Integrand size = 22, antiderivative size = 69 \[ \int \frac {x^{3/2} (A+B x)}{b x+c x^2} \, dx=-\frac {2 (b B-A c) \sqrt {x}}{c^2}+\frac {2 B x^{3/2}}{3 c}+\frac {2 \sqrt {b} (b B-A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{5/2}} \] Output:
-2*(-A*c+B*b)*x^(1/2)/c^2+2/3*B*x^(3/2)/c+2*b^(1/2)*(-A*c+B*b)*arctan(c^(1 /2)*x^(1/2)/b^(1/2))/c^(5/2)
Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.91 \[ \int \frac {x^{3/2} (A+B x)}{b x+c x^2} \, dx=\frac {2 \sqrt {x} (-3 b B+3 A c+B c x)}{3 c^2}+\frac {2 \sqrt {b} (b B-A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{5/2}} \] Input:
Integrate[(x^(3/2)*(A + B*x))/(b*x + c*x^2),x]
Output:
(2*Sqrt[x]*(-3*b*B + 3*A*c + B*c*x))/(3*c^2) + (2*Sqrt[b]*(b*B - A*c)*ArcT an[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/c^(5/2)
Time = 0.29 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {9, 90, 60, 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.
\(\displaystyle \int \frac {x^{3/2} (A+B x)}{b x+c x^2} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {\sqrt {x} (A+B x)}{b+c x}dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {2 B x^{3/2}}{3 c}-\frac {(b B-A c) \int \frac {\sqrt {x}}{b+c x}dx}{c}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2 B x^{3/2}}{3 c}-\frac {(b B-A c) \left (\frac {2 \sqrt {x}}{c}-\frac {b \int \frac {1}{\sqrt {x} (b+c x)}dx}{c}\right )}{c}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2 B x^{3/2}}{3 c}-\frac {(b B-A c) \left (\frac {2 \sqrt {x}}{c}-\frac {2 b \int \frac {1}{b+c x}d\sqrt {x}}{c}\right )}{c}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {2 B x^{3/2}}{3 c}-\frac {(b B-A c) \left (\frac {2 \sqrt {x}}{c}-\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{3/2}}\right )}{c}\) |
Input:
Int[(x^(3/2)*(A + B*x))/(b*x + c*x^2),x]
Output:
(2*B*x^(3/2))/(3*c) - ((b*B - A*c)*((2*Sqrt[x])/c - (2*Sqrt[b]*ArcTan[(Sqr t[c]*Sqrt[x])/Sqrt[b]])/c^(3/2)))/c
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, 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[((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)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Time = 0.79 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.77
method | result | size |
risch | \(\frac {2 \left (B c x +3 A c -3 B b \right ) \sqrt {x}}{3 c^{2}}-\frac {2 b \left (A c -B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{c^{2} \sqrt {b c}}\) | \(53\) |
derivativedivides | \(\frac {\frac {2 B c \,x^{\frac {3}{2}}}{3}+2 A c \sqrt {x}-2 B b \sqrt {x}}{c^{2}}-\frac {2 b \left (A c -B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{c^{2} \sqrt {b c}}\) | \(58\) |
default | \(\frac {\frac {2 B c \,x^{\frac {3}{2}}}{3}+2 A c \sqrt {x}-2 B b \sqrt {x}}{c^{2}}-\frac {2 b \left (A c -B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{c^{2} \sqrt {b c}}\) | \(58\) |
Input:
int(x^(3/2)*(B*x+A)/(c*x^2+b*x),x,method=_RETURNVERBOSE)
Output:
2/3*(B*c*x+3*A*c-3*B*b)*x^(1/2)/c^2-2*b*(A*c-B*b)/c^2/(b*c)^(1/2)*arctan(c *x^(1/2)/(b*c)^(1/2))
Time = 0.09 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.87 \[ \int \frac {x^{3/2} (A+B x)}{b x+c x^2} \, dx=\left [-\frac {3 \, {\left (B b - A c\right )} \sqrt {-\frac {b}{c}} \log \left (\frac {c x - 2 \, c \sqrt {x} \sqrt {-\frac {b}{c}} - b}{c x + b}\right ) - 2 \, {\left (B c x - 3 \, B b + 3 \, A c\right )} \sqrt {x}}{3 \, c^{2}}, \frac {2 \, {\left (3 \, {\left (B b - A c\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c \sqrt {x} \sqrt {\frac {b}{c}}}{b}\right ) + {\left (B c x - 3 \, B b + 3 \, A c\right )} \sqrt {x}\right )}}{3 \, c^{2}}\right ] \] Input:
integrate(x^(3/2)*(B*x+A)/(c*x^2+b*x),x, algorithm="fricas")
Output:
[-1/3*(3*(B*b - A*c)*sqrt(-b/c)*log((c*x - 2*c*sqrt(x)*sqrt(-b/c) - b)/(c* x + b)) - 2*(B*c*x - 3*B*b + 3*A*c)*sqrt(x))/c^2, 2/3*(3*(B*b - A*c)*sqrt( b/c)*arctan(c*sqrt(x)*sqrt(b/c)/b) + (B*c*x - 3*B*b + 3*A*c)*sqrt(x))/c^2]
Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (65) = 130\).
Time = 1.73 (sec) , antiderivative size = 221, normalized size of antiderivative = 3.20 \[ \int \frac {x^{3/2} (A+B x)}{b x+c x^2} \, dx=\begin {cases} \tilde {\infty } \left (2 A \sqrt {x} + \frac {2 B x^{\frac {3}{2}}}{3}\right ) & \text {for}\: b = 0 \wedge c = 0 \\\frac {\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {5}{2}}}{5}}{b} & \text {for}\: c = 0 \\\frac {2 A \sqrt {x} + \frac {2 B x^{\frac {3}{2}}}{3}}{c} & \text {for}\: b = 0 \\- \frac {A b \log {\left (\sqrt {x} - \sqrt {- \frac {b}{c}} \right )}}{c^{2} \sqrt {- \frac {b}{c}}} + \frac {A b \log {\left (\sqrt {x} + \sqrt {- \frac {b}{c}} \right )}}{c^{2} \sqrt {- \frac {b}{c}}} + \frac {2 A \sqrt {x}}{c} + \frac {B b^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{c}} \right )}}{c^{3} \sqrt {- \frac {b}{c}}} - \frac {B b^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{c}} \right )}}{c^{3} \sqrt {- \frac {b}{c}}} - \frac {2 B b \sqrt {x}}{c^{2}} + \frac {2 B x^{\frac {3}{2}}}{3 c} & \text {otherwise} \end {cases} \] Input:
integrate(x**(3/2)*(B*x+A)/(c*x**2+b*x),x)
Output:
Piecewise((zoo*(2*A*sqrt(x) + 2*B*x**(3/2)/3), Eq(b, 0) & Eq(c, 0)), ((2*A *x**(3/2)/3 + 2*B*x**(5/2)/5)/b, Eq(c, 0)), ((2*A*sqrt(x) + 2*B*x**(3/2)/3 )/c, Eq(b, 0)), (-A*b*log(sqrt(x) - sqrt(-b/c))/(c**2*sqrt(-b/c)) + A*b*lo g(sqrt(x) + sqrt(-b/c))/(c**2*sqrt(-b/c)) + 2*A*sqrt(x)/c + B*b**2*log(sqr t(x) - sqrt(-b/c))/(c**3*sqrt(-b/c)) - B*b**2*log(sqrt(x) + sqrt(-b/c))/(c **3*sqrt(-b/c)) - 2*B*b*sqrt(x)/c**2 + 2*B*x**(3/2)/(3*c), True))
Time = 0.11 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.84 \[ \int \frac {x^{3/2} (A+B x)}{b x+c x^2} \, dx=\frac {2 \, {\left (B b^{2} - A b c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} c^{2}} + \frac {2 \, {\left (B c x^{\frac {3}{2}} - 3 \, {\left (B b - A c\right )} \sqrt {x}\right )}}{3 \, c^{2}} \] Input:
integrate(x^(3/2)*(B*x+A)/(c*x^2+b*x),x, algorithm="maxima")
Output:
2*(B*b^2 - A*b*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*c^2) + 2/3*(B*c*x ^(3/2) - 3*(B*b - A*c)*sqrt(x))/c^2
Time = 0.15 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.93 \[ \int \frac {x^{3/2} (A+B x)}{b x+c x^2} \, dx=\frac {2 \, {\left (B b^{2} - A b c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} c^{2}} + \frac {2 \, {\left (B c^{2} x^{\frac {3}{2}} - 3 \, B b c \sqrt {x} + 3 \, A c^{2} \sqrt {x}\right )}}{3 \, c^{3}} \] Input:
integrate(x^(3/2)*(B*x+A)/(c*x^2+b*x),x, algorithm="giac")
Output:
2*(B*b^2 - A*b*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*c^2) + 2/3*(B*c^2 *x^(3/2) - 3*B*b*c*sqrt(x) + 3*A*c^2*sqrt(x))/c^3
Time = 0.05 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.10 \[ \int \frac {x^{3/2} (A+B x)}{b x+c x^2} \, dx=\sqrt {x}\,\left (\frac {2\,A}{c}-\frac {2\,B\,b}{c^2}\right )+\frac {2\,B\,x^{3/2}}{3\,c}+\frac {2\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c}\,\sqrt {x}\,\left (A\,c-B\,b\right )}{B\,b^2-A\,b\,c}\right )\,\left (A\,c-B\,b\right )}{c^{5/2}} \] Input:
int((x^(3/2)*(A + B*x))/(b*x + c*x^2),x)
Output:
x^(1/2)*((2*A)/c - (2*B*b)/c^2) + (2*B*x^(3/2))/(3*c) + (2*b^(1/2)*atan((b ^(1/2)*c^(1/2)*x^(1/2)*(A*c - B*b))/(B*b^2 - A*b*c))*(A*c - B*b))/c^(5/2)
Time = 0.21 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.06 \[ \int \frac {x^{3/2} (A+B x)}{b x+c x^2} \, dx=\frac {-2 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) a c +2 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) b^{2}+2 \sqrt {x}\, a \,c^{2}-2 \sqrt {x}\, b^{2} c +\frac {2 \sqrt {x}\, b \,c^{2} x}{3}}{c^{3}} \] Input:
int(x^(3/2)*(B*x+A)/(c*x^2+b*x),x)
Output:
(2*( - 3*sqrt(c)*sqrt(b)*atan((sqrt(x)*c)/(sqrt(c)*sqrt(b)))*a*c + 3*sqrt( c)*sqrt(b)*atan((sqrt(x)*c)/(sqrt(c)*sqrt(b)))*b**2 + 3*sqrt(x)*a*c**2 - 3 *sqrt(x)*b**2*c + sqrt(x)*b*c**2*x))/(3*c**3)