Integrand size = 13, antiderivative size = 68 \[ \int \frac {x^{5/2}}{a+b x} \, dx=\frac {2 a^2 \sqrt {x}}{b^3}-\frac {2 a x^{3/2}}{3 b^2}+\frac {2 x^{5/2}}{5 b}-\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}} \] Output:
2*a^2*x^(1/2)/b^3-2/3*a*x^(3/2)/b^2+2/5*x^(5/2)/b-2*a^(5/2)*arctan(b^(1/2) *x^(1/2)/a^(1/2))/b^(7/2)
Time = 0.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.90 \[ \int \frac {x^{5/2}}{a+b x} \, dx=\frac {2 \sqrt {x} \left (15 a^2-5 a b x+3 b^2 x^2\right )}{15 b^3}-\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}} \] Input:
Integrate[x^(5/2)/(a + b*x),x]
Output:
(2*Sqrt[x]*(15*a^2 - 5*a*b*x + 3*b^2*x^2))/(15*b^3) - (2*a^(5/2)*ArcTan[(S qrt[b]*Sqrt[x])/Sqrt[a]])/b^(7/2)
Time = 0.19 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {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.
\(\displaystyle \int \frac {x^{5/2}}{a+b x} \, dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2 x^{5/2}}{5 b}-\frac {a \int \frac {x^{3/2}}{a+b x}dx}{b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2 x^{5/2}}{5 b}-\frac {a \left (\frac {2 x^{3/2}}{3 b}-\frac {a \int \frac {\sqrt {x}}{a+b x}dx}{b}\right )}{b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2 x^{5/2}}{5 b}-\frac {a \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {a \int \frac {1}{\sqrt {x} (a+b x)}dx}{b}\right )}{b}\right )}{b}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2 x^{5/2}}{5 b}-\frac {a \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \int \frac {1}{a+b x}d\sqrt {x}}{b}\right )}{b}\right )}{b}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {2 x^{5/2}}{5 b}-\frac {a \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}}\right )}{b}\right )}{b}\) |
Input:
Int[x^(5/2)/(a + b*x),x]
Output:
(2*x^(5/2))/(5*b) - (a*((2*x^(3/2))/(3*b) - (a*((2*Sqrt[x])/b - (2*Sqrt[a] *ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(3/2)))/b))/b
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]
Time = 0.10 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.78
method | result | size |
risch | \(\frac {2 \left (3 b^{2} x^{2}-5 a b x +15 a^{2}\right ) \sqrt {x}}{15 b^{3}}-\frac {2 a^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b^{3} \sqrt {a b}}\) | \(53\) |
derivativedivides | \(\frac {\frac {2 b^{2} x^{\frac {5}{2}}}{5}-\frac {2 a b \,x^{\frac {3}{2}}}{3}+2 a^{2} \sqrt {x}}{b^{3}}-\frac {2 a^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b^{3} \sqrt {a b}}\) | \(54\) |
default | \(\frac {\frac {2 b^{2} x^{\frac {5}{2}}}{5}-\frac {2 a b \,x^{\frac {3}{2}}}{3}+2 a^{2} \sqrt {x}}{b^{3}}-\frac {2 a^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b^{3} \sqrt {a b}}\) | \(54\) |
Input:
int(x^(5/2)/(b*x+a),x,method=_RETURNVERBOSE)
Output:
2/15*(3*b^2*x^2-5*a*b*x+15*a^2)*x^(1/2)/b^3-2*a^3/b^3/(a*b)^(1/2)*arctan(b *x^(1/2)/(a*b)^(1/2))
Time = 0.08 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.94 \[ \int \frac {x^{5/2}}{a+b x} \, dx=\left [\frac {15 \, a^{2} \sqrt {-\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) + 2 \, {\left (3 \, b^{2} x^{2} - 5 \, a b x + 15 \, a^{2}\right )} \sqrt {x}}{15 \, b^{3}}, -\frac {2 \, {\left (15 \, a^{2} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (3 \, b^{2} x^{2} - 5 \, a b x + 15 \, a^{2}\right )} \sqrt {x}\right )}}{15 \, b^{3}}\right ] \] Input:
integrate(x^(5/2)/(b*x+a),x, algorithm="fricas")
Output:
[1/15*(15*a^2*sqrt(-a/b)*log((b*x - 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) + 2*(3*b^2*x^2 - 5*a*b*x + 15*a^2)*sqrt(x))/b^3, -2/15*(15*a^2*sqrt(a/b)* arctan(b*sqrt(x)*sqrt(a/b)/a) - (3*b^2*x^2 - 5*a*b*x + 15*a^2)*sqrt(x))/b^ 3]
Time = 1.85 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.79 \[ \int \frac {x^{5/2}}{a+b x} \, dx=\begin {cases} \tilde {\infty } x^{\frac {5}{2}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {7}{2}}}{7 a} & \text {for}\: b = 0 \\\frac {2 x^{\frac {5}{2}}}{5 b} & \text {for}\: a = 0 \\- \frac {a^{3} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{b^{4} \sqrt {- \frac {a}{b}}} + \frac {a^{3} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{b^{4} \sqrt {- \frac {a}{b}}} + \frac {2 a^{2} \sqrt {x}}{b^{3}} - \frac {2 a x^{\frac {3}{2}}}{3 b^{2}} + \frac {2 x^{\frac {5}{2}}}{5 b} & \text {otherwise} \end {cases} \] Input:
integrate(x**(5/2)/(b*x+a),x)
Output:
Piecewise((zoo*x**(5/2), Eq(a, 0) & Eq(b, 0)), (2*x**(7/2)/(7*a), Eq(b, 0) ), (2*x**(5/2)/(5*b), Eq(a, 0)), (-a**3*log(sqrt(x) - sqrt(-a/b))/(b**4*sq rt(-a/b)) + a**3*log(sqrt(x) + sqrt(-a/b))/(b**4*sqrt(-a/b)) + 2*a**2*sqrt (x)/b**3 - 2*a*x**(3/2)/(3*b**2) + 2*x**(5/2)/(5*b), True))
Time = 0.11 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.79 \[ \int \frac {x^{5/2}}{a+b x} \, dx=-\frac {2 \, a^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {2 \, {\left (3 \, b^{2} x^{\frac {5}{2}} - 5 \, a b x^{\frac {3}{2}} + 15 \, a^{2} \sqrt {x}\right )}}{15 \, b^{3}} \] Input:
integrate(x^(5/2)/(b*x+a),x, algorithm="maxima")
Output:
-2*a^3*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) + 2/15*(3*b^2*x^(5/2) - 5*a*b*x^(3/2) + 15*a^2*sqrt(x))/b^3
Time = 0.12 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.87 \[ \int \frac {x^{5/2}}{a+b x} \, dx=-\frac {2 \, a^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {2 \, {\left (3 \, b^{4} x^{\frac {5}{2}} - 5 \, a b^{3} x^{\frac {3}{2}} + 15 \, a^{2} b^{2} \sqrt {x}\right )}}{15 \, b^{5}} \] Input:
integrate(x^(5/2)/(b*x+a),x, algorithm="giac")
Output:
-2*a^3*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) + 2/15*(3*b^4*x^(5/2) - 5*a*b^3*x^(3/2) + 15*a^2*b^2*sqrt(x))/b^5
Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.71 \[ \int \frac {x^{5/2}}{a+b x} \, dx=\frac {2\,x^{5/2}}{5\,b}-\frac {2\,a\,x^{3/2}}{3\,b^2}+\frac {2\,a^2\,\sqrt {x}}{b^3}-\frac {2\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}} \] Input:
int(x^(5/2)/(a + b*x),x)
Output:
(2*x^(5/2))/(5*b) - (2*a*x^(3/2))/(3*b^2) + (2*a^2*x^(1/2))/b^3 - (2*a^(5/ 2)*atan((b^(1/2)*x^(1/2))/a^(1/2)))/b^(7/2)
Time = 0.16 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.81 \[ \int \frac {x^{5/2}}{a+b x} \, dx=\frac {-2 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{2}+2 \sqrt {x}\, a^{2} b -\frac {2 \sqrt {x}\, a \,b^{2} x}{3}+\frac {2 \sqrt {x}\, b^{3} x^{2}}{5}}{b^{4}} \] Input:
int(x^(5/2)/(b*x+a),x)
Output:
(2*( - 15*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**2 + 15*sq rt(x)*a**2*b - 5*sqrt(x)*a*b**2*x + 3*sqrt(x)*b**3*x**2))/(15*b**4)