Integrand size = 18, antiderivative size = 118 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^2} \, dx=a (5 A b+2 a B) \sqrt {a+b x}+\frac {1}{3} (5 A b+2 a B) (a+b x)^{3/2}+\frac {(5 A b+2 a B) (a+b x)^{5/2}}{5 a}-\frac {A (a+b x)^{7/2}}{a x}-a^{3/2} (5 A b+2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]
1/3*(5*A*b+2*B*a)*(b*x+a)^(3/2)+1/5*(5*A*b+2*B*a)*(b*x+a)^(5/2)/a-A*(b*x+a )^(7/2)/a/x-a^(3/2)*(5*A*b+2*B*a)*arctanh((b*x+a)^(1/2)/a^(1/2))+a*(5*A*b+ 2*B*a)*(b*x+a)^(1/2)
Time = 0.16 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.77 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^2} \, dx=\frac {\sqrt {a+b x} \left (2 b^2 x^2 (5 A+3 B x)+2 a b x (35 A+11 B x)+a^2 (-15 A+46 B x)\right )}{15 x}-a^{3/2} (5 A b+2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]
(Sqrt[a + b*x]*(2*b^2*x^2*(5*A + 3*B*x) + 2*a*b*x*(35*A + 11*B*x) + a^2*(- 15*A + 46*B*x)))/(15*x) - a^(3/2)*(5*A*b + 2*a*B)*ArcTanh[Sqrt[a + b*x]/Sq rt[a]]
Time = 0.21 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.86, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {87, 60, 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.
| \(\displaystyle \int \frac {(a+b x)^{5/2} (A+B x)}{x^2} \, dx\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {(2 a B+5 A b) \int \frac {(a+b x)^{5/2}}{x}dx}{2 a}-\frac {A (a+b x)^{7/2}}{a x}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(2 a B+5 A b) \left (a \int \frac {(a+b x)^{3/2}}{x}dx+\frac {2}{5} (a+b x)^{5/2}\right )}{2 a}-\frac {A (a+b x)^{7/2}}{a x}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(2 a B+5 A b) \left (a \left (a \int \frac {\sqrt {a+b x}}{x}dx+\frac {2}{3} (a+b x)^{3/2}\right )+\frac {2}{5} (a+b x)^{5/2}\right )}{2 a}-\frac {A (a+b x)^{7/2}}{a x}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(2 a B+5 A b) \left (a \left (a \left (a \int \frac {1}{x \sqrt {a+b x}}dx+2 \sqrt {a+b x}\right )+\frac {2}{3} (a+b x)^{3/2}\right )+\frac {2}{5} (a+b x)^{5/2}\right )}{2 a}-\frac {A (a+b x)^{7/2}}{a x}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(2 a B+5 A b) \left (a \left (a \left (\frac {2 a \int \frac {1}{\frac {a+b x}{b}-\frac {a}{b}}d\sqrt {a+b x}}{b}+2 \sqrt {a+b x}\right )+\frac {2}{3} (a+b x)^{3/2}\right )+\frac {2}{5} (a+b x)^{5/2}\right )}{2 a}-\frac {A (a+b x)^{7/2}}{a x}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(2 a B+5 A b) \left (a \left (a \left (2 \sqrt {a+b x}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )+\frac {2}{3} (a+b x)^{3/2}\right )+\frac {2}{5} (a+b x)^{5/2}\right )}{2 a}-\frac {A (a+b x)^{7/2}}{a x}\) |
-((A*(a + b*x)^(7/2))/(a*x)) + ((5*A*b + 2*a*B)*((2*(a + b*x)^(5/2))/5 + a *((2*(a + b*x)^(3/2))/3 + a*(2*Sqrt[a + b*x] - 2*Sqrt[a]*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]))))/(2*a)
3.5.16.3.1 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*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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 = 0.53 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.75
| method | result | size |
| pseudoelliptic | \(\frac {\frac {2 \left (-\frac {15}{2} a^{2} b A -3 a^{3} B \right ) x \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{3}+\frac {2 \sqrt {b x +a}\, \left (7 x \left (\frac {11 B x}{35}+A \right ) b \,a^{\frac {3}{2}}+\left (\frac {23 B x}{5}-\frac {3 A}{2}\right ) a^{\frac {5}{2}}+b^{2} x^{2} \sqrt {a}\, \left (\frac {3 B x}{5}+A \right )\right )}{3}}{x \sqrt {a}}\) | \(88\) |
| risch | \(-\frac {a^{2} A \sqrt {b x +a}}{x}+\frac {2 B \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 A b \left (b x +a \right )^{\frac {3}{2}}}{3}+\frac {2 B a \left (b x +a \right )^{\frac {3}{2}}}{3}+4 A a b \sqrt {b x +a}+2 B \,a^{2} \sqrt {b x +a}-a^{\frac {3}{2}} \left (5 A b +2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )\) | \(101\) |
| derivativedivides | \(\frac {2 B \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 A b \left (b x +a \right )^{\frac {3}{2}}}{3}+\frac {2 B a \left (b x +a \right )^{\frac {3}{2}}}{3}+4 A a b \sqrt {b x +a}+2 B \,a^{2} \sqrt {b x +a}-2 a^{2} \left (\frac {A \sqrt {b x +a}}{2 x}+\frac {\left (5 A b +2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\) | \(104\) |
| default | \(\frac {2 B \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 A b \left (b x +a \right )^{\frac {3}{2}}}{3}+\frac {2 B a \left (b x +a \right )^{\frac {3}{2}}}{3}+4 A a b \sqrt {b x +a}+2 B \,a^{2} \sqrt {b x +a}-2 a^{2} \left (\frac {A \sqrt {b x +a}}{2 x}+\frac {\left (5 A b +2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\) | \(104\) |
2/3/a^(1/2)*((-15/2*a^2*b*A-3*a^3*B)*x*arctanh((b*x+a)^(1/2)/a^(1/2))+(b*x +a)^(1/2)*(7*x*(11/35*B*x+A)*b*a^(3/2)+(23/5*B*x-3/2*A)*a^(5/2)+b^2*x^2*a^ (1/2)*(3/5*B*x+A)))/x
Time = 0.23 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.74 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^2} \, dx=\left [\frac {15 \, {\left (2 \, B a^{2} + 5 \, A a b\right )} \sqrt {a} x \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (6 \, B b^{2} x^{3} - 15 \, A a^{2} + 2 \, {\left (11 \, B a b + 5 \, A b^{2}\right )} x^{2} + 2 \, {\left (23 \, B a^{2} + 35 \, A a b\right )} x\right )} \sqrt {b x + a}}{30 \, x}, \frac {15 \, {\left (2 \, B a^{2} + 5 \, A a b\right )} \sqrt {-a} x \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (6 \, B b^{2} x^{3} - 15 \, A a^{2} + 2 \, {\left (11 \, B a b + 5 \, A b^{2}\right )} x^{2} + 2 \, {\left (23 \, B a^{2} + 35 \, A a b\right )} x\right )} \sqrt {b x + a}}{15 \, x}\right ] \]
[1/30*(15*(2*B*a^2 + 5*A*a*b)*sqrt(a)*x*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(6*B*b^2*x^3 - 15*A*a^2 + 2*(11*B*a*b + 5*A*b^2)*x^2 + 2*(2 3*B*a^2 + 35*A*a*b)*x)*sqrt(b*x + a))/x, 1/15*(15*(2*B*a^2 + 5*A*a*b)*sqrt (-a)*x*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (6*B*b^2*x^3 - 15*A*a^2 + 2*(11* B*a*b + 5*A*b^2)*x^2 + 2*(23*B*a^2 + 35*A*a*b)*x)*sqrt(b*x + a))/x]
Time = 12.39 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.14 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^2} \, dx=- A a^{\frac {3}{2}} b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )} - \frac {A a^{2} \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{\sqrt {x}} + 2 A a b \left (\begin {cases} \frac {2 a \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 \sqrt {a + b x} & \text {for}\: b \neq 0 \\\sqrt {a} \log {\left (x \right )} & \text {otherwise} \end {cases}\right ) + A b^{2} \left (\begin {cases} \frac {2 \left (a + b x\right )^{\frac {3}{2}}}{3 b} & \text {for}\: b \neq 0 \\\sqrt {a} x & \text {otherwise} \end {cases}\right ) + B a^{2} \left (\begin {cases} \frac {2 a \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 \sqrt {a + b x} & \text {for}\: b \neq 0 \\\sqrt {a} \log {\left (x \right )} & \text {otherwise} \end {cases}\right ) + 2 B a b \left (\begin {cases} \frac {2 \left (a + b x\right )^{\frac {3}{2}}}{3 b} & \text {for}\: b \neq 0 \\\sqrt {a} x & \text {otherwise} \end {cases}\right ) + B b^{2} \left (\begin {cases} - \frac {2 a \left (a + b x\right )^{\frac {3}{2}}}{3 b^{2}} + \frac {2 \left (a + b x\right )^{\frac {5}{2}}}{5 b^{2}} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{2}}{2} & \text {otherwise} \end {cases}\right ) \]
-A*a**(3/2)*b*asinh(sqrt(a)/(sqrt(b)*sqrt(x))) - A*a**2*sqrt(b)*sqrt(a/(b* x) + 1)/sqrt(x) + 2*A*a*b*Piecewise((2*a*atan(sqrt(a + b*x)/sqrt(-a))/sqrt (-a) + 2*sqrt(a + b*x), Ne(b, 0)), (sqrt(a)*log(x), True)) + A*b**2*Piecew ise((2*(a + b*x)**(3/2)/(3*b), Ne(b, 0)), (sqrt(a)*x, True)) + B*a**2*Piec ewise((2*a*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) + 2*sqrt(a + b*x), Ne(b, 0)), (sqrt(a)*log(x), True)) + 2*B*a*b*Piecewise((2*(a + b*x)**(3/2)/(3*b) , Ne(b, 0)), (sqrt(a)*x, True)) + B*b**2*Piecewise((-2*a*(a + b*x)**(3/2)/ (3*b**2) + 2*(a + b*x)**(5/2)/(5*b**2), Ne(b, 0)), (sqrt(a)*x**2/2, True))
Time = 0.28 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.02 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^2} \, dx=\frac {1}{30} \, {\left (\frac {15 \, {\left (2 \, B a + 5 \, A b\right )} a^{\frac {3}{2}} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{b} - \frac {30 \, \sqrt {b x + a} A a^{2}}{b x} + \frac {4 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} B + 5 \, {\left (B a + A b\right )} {\left (b x + a\right )}^{\frac {3}{2}} + 15 \, {\left (B a^{2} + 2 \, A a b\right )} \sqrt {b x + a}\right )}}{b}\right )} b \]
1/30*(15*(2*B*a + 5*A*b)*a^(3/2)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/b - 30*sqrt(b*x + a)*A*a^2/(b*x) + 4*(3*(b*x + a)^(5/2)*B + 5*(B*a + A*b)*(b*x + a)^(3/2) + 15*(B*a^2 + 2*A*a*b)*sqrt(b*x + a))/b)*b
Time = 0.28 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.06 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^2} \, dx=\frac {6 \, {\left (b x + a\right )}^{\frac {5}{2}} B b + 10 \, {\left (b x + a\right )}^{\frac {3}{2}} B a b + 30 \, \sqrt {b x + a} B a^{2} b + 10 \, {\left (b x + a\right )}^{\frac {3}{2}} A b^{2} + 60 \, \sqrt {b x + a} A a b^{2} - \frac {15 \, \sqrt {b x + a} A a^{2} b}{x} + \frac {15 \, {\left (2 \, B a^{3} b + 5 \, A a^{2} b^{2}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}}}{15 \, b} \]
1/15*(6*(b*x + a)^(5/2)*B*b + 10*(b*x + a)^(3/2)*B*a*b + 30*sqrt(b*x + a)* B*a^2*b + 10*(b*x + a)^(3/2)*A*b^2 + 60*sqrt(b*x + a)*A*a*b^2 - 15*sqrt(b* x + a)*A*a^2*b/x + 15*(2*B*a^3*b + 5*A*a^2*b^2)*arctan(sqrt(b*x + a)/sqrt( -a))/sqrt(-a))/b
Time = 0.09 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.85 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^2} \, dx=\left (\frac {2\,A\,b}{3}+\frac {2\,B\,a}{3}\right )\,{\left (a+b\,x\right )}^{3/2}+\left (2\,a\,\left (2\,A\,b+2\,B\,a\right )-2\,B\,a^2\right )\,\sqrt {a+b\,x}+\frac {2\,B\,{\left (a+b\,x\right )}^{5/2}}{5}-\frac {A\,a^2\,\sqrt {a+b\,x}}{x}+a^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,\left (5\,A\,b+2\,B\,a\right )\,1{}\mathrm {i} \]