}}{\left (a x +b \right ) a^{5}}-\frac {8 b^{3} \sqrt {x}}{a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(a+b/x)^2,x)

[Out]

2/7*x^(7/2)/a^2-4/5*b*x^(5/2)/a^3+2*b^2*x^(3/2)/a^4-8*b^3*x^(1/2)/a^5-1/a^5*b^4*x^(1/2)/(a*x+b)+9/a^5*b^4/(a*b
)^(1/2)*arctan(1/(a*b)^(1/2)*a*x^(1/2))

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maxima [A]  time = 2.39, size = 88, normalized size = 0.90 \[ \frac {10 \, a^{4} - \frac {18 \, a^{3} b}{x} + \frac {42 \, a^{2} b^{2}}{x^{2}} - \frac {210 \, a b^{3}}{x^{3}} - \frac {315 \, b^{4}}{x^{4}}}{35 \, {\left (\frac {a^{6}}{x^{\frac {7}{2}}} + \frac {a^{5} b}{x^{\frac {9}{2}}}\right )}} - \frac {9 \, b^{4} \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{\sqrt {a b} a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(a+b/x)^2,x, algorithm="maxima")

[Out]

1/35*(10*a^4 - 18*a^3*b/x + 42*a^2*b^2/x^2 - 210*a*b^3/x^3 - 315*b^4/x^4)/(a^6/x^(7/2) + a^5*b/x^(9/2)) - 9*b^
4*arctan(b/(sqrt(a*b)*sqrt(x)))/(sqrt(a*b)*a^5)

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mupad [B]  time = 1.08, size = 80, normalized size = 0.82 \[ \frac {2\,x^{7/2}}{7\,a^2}-\frac {4\,b\,x^{5/2}}{5\,a^3}+\frac {2\,b^2\,x^{3/2}}{a^4}-\frac {8\,b^3\,\sqrt {x}}{a^5}-\frac {b^4\,\sqrt {x}}{x\,a^6+b\,a^5}+\frac {9\,b^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{a^{11/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(a + b/x)^2,x)

[Out]

(2*x^(7/2))/(7*a^2) - (4*b*x^(5/2))/(5*a^3) + (2*b^2*x^(3/2))/a^4 - (8*b^3*x^(1/2))/a^5 - (b^4*x^(1/2))/(a^5*b
+ a^6*x) + (9*b^(7/2)*atan((a^(1/2)*x^(1/2))/b^(1/2)))/a^(11/2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)/(a+b/x)**2,x)

[Out]

Timed out

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