Integrand size = 18, antiderivative size = 69 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x} \, dx=2 a A \sqrt {a+b x}+\frac {2}{3} A (a+b x)^{3/2}+\frac {2 B (a+b x)^{5/2}}{5 b}-2 a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]
2/3*A*(b*x+a)^(3/2)+2/5*B*(b*x+a)^(5/2)/b-2*a^(3/2)*A*arctanh((b*x+a)^(1/2 )/a^(1/2))+2*a*A*(b*x+a)^(1/2)
Time = 0.08 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x} \, dx=\frac {2 \sqrt {a+b x} \left (15 a A b+5 A b (a+b x)+3 B (a+b x)^2\right )}{15 b}-2 a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]
(2*Sqrt[a + b*x]*(15*a*A*b + 5*A*b*(a + b*x) + 3*B*(a + b*x)^2))/(15*b) - 2*a^(3/2)*A*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]
Time = 0.18 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {90, 60, 60, 73, 221}
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.
\(\displaystyle \int \frac {(a+b x)^{3/2} (A+B x)}{x} \, dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle A \int \frac {(a+b x)^{3/2}}{x}dx+\frac {2 B (a+b x)^{5/2}}{5 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle A \left (a \int \frac {\sqrt {a+b x}}{x}dx+\frac {2}{3} (a+b x)^{3/2}\right )+\frac {2 B (a+b x)^{5/2}}{5 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle A \left (a \left (a \int \frac {1}{x \sqrt {a+b x}}dx+2 \sqrt {a+b x}\right )+\frac {2}{3} (a+b x)^{3/2}\right )+\frac {2 B (a+b x)^{5/2}}{5 b}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle A \left (a \left (\frac {2 a \int \frac {1}{\frac {a+b x}{b}-\frac {a}{b}}d\sqrt {a+b x}}{b}+2 \sqrt {a+b x}\right )+\frac {2}{3} (a+b x)^{3/2}\right )+\frac {2 B (a+b x)^{5/2}}{5 b}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle A \left (a \left (2 \sqrt {a+b x}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )+\frac {2}{3} (a+b x)^{3/2}\right )+\frac {2 B (a+b x)^{5/2}}{5 b}\) |
(2*B*(a + b*x)^(5/2))/(5*b) + A*((2*(a + b*x)^(3/2))/3 + a*(2*Sqrt[a + b*x ] - 2*Sqrt[a]*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]))
3.5.3.3.1 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[((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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Time = 0.50 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.84
method | result | size |
derivativedivides | \(\frac {\frac {2 B \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 A b \left (b x +a \right )^{\frac {3}{2}}}{3}+2 A a b \sqrt {b x +a}-2 A \,a^{\frac {3}{2}} b \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{b}\) | \(58\) |
default | \(\frac {\frac {2 B \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 A b \left (b x +a \right )^{\frac {3}{2}}}{3}+2 A a b \sqrt {b x +a}-2 A \,a^{\frac {3}{2}} b \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{b}\) | \(58\) |
pseudoelliptic | \(\frac {-2 A \,a^{\frac {3}{2}} b \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\frac {8 \left (\frac {\left (\frac {3 B x}{5}+A \right ) x \,b^{2}}{4}+a \left (\frac {3 B x}{10}+A \right ) b +\frac {3 a^{2} B}{20}\right ) \sqrt {b x +a}}{3}}{b}\) | \(63\) |
2/b*(1/5*B*(b*x+a)^(5/2)+1/3*A*b*(b*x+a)^(3/2)+A*a*b*(b*x+a)^(1/2)-A*a^(3/ 2)*b*arctanh((b*x+a)^(1/2)/a^(1/2)))
Time = 0.23 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.29 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x} \, dx=\left [\frac {15 \, A a^{\frac {3}{2}} b \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (3 \, B b^{2} x^{2} + 3 \, B a^{2} + 20 \, A a b + {\left (6 \, B a b + 5 \, A b^{2}\right )} x\right )} \sqrt {b x + a}}{15 \, b}, \frac {2 \, {\left (15 \, A \sqrt {-a} a b \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (3 \, B b^{2} x^{2} + 3 \, B a^{2} + 20 \, A a b + {\left (6 \, B a b + 5 \, A b^{2}\right )} x\right )} \sqrt {b x + a}\right )}}{15 \, b}\right ] \]
[1/15*(15*A*a^(3/2)*b*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(3* B*b^2*x^2 + 3*B*a^2 + 20*A*a*b + (6*B*a*b + 5*A*b^2)*x)*sqrt(b*x + a))/b, 2/15*(15*A*sqrt(-a)*a*b*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (3*B*b^2*x^2 + 3*B*a^2 + 20*A*a*b + (6*B*a*b + 5*A*b^2)*x)*sqrt(b*x + a))/b]
Time = 1.46 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.26 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x} \, dx=\begin {cases} \frac {2 A a^{2} \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 A a \sqrt {a + b x} + \frac {2 A \left (a + b x\right )^{\frac {3}{2}}}{3} + \frac {2 B \left (a + b x\right )^{\frac {5}{2}}}{5 b} & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (A \log {\left (B x \right )} + B x\right ) & \text {otherwise} \end {cases} \]
Piecewise((2*A*a**2*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) + 2*A*a*sqrt(a + b*x) + 2*A*(a + b*x)**(3/2)/3 + 2*B*(a + b*x)**(5/2)/(5*b), Ne(b, 0)), (a **(3/2)*(A*log(B*x) + B*x), True))
Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.06 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x} \, dx=A a^{\frac {3}{2}} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + \frac {2 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} B + 5 \, {\left (b x + a\right )}^{\frac {3}{2}} A b + 15 \, \sqrt {b x + a} A a b\right )}}{15 \, b} \]
A*a^(3/2)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a))) + 2/15* (3*(b*x + a)^(5/2)*B + 5*(b*x + a)^(3/2)*A*b + 15*sqrt(b*x + a)*A*a*b)/b
Time = 0.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.04 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x} \, dx=\frac {2 \, A a^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} B b^{4} + 5 \, {\left (b x + a\right )}^{\frac {3}{2}} A b^{5} + 15 \, \sqrt {b x + a} A a b^{5}\right )}}{15 \, b^{5}} \]
2*A*a^2*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) + 2/15*(3*(b*x + a)^(5/2)* B*b^4 + 5*(b*x + a)^(3/2)*A*b^5 + 15*sqrt(b*x + a)*A*a*b^5)/b^5
Time = 0.42 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.38 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x} \, dx=\left (\frac {2\,A\,b-2\,B\,a}{3\,b}+\frac {2\,B\,a}{3\,b}\right )\,{\left (a+b\,x\right )}^{3/2}+\frac {2\,B\,{\left (a+b\,x\right )}^{5/2}}{5\,b}+a\,\left (\frac {2\,A\,b-2\,B\,a}{b}+\frac {2\,B\,a}{b}\right )\,\sqrt {a+b\,x}+A\,a^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,2{}\mathrm {i} \]