Integrand size = 18, antiderivative size = 147 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^3} \, dx=\frac {5 (3 A b-7 a B) \sqrt {x}}{4 b^4}-\frac {5 (3 A b-7 a B) x^{3/2}}{12 a b^3}+\frac {(A b-a B) x^{7/2}}{2 a b (a+b x)^2}+\frac {(3 A b-7 a B) x^{5/2}}{4 a b^2 (a+b x)}-\frac {5 \sqrt {a} (3 A b-7 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{9/2}} \]
-5/12*(3*A*b-7*B*a)*x^(3/2)/a/b^3+1/2*(A*b-B*a)*x^(7/2)/a/b/(b*x+a)^2+1/4* (3*A*b-7*B*a)*x^(5/2)/a/b^2/(b*x+a)-5/4*(3*A*b-7*B*a)*arctan(b^(1/2)*x^(1/ 2)/a^(1/2))*a^(1/2)/b^(9/2)+5/4*(3*A*b-7*B*a)*x^(1/2)/b^4
Time = 0.17 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.75 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^3} \, dx=\frac {\sqrt {x} \left (-105 a^3 B+a b^2 x (75 A-56 B x)+5 a^2 b (9 A-35 B x)+8 b^3 x^2 (3 A+B x)\right )}{12 b^4 (a+b x)^2}+\frac {5 \sqrt {a} (-3 A b+7 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{9/2}} \]
(Sqrt[x]*(-105*a^3*B + a*b^2*x*(75*A - 56*B*x) + 5*a^2*b*(9*A - 35*B*x) + 8*b^3*x^2*(3*A + B*x)))/(12*b^4*(a + b*x)^2) + (5*Sqrt[a]*(-3*A*b + 7*a*B) *ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*b^(9/2))
Time = 0.23 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.91, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {87, 51, 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.
\(\displaystyle \int \frac {x^{5/2} (A+B x)}{(a+b x)^3} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {x^{7/2} (A b-a B)}{2 a b (a+b x)^2}-\frac {(3 A b-7 a B) \int \frac {x^{5/2}}{(a+b x)^2}dx}{4 a b}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {x^{7/2} (A b-a B)}{2 a b (a+b x)^2}-\frac {(3 A b-7 a B) \left (\frac {5 \int \frac {x^{3/2}}{a+b x}dx}{2 b}-\frac {x^{5/2}}{b (a+b x)}\right )}{4 a b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {x^{7/2} (A b-a B)}{2 a b (a+b x)^2}-\frac {(3 A b-7 a B) \left (\frac {5 \left (\frac {2 x^{3/2}}{3 b}-\frac {a \int \frac {\sqrt {x}}{a+b x}dx}{b}\right )}{2 b}-\frac {x^{5/2}}{b (a+b x)}\right )}{4 a b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {x^{7/2} (A b-a B)}{2 a b (a+b x)^2}-\frac {(3 A b-7 a B) \left (\frac {5 \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {a \int \frac {1}{\sqrt {x} (a+b x)}dx}{b}\right )}{b}\right )}{2 b}-\frac {x^{5/2}}{b (a+b x)}\right )}{4 a b}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {x^{7/2} (A b-a B)}{2 a b (a+b x)^2}-\frac {(3 A b-7 a B) \left (\frac {5 \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \int \frac {1}{a+b x}d\sqrt {x}}{b}\right )}{b}\right )}{2 b}-\frac {x^{5/2}}{b (a+b x)}\right )}{4 a b}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {x^{7/2} (A b-a B)}{2 a b (a+b x)^2}-\frac {(3 A b-7 a B) \left (\frac {5 \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}}\right )}{b}\right )}{2 b}-\frac {x^{5/2}}{b (a+b x)}\right )}{4 a b}\) |
((A*b - a*B)*x^(7/2))/(2*a*b*(a + b*x)^2) - ((3*A*b - 7*a*B)*(-(x^(5/2)/(b *(a + b*x))) + (5*((2*x^(3/2))/(3*b) - (a*((2*Sqrt[x])/b - (2*Sqrt[a]*ArcT an[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(3/2)))/b))/(2*b)))/(4*a*b)
3.4.64.3.1 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] :> 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)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Time = 0.50 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.67
method | result | size |
risch | \(\frac {2 \left (b B x +3 A b -9 B a \right ) \sqrt {x}}{3 b^{4}}-\frac {a \left (\frac {2 \left (-\frac {9}{8} b^{2} A +\frac {13}{8} a b B \right ) x^{\frac {3}{2}}-\frac {a \left (7 A b -11 B a \right ) \sqrt {x}}{4}}{\left (b x +a \right )^{2}}+\frac {5 \left (3 A b -7 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}\right )}{b^{4}}\) | \(98\) |
derivativedivides | \(\frac {\frac {2 b B \,x^{\frac {3}{2}}}{3}+2 A b \sqrt {x}-6 B a \sqrt {x}}{b^{4}}-\frac {2 a \left (\frac {\left (-\frac {9}{8} b^{2} A +\frac {13}{8} a b B \right ) x^{\frac {3}{2}}-\frac {a \left (7 A b -11 B a \right ) \sqrt {x}}{8}}{\left (b x +a \right )^{2}}+\frac {5 \left (3 A b -7 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{4}}\) | \(102\) |
default | \(\frac {\frac {2 b B \,x^{\frac {3}{2}}}{3}+2 A b \sqrt {x}-6 B a \sqrt {x}}{b^{4}}-\frac {2 a \left (\frac {\left (-\frac {9}{8} b^{2} A +\frac {13}{8} a b B \right ) x^{\frac {3}{2}}-\frac {a \left (7 A b -11 B a \right ) \sqrt {x}}{8}}{\left (b x +a \right )^{2}}+\frac {5 \left (3 A b -7 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{4}}\) | \(102\) |
2/3*(B*b*x+3*A*b-9*B*a)*x^(1/2)/b^4-a/b^4*(2*((-9/8*b^2*A+13/8*a*b*B)*x^(3 /2)-1/8*a*(7*A*b-11*B*a)*x^(1/2))/(b*x+a)^2+5/4*(3*A*b-7*B*a)/(a*b)^(1/2)* arctan(b*x^(1/2)/(a*b)^(1/2)))
Time = 0.25 (sec) , antiderivative size = 349, normalized size of antiderivative = 2.37 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^3} \, dx=\left [-\frac {15 \, {\left (7 \, B a^{3} - 3 \, A a^{2} b + {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} + 2 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) - 2 \, {\left (8 \, B b^{3} x^{3} - 105 \, B a^{3} + 45 \, A a^{2} b - 8 \, {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} - 25 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt {x}}{24 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}, \frac {15 \, {\left (7 \, B a^{3} - 3 \, A a^{2} b + {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} + 2 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) + {\left (8 \, B b^{3} x^{3} - 105 \, B a^{3} + 45 \, A a^{2} b - 8 \, {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} - 25 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt {x}}{12 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}\right ] \]
[-1/24*(15*(7*B*a^3 - 3*A*a^2*b + (7*B*a*b^2 - 3*A*b^3)*x^2 + 2*(7*B*a^2*b - 3*A*a*b^2)*x)*sqrt(-a/b)*log((b*x - 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) - 2*(8*B*b^3*x^3 - 105*B*a^3 + 45*A*a^2*b - 8*(7*B*a*b^2 - 3*A*b^3)*x^ 2 - 25*(7*B*a^2*b - 3*A*a*b^2)*x)*sqrt(x))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4) , 1/12*(15*(7*B*a^3 - 3*A*a^2*b + (7*B*a*b^2 - 3*A*b^3)*x^2 + 2*(7*B*a^2*b - 3*A*a*b^2)*x)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) + (8*B*b^3*x^3 - 105*B*a^3 + 45*A*a^2*b - 8*(7*B*a*b^2 - 3*A*b^3)*x^2 - 25*(7*B*a^2*b - 3*A *a*b^2)*x)*sqrt(x))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4)]
Leaf count of result is larger than twice the leaf count of optimal. 1496 vs. \(2 (139) = 278\).
Time = 21.31 (sec) , antiderivative size = 1496, normalized size of antiderivative = 10.18 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^3} \, dx=\text {Too large to display} \]
Piecewise((zoo*(2*A*sqrt(x) + 2*B*x**(3/2)/3), Eq(a, 0) & Eq(b, 0)), ((2*A *x**(7/2)/7 + 2*B*x**(9/2)/9)/a**3, Eq(b, 0)), ((2*A*sqrt(x) + 2*B*x**(3/2 )/3)/b**3, Eq(a, 0)), (-45*A*a**3*b*log(sqrt(x) - sqrt(-a/b))/(24*a**2*b** 5*sqrt(-a/b) + 48*a*b**6*x*sqrt(-a/b) + 24*b**7*x**2*sqrt(-a/b)) + 45*A*a* *3*b*log(sqrt(x) + sqrt(-a/b))/(24*a**2*b**5*sqrt(-a/b) + 48*a*b**6*x*sqrt (-a/b) + 24*b**7*x**2*sqrt(-a/b)) + 90*A*a**2*b**2*sqrt(x)*sqrt(-a/b)/(24* a**2*b**5*sqrt(-a/b) + 48*a*b**6*x*sqrt(-a/b) + 24*b**7*x**2*sqrt(-a/b)) - 90*A*a**2*b**2*x*log(sqrt(x) - sqrt(-a/b))/(24*a**2*b**5*sqrt(-a/b) + 48* a*b**6*x*sqrt(-a/b) + 24*b**7*x**2*sqrt(-a/b)) + 90*A*a**2*b**2*x*log(sqrt (x) + sqrt(-a/b))/(24*a**2*b**5*sqrt(-a/b) + 48*a*b**6*x*sqrt(-a/b) + 24*b **7*x**2*sqrt(-a/b)) + 150*A*a*b**3*x**(3/2)*sqrt(-a/b)/(24*a**2*b**5*sqrt (-a/b) + 48*a*b**6*x*sqrt(-a/b) + 24*b**7*x**2*sqrt(-a/b)) - 45*A*a*b**3*x **2*log(sqrt(x) - sqrt(-a/b))/(24*a**2*b**5*sqrt(-a/b) + 48*a*b**6*x*sqrt( -a/b) + 24*b**7*x**2*sqrt(-a/b)) + 45*A*a*b**3*x**2*log(sqrt(x) + sqrt(-a/ b))/(24*a**2*b**5*sqrt(-a/b) + 48*a*b**6*x*sqrt(-a/b) + 24*b**7*x**2*sqrt( -a/b)) + 48*A*b**4*x**(5/2)*sqrt(-a/b)/(24*a**2*b**5*sqrt(-a/b) + 48*a*b** 6*x*sqrt(-a/b) + 24*b**7*x**2*sqrt(-a/b)) + 105*B*a**4*log(sqrt(x) - sqrt( -a/b))/(24*a**2*b**5*sqrt(-a/b) + 48*a*b**6*x*sqrt(-a/b) + 24*b**7*x**2*sq rt(-a/b)) - 105*B*a**4*log(sqrt(x) + sqrt(-a/b))/(24*a**2*b**5*sqrt(-a/b) + 48*a*b**6*x*sqrt(-a/b) + 24*b**7*x**2*sqrt(-a/b)) - 210*B*a**3*b*sqrt...
Time = 0.28 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.84 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^3} \, dx=-\frac {{\left (13 \, B a^{2} b - 9 \, A a b^{2}\right )} x^{\frac {3}{2}} + {\left (11 \, B a^{3} - 7 \, A a^{2} b\right )} \sqrt {x}}{4 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} + \frac {5 \, {\left (7 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{4}} + \frac {2 \, {\left (B b x^{\frac {3}{2}} - 3 \, {\left (3 \, B a - A b\right )} \sqrt {x}\right )}}{3 \, b^{4}} \]
-1/4*((13*B*a^2*b - 9*A*a*b^2)*x^(3/2) + (11*B*a^3 - 7*A*a^2*b)*sqrt(x))/( b^6*x^2 + 2*a*b^5*x + a^2*b^4) + 5/4*(7*B*a^2 - 3*A*a*b)*arctan(b*sqrt(x)/ sqrt(a*b))/(sqrt(a*b)*b^4) + 2/3*(B*b*x^(3/2) - 3*(3*B*a - A*b)*sqrt(x))/b ^4
Time = 0.29 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.81 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^3} \, dx=\frac {5 \, {\left (7 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{4}} - \frac {13 \, B a^{2} b x^{\frac {3}{2}} - 9 \, A a b^{2} x^{\frac {3}{2}} + 11 \, B a^{3} \sqrt {x} - 7 \, A a^{2} b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} b^{4}} + \frac {2 \, {\left (B b^{6} x^{\frac {3}{2}} - 9 \, B a b^{5} \sqrt {x} + 3 \, A b^{6} \sqrt {x}\right )}}{3 \, b^{9}} \]
5/4*(7*B*a^2 - 3*A*a*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4) - 1/4* (13*B*a^2*b*x^(3/2) - 9*A*a*b^2*x^(3/2) + 11*B*a^3*sqrt(x) - 7*A*a^2*b*sqr t(x))/((b*x + a)^2*b^4) + 2/3*(B*b^6*x^(3/2) - 9*B*a*b^5*sqrt(x) + 3*A*b^6 *sqrt(x))/b^9
Time = 0.58 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.97 \[ \int \frac {x^{5/2} (A+B x)}{(a+b x)^3} \, dx=\frac {x^{3/2}\,\left (\frac {9\,A\,a\,b^2}{4}-\frac {13\,B\,a^2\,b}{4}\right )-\sqrt {x}\,\left (\frac {11\,B\,a^3}{4}-\frac {7\,A\,a^2\,b}{4}\right )}{a^2\,b^4+2\,a\,b^5\,x+b^6\,x^2}+\sqrt {x}\,\left (\frac {2\,A}{b^3}-\frac {6\,B\,a}{b^4}\right )+\frac {2\,B\,x^{3/2}}{3\,b^3}+\frac {5\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,\sqrt {x}\,\left (3\,A\,b-7\,B\,a\right )}{7\,B\,a^2-3\,A\,a\,b}\right )\,\left (3\,A\,b-7\,B\,a\right )}{4\,b^{9/2}} \]