Integrand size = 18, antiderivative size = 136 \[ \int \frac {x^{7/2} (A+B x)}{a+b x} \, dx=-\frac {2 a^3 (A b-a B) \sqrt {x}}{b^5}+\frac {2 a^2 (A b-a B) x^{3/2}}{3 b^4}-\frac {2 a (A b-a B) x^{5/2}}{5 b^3}+\frac {2 (A b-a B) x^{7/2}}{7 b^2}+\frac {2 B x^{9/2}}{9 b}+\frac {2 a^{7/2} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{11/2}} \]
2/3*a^2*(A*b-B*a)*x^(3/2)/b^4-2/5*a*(A*b-B*a)*x^(5/2)/b^3+2/7*(A*b-B*a)*x^ (7/2)/b^2+2/9*B*x^(9/2)/b+2*a^(7/2)*(A*b-B*a)*arctan(b^(1/2)*x^(1/2)/a^(1/ 2))/b^(11/2)-2*a^3*(A*b-B*a)*x^(1/2)/b^5
Time = 0.15 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.88 \[ \int \frac {x^{7/2} (A+B x)}{a+b x} \, dx=\frac {2 \sqrt {x} \left (315 a^4 B-105 a^3 b (3 A+B x)+21 a^2 b^2 x (5 A+3 B x)-9 a b^3 x^2 (7 A+5 B x)+5 b^4 x^3 (9 A+7 B x)\right )}{315 b^5}-\frac {2 a^{7/2} (-A b+a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{11/2}} \]
(2*Sqrt[x]*(315*a^4*B - 105*a^3*b*(3*A + B*x) + 21*a^2*b^2*x*(5*A + 3*B*x) - 9*a*b^3*x^2*(7*A + 5*B*x) + 5*b^4*x^3*(9*A + 7*B*x)))/(315*b^5) - (2*a^ (7/2)*(-(A*b) + a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(11/2)
Time = 0.22 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.90, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {90, 60, 60, 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.
| \(\displaystyle \int \frac {x^{7/2} (A+B x)}{a+b x} \, dx\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {(A b-a B) \int \frac {x^{7/2}}{a+b x}dx}{b}+\frac {2 B x^{9/2}}{9 b}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(A b-a B) \left (\frac {2 x^{7/2}}{7 b}-\frac {a \int \frac {x^{5/2}}{a+b x}dx}{b}\right )}{b}+\frac {2 B x^{9/2}}{9 b}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(A b-a B) \left (\frac {2 x^{7/2}}{7 b}-\frac {a \left (\frac {2 x^{5/2}}{5 b}-\frac {a \int \frac {x^{3/2}}{a+b x}dx}{b}\right )}{b}\right )}{b}+\frac {2 B x^{9/2}}{9 b}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(A b-a B) \left (\frac {2 x^{7/2}}{7 b}-\frac {a \left (\frac {2 x^{5/2}}{5 b}-\frac {a \left (\frac {2 x^{3/2}}{3 b}-\frac {a \int \frac {\sqrt {x}}{a+b x}dx}{b}\right )}{b}\right )}{b}\right )}{b}+\frac {2 B x^{9/2}}{9 b}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(A b-a B) \left (\frac {2 x^{7/2}}{7 b}-\frac {a \left (\frac {2 x^{5/2}}{5 b}-\frac {a \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {a \int \frac {1}{\sqrt {x} (a+b x)}dx}{b}\right )}{b}\right )}{b}\right )}{b}\right )}{b}+\frac {2 B x^{9/2}}{9 b}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(A b-a B) \left (\frac {2 x^{7/2}}{7 b}-\frac {a \left (\frac {2 x^{5/2}}{5 b}-\frac {a \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \int \frac {1}{a+b x}d\sqrt {x}}{b}\right )}{b}\right )}{b}\right )}{b}\right )}{b}+\frac {2 B x^{9/2}}{9 b}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {(A b-a B) \left (\frac {2 x^{7/2}}{7 b}-\frac {a \left (\frac {2 x^{5/2}}{5 b}-\frac {a \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}}\right )}{b}\right )}{b}\right )}{b}\right )}{b}+\frac {2 B x^{9/2}}{9 b}\) |
(2*B*x^(9/2))/(9*b) + ((A*b - a*B)*((2*x^(7/2))/(7*b) - (a*((2*x^(5/2))/(5 *b) - (a*((2*x^(3/2))/(3*b) - (a*((2*Sqrt[x])/b - (2*Sqrt[a]*ArcTan[(Sqrt[ b]*Sqrt[x])/Sqrt[a]])/b^(3/2)))/b))/b))/b))/b
3.4.44.3.1 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)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Time = 1.31 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.91
| method | result | size |
| risch | \(-\frac {2 \left (-35 B \,x^{4} b^{4}-45 A \,x^{3} b^{4}+45 B \,x^{3} a \,b^{3}+63 A \,x^{2} a \,b^{3}-63 B \,x^{2} a^{2} b^{2}-105 A x \,a^{2} b^{2}+105 B x \,a^{3} b +315 A \,a^{3} b -315 B \,a^{4}\right ) \sqrt {x}}{315 b^{5}}+\frac {2 a^{4} \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b^{5} \sqrt {a b}}\) | \(124\) |
| derivativedivides | \(-\frac {2 \left (-\frac {B \,x^{\frac {9}{2}} b^{4}}{9}-\frac {A \,b^{4} x^{\frac {7}{2}}}{7}+\frac {B a \,b^{3} x^{\frac {7}{2}}}{7}+\frac {A a \,b^{3} x^{\frac {5}{2}}}{5}-\frac {B \,a^{2} b^{2} x^{\frac {5}{2}}}{5}-\frac {A \,a^{2} b^{2} x^{\frac {3}{2}}}{3}+\frac {B \,a^{3} b \,x^{\frac {3}{2}}}{3}+A \,a^{3} b \sqrt {x}-B \,a^{4} \sqrt {x}\right )}{b^{5}}+\frac {2 a^{4} \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b^{5} \sqrt {a b}}\) | \(130\) |
| default | \(-\frac {2 \left (-\frac {B \,x^{\frac {9}{2}} b^{4}}{9}-\frac {A \,b^{4} x^{\frac {7}{2}}}{7}+\frac {B a \,b^{3} x^{\frac {7}{2}}}{7}+\frac {A a \,b^{3} x^{\frac {5}{2}}}{5}-\frac {B \,a^{2} b^{2} x^{\frac {5}{2}}}{5}-\frac {A \,a^{2} b^{2} x^{\frac {3}{2}}}{3}+\frac {B \,a^{3} b \,x^{\frac {3}{2}}}{3}+A \,a^{3} b \sqrt {x}-B \,a^{4} \sqrt {x}\right )}{b^{5}}+\frac {2 a^{4} \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b^{5} \sqrt {a b}}\) | \(130\) |
-2/315*(-35*B*b^4*x^4-45*A*b^4*x^3+45*B*a*b^3*x^3+63*A*a*b^3*x^2-63*B*a^2* b^2*x^2-105*A*a^2*b^2*x+105*B*a^3*b*x+315*A*a^3*b-315*B*a^4)*x^(1/2)/b^5+2 *a^4*(A*b-B*a)/b^5/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(1/2))
Time = 0.24 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.03 \[ \int \frac {x^{7/2} (A+B x)}{a+b x} \, dx=\left [-\frac {315 \, {\left (B a^{4} - A a^{3} b\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x + 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) - 2 \, {\left (35 \, B b^{4} x^{4} + 315 \, B a^{4} - 315 \, A a^{3} b - 45 \, {\left (B a b^{3} - A b^{4}\right )} x^{3} + 63 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} x^{2} - 105 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{315 \, b^{5}}, -\frac {2 \, {\left (315 \, {\left (B a^{4} - A a^{3} b\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (35 \, B b^{4} x^{4} + 315 \, B a^{4} - 315 \, A a^{3} b - 45 \, {\left (B a b^{3} - A b^{4}\right )} x^{3} + 63 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} x^{2} - 105 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt {x}\right )}}{315 \, b^{5}}\right ] \]
[-1/315*(315*(B*a^4 - A*a^3*b)*sqrt(-a/b)*log((b*x + 2*b*sqrt(x)*sqrt(-a/b ) - a)/(b*x + a)) - 2*(35*B*b^4*x^4 + 315*B*a^4 - 315*A*a^3*b - 45*(B*a*b^ 3 - A*b^4)*x^3 + 63*(B*a^2*b^2 - A*a*b^3)*x^2 - 105*(B*a^3*b - A*a^2*b^2)* x)*sqrt(x))/b^5, -2/315*(315*(B*a^4 - A*a^3*b)*sqrt(a/b)*arctan(b*sqrt(x)* sqrt(a/b)/a) - (35*B*b^4*x^4 + 315*B*a^4 - 315*A*a^3*b - 45*(B*a*b^3 - A*b ^4)*x^3 + 63*(B*a^2*b^2 - A*a*b^3)*x^2 - 105*(B*a^3*b - A*a^2*b^2)*x)*sqrt (x))/b^5]
Leaf count of result is larger than twice the leaf count of optimal. 328 vs. \(2 (131) = 262\).
Time = 10.89 (sec) , antiderivative size = 328, normalized size of antiderivative = 2.41 \[ \int \frac {x^{7/2} (A+B x)}{a+b x} \, dx=\begin {cases} \tilde {\infty } \left (\frac {2 A x^{\frac {7}{2}}}{7} + \frac {2 B x^{\frac {9}{2}}}{9}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {\frac {2 A x^{\frac {9}{2}}}{9} + \frac {2 B x^{\frac {11}{2}}}{11}}{a} & \text {for}\: b = 0 \\\frac {\frac {2 A x^{\frac {7}{2}}}{7} + \frac {2 B x^{\frac {9}{2}}}{9}}{b} & \text {for}\: a = 0 \\\frac {A a^{4} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{b^{5} \sqrt {- \frac {a}{b}}} - \frac {A a^{4} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{b^{5} \sqrt {- \frac {a}{b}}} - \frac {2 A a^{3} \sqrt {x}}{b^{4}} + \frac {2 A a^{2} x^{\frac {3}{2}}}{3 b^{3}} - \frac {2 A a x^{\frac {5}{2}}}{5 b^{2}} + \frac {2 A x^{\frac {7}{2}}}{7 b} - \frac {B a^{5} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{b^{6} \sqrt {- \frac {a}{b}}} + \frac {B a^{5} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{b^{6} \sqrt {- \frac {a}{b}}} + \frac {2 B a^{4} \sqrt {x}}{b^{5}} - \frac {2 B a^{3} x^{\frac {3}{2}}}{3 b^{4}} + \frac {2 B a^{2} x^{\frac {5}{2}}}{5 b^{3}} - \frac {2 B a x^{\frac {7}{2}}}{7 b^{2}} + \frac {2 B x^{\frac {9}{2}}}{9 b} & \text {otherwise} \end {cases} \]
Piecewise((zoo*(2*A*x**(7/2)/7 + 2*B*x**(9/2)/9), Eq(a, 0) & Eq(b, 0)), (( 2*A*x**(9/2)/9 + 2*B*x**(11/2)/11)/a, Eq(b, 0)), ((2*A*x**(7/2)/7 + 2*B*x* *(9/2)/9)/b, Eq(a, 0)), (A*a**4*log(sqrt(x) - sqrt(-a/b))/(b**5*sqrt(-a/b) ) - A*a**4*log(sqrt(x) + sqrt(-a/b))/(b**5*sqrt(-a/b)) - 2*A*a**3*sqrt(x)/ b**4 + 2*A*a**2*x**(3/2)/(3*b**3) - 2*A*a*x**(5/2)/(5*b**2) + 2*A*x**(7/2) /(7*b) - B*a**5*log(sqrt(x) - sqrt(-a/b))/(b**6*sqrt(-a/b)) + B*a**5*log(s qrt(x) + sqrt(-a/b))/(b**6*sqrt(-a/b)) + 2*B*a**4*sqrt(x)/b**5 - 2*B*a**3* x**(3/2)/(3*b**4) + 2*B*a**2*x**(5/2)/(5*b**3) - 2*B*a*x**(7/2)/(7*b**2) + 2*B*x**(9/2)/(9*b), True))
Time = 0.30 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.94 \[ \int \frac {x^{7/2} (A+B x)}{a+b x} \, dx=-\frac {2 \, {\left (B a^{5} - A a^{4} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{5}} + \frac {2 \, {\left (35 \, B b^{4} x^{\frac {9}{2}} - 45 \, {\left (B a b^{3} - A b^{4}\right )} x^{\frac {7}{2}} + 63 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} x^{\frac {5}{2}} - 105 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} x^{\frac {3}{2}} + 315 \, {\left (B a^{4} - A a^{3} b\right )} \sqrt {x}\right )}}{315 \, b^{5}} \]
-2*(B*a^5 - A*a^4*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^5) + 2/315*( 35*B*b^4*x^(9/2) - 45*(B*a*b^3 - A*b^4)*x^(7/2) + 63*(B*a^2*b^2 - A*a*b^3) *x^(5/2) - 105*(B*a^3*b - A*a^2*b^2)*x^(3/2) + 315*(B*a^4 - A*a^3*b)*sqrt( x))/b^5
Time = 0.31 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.02 \[ \int \frac {x^{7/2} (A+B x)}{a+b x} \, dx=-\frac {2 \, {\left (B a^{5} - A a^{4} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{5}} + \frac {2 \, {\left (35 \, B b^{8} x^{\frac {9}{2}} - 45 \, B a b^{7} x^{\frac {7}{2}} + 45 \, A b^{8} x^{\frac {7}{2}} + 63 \, B a^{2} b^{6} x^{\frac {5}{2}} - 63 \, A a b^{7} x^{\frac {5}{2}} - 105 \, B a^{3} b^{5} x^{\frac {3}{2}} + 105 \, A a^{2} b^{6} x^{\frac {3}{2}} + 315 \, B a^{4} b^{4} \sqrt {x} - 315 \, A a^{3} b^{5} \sqrt {x}\right )}}{315 \, b^{9}} \]
-2*(B*a^5 - A*a^4*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^5) + 2/315*( 35*B*b^8*x^(9/2) - 45*B*a*b^7*x^(7/2) + 45*A*b^8*x^(7/2) + 63*B*a^2*b^6*x^ (5/2) - 63*A*a*b^7*x^(5/2) - 105*B*a^3*b^5*x^(3/2) + 105*A*a^2*b^6*x^(3/2) + 315*B*a^4*b^4*sqrt(x) - 315*A*a^3*b^5*sqrt(x))/b^9
Time = 0.06 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.11 \[ \int \frac {x^{7/2} (A+B x)}{a+b x} \, dx=x^{7/2}\,\left (\frac {2\,A}{7\,b}-\frac {2\,B\,a}{7\,b^2}\right )+\frac {2\,B\,x^{9/2}}{9\,b}+\frac {a^2\,x^{3/2}\,\left (\frac {2\,A}{b}-\frac {2\,B\,a}{b^2}\right )}{3\,b^2}-\frac {a^3\,\sqrt {x}\,\left (\frac {2\,A}{b}-\frac {2\,B\,a}{b^2}\right )}{b^3}-\frac {2\,a^{7/2}\,\mathrm {atan}\left (\frac {a^{7/2}\,\sqrt {b}\,\sqrt {x}\,\left (A\,b-B\,a\right )}{B\,a^5-A\,a^4\,b}\right )\,\left (A\,b-B\,a\right )}{b^{11/2}}-\frac {a\,x^{5/2}\,\left (\frac {2\,A}{b}-\frac {2\,B\,a}{b^2}\right )}{5\,b} \]
x^(7/2)*((2*A)/(7*b) - (2*B*a)/(7*b^2)) + (2*B*x^(9/2))/(9*b) + (a^2*x^(3/ 2)*((2*A)/b - (2*B*a)/b^2))/(3*b^2) - (a^3*x^(1/2)*((2*A)/b - (2*B*a)/b^2) )/b^3 - (2*a^(7/2)*atan((a^(7/2)*b^(1/2)*x^(1/2)*(A*b - B*a))/(B*a^5 - A*a ^4*b))*(A*b - B*a))/b^(11/2) - (a*x^(5/2)*((2*A)/b - (2*B*a)/b^2))/(5*b)