Integrand size = 18, antiderivative size = 100 \[ \int \frac {\sqrt {x} (A+B x)}{(a+b x)^3} \, dx=\frac {(A b-a B) x^{3/2}}{2 a b (a+b x)^2}-\frac {(A b+3 a B) \sqrt {x}}{4 a b^2 (a+b x)}+\frac {(A b+3 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{3/2} b^{5/2}} \]
1/2*(A*b-B*a)*x^(3/2)/a/b/(b*x+a)^2+1/4*(A*b+3*B*a)*arctan(b^(1/2)*x^(1/2) /a^(1/2))/a^(3/2)/b^(5/2)-1/4*(A*b+3*B*a)*x^(1/2)/a/b^2/(b*x+a)
Time = 0.14 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt {x} (A+B x)}{(a+b x)^3} \, dx=-\frac {\sqrt {x} \left (a A b+3 a^2 B-A b^2 x+5 a b B x\right )}{4 a b^2 (a+b x)^2}+\frac {(A b+3 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{3/2} b^{5/2}} \]
-1/4*(Sqrt[x]*(a*A*b + 3*a^2*B - A*b^2*x + 5*a*b*B*x))/(a*b^2*(a + b*x)^2) + ((A*b + 3*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*a^(3/2)*b^(5/2))
Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {87, 51, 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.
\(\displaystyle \int \frac {\sqrt {x} (A+B x)}{(a+b x)^3} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {(3 a B+A b) \int \frac {\sqrt {x}}{(a+b x)^2}dx}{4 a b}+\frac {x^{3/2} (A b-a B)}{2 a b (a+b x)^2}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {(3 a B+A b) \left (\frac {\int \frac {1}{\sqrt {x} (a+b x)}dx}{2 b}-\frac {\sqrt {x}}{b (a+b x)}\right )}{4 a b}+\frac {x^{3/2} (A b-a B)}{2 a b (a+b x)^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(3 a B+A b) \left (\frac {\int \frac {1}{a+b x}d\sqrt {x}}{b}-\frac {\sqrt {x}}{b (a+b x)}\right )}{4 a b}+\frac {x^{3/2} (A b-a B)}{2 a b (a+b x)^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {(3 a B+A b) \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}}-\frac {\sqrt {x}}{b (a+b x)}\right )}{4 a b}+\frac {x^{3/2} (A b-a B)}{2 a b (a+b x)^2}\) |
((A*b - a*B)*x^(3/2))/(2*a*b*(a + b*x)^2) + ((A*b + 3*a*B)*(-(Sqrt[x]/(b*( a + b*x))) + ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]]/(Sqrt[a]*b^(3/2))))/(4*a*b)
3.4.66.3.1 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, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Time = 0.46 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.79
method | result | size |
derivativedivides | \(\frac {\frac {\left (A b -5 B a \right ) x^{\frac {3}{2}}}{4 a b}-\frac {\left (A b +3 B a \right ) \sqrt {x}}{4 b^{2}}}{\left (b x +a \right )^{2}}+\frac {\left (A b +3 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 b^{2} a \sqrt {a b}}\) | \(79\) |
default | \(\frac {\frac {\left (A b -5 B a \right ) x^{\frac {3}{2}}}{4 a b}-\frac {\left (A b +3 B a \right ) \sqrt {x}}{4 b^{2}}}{\left (b x +a \right )^{2}}+\frac {\left (A b +3 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 b^{2} a \sqrt {a b}}\) | \(79\) |
2*(1/8*(A*b-5*B*a)/a/b*x^(3/2)-1/8*(A*b+3*B*a)/b^2*x^(1/2))/(b*x+a)^2+1/4* (A*b+3*B*a)/b^2/a/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(1/2))
Time = 0.23 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.91 \[ \int \frac {\sqrt {x} (A+B x)}{(a+b x)^3} \, dx=\left [-\frac {{\left (3 \, B a^{3} + A a^{2} b + {\left (3 \, B a b^{2} + A b^{3}\right )} x^{2} + 2 \, {\left (3 \, B a^{2} b + A a b^{2}\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) + 2 \, {\left (3 \, B a^{3} b + A a^{2} b^{2} + {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} \sqrt {x}}{8 \, {\left (a^{2} b^{5} x^{2} + 2 \, a^{3} b^{4} x + a^{4} b^{3}\right )}}, -\frac {{\left (3 \, B a^{3} + A a^{2} b + {\left (3 \, B a b^{2} + A b^{3}\right )} x^{2} + 2 \, {\left (3 \, B a^{2} b + A a b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (3 \, B a^{3} b + A a^{2} b^{2} + {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} \sqrt {x}}{4 \, {\left (a^{2} b^{5} x^{2} + 2 \, a^{3} b^{4} x + a^{4} b^{3}\right )}}\right ] \]
[-1/8*((3*B*a^3 + A*a^2*b + (3*B*a*b^2 + A*b^3)*x^2 + 2*(3*B*a^2*b + A*a*b ^2)*x)*sqrt(-a*b)*log((b*x - a - 2*sqrt(-a*b)*sqrt(x))/(b*x + a)) + 2*(3*B *a^3*b + A*a^2*b^2 + (5*B*a^2*b^2 - A*a*b^3)*x)*sqrt(x))/(a^2*b^5*x^2 + 2* a^3*b^4*x + a^4*b^3), -1/4*((3*B*a^3 + A*a^2*b + (3*B*a*b^2 + A*b^3)*x^2 + 2*(3*B*a^2*b + A*a*b^2)*x)*sqrt(a*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (3*B *a^3*b + A*a^2*b^2 + (5*B*a^2*b^2 - A*a*b^3)*x)*sqrt(x))/(a^2*b^5*x^2 + 2* a^3*b^4*x + a^4*b^3)]
Leaf count of result is larger than twice the leaf count of optimal. 1316 vs. \(2 (90) = 180\).
Time = 5.71 (sec) , antiderivative size = 1316, normalized size of antiderivative = 13.16 \[ \int \frac {\sqrt {x} (A+B x)}{(a+b x)^3} \, dx=\text {Too large to display} \]
Piecewise((zoo*(-2*A/(3*x**(3/2)) - 2*B/sqrt(x)), Eq(a, 0) & Eq(b, 0)), (( 2*A*x**(3/2)/3 + 2*B*x**(5/2)/5)/a**3, Eq(b, 0)), ((-2*A/(3*x**(3/2)) - 2* B/sqrt(x))/b**3, Eq(a, 0)), (A*a**2*b*log(sqrt(x) - sqrt(-a/b))/(8*a**3*b* *3*sqrt(-a/b) + 16*a**2*b**4*x*sqrt(-a/b) + 8*a*b**5*x**2*sqrt(-a/b)) - A* a**2*b*log(sqrt(x) + sqrt(-a/b))/(8*a**3*b**3*sqrt(-a/b) + 16*a**2*b**4*x* sqrt(-a/b) + 8*a*b**5*x**2*sqrt(-a/b)) - 2*A*a*b**2*sqrt(x)*sqrt(-a/b)/(8* a**3*b**3*sqrt(-a/b) + 16*a**2*b**4*x*sqrt(-a/b) + 8*a*b**5*x**2*sqrt(-a/b )) + 2*A*a*b**2*x*log(sqrt(x) - sqrt(-a/b))/(8*a**3*b**3*sqrt(-a/b) + 16*a **2*b**4*x*sqrt(-a/b) + 8*a*b**5*x**2*sqrt(-a/b)) - 2*A*a*b**2*x*log(sqrt( x) + sqrt(-a/b))/(8*a**3*b**3*sqrt(-a/b) + 16*a**2*b**4*x*sqrt(-a/b) + 8*a *b**5*x**2*sqrt(-a/b)) + 2*A*b**3*x**(3/2)*sqrt(-a/b)/(8*a**3*b**3*sqrt(-a /b) + 16*a**2*b**4*x*sqrt(-a/b) + 8*a*b**5*x**2*sqrt(-a/b)) + A*b**3*x**2* log(sqrt(x) - sqrt(-a/b))/(8*a**3*b**3*sqrt(-a/b) + 16*a**2*b**4*x*sqrt(-a /b) + 8*a*b**5*x**2*sqrt(-a/b)) - A*b**3*x**2*log(sqrt(x) + sqrt(-a/b))/(8 *a**3*b**3*sqrt(-a/b) + 16*a**2*b**4*x*sqrt(-a/b) + 8*a*b**5*x**2*sqrt(-a/ b)) + 3*B*a**3*log(sqrt(x) - sqrt(-a/b))/(8*a**3*b**3*sqrt(-a/b) + 16*a**2 *b**4*x*sqrt(-a/b) + 8*a*b**5*x**2*sqrt(-a/b)) - 3*B*a**3*log(sqrt(x) + sq rt(-a/b))/(8*a**3*b**3*sqrt(-a/b) + 16*a**2*b**4*x*sqrt(-a/b) + 8*a*b**5*x **2*sqrt(-a/b)) - 6*B*a**2*b*sqrt(x)*sqrt(-a/b)/(8*a**3*b**3*sqrt(-a/b) + 16*a**2*b**4*x*sqrt(-a/b) + 8*a*b**5*x**2*sqrt(-a/b)) + 6*B*a**2*b*x*lo...
Time = 0.29 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {x} (A+B x)}{(a+b x)^3} \, dx=-\frac {{\left (5 \, B a b - A b^{2}\right )} x^{\frac {3}{2}} + {\left (3 \, B a^{2} + A a b\right )} \sqrt {x}}{4 \, {\left (a b^{4} x^{2} + 2 \, a^{2} b^{3} x + a^{3} b^{2}\right )}} + \frac {{\left (3 \, B a + A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a b^{2}} \]
-1/4*((5*B*a*b - A*b^2)*x^(3/2) + (3*B*a^2 + A*a*b)*sqrt(x))/(a*b^4*x^2 + 2*a^2*b^3*x + a^3*b^2) + 1/4*(3*B*a + A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sq rt(a*b)*a*b^2)
Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {x} (A+B x)}{(a+b x)^3} \, dx=\frac {{\left (3 \, B a + A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a b^{2}} - \frac {5 \, B a b x^{\frac {3}{2}} - A b^{2} x^{\frac {3}{2}} + 3 \, B a^{2} \sqrt {x} + A a b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a b^{2}} \]
1/4*(3*B*a + A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a*b^2) - 1/4*(5*B *a*b*x^(3/2) - A*b^2*x^(3/2) + 3*B*a^2*sqrt(x) + A*a*b*sqrt(x))/((b*x + a) ^2*a*b^2)
Time = 0.51 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {x} (A+B x)}{(a+b x)^3} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (A\,b+3\,B\,a\right )}{4\,a^{3/2}\,b^{5/2}}-\frac {\frac {\sqrt {x}\,\left (A\,b+3\,B\,a\right )}{4\,b^2}-\frac {x^{3/2}\,\left (A\,b-5\,B\,a\right )}{4\,a\,b}}{a^2+2\,a\,b\,x+b^2\,x^2} \]