Integrand size = 24, antiderivative size = 216 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{17/2}} \, dx=-\frac {c^2 (10 b B-3 A c) \sqrt {b x+c x^2}}{64 b x^{5/2}}-\frac {c^3 (10 b B-3 A c) \sqrt {b x+c x^2}}{128 b^2 x^{3/2}}-\frac {c (10 b B-3 A c) \left (b x+c x^2\right )^{3/2}}{48 b x^{9/2}}-\frac {(10 b B-3 A c) \left (b x+c x^2\right )^{5/2}}{40 b x^{13/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}+\frac {c^4 (10 b B-3 A c) \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{128 b^{5/2}} \]
-1/48*c*(-3*A*c+10*B*b)*(c*x^2+b*x)^(3/2)/b/x^(9/2)-1/40*(-3*A*c+10*B*b)*( c*x^2+b*x)^(5/2)/b/x^(13/2)-1/5*A*(c*x^2+b*x)^(7/2)/b/x^(17/2)+1/128*c^4*( -3*A*c+10*B*b)*arctanh((c*x^2+b*x)^(1/2)/b^(1/2)/x^(1/2))/b^(5/2)-1/64*c^2 *(-3*A*c+10*B*b)*(c*x^2+b*x)^(1/2)/b/x^(5/2)-1/128*c^3*(-3*A*c+10*B*b)*(c* x^2+b*x)^(1/2)/b^2/x^(3/2)
Time = 0.36 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.74 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{17/2}} \, dx=\frac {-\sqrt {b} (b+c x) \left (10 b B x \left (48 b^3+136 b^2 c x+118 b c^2 x^2+15 c^3 x^3\right )+3 A \left (128 b^4+336 b^3 c x+248 b^2 c^2 x^2+10 b c^3 x^3-15 c^4 x^4\right )\right )+15 c^4 (10 b B-3 A c) x^5 \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )}{1920 b^{5/2} x^{9/2} \sqrt {x (b+c x)}} \]
(-(Sqrt[b]*(b + c*x)*(10*b*B*x*(48*b^3 + 136*b^2*c*x + 118*b*c^2*x^2 + 15* c^3*x^3) + 3*A*(128*b^4 + 336*b^3*c*x + 248*b^2*c^2*x^2 + 10*b*c^3*x^3 - 1 5*c^4*x^4))) + 15*c^4*(10*b*B - 3*A*c)*x^5*Sqrt[b + c*x]*ArcTanh[Sqrt[b + c*x]/Sqrt[b]])/(1920*b^(5/2)*x^(9/2)*Sqrt[x*(b + c*x)])
Time = 0.35 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.85, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1220, 1130, 1130, 1130, 1135, 1136, 220}
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.
| \(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{17/2}} \, dx\) |
| \(\Big \downarrow \) 1220 |
| \(\displaystyle \frac {(10 b B-3 A c) \int \frac {\left (c x^2+b x\right )^{5/2}}{x^{15/2}}dx}{10 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}\) |
| \(\Big \downarrow \) 1130 |
| \(\displaystyle \frac {(10 b B-3 A c) \left (\frac {5}{8} c \int \frac {\left (c x^2+b x\right )^{3/2}}{x^{11/2}}dx-\frac {\left (b x+c x^2\right )^{5/2}}{4 x^{13/2}}\right )}{10 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}\) |
| \(\Big \downarrow \) 1130 |
| \(\displaystyle \frac {(10 b B-3 A c) \left (\frac {5}{8} c \left (\frac {1}{2} c \int \frac {\sqrt {c x^2+b x}}{x^{7/2}}dx-\frac {\left (b x+c x^2\right )^{3/2}}{3 x^{9/2}}\right )-\frac {\left (b x+c x^2\right )^{5/2}}{4 x^{13/2}}\right )}{10 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}\) |
| \(\Big \downarrow \) 1130 |
| \(\displaystyle \frac {(10 b B-3 A c) \left (\frac {5}{8} c \left (\frac {1}{2} c \left (\frac {1}{4} c \int \frac {1}{x^{3/2} \sqrt {c x^2+b x}}dx-\frac {\sqrt {b x+c x^2}}{2 x^{5/2}}\right )-\frac {\left (b x+c x^2\right )^{3/2}}{3 x^{9/2}}\right )-\frac {\left (b x+c x^2\right )^{5/2}}{4 x^{13/2}}\right )}{10 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}\) |
| \(\Big \downarrow \) 1135 |
| \(\displaystyle \frac {(10 b B-3 A c) \left (\frac {5}{8} c \left (\frac {1}{2} c \left (\frac {1}{4} c \left (-\frac {c \int \frac {1}{\sqrt {x} \sqrt {c x^2+b x}}dx}{2 b}-\frac {\sqrt {b x+c x^2}}{b x^{3/2}}\right )-\frac {\sqrt {b x+c x^2}}{2 x^{5/2}}\right )-\frac {\left (b x+c x^2\right )^{3/2}}{3 x^{9/2}}\right )-\frac {\left (b x+c x^2\right )^{5/2}}{4 x^{13/2}}\right )}{10 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}\) |
| \(\Big \downarrow \) 1136 |
| \(\displaystyle \frac {(10 b B-3 A c) \left (\frac {5}{8} c \left (\frac {1}{2} c \left (\frac {1}{4} c \left (-\frac {c \int \frac {1}{\frac {c x^2+b x}{x}-b}d\frac {\sqrt {c x^2+b x}}{\sqrt {x}}}{b}-\frac {\sqrt {b x+c x^2}}{b x^{3/2}}\right )-\frac {\sqrt {b x+c x^2}}{2 x^{5/2}}\right )-\frac {\left (b x+c x^2\right )^{3/2}}{3 x^{9/2}}\right )-\frac {\left (b x+c x^2\right )^{5/2}}{4 x^{13/2}}\right )}{10 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle \frac {(10 b B-3 A c) \left (\frac {5}{8} c \left (\frac {1}{2} c \left (\frac {1}{4} c \left (\frac {c \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{3/2}}-\frac {\sqrt {b x+c x^2}}{b x^{3/2}}\right )-\frac {\sqrt {b x+c x^2}}{2 x^{5/2}}\right )-\frac {\left (b x+c x^2\right )^{3/2}}{3 x^{9/2}}\right )-\frac {\left (b x+c x^2\right )^{5/2}}{4 x^{13/2}}\right )}{10 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}\) |
-1/5*(A*(b*x + c*x^2)^(7/2))/(b*x^(17/2)) + ((10*b*B - 3*A*c)*(-1/4*(b*x + c*x^2)^(5/2)/x^(13/2) + (5*c*(-1/3*(b*x + c*x^2)^(3/2)/x^(9/2) + (c*(-1/2 *Sqrt[b*x + c*x^2]/x^(5/2) + (c*(-(Sqrt[b*x + c*x^2]/(b*x^(3/2))) + (c*Arc Tanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/b^(3/2)))/4))/2))/8))/(10*b)
3.3.24.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + p + 1))), x] - Simp[c*(p/(e^2*(m + p + 1))) Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] & & IntegerQ[2*p]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(-e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((m + p + 1)*(2* c*d - b*e))), x] + Simp[c*((m + 2*p + 2)/((m + p + 1)*(2*c*d - b*e))) Int [(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && I ntegerQ[2*p]
Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x _Symbol] :> Simp[2*e Subst[Int[1/(2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x ^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e *f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)) Int[(d + e*x )^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((LtQ[m, -1] && !IGtQ[m + p + 1, 0 ]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0 ]
Time = 0.08 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.73
| method | result | size |
| risch | \(-\frac {\left (c x +b \right ) \left (-45 A \,c^{4} x^{4}+150 B b \,c^{3} x^{4}+30 A b \,c^{3} x^{3}+1180 B \,b^{2} c^{2} x^{3}+744 A \,b^{2} c^{2} x^{2}+1360 B \,b^{3} c \,x^{2}+1008 A \,b^{3} c x +480 b^{4} B x +384 A \,b^{4}\right )}{1920 x^{\frac {9}{2}} b^{2} \sqrt {x \left (c x +b \right )}}-\frac {c^{4} \left (3 A c -10 B b \right ) \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{128 b^{\frac {5}{2}} \sqrt {x \left (c x +b \right )}}\) | \(157\) |
| default | \(-\frac {\sqrt {x \left (c x +b \right )}\, \left (45 A \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c^{5} x^{5}-150 B \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b \,c^{4} x^{5}-45 A \,c^{4} x^{4} \sqrt {c x +b}\, \sqrt {b}+150 B \,b^{\frac {3}{2}} c^{3} x^{4} \sqrt {c x +b}+30 A \,b^{\frac {3}{2}} c^{3} x^{3} \sqrt {c x +b}+1180 B \,b^{\frac {5}{2}} c^{2} x^{3} \sqrt {c x +b}+744 A \,b^{\frac {5}{2}} c^{2} x^{2} \sqrt {c x +b}+1360 B \,b^{\frac {7}{2}} c \,x^{2} \sqrt {c x +b}+1008 A \,b^{\frac {7}{2}} c x \sqrt {c x +b}+480 B \,b^{\frac {9}{2}} x \sqrt {c x +b}+384 A \,b^{\frac {9}{2}} \sqrt {c x +b}\right )}{1920 b^{\frac {5}{2}} x^{\frac {11}{2}} \sqrt {c x +b}}\) | \(223\) |
-1/1920*(c*x+b)*(-45*A*c^4*x^4+150*B*b*c^3*x^4+30*A*b*c^3*x^3+1180*B*b^2*c ^2*x^3+744*A*b^2*c^2*x^2+1360*B*b^3*c*x^2+1008*A*b^3*c*x+480*B*b^4*x+384*A *b^4)/x^(9/2)/b^2/(x*(c*x+b))^(1/2)-1/128*c^4*(3*A*c-10*B*b)/b^(5/2)*arcta nh((c*x+b)^(1/2)/b^(1/2))*(c*x+b)^(1/2)*x^(1/2)/(x*(c*x+b))^(1/2)
Time = 0.28 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.56 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{17/2}} \, dx=\left [-\frac {15 \, {\left (10 \, B b c^{4} - 3 \, A c^{5}\right )} \sqrt {b} x^{6} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (384 \, A b^{5} + 15 \, {\left (10 \, B b^{2} c^{3} - 3 \, A b c^{4}\right )} x^{4} + 10 \, {\left (118 \, B b^{3} c^{2} + 3 \, A b^{2} c^{3}\right )} x^{3} + 8 \, {\left (170 \, B b^{4} c + 93 \, A b^{3} c^{2}\right )} x^{2} + 48 \, {\left (10 \, B b^{5} + 21 \, A b^{4} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{3840 \, b^{3} x^{6}}, -\frac {15 \, {\left (10 \, B b c^{4} - 3 \, A c^{5}\right )} \sqrt {-b} x^{6} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (384 \, A b^{5} + 15 \, {\left (10 \, B b^{2} c^{3} - 3 \, A b c^{4}\right )} x^{4} + 10 \, {\left (118 \, B b^{3} c^{2} + 3 \, A b^{2} c^{3}\right )} x^{3} + 8 \, {\left (170 \, B b^{4} c + 93 \, A b^{3} c^{2}\right )} x^{2} + 48 \, {\left (10 \, B b^{5} + 21 \, A b^{4} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{1920 \, b^{3} x^{6}}\right ] \]
[-1/3840*(15*(10*B*b*c^4 - 3*A*c^5)*sqrt(b)*x^6*log(-(c*x^2 + 2*b*x - 2*sq rt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(384*A*b^5 + 15*(10*B*b^2*c^3 - 3*A*b*c^4)*x^4 + 10*(118*B*b^3*c^2 + 3*A*b^2*c^3)*x^3 + 8*(170*B*b^4*c + 9 3*A*b^3*c^2)*x^2 + 48*(10*B*b^5 + 21*A*b^4*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x) )/(b^3*x^6), -1/1920*(15*(10*B*b*c^4 - 3*A*c^5)*sqrt(-b)*x^6*arctan(sqrt(- b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (384*A*b^5 + 15*(10*B*b^2*c^3 - 3*A*b*c^4) *x^4 + 10*(118*B*b^3*c^2 + 3*A*b^2*c^3)*x^3 + 8*(170*B*b^4*c + 93*A*b^3*c^ 2)*x^2 + 48*(10*B*b^5 + 21*A*b^4*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*x^6 )]
Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{17/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{17/2}} \, dx=\int { \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} {\left (B x + A\right )}}{x^{\frac {17}{2}}} \,d x } \]
Time = 0.34 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{17/2}} \, dx=-\frac {\frac {15 \, {\left (10 \, B b c^{5} - 3 \, A c^{6}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} + \frac {150 \, {\left (c x + b\right )}^{\frac {9}{2}} B b c^{5} + 580 \, {\left (c x + b\right )}^{\frac {7}{2}} B b^{2} c^{5} - 1280 \, {\left (c x + b\right )}^{\frac {5}{2}} B b^{3} c^{5} + 700 \, {\left (c x + b\right )}^{\frac {3}{2}} B b^{4} c^{5} - 150 \, \sqrt {c x + b} B b^{5} c^{5} - 45 \, {\left (c x + b\right )}^{\frac {9}{2}} A c^{6} + 210 \, {\left (c x + b\right )}^{\frac {7}{2}} A b c^{6} + 384 \, {\left (c x + b\right )}^{\frac {5}{2}} A b^{2} c^{6} - 210 \, {\left (c x + b\right )}^{\frac {3}{2}} A b^{3} c^{6} + 45 \, \sqrt {c x + b} A b^{4} c^{6}}{b^{2} c^{5} x^{5}}}{1920 \, c} \]
-1/1920*(15*(10*B*b*c^5 - 3*A*c^6)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b )*b^2) + (150*(c*x + b)^(9/2)*B*b*c^5 + 580*(c*x + b)^(7/2)*B*b^2*c^5 - 12 80*(c*x + b)^(5/2)*B*b^3*c^5 + 700*(c*x + b)^(3/2)*B*b^4*c^5 - 150*sqrt(c* x + b)*B*b^5*c^5 - 45*(c*x + b)^(9/2)*A*c^6 + 210*(c*x + b)^(7/2)*A*b*c^6 + 384*(c*x + b)^(5/2)*A*b^2*c^6 - 210*(c*x + b)^(3/2)*A*b^3*c^6 + 45*sqrt( c*x + b)*A*b^4*c^6)/(b^2*c^5*x^5))/c
Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{17/2}} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{5/2}\,\left (A+B\,x\right )}{x^{17/2}} \,d x \]