Integrand size = 20, antiderivative size = 117 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{15/2}} \, dx=-\frac {2 A (a+b x)^{7/2}}{13 a x^{13/2}}+\frac {2 (6 A b-13 a B) (a+b x)^{7/2}}{143 a^2 x^{11/2}}-\frac {8 b (6 A b-13 a B) (a+b x)^{7/2}}{1287 a^3 x^{9/2}}+\frac {16 b^2 (6 A b-13 a B) (a+b x)^{7/2}}{9009 a^4 x^{7/2}} \]
-2/13*A*(b*x+a)^(7/2)/a/x^(13/2)+2/143*(6*A*b-13*B*a)*(b*x+a)^(7/2)/a^2/x^ (11/2)-8/1287*b*(6*A*b-13*B*a)*(b*x+a)^(7/2)/a^3/x^(9/2)+16/9009*b^2*(6*A* b-13*B*a)*(b*x+a)^(7/2)/a^4/x^(7/2)
Time = 0.34 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.65 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{15/2}} \, dx=-\frac {2 (a+b x)^{7/2} \left (-48 A b^3 x^3+63 a^3 (11 A+13 B x)+8 a b^2 x^2 (21 A+13 B x)-14 a^2 b x (27 A+26 B x)\right )}{9009 a^4 x^{13/2}} \]
(-2*(a + b*x)^(7/2)*(-48*A*b^3*x^3 + 63*a^3*(11*A + 13*B*x) + 8*a*b^2*x^2* (21*A + 13*B*x) - 14*a^2*b*x*(27*A + 26*B*x)))/(9009*a^4*x^(13/2))
Time = 0.19 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {87, 55, 55, 48}
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.
\(\displaystyle \int \frac {(a+b x)^{5/2} (A+B x)}{x^{15/2}} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle -\frac {(6 A b-13 a B) \int \frac {(a+b x)^{5/2}}{x^{13/2}}dx}{13 a}-\frac {2 A (a+b x)^{7/2}}{13 a x^{13/2}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {(6 A b-13 a B) \left (-\frac {4 b \int \frac {(a+b x)^{5/2}}{x^{11/2}}dx}{11 a}-\frac {2 (a+b x)^{7/2}}{11 a x^{11/2}}\right )}{13 a}-\frac {2 A (a+b x)^{7/2}}{13 a x^{13/2}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {(6 A b-13 a B) \left (-\frac {4 b \left (-\frac {2 b \int \frac {(a+b x)^{5/2}}{x^{9/2}}dx}{9 a}-\frac {2 (a+b x)^{7/2}}{9 a x^{9/2}}\right )}{11 a}-\frac {2 (a+b x)^{7/2}}{11 a x^{11/2}}\right )}{13 a}-\frac {2 A (a+b x)^{7/2}}{13 a x^{13/2}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {\left (-\frac {4 b \left (\frac {4 b (a+b x)^{7/2}}{63 a^2 x^{7/2}}-\frac {2 (a+b x)^{7/2}}{9 a x^{9/2}}\right )}{11 a}-\frac {2 (a+b x)^{7/2}}{11 a x^{11/2}}\right ) (6 A b-13 a B)}{13 a}-\frac {2 A (a+b x)^{7/2}}{13 a x^{13/2}}\) |
(-2*A*(a + b*x)^(7/2))/(13*a*x^(13/2)) - ((6*A*b - 13*a*B)*((-2*(a + b*x)^ (7/2))/(11*a*x^(11/2)) - (4*b*((-2*(a + b*x)^(7/2))/(9*a*x^(9/2)) + (4*b*( a + b*x)^(7/2))/(63*a^2*x^(7/2))))/(11*a)))/(13*a)
3.6.11.3.1 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 + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
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]))))
Time = 0.52 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.66
method | result | size |
gosper | \(-\frac {2 \left (b x +a \right )^{\frac {7}{2}} \left (-48 A \,b^{3} x^{3}+104 B a \,b^{2} x^{3}+168 a A \,b^{2} x^{2}-364 B \,a^{2} b \,x^{2}-378 a^{2} A b x +819 a^{3} B x +693 a^{3} A \right )}{9009 x^{\frac {13}{2}} a^{4}}\) | \(77\) |
default | \(-\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (-48 A \,b^{5} x^{5}+104 B a \,b^{4} x^{5}+72 a A \,b^{4} x^{4}-156 B \,a^{2} b^{3} x^{4}-90 a^{2} A \,b^{3} x^{3}+195 B \,a^{3} b^{2} x^{3}+105 a^{3} A \,b^{2} x^{2}+1274 B \,a^{4} b \,x^{2}+1008 a^{4} A b x +819 a^{5} B x +693 a^{5} A \right )}{9009 x^{\frac {13}{2}} a^{4}}\) | \(125\) |
risch | \(-\frac {2 \sqrt {b x +a}\, \left (-48 A \,b^{6} x^{6}+104 B a \,b^{5} x^{6}+24 A a \,b^{5} x^{5}-52 B \,a^{2} b^{4} x^{5}-18 A \,a^{2} b^{4} x^{4}+39 B \,a^{3} b^{3} x^{4}+15 A \,a^{3} b^{3} x^{3}+1469 B \,a^{4} b^{2} x^{3}+1113 A \,a^{4} b^{2} x^{2}+2093 B \,a^{5} b \,x^{2}+1701 A \,a^{5} b x +819 B \,a^{6} x +693 A \,a^{6}\right )}{9009 x^{\frac {13}{2}} a^{4}}\) | \(149\) |
-2/9009*(b*x+a)^(7/2)*(-48*A*b^3*x^3+104*B*a*b^2*x^3+168*A*a*b^2*x^2-364*B *a^2*b*x^2-378*A*a^2*b*x+819*B*a^3*x+693*A*a^3)/x^(13/2)/a^4
Time = 0.23 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.27 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{15/2}} \, dx=-\frac {2 \, {\left (693 \, A a^{6} + 8 \, {\left (13 \, B a b^{5} - 6 \, A b^{6}\right )} x^{6} - 4 \, {\left (13 \, B a^{2} b^{4} - 6 \, A a b^{5}\right )} x^{5} + 3 \, {\left (13 \, B a^{3} b^{3} - 6 \, A a^{2} b^{4}\right )} x^{4} + {\left (1469 \, B a^{4} b^{2} + 15 \, A a^{3} b^{3}\right )} x^{3} + 7 \, {\left (299 \, B a^{5} b + 159 \, A a^{4} b^{2}\right )} x^{2} + 63 \, {\left (13 \, B a^{6} + 27 \, A a^{5} b\right )} x\right )} \sqrt {b x + a}}{9009 \, a^{4} x^{\frac {13}{2}}} \]
-2/9009*(693*A*a^6 + 8*(13*B*a*b^5 - 6*A*b^6)*x^6 - 4*(13*B*a^2*b^4 - 6*A* a*b^5)*x^5 + 3*(13*B*a^3*b^3 - 6*A*a^2*b^4)*x^4 + (1469*B*a^4*b^2 + 15*A*a ^3*b^3)*x^3 + 7*(299*B*a^5*b + 159*A*a^4*b^2)*x^2 + 63*(13*B*a^6 + 27*A*a^ 5*b)*x)*sqrt(b*x + a)/(a^4*x^(13/2))
Timed out. \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{15/2}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 350 vs. \(2 (93) = 186\).
Time = 0.22 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.99 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{15/2}} \, dx=-\frac {16 \, \sqrt {b x^{2} + a x} B b^{5}}{693 \, a^{3} x} + \frac {32 \, \sqrt {b x^{2} + a x} A b^{6}}{3003 \, a^{4} x} + \frac {8 \, \sqrt {b x^{2} + a x} B b^{4}}{693 \, a^{2} x^{2}} - \frac {16 \, \sqrt {b x^{2} + a x} A b^{5}}{3003 \, a^{3} x^{2}} - \frac {2 \, \sqrt {b x^{2} + a x} B b^{3}}{231 \, a x^{3}} + \frac {4 \, \sqrt {b x^{2} + a x} A b^{4}}{1001 \, a^{2} x^{3}} + \frac {5 \, \sqrt {b x^{2} + a x} B b^{2}}{693 \, x^{4}} - \frac {10 \, \sqrt {b x^{2} + a x} A b^{3}}{3003 \, a x^{4}} - \frac {5 \, \sqrt {b x^{2} + a x} B a b}{792 \, x^{5}} + \frac {5 \, \sqrt {b x^{2} + a x} A b^{2}}{1716 \, x^{5}} - \frac {5 \, \sqrt {b x^{2} + a x} B a^{2}}{88 \, x^{6}} - \frac {3 \, \sqrt {b x^{2} + a x} A a b}{1144 \, x^{6}} + \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a}{24 \, x^{7}} - \frac {3 \, \sqrt {b x^{2} + a x} A a^{2}}{104 \, x^{7}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} B}{3 \, x^{8}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A a}{8 \, x^{8}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} A}{4 \, x^{9}} \]
-16/693*sqrt(b*x^2 + a*x)*B*b^5/(a^3*x) + 32/3003*sqrt(b*x^2 + a*x)*A*b^6/ (a^4*x) + 8/693*sqrt(b*x^2 + a*x)*B*b^4/(a^2*x^2) - 16/3003*sqrt(b*x^2 + a *x)*A*b^5/(a^3*x^2) - 2/231*sqrt(b*x^2 + a*x)*B*b^3/(a*x^3) + 4/1001*sqrt( b*x^2 + a*x)*A*b^4/(a^2*x^3) + 5/693*sqrt(b*x^2 + a*x)*B*b^2/x^4 - 10/3003 *sqrt(b*x^2 + a*x)*A*b^3/(a*x^4) - 5/792*sqrt(b*x^2 + a*x)*B*a*b/x^5 + 5/1 716*sqrt(b*x^2 + a*x)*A*b^2/x^5 - 5/88*sqrt(b*x^2 + a*x)*B*a^2/x^6 - 3/114 4*sqrt(b*x^2 + a*x)*A*a*b/x^6 + 5/24*(b*x^2 + a*x)^(3/2)*B*a/x^7 - 3/104*s qrt(b*x^2 + a*x)*A*a^2/x^7 - 1/3*(b*x^2 + a*x)^(5/2)*B/x^8 + 1/8*(b*x^2 + a*x)^(3/2)*A*a/x^8 - 1/4*(b*x^2 + a*x)^(5/2)*A/x^9
Time = 0.37 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.23 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{15/2}} \, dx=-\frac {2 \, {\left ({\left (b x + a\right )} {\left (4 \, {\left (b x + a\right )} {\left (\frac {2 \, {\left (13 \, B a^{3} b^{12} - 6 \, A a^{2} b^{13}\right )} {\left (b x + a\right )}}{a^{6}} - \frac {13 \, {\left (13 \, B a^{4} b^{12} - 6 \, A a^{3} b^{13}\right )}}{a^{6}}\right )} + \frac {143 \, {\left (13 \, B a^{5} b^{12} - 6 \, A a^{4} b^{13}\right )}}{a^{6}}\right )} - \frac {1287 \, {\left (B a^{6} b^{12} - A a^{5} b^{13}\right )}}{a^{6}}\right )} {\left (b x + a\right )}^{\frac {7}{2}} b}{9009 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {13}{2}} {\left | b \right |}} \]
-2/9009*((b*x + a)*(4*(b*x + a)*(2*(13*B*a^3*b^12 - 6*A*a^2*b^13)*(b*x + a )/a^6 - 13*(13*B*a^4*b^12 - 6*A*a^3*b^13)/a^6) + 143*(13*B*a^5*b^12 - 6*A* a^4*b^13)/a^6) - 1287*(B*a^6*b^12 - A*a^5*b^13)/a^6)*(b*x + a)^(7/2)*b/((( b*x + a)*b - a*b)^(13/2)*abs(b))
Time = 0.90 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.16 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{15/2}} \, dx=-\frac {\sqrt {a+b\,x}\,\left (\frac {2\,A\,a^2}{13}+\frac {x\,\left (1638\,B\,a^6+3402\,A\,b\,a^5\right )}{9009\,a^4}-\frac {x^6\,\left (96\,A\,b^6-208\,B\,a\,b^5\right )}{9009\,a^4}+\frac {2\,b\,x^2\,\left (159\,A\,b+299\,B\,a\right )}{1287}-\frac {2\,b^3\,x^4\,\left (6\,A\,b-13\,B\,a\right )}{3003\,a^2}+\frac {8\,b^4\,x^5\,\left (6\,A\,b-13\,B\,a\right )}{9009\,a^3}+\frac {2\,b^2\,x^3\,\left (15\,A\,b+1469\,B\,a\right )}{9009\,a}\right )}{x^{13/2}} \]
-((a + b*x)^(1/2)*((2*A*a^2)/13 + (x*(1638*B*a^6 + 3402*A*a^5*b))/(9009*a^ 4) - (x^6*(96*A*b^6 - 208*B*a*b^5))/(9009*a^4) + (2*b*x^2*(159*A*b + 299*B *a))/1287 - (2*b^3*x^4*(6*A*b - 13*B*a))/(3003*a^2) + (8*b^4*x^5*(6*A*b - 13*B*a))/(9009*a^3) + (2*b^2*x^3*(15*A*b + 1469*B*a))/(9009*a)))/x^(13/2)