Integrand size = 22, antiderivative size = 90 \[ \int \frac {x^{5/2} (A+B x)}{b x+c x^2} \, dx=\frac {2 b (b B-A c) \sqrt {x}}{c^3}-\frac {2 (b B-A c) x^{3/2}}{3 c^2}+\frac {2 B x^{5/2}}{5 c}-\frac {2 b^{3/2} (b B-A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{7/2}} \] Output:
2*b*(-A*c+B*b)*x^(1/2)/c^3-2/3*(-A*c+B*b)*x^(3/2)/c^2+2/5*B*x^(5/2)/c-2*b^ (3/2)*(-A*c+B*b)*arctan(c^(1/2)*x^(1/2)/b^(1/2))/c^(7/2)
Time = 0.12 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.90 \[ \int \frac {x^{5/2} (A+B x)}{b x+c x^2} \, dx=\frac {2 \sqrt {x} \left (15 b^2 B-5 b c (3 A+B x)+c^2 x (5 A+3 B x)\right )}{15 c^3}-\frac {2 b^{3/2} (b B-A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{7/2}} \] Input:
Integrate[(x^(5/2)*(A + B*x))/(b*x + c*x^2),x]
Output:
(2*Sqrt[x]*(15*b^2*B - 5*b*c*(3*A + B*x) + c^2*x*(5*A + 3*B*x)))/(15*c^3) - (2*b^(3/2)*(b*B - A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/c^(7/2)
Time = 0.30 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {9, 90, 60, 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.
\(\displaystyle \int \frac {x^{5/2} (A+B x)}{b x+c x^2} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {x^{3/2} (A+B x)}{b+c x}dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {2 B x^{5/2}}{5 c}-\frac {(b B-A c) \int \frac {x^{3/2}}{b+c x}dx}{c}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2 B x^{5/2}}{5 c}-\frac {(b B-A c) \left (\frac {2 x^{3/2}}{3 c}-\frac {b \int \frac {\sqrt {x}}{b+c x}dx}{c}\right )}{c}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2 B x^{5/2}}{5 c}-\frac {(b B-A c) \left (\frac {2 x^{3/2}}{3 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {b \int \frac {1}{\sqrt {x} (b+c x)}dx}{c}\right )}{c}\right )}{c}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2 B x^{5/2}}{5 c}-\frac {(b B-A c) \left (\frac {2 x^{3/2}}{3 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {2 b \int \frac {1}{b+c x}d\sqrt {x}}{c}\right )}{c}\right )}{c}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {2 B x^{5/2}}{5 c}-\frac {(b B-A c) \left (\frac {2 x^{3/2}}{3 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{3/2}}\right )}{c}\right )}{c}\) |
Input:
Int[(x^(5/2)*(A + B*x))/(b*x + c*x^2),x]
Output:
(2*B*x^(5/2))/(5*c) - ((b*B - A*c)*((2*x^(3/2))/(3*c) - (b*((2*Sqrt[x])/c - (2*Sqrt[b]*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/c^(3/2)))/c))/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 = 0.86 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84
method | result | size |
risch | \(-\frac {2 \left (-3 B \,c^{2} x^{2}-5 A \,c^{2} x +5 B b c x +15 A b c -15 B \,b^{2}\right ) \sqrt {x}}{15 c^{3}}+\frac {2 b^{2} \left (A c -B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{c^{3} \sqrt {b c}}\) | \(76\) |
derivativedivides | \(-\frac {2 \left (-\frac {B \,c^{2} x^{\frac {5}{2}}}{5}-\frac {A \,c^{2} x^{\frac {3}{2}}}{3}+\frac {B b c \,x^{\frac {3}{2}}}{3}+A b c \sqrt {x}-B \,b^{2} \sqrt {x}\right )}{c^{3}}+\frac {2 b^{2} \left (A c -B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{c^{3} \sqrt {b c}}\) | \(82\) |
default | \(-\frac {2 \left (-\frac {B \,c^{2} x^{\frac {5}{2}}}{5}-\frac {A \,c^{2} x^{\frac {3}{2}}}{3}+\frac {B b c \,x^{\frac {3}{2}}}{3}+A b c \sqrt {x}-B \,b^{2} \sqrt {x}\right )}{c^{3}}+\frac {2 b^{2} \left (A c -B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{c^{3} \sqrt {b c}}\) | \(82\) |
Input:
int(x^(5/2)*(B*x+A)/(c*x^2+b*x),x,method=_RETURNVERBOSE)
Output:
-2/15*(-3*B*c^2*x^2-5*A*c^2*x+5*B*b*c*x+15*A*b*c-15*B*b^2)*x^(1/2)/c^3+2*b ^2*(A*c-B*b)/c^3/(b*c)^(1/2)*arctan(c*x^(1/2)/(b*c)^(1/2))
Time = 0.10 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.00 \[ \int \frac {x^{5/2} (A+B x)}{b x+c x^2} \, dx=\left [-\frac {15 \, {\left (B b^{2} - A b c\right )} \sqrt {-\frac {b}{c}} \log \left (\frac {c x + 2 \, c \sqrt {x} \sqrt {-\frac {b}{c}} - b}{c x + b}\right ) - 2 \, {\left (3 \, B c^{2} x^{2} + 15 \, B b^{2} - 15 \, A b c - 5 \, {\left (B b c - A c^{2}\right )} x\right )} \sqrt {x}}{15 \, c^{3}}, -\frac {2 \, {\left (15 \, {\left (B b^{2} - A b c\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c \sqrt {x} \sqrt {\frac {b}{c}}}{b}\right ) - {\left (3 \, B c^{2} x^{2} + 15 \, B b^{2} - 15 \, A b c - 5 \, {\left (B b c - A c^{2}\right )} x\right )} \sqrt {x}\right )}}{15 \, c^{3}}\right ] \] Input:
integrate(x^(5/2)*(B*x+A)/(c*x^2+b*x),x, algorithm="fricas")
Output:
[-1/15*(15*(B*b^2 - A*b*c)*sqrt(-b/c)*log((c*x + 2*c*sqrt(x)*sqrt(-b/c) - b)/(c*x + b)) - 2*(3*B*c^2*x^2 + 15*B*b^2 - 15*A*b*c - 5*(B*b*c - A*c^2)*x )*sqrt(x))/c^3, -2/15*(15*(B*b^2 - A*b*c)*sqrt(b/c)*arctan(c*sqrt(x)*sqrt( b/c)/b) - (3*B*c^2*x^2 + 15*B*b^2 - 15*A*b*c - 5*(B*b*c - A*c^2)*x)*sqrt(x ))/c^3]
Leaf count of result is larger than twice the leaf count of optimal. 260 vs. \(2 (87) = 174\).
Time = 4.45 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.89 \[ \int \frac {x^{5/2} (A+B x)}{b x+c x^2} \, dx=\begin {cases} \tilde {\infty } \left (\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {5}{2}}}{5}\right ) & \text {for}\: b = 0 \wedge c = 0 \\\frac {\frac {2 A x^{\frac {5}{2}}}{5} + \frac {2 B x^{\frac {7}{2}}}{7}}{b} & \text {for}\: c = 0 \\\frac {\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {5}{2}}}{5}}{c} & \text {for}\: b = 0 \\\frac {A b^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{c}} \right )}}{c^{3} \sqrt {- \frac {b}{c}}} - \frac {A b^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{c}} \right )}}{c^{3} \sqrt {- \frac {b}{c}}} - \frac {2 A b \sqrt {x}}{c^{2}} + \frac {2 A x^{\frac {3}{2}}}{3 c} - \frac {B b^{3} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{c}} \right )}}{c^{4} \sqrt {- \frac {b}{c}}} + \frac {B b^{3} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{c}} \right )}}{c^{4} \sqrt {- \frac {b}{c}}} + \frac {2 B b^{2} \sqrt {x}}{c^{3}} - \frac {2 B b x^{\frac {3}{2}}}{3 c^{2}} + \frac {2 B x^{\frac {5}{2}}}{5 c} & \text {otherwise} \end {cases} \] Input:
integrate(x**(5/2)*(B*x+A)/(c*x**2+b*x),x)
Output:
Piecewise((zoo*(2*A*x**(3/2)/3 + 2*B*x**(5/2)/5), Eq(b, 0) & Eq(c, 0)), (( 2*A*x**(5/2)/5 + 2*B*x**(7/2)/7)/b, Eq(c, 0)), ((2*A*x**(3/2)/3 + 2*B*x**( 5/2)/5)/c, Eq(b, 0)), (A*b**2*log(sqrt(x) - sqrt(-b/c))/(c**3*sqrt(-b/c)) - A*b**2*log(sqrt(x) + sqrt(-b/c))/(c**3*sqrt(-b/c)) - 2*A*b*sqrt(x)/c**2 + 2*A*x**(3/2)/(3*c) - B*b**3*log(sqrt(x) - sqrt(-b/c))/(c**4*sqrt(-b/c)) + B*b**3*log(sqrt(x) + sqrt(-b/c))/(c**4*sqrt(-b/c)) + 2*B*b**2*sqrt(x)/c* *3 - 2*B*b*x**(3/2)/(3*c**2) + 2*B*x**(5/2)/(5*c), True))
Time = 0.11 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.91 \[ \int \frac {x^{5/2} (A+B x)}{b x+c x^2} \, dx=-\frac {2 \, {\left (B b^{3} - A b^{2} c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} c^{3}} + \frac {2 \, {\left (3 \, B c^{2} x^{\frac {5}{2}} - 5 \, {\left (B b c - A c^{2}\right )} x^{\frac {3}{2}} + 15 \, {\left (B b^{2} - A b c\right )} \sqrt {x}\right )}}{15 \, c^{3}} \] Input:
integrate(x^(5/2)*(B*x+A)/(c*x^2+b*x),x, algorithm="maxima")
Output:
-2*(B*b^3 - A*b^2*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*c^3) + 2/15*(3 *B*c^2*x^(5/2) - 5*(B*b*c - A*c^2)*x^(3/2) + 15*(B*b^2 - A*b*c)*sqrt(x))/c ^3
Time = 0.13 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.01 \[ \int \frac {x^{5/2} (A+B x)}{b x+c x^2} \, dx=-\frac {2 \, {\left (B b^{3} - A b^{2} c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} c^{3}} + \frac {2 \, {\left (3 \, B c^{4} x^{\frac {5}{2}} - 5 \, B b c^{3} x^{\frac {3}{2}} + 5 \, A c^{4} x^{\frac {3}{2}} + 15 \, B b^{2} c^{2} \sqrt {x} - 15 \, A b c^{3} \sqrt {x}\right )}}{15 \, c^{5}} \] Input:
integrate(x^(5/2)*(B*x+A)/(c*x^2+b*x),x, algorithm="giac")
Output:
-2*(B*b^3 - A*b^2*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*c^3) + 2/15*(3 *B*c^4*x^(5/2) - 5*B*b*c^3*x^(3/2) + 5*A*c^4*x^(3/2) + 15*B*b^2*c^2*sqrt(x ) - 15*A*b*c^3*sqrt(x))/c^5
Time = 0.04 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.12 \[ \int \frac {x^{5/2} (A+B x)}{b x+c x^2} \, dx=x^{3/2}\,\left (\frac {2\,A}{3\,c}-\frac {2\,B\,b}{3\,c^2}\right )+\frac {2\,B\,x^{5/2}}{5\,c}-\frac {2\,b^{3/2}\,\mathrm {atan}\left (\frac {b^{3/2}\,\sqrt {c}\,\sqrt {x}\,\left (A\,c-B\,b\right )}{B\,b^3-A\,b^2\,c}\right )\,\left (A\,c-B\,b\right )}{c^{7/2}}-\frac {b\,\sqrt {x}\,\left (\frac {2\,A}{c}-\frac {2\,B\,b}{c^2}\right )}{c} \] Input:
int((x^(5/2)*(A + B*x))/(b*x + c*x^2),x)
Output:
x^(3/2)*((2*A)/(3*c) - (2*B*b)/(3*c^2)) + (2*B*x^(5/2))/(5*c) - (2*b^(3/2) *atan((b^(3/2)*c^(1/2)*x^(1/2)*(A*c - B*b))/(B*b^3 - A*b^2*c))*(A*c - B*b) )/c^(7/2) - (b*x^(1/2)*((2*A)/c - (2*B*b)/c^2))/c
Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.09 \[ \int \frac {x^{5/2} (A+B x)}{b x+c x^2} \, dx=\frac {2 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) a b c -2 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) b^{3}-2 \sqrt {x}\, a b \,c^{2}+\frac {2 \sqrt {x}\, a \,c^{3} x}{3}+2 \sqrt {x}\, b^{3} c -\frac {2 \sqrt {x}\, b^{2} c^{2} x}{3}+\frac {2 \sqrt {x}\, b \,c^{3} x^{2}}{5}}{c^{4}} \] Input:
int(x^(5/2)*(B*x+A)/(c*x^2+b*x),x)
Output:
(2*(15*sqrt(c)*sqrt(b)*atan((sqrt(x)*c)/(sqrt(c)*sqrt(b)))*a*b*c - 15*sqrt (c)*sqrt(b)*atan((sqrt(x)*c)/(sqrt(c)*sqrt(b)))*b**3 - 15*sqrt(x)*a*b*c**2 + 5*sqrt(x)*a*c**3*x + 15*sqrt(x)*b**3*c - 5*sqrt(x)*b**2*c**2*x + 3*sqrt (x)*b*c**3*x**2))/(15*c**4)