3.17.65 \(\int \frac {x^{5/2}}{a+\frac {b}{x}} \, dx\) [1665]

Optimal. Leaf size=83 \[ -\frac {2 b^3 \sqrt {x}}{a^4}+\frac {2 b^2 x^{3/2}}{3 a^3}-\frac {2 b x^{5/2}}{5 a^2}+\frac {2 x^{7/2}}{7 a}+\frac {2 b^{7/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{9/2}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {269, 52, 65, 211} \begin {gather*} \frac {2 b^{7/2} \text {ArcTan}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{9/2}}-\frac {2 b^3 \sqrt {x}}{a^4}+\frac {2 b^2 x^{3/2}}{3 a^3}-\frac {2 b x^{5/2}}{5 a^2}+\frac {2 x^{7/2}}{7 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(a + b/x),x]

[Out]

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

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

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Mathematica [A]
time = 0.06, size = 72, normalized size = 0.87 \begin {gather*} \frac {2 \sqrt {x} \left (-105 b^3+35 a b^2 x-21 a^2 b x^2+15 a^3 x^3\right )}{105 a^4}+\frac {2 b^{7/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(a + b/x),x]

[Out]

(2*Sqrt[x]*(-105*b^3 + 35*a*b^2*x - 21*a^2*b*x^2 + 15*a^3*x^3))/(105*a^4) + (2*b^(7/2)*ArcTan[(Sqrt[a]*Sqrt[x]
)/Sqrt[b]])/a^(9/2)

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Maple [A]
time = 0.08, size = 66, normalized size = 0.80

method	result	size
risch	\(\frac {2 \left (15 a^{3} x^{3}-21 a^{2} b \,x^{2}+35 a \,b^{2} x -105 b^{3}\right ) \sqrt {x}}{105 a^{4}}+\frac {2 b^{4} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{a^{4} \sqrt {a b}}\)	\(64\)
derivativedivides	\(\frac {\frac {2 a^{3} x^{\frac {7}{2}}}{7}-\frac {2 a^{2} b \,x^{\frac {5}{2}}}{5}+\frac {2 a \,b^{2} x^{\frac {3}{2}}}{3}-2 b^{3} \sqrt {x}}{a^{4}}+\frac {2 b^{4} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{a^{4} \sqrt {a b}}\)	\(66\)
default	\(\frac {\frac {2 a^{3} x^{\frac {7}{2}}}{7}-\frac {2 a^{2} b \,x^{\frac {5}{2}}}{5}+\frac {2 a \,b^{2} x^{\frac {3}{2}}}{3}-2 b^{3} \sqrt {x}}{a^{4}}+\frac {2 b^{4} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{a^{4} \sqrt {a b}}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(a+1/x*b),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.51, size = 65, normalized size = 0.78 \begin {gather*} \frac {2 \, {\left (15 \, a^{3} - \frac {21 \, a^{2} b}{x} + \frac {35 \, a b^{2}}{x^{2}} - \frac {105 \, b^{3}}{x^{3}}\right )} x^{\frac {7}{2}}}{105 \, a^{4}} - \frac {2 \, b^{4} \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{\sqrt {a b} a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*a^3 - 21*a^2*b/x + 35*a*b^2/x^2 - 105*b^3/x^3)*x^(7/2)/a^4 - 2*b^4*arctan(b/(sqrt(a*b)*sqrt(x)))/(sq
rt(a*b)*a^4)

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Fricas [A]
time = 0.37, size = 153, normalized size = 1.84 \begin {gather*} \left [\frac {105 \, b^{3} \sqrt {-\frac {b}{a}} \log \left (\frac {a x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - b}{a x + b}\right ) + 2 \, {\left (15 \, a^{3} x^{3} - 21 \, a^{2} b x^{2} + 35 \, a b^{2} x - 105 \, b^{3}\right )} \sqrt {x}}{105 \, a^{4}}, \frac {2 \, {\left (105 \, b^{3} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {x} \sqrt {\frac {b}{a}}}{b}\right ) + {\left (15 \, a^{3} x^{3} - 21 \, a^{2} b x^{2} + 35 \, a b^{2} x - 105 \, b^{3}\right )} \sqrt {x}\right )}}{105 \, a^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/105*(105*b^3*sqrt(-b/a)*log((a*x + 2*a*sqrt(x)*sqrt(-b/a) - b)/(a*x + b)) + 2*(15*a^3*x^3 - 21*a^2*b*x^2 +
35*a*b^2*x - 105*b^3)*sqrt(x))/a^4, 2/105*(105*b^3*sqrt(b/a)*arctan(a*sqrt(x)*sqrt(b/a)/b) + (15*a^3*x^3 - 21*
a^2*b*x^2 + 35*a*b^2*x - 105*b^3)*sqrt(x))/a^4]

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Sympy [A]
time = 4.00, size = 138, normalized size = 1.66 \begin {gather*} \begin {cases} \tilde {\infty } x^{\frac {9}{2}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {9}{2}}}{9 b} & \text {for}\: a = 0 \\\frac {2 x^{\frac {7}{2}}}{7 a} & \text {for}\: b = 0 \\\frac {2 x^{\frac {7}{2}}}{7 a} - \frac {2 b x^{\frac {5}{2}}}{5 a^{2}} + \frac {2 b^{2} x^{\frac {3}{2}}}{3 a^{3}} - \frac {2 b^{3} \sqrt {x}}{a^{4}} + \frac {b^{4} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{a^{5} \sqrt {- \frac {b}{a}}} - \frac {b^{4} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{a^{5} \sqrt {- \frac {b}{a}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x**(9/2), Eq(a, 0) & Eq(b, 0)), (2*x**(9/2)/(9*b), Eq(a, 0)), (2*x**(7/2)/(7*a), Eq(b, 0)), (2*
x**(7/2)/(7*a) - 2*b*x**(5/2)/(5*a**2) + 2*b**2*x**(3/2)/(3*a**3) - 2*b**3*sqrt(x)/a**4 + b**4*log(sqrt(x) - s
qrt(-b/a))/(a**5*sqrt(-b/a)) - b**4*log(sqrt(x) + sqrt(-b/a))/(a**5*sqrt(-b/a)), True))

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Giac [A]
time = 0.53, size = 70, normalized size = 0.84 \begin {gather*} \frac {2 \, b^{4} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{4}} + \frac {2 \, {\left (15 \, a^{6} x^{\frac {7}{2}} - 21 \, a^{5} b x^{\frac {5}{2}} + 35 \, a^{4} b^{2} x^{\frac {3}{2}} - 105 \, a^{3} b^{3} \sqrt {x}\right )}}{105 \, a^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^4*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4) + 2/105*(15*a^6*x^(7/2) - 21*a^5*b*x^(5/2) + 35*a^4*b^2*x^(3
/2) - 105*a^3*b^3*sqrt(x))/a^7

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Mupad [B]
time = 0.04, size = 59, normalized size = 0.71 \begin {gather*} \frac {2\,x^{7/2}}{7\,a}-\frac {2\,b\,x^{5/2}}{5\,a^2}+\frac {2\,b^2\,x^{3/2}}{3\,a^3}-\frac {2\,b^3\,\sqrt {x}}{a^4}+\frac {2\,b^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{a^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(a + b/x),x)

[Out]

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

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