Integrand size = 15, antiderivative size = 86 \[ \int \frac {(a+b x)^{5/2}}{x^{5/2}} \, dx=5 b^2 \sqrt {x} \sqrt {a+b x}-\frac {10 b (a+b x)^{3/2}}{3 \sqrt {x}}-\frac {2 (a+b x)^{5/2}}{3 x^{3/2}}+5 a b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right ) \]
-2/3*(b*x+a)^(5/2)/x^(3/2)+5*a*b^(3/2)*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/ 2))-10/3*b*(b*x+a)^(3/2)/x^(1/2)+5*b^2*x^(1/2)*(b*x+a)^(1/2)
Time = 0.19 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.90 \[ \int \frac {(a+b x)^{5/2}}{x^{5/2}} \, dx=\frac {\sqrt {a+b x} \left (-2 a^2-14 a b x+3 b^2 x^2\right )}{3 x^{3/2}}+10 a b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right ) \]
(Sqrt[a + b*x]*(-2*a^2 - 14*a*b*x + 3*b^2*x^2))/(3*x^(3/2)) + 10*a*b^(3/2) *ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x])]
Time = 0.18 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {57, 57, 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.
\(\displaystyle \int \frac {(a+b x)^{5/2}}{x^{5/2}} \, dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {5}{3} b \int \frac {(a+b x)^{3/2}}{x^{3/2}}dx-\frac {2 (a+b x)^{5/2}}{3 x^{3/2}}\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {5}{3} b \left (3 b \int \frac {\sqrt {a+b x}}{\sqrt {x}}dx-\frac {2 (a+b x)^{3/2}}{\sqrt {x}}\right )-\frac {2 (a+b x)^{5/2}}{3 x^{3/2}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5}{3} b \left (3 b \left (\frac {1}{2} a \int \frac {1}{\sqrt {x} \sqrt {a+b x}}dx+\sqrt {x} \sqrt {a+b x}\right )-\frac {2 (a+b x)^{3/2}}{\sqrt {x}}\right )-\frac {2 (a+b x)^{5/2}}{3 x^{3/2}}\) |
\(\Big \downarrow \) 65 |
\(\displaystyle \frac {5}{3} b \left (3 b \left (a \int \frac {1}{1-\frac {b x}{a+b x}}d\frac {\sqrt {x}}{\sqrt {a+b x}}+\sqrt {x} \sqrt {a+b x}\right )-\frac {2 (a+b x)^{3/2}}{\sqrt {x}}\right )-\frac {2 (a+b x)^{5/2}}{3 x^{3/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {5}{3} b \left (3 b \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{\sqrt {b}}+\sqrt {x} \sqrt {a+b x}\right )-\frac {2 (a+b x)^{3/2}}{\sqrt {x}}\right )-\frac {2 (a+b x)^{5/2}}{3 x^{3/2}}\) |
(-2*(a + b*x)^(5/2))/(3*x^(3/2)) + (5*b*((-2*(a + b*x)^(3/2))/Sqrt[x] + 3* b*(Sqrt[x]*Sqrt[a + b*x] + (a*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/Sq rt[b])))/3
3.6.50.3.1 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 + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & 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]
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 = 0.09 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.95
method | result | size |
risch | \(-\frac {\sqrt {b x +a}\, \left (-3 b^{2} x^{2}+14 a b x +2 a^{2}\right )}{3 x^{\frac {3}{2}}}+\frac {5 a \,b^{\frac {3}{2}} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{2 \sqrt {x}\, \sqrt {b x +a}}\) | \(82\) |
-1/3*(b*x+a)^(1/2)*(-3*b^2*x^2+14*a*b*x+2*a^2)/x^(3/2)+5/2*a*b^(3/2)*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 = 0.23 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.60 \[ \int \frac {(a+b x)^{5/2}}{x^{5/2}} \, dx=\left [\frac {15 \, a b^{\frac {3}{2}} x^{2} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (3 \, b^{2} x^{2} - 14 \, a b x - 2 \, a^{2}\right )} \sqrt {b x + a} \sqrt {x}}{6 \, x^{2}}, -\frac {15 \, a \sqrt {-b} b x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (3 \, b^{2} x^{2} - 14 \, a b x - 2 \, a^{2}\right )} \sqrt {b x + a} \sqrt {x}}{3 \, x^{2}}\right ] \]
[1/6*(15*a*b^(3/2)*x^2*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(3*b^2*x^2 - 14*a*b*x - 2*a^2)*sqrt(b*x + a)*sqrt(x))/x^2, -1/3*(15*a*sq rt(-b)*b*x^2*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (3*b^2*x^2 - 14* a*b*x - 2*a^2)*sqrt(b*x + a)*sqrt(x))/x^2]
Time = 3.89 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.15 \[ \int \frac {(a+b x)^{5/2}}{x^{5/2}} \, dx=- \frac {2 a^{2} \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{3 x} - \frac {14 a b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{3} - \frac {5 a b^{\frac {3}{2}} \log {\left (\frac {a}{b x} \right )}}{2} + 5 a b^{\frac {3}{2}} \log {\left (\sqrt {\frac {a}{b x} + 1} + 1 \right )} + b^{\frac {5}{2}} x \sqrt {\frac {a}{b x} + 1} \]
-2*a**2*sqrt(b)*sqrt(a/(b*x) + 1)/(3*x) - 14*a*b**(3/2)*sqrt(a/(b*x) + 1)/ 3 - 5*a*b**(3/2)*log(a/(b*x))/2 + 5*a*b**(3/2)*log(sqrt(a/(b*x) + 1) + 1) + b**(5/2)*x*sqrt(a/(b*x) + 1)
Time = 0.32 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.16 \[ \int \frac {(a+b x)^{5/2}}{x^{5/2}} \, dx=-\frac {5}{2} \, a b^{\frac {3}{2}} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + a}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + a}}{\sqrt {x}}}\right ) - \frac {4 \, \sqrt {b x + a} a b}{\sqrt {x}} - \frac {\sqrt {b x + a} a b^{2}}{{\left (b - \frac {b x + a}{x}\right )} \sqrt {x}} - \frac {2 \, {\left (b x + a\right )}^{\frac {3}{2}} a}{3 \, x^{\frac {3}{2}}} \]
-5/2*a*b^(3/2)*log(-(sqrt(b) - sqrt(b*x + a)/sqrt(x))/(sqrt(b) + sqrt(b*x + a)/sqrt(x))) - 4*sqrt(b*x + a)*a*b/sqrt(x) - sqrt(b*x + a)*a*b^2/((b - ( b*x + a)/x)*sqrt(x)) - 2/3*(b*x + a)^(3/2)*a/x^(3/2)
Time = 77.24 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.14 \[ \int \frac {(a+b x)^{5/2}}{x^{5/2}} \, dx=-\frac {{\left (15 \, a b^{\frac {3}{2}} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right ) - \frac {{\left (15 \, a^{2} b^{3} + {\left (3 \, {\left (b x + a\right )} b^{3} - 20 \, a b^{3}\right )} {\left (b x + a\right )}\right )} \sqrt {b x + a}}{{\left ({\left (b x + a\right )} b - a b\right )}^{\frac {3}{2}}}\right )} b}{3 \, {\left | b \right |}} \]
-1/3*(15*a*b^(3/2)*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b*x + a)*b - a*b ))) - (15*a^2*b^3 + (3*(b*x + a)*b^3 - 20*a*b^3)*(b*x + a))*sqrt(b*x + a)/ ((b*x + a)*b - a*b)^(3/2))*b/abs(b)
Timed out. \[ \int \frac {(a+b x)^{5/2}}{x^{5/2}} \, dx=\int \frac {{\left (a+b\,x\right )}^{5/2}}{x^{5/2}} \,d x \]