Integrand size = 15, antiderivative size = 83 \[ \int \frac {x^{5/2}}{a+\frac {b}{x}} \, dx=-\frac {2 b^3 \sqrt {x}}{a^4}+\frac {2 b^2 x^{3/2}}{3 a^3}-\frac {2 b x^{5/2}}{5 a^2}+\frac {2 x^{7/2}}{7 a}+\frac {2 b^{7/2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{9/2}} \] Output:
-2*b^3*x^(1/2)/a^4+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)
Time = 0.06 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.87 \[ \int \frac {x^{5/2}}{a+\frac {b}{x}} \, dx=\frac {2 \sqrt {x} \left (-105 b^3+35 a b^2 x-21 a^2 b x^2+15 a^3 x^3\right )}{105 a^4}+\frac {2 b^{7/2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{9/2}} \] Input:
Integrate[x^(5/2)/(a + b/x),x]
Output:
(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)
Time = 0.32 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.17, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {795, 60, 60, 60, 60, 73, 218}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^{5/2}}{a+\frac {b}{x}} \, dx\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \int \frac {x^{7/2}}{a x+b}dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2 x^{7/2}}{7 a}-\frac {b \int \frac {x^{5/2}}{b+a x}dx}{a}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2 x^{7/2}}{7 a}-\frac {b \left (\frac {2 x^{5/2}}{5 a}-\frac {b \int \frac {x^{3/2}}{b+a x}dx}{a}\right )}{a}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2 x^{7/2}}{7 a}-\frac {b \left (\frac {2 x^{5/2}}{5 a}-\frac {b \left (\frac {2 x^{3/2}}{3 a}-\frac {b \int \frac {\sqrt {x}}{b+a x}dx}{a}\right )}{a}\right )}{a}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2 x^{7/2}}{7 a}-\frac {b \left (\frac {2 x^{5/2}}{5 a}-\frac {b \left (\frac {2 x^{3/2}}{3 a}-\frac {b \left (\frac {2 \sqrt {x}}{a}-\frac {b \int \frac {1}{\sqrt {x} (b+a x)}dx}{a}\right )}{a}\right )}{a}\right )}{a}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2 x^{7/2}}{7 a}-\frac {b \left (\frac {2 x^{5/2}}{5 a}-\frac {b \left (\frac {2 x^{3/2}}{3 a}-\frac {b \left (\frac {2 \sqrt {x}}{a}-\frac {2 b \int \frac {1}{b+a x}d\sqrt {x}}{a}\right )}{a}\right )}{a}\right )}{a}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {2 x^{7/2}}{7 a}-\frac {b \left (\frac {2 x^{5/2}}{5 a}-\frac {b \left (\frac {2 x^{3/2}}{3 a}-\frac {b \left (\frac {2 \sqrt {x}}{a}-\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{3/2}}\right )}{a}\right )}{a}\right )}{a}\) |
Input:
Int[x^(5/2)/(a + b/x),x]
Output:
(2*x^(7/2))/(7*a) - (b*((2*x^(5/2))/(5*a) - (b*((2*x^(3/2))/(3*a) - (b*((2 *Sqrt[x])/a - (2*Sqrt[b]*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/a^(3/2)))/a))/ a))/a
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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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]
Time = 0.37 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.77
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 | \(\frac {\frac {2 a^{3} x^{\frac {7}{2}}}{7}-\frac {2 a^{2} b \,x^{\frac {5}{2}}}{5}+\frac {2 a \,b^{2} x^{\frac {3}{2}}}{3}-2 b^{3} \sqrt {x}}{a^{4}}+\frac {2 b^{4} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{a^{4} \sqrt {a b}}\) | \(66\) |
Input:
int(x^(5/2)/(a+b/x),x,method=_RETURNVERBOSE)
Output:
2/105*(15*a^3*x^3-21*a^2*b*x^2+35*a*b^2*x-105*b^3)*x^(1/2)/a^4+2*b^4/a^4/( a*b)^(1/2)*arctan(a*x^(1/2)/(a*b)^(1/2))
Time = 0.07 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.84 \[ \int \frac {x^{5/2}}{a+\frac {b}{x}} \, dx=\left [\frac {105 \, b^{3} \sqrt {-\frac {b}{a}} \log \left (\frac {a x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - b}{a x + b}\right ) + 2 \, {\left (15 \, a^{3} x^{3} - 21 \, a^{2} b x^{2} + 35 \, a b^{2} x - 105 \, b^{3}\right )} \sqrt {x}}{105 \, a^{4}}, \frac {2 \, {\left (105 \, b^{3} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {x} \sqrt {\frac {b}{a}}}{b}\right ) + {\left (15 \, a^{3} x^{3} - 21 \, a^{2} b x^{2} + 35 \, a b^{2} x - 105 \, b^{3}\right )} \sqrt {x}\right )}}{105 \, a^{4}}\right ] \] Input:
integrate(x^(5/2)/(a+b/x),x, algorithm="fricas")
Output:
[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]
Time = 3.87 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.66 \[ \int \frac {x^{5/2}}{a+\frac {b}{x}} \, dx=\begin {cases} \tilde {\infty } x^{\frac {9}{2}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {9}{2}}}{9 b} & \text {for}\: a = 0 \\\frac {2 x^{\frac {7}{2}}}{7 a} & \text {for}\: b = 0 \\\frac {2 x^{\frac {7}{2}}}{7 a} - \frac {2 b x^{\frac {5}{2}}}{5 a^{2}} + \frac {2 b^{2} x^{\frac {3}{2}}}{3 a^{3}} - \frac {2 b^{3} \sqrt {x}}{a^{4}} + \frac {b^{4} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{a^{5} \sqrt {- \frac {b}{a}}} - \frac {b^{4} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{a^{5} \sqrt {- \frac {b}{a}}} & \text {otherwise} \end {cases} \] Input:
integrate(x**(5/2)/(a+b/x),x)
Output:
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) - sqr t(-b/a))/(a**5*sqrt(-b/a)) - b**4*log(sqrt(x) + sqrt(-b/a))/(a**5*sqrt(-b/ a)), True))
Time = 0.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.78 \[ \int \frac {x^{5/2}}{a+\frac {b}{x}} \, dx=\frac {2 \, {\left (15 \, a^{3} - \frac {21 \, a^{2} b}{x} + \frac {35 \, a b^{2}}{x^{2}} - \frac {105 \, b^{3}}{x^{3}}\right )} x^{\frac {7}{2}}}{105 \, a^{4}} - \frac {2 \, b^{4} \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{\sqrt {a b} a^{4}} \] Input:
integrate(x^(5/2)/(a+b/x),x, algorithm="maxima")
Output:
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)))/(sqrt(a*b)*a^4)
Time = 0.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.84 \[ \int \frac {x^{5/2}}{a+\frac {b}{x}} \, dx=\frac {2 \, b^{4} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{4}} + \frac {2 \, {\left (15 \, a^{6} x^{\frac {7}{2}} - 21 \, a^{5} b x^{\frac {5}{2}} + 35 \, a^{4} b^{2} x^{\frac {3}{2}} - 105 \, a^{3} b^{3} \sqrt {x}\right )}}{105 \, a^{7}} \] Input:
integrate(x^(5/2)/(a+b/x),x, algorithm="giac")
Output:
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
Time = 0.24 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.71 \[ \int \frac {x^{5/2}}{a+\frac {b}{x}} \, dx=\frac {2\,x^{7/2}}{7\,a}-\frac {2\,b\,x^{5/2}}{5\,a^2}+\frac {2\,b^2\,x^{3/2}}{3\,a^3}-\frac {2\,b^3\,\sqrt {x}}{a^4}+\frac {2\,b^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{a^{9/2}} \] Input:
int(x^(5/2)/(a + b/x),x)
Output:
(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)
Time = 0.24 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.82 \[ \int \frac {x^{5/2}}{a+\frac {b}{x}} \, dx=\frac {2 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, a}{\sqrt {b}\, \sqrt {a}}\right ) b^{3}+\frac {2 \sqrt {x}\, a^{4} x^{3}}{7}-\frac {2 \sqrt {x}\, a^{3} b \,x^{2}}{5}+\frac {2 \sqrt {x}\, a^{2} b^{2} x}{3}-2 \sqrt {x}\, a \,b^{3}}{a^{5}} \] Input:
int(x^(5/2)/(a+b/x),x)
Output:
(2*(105*sqrt(b)*sqrt(a)*atan((sqrt(x)*a)/(sqrt(b)*sqrt(a)))*b**3 + 15*sqrt (x)*a**4*x**3 - 21*sqrt(x)*a**3*b*x**2 + 35*sqrt(x)*a**2*b**2*x - 105*sqrt (x)*a*b**3))/(105*a**5)