Integrand size = 15, antiderivative size = 101 \[ \int \frac {x^{5/2}}{\sqrt {a+b x}} \, dx=\frac {5 a^2 \sqrt {x} \sqrt {a+b x}}{8 b^3}-\frac {5 a x^{3/2} \sqrt {a+b x}}{12 b^2}+\frac {x^{5/2} \sqrt {a+b x}}{3 b}-\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{7/2}} \]
-5/8*a^3*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2))/b^(7/2)-5/12*a*x^(3/2)*(b* x+a)^(1/2)/b^2+1/3*x^(5/2)*(b*x+a)^(1/2)/b+5/8*a^2*x^(1/2)*(b*x+a)^(1/2)/b ^3
Time = 0.23 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.83 \[ \int \frac {x^{5/2}}{\sqrt {a+b x}} \, dx=\frac {\sqrt {x} \sqrt {a+b x} \left (15 a^2-10 a b x+8 b^2 x^2\right )}{24 b^3}+\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )}{4 b^{7/2}} \]
(Sqrt[x]*Sqrt[a + b*x]*(15*a^2 - 10*a*b*x + 8*b^2*x^2))/(24*b^3) + (5*a^3* ArcTanh[(Sqrt[b]*Sqrt[x])/(Sqrt[a] - Sqrt[a + b*x])])/(4*b^(7/2))
Time = 0.18 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {60, 60, 60, 65, 219}
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.
\(\displaystyle \int \frac {x^{5/2}}{\sqrt {a+b x}} \, dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {x^{5/2} \sqrt {a+b x}}{3 b}-\frac {5 a \int \frac {x^{3/2}}{\sqrt {a+b x}}dx}{6 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {x^{5/2} \sqrt {a+b x}}{3 b}-\frac {5 a \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \int \frac {\sqrt {x}}{\sqrt {a+b x}}dx}{4 b}\right )}{6 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {x^{5/2} \sqrt {a+b x}}{3 b}-\frac {5 a \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \left (\frac {\sqrt {x} \sqrt {a+b x}}{b}-\frac {a \int \frac {1}{\sqrt {x} \sqrt {a+b x}}dx}{2 b}\right )}{4 b}\right )}{6 b}\) |
\(\Big \downarrow \) 65 |
\(\displaystyle \frac {x^{5/2} \sqrt {a+b x}}{3 b}-\frac {5 a \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \left (\frac {\sqrt {x} \sqrt {a+b x}}{b}-\frac {a \int \frac {1}{1-\frac {b x}{a+b x}}d\frac {\sqrt {x}}{\sqrt {a+b x}}}{b}\right )}{4 b}\right )}{6 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {x^{5/2} \sqrt {a+b x}}{3 b}-\frac {5 a \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \left (\frac {\sqrt {x} \sqrt {a+b x}}{b}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{3/2}}\right )}{4 b}\right )}{6 b}\) |
(x^(5/2)*Sqrt[a + b*x])/(3*b) - (5*a*((x^(3/2)*Sqrt[a + b*x])/(2*b) - (3*a *((Sqrt[x]*Sqrt[a + b*x])/b - (a*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]]) /b^(3/2)))/(4*b)))/(6*b)
3.6.69.3.1 Defintions of rubi rules used
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[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2 Sub st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d }, x] && !GtQ[c, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.86
method | result | size |
risch | \(\frac {\left (8 b^{2} x^{2}-10 a b x +15 a^{2}\right ) \sqrt {x}\, \sqrt {b x +a}}{24 b^{3}}-\frac {5 a^{3} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{16 b^{\frac {7}{2}} \sqrt {x}\, \sqrt {b x +a}}\) | \(87\) |
default | \(\frac {x^{\frac {5}{2}} \sqrt {b x +a}}{3 b}-\frac {5 a \left (\frac {x^{\frac {3}{2}} \sqrt {b x +a}}{2 b}-\frac {3 a \left (\frac {\sqrt {x}\, \sqrt {b x +a}}{b}-\frac {a \sqrt {x \left (b x +a \right )}\, \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2 b^{\frac {3}{2}} \sqrt {x}\, \sqrt {b x +a}}\right )}{4 b}\right )}{6 b}\) | \(109\) |
1/24*(8*b^2*x^2-10*a*b*x+15*a^2)*x^(1/2)*(b*x+a)^(1/2)/b^3-5/16*a^3/b^(7/2 )*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))*(x*(b*x+a))^(1/2)/x^(1/2)/(b*x +a)^(1/2)
Time = 0.24 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.39 \[ \int \frac {x^{5/2}}{\sqrt {a+b x}} \, dx=\left [\frac {15 \, a^{3} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (8 \, b^{3} x^{2} - 10 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{48 \, b^{4}}, \frac {15 \, a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (8 \, b^{3} x^{2} - 10 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{24 \, b^{4}}\right ] \]
[1/48*(15*a^3*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2 *(8*b^3*x^2 - 10*a*b^2*x + 15*a^2*b)*sqrt(b*x + a)*sqrt(x))/b^4, 1/24*(15* a^3*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (8*b^3*x^2 - 10* a*b^2*x + 15*a^2*b)*sqrt(b*x + a)*sqrt(x))/b^4]
Time = 10.02 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.27 \[ \int \frac {x^{5/2}}{\sqrt {a+b x}} \, dx=\frac {5 a^{\frac {5}{2}} \sqrt {x}}{8 b^{3} \sqrt {1 + \frac {b x}{a}}} + \frac {5 a^{\frac {3}{2}} x^{\frac {3}{2}}}{24 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {\sqrt {a} x^{\frac {5}{2}}}{12 b \sqrt {1 + \frac {b x}{a}}} - \frac {5 a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{8 b^{\frac {7}{2}}} + \frac {x^{\frac {7}{2}}}{3 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} \]
5*a**(5/2)*sqrt(x)/(8*b**3*sqrt(1 + b*x/a)) + 5*a**(3/2)*x**(3/2)/(24*b**2 *sqrt(1 + b*x/a)) - sqrt(a)*x**(5/2)/(12*b*sqrt(1 + b*x/a)) - 5*a**3*asinh (sqrt(b)*sqrt(x)/sqrt(a))/(8*b**(7/2)) + x**(7/2)/(3*sqrt(a)*sqrt(1 + b*x/ a))
Time = 0.30 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.45 \[ \int \frac {x^{5/2}}{\sqrt {a+b x}} \, dx=\frac {5 \, a^{3} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + a}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + a}}{\sqrt {x}}}\right )}{16 \, b^{\frac {7}{2}}} - \frac {\frac {33 \, \sqrt {b x + a} a^{3} b^{2}}{\sqrt {x}} - \frac {40 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} b}{x^{\frac {3}{2}}} + \frac {15 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3}}{x^{\frac {5}{2}}}}{24 \, {\left (b^{6} - \frac {3 \, {\left (b x + a\right )} b^{5}}{x} + \frac {3 \, {\left (b x + a\right )}^{2} b^{4}}{x^{2}} - \frac {{\left (b x + a\right )}^{3} b^{3}}{x^{3}}\right )}} \]
5/16*a^3*log(-(sqrt(b) - sqrt(b*x + a)/sqrt(x))/(sqrt(b) + sqrt(b*x + a)/s qrt(x)))/b^(7/2) - 1/24*(33*sqrt(b*x + a)*a^3*b^2/sqrt(x) - 40*(b*x + a)^( 3/2)*a^3*b/x^(3/2) + 15*(b*x + a)^(5/2)*a^3/x^(5/2))/(b^6 - 3*(b*x + a)*b^ 5/x + 3*(b*x + a)^2*b^4/x^2 - (b*x + a)^3*b^3/x^3)
Time = 77.35 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.99 \[ \int \frac {x^{5/2}}{\sqrt {a+b x}} \, dx=\frac {{\left (\frac {15 \, a^{3} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{b^{\frac {3}{2}}} + \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )}}{b^{2}} - \frac {13 \, a}{b^{2}}\right )} + \frac {33 \, a^{2}}{b^{2}}\right )}\right )} {\left | b \right |}}{24 \, b^{3}} \]
1/24*(15*a^3*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b*x + a)*b - a*b)))/b^ (3/2) + sqrt((b*x + a)*b - a*b)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)/b^ 2 - 13*a/b^2) + 33*a^2/b^2))*abs(b)/b^3
Timed out. \[ \int \frac {x^{5/2}}{\sqrt {a+b x}} \, dx=\int \frac {x^{5/2}}{\sqrt {a+b\,x}} \,d x \]