Integrand size = 18, antiderivative size = 177 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^6} \, dx=\frac {b^2 (3 A b-10 a B) \sqrt {a+b x}}{64 a x^2}+\frac {b^3 (3 A b-10 a B) \sqrt {a+b x}}{128 a^2 x}+\frac {b (3 A b-10 a B) (a+b x)^{3/2}}{48 a x^3}+\frac {(3 A b-10 a B) (a+b x)^{5/2}}{40 a x^4}-\frac {A (a+b x)^{7/2}}{5 a x^5}-\frac {b^4 (3 A b-10 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{128 a^{5/2}} \]
1/48*b*(3*A*b-10*B*a)*(b*x+a)^(3/2)/a/x^3+1/40*(3*A*b-10*B*a)*(b*x+a)^(5/2 )/a/x^4-1/5*A*(b*x+a)^(7/2)/a/x^5-1/128*b^4*(3*A*b-10*B*a)*arctanh((b*x+a) ^(1/2)/a^(1/2))/a^(5/2)+1/64*b^2*(3*A*b-10*B*a)*(b*x+a)^(1/2)/a/x^2+1/128* b^3*(3*A*b-10*B*a)*(b*x+a)^(1/2)/a^2/x
Time = 0.31 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.73 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^6} \, dx=-\frac {\sqrt {a+b x} \left (-45 A b^4 x^4+30 a b^3 x^3 (A+5 B x)+96 a^4 (4 A+5 B x)+16 a^3 b x (63 A+85 B x)+4 a^2 b^2 x^2 (186 A+295 B x)\right )}{1920 a^2 x^5}+\frac {b^4 (-3 A b+10 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{128 a^{5/2}} \]
-1/1920*(Sqrt[a + b*x]*(-45*A*b^4*x^4 + 30*a*b^3*x^3*(A + 5*B*x) + 96*a^4* (4*A + 5*B*x) + 16*a^3*b*x*(63*A + 85*B*x) + 4*a^2*b^2*x^2*(186*A + 295*B* x)))/(a^2*x^5) + (b^4*(-3*A*b + 10*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(1 28*a^(5/2))
Time = 0.22 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.81, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {87, 51, 51, 51, 52, 73, 221}
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.
| \(\displaystyle \int \frac {(a+b x)^{5/2} (A+B x)}{x^6} \, dx\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle -\frac {(3 A b-10 a B) \int \frac {(a+b x)^{5/2}}{x^5}dx}{10 a}-\frac {A (a+b x)^{7/2}}{5 a x^5}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle -\frac {(3 A b-10 a B) \left (\frac {5}{8} b \int \frac {(a+b x)^{3/2}}{x^4}dx-\frac {(a+b x)^{5/2}}{4 x^4}\right )}{10 a}-\frac {A (a+b x)^{7/2}}{5 a x^5}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle -\frac {(3 A b-10 a B) \left (\frac {5}{8} b \left (\frac {1}{2} b \int \frac {\sqrt {a+b x}}{x^3}dx-\frac {(a+b x)^{3/2}}{3 x^3}\right )-\frac {(a+b x)^{5/2}}{4 x^4}\right )}{10 a}-\frac {A (a+b x)^{7/2}}{5 a x^5}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle -\frac {(3 A b-10 a B) \left (\frac {5}{8} b \left (\frac {1}{2} b \left (\frac {1}{4} b \int \frac {1}{x^2 \sqrt {a+b x}}dx-\frac {\sqrt {a+b x}}{2 x^2}\right )-\frac {(a+b x)^{3/2}}{3 x^3}\right )-\frac {(a+b x)^{5/2}}{4 x^4}\right )}{10 a}-\frac {A (a+b x)^{7/2}}{5 a x^5}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {(3 A b-10 a B) \left (\frac {5}{8} b \left (\frac {1}{2} b \left (\frac {1}{4} b \left (-\frac {b \int \frac {1}{x \sqrt {a+b x}}dx}{2 a}-\frac {\sqrt {a+b x}}{a x}\right )-\frac {\sqrt {a+b x}}{2 x^2}\right )-\frac {(a+b x)^{3/2}}{3 x^3}\right )-\frac {(a+b x)^{5/2}}{4 x^4}\right )}{10 a}-\frac {A (a+b x)^{7/2}}{5 a x^5}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {(3 A b-10 a B) \left (\frac {5}{8} b \left (\frac {1}{2} b \left (\frac {1}{4} b \left (-\frac {\int \frac {1}{\frac {a+b x}{b}-\frac {a}{b}}d\sqrt {a+b x}}{a}-\frac {\sqrt {a+b x}}{a x}\right )-\frac {\sqrt {a+b x}}{2 x^2}\right )-\frac {(a+b x)^{3/2}}{3 x^3}\right )-\frac {(a+b x)^{5/2}}{4 x^4}\right )}{10 a}-\frac {A (a+b x)^{7/2}}{5 a x^5}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {(3 A b-10 a B) \left (\frac {5}{8} b \left (\frac {1}{2} b \left (\frac {1}{4} b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b x}}{a x}\right )-\frac {\sqrt {a+b x}}{2 x^2}\right )-\frac {(a+b x)^{3/2}}{3 x^3}\right )-\frac {(a+b x)^{5/2}}{4 x^4}\right )}{10 a}-\frac {A (a+b x)^{7/2}}{5 a x^5}\) |
-1/5*(A*(a + b*x)^(7/2))/(a*x^5) - ((3*A*b - 10*a*B)*(-1/4*(a + b*x)^(5/2) /x^4 + (5*b*(-1/3*(a + b*x)^(3/2)/x^3 + (b*(-1/2*Sqrt[a + b*x]/x^2 + (b*(- (Sqrt[a + b*x]/(a*x)) + (b*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/a^(3/2)))/4))/2 ))/8))/(10*a)
3.5.20.3.1 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 + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
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)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Time = 0.55 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.67
| method | result | size |
| pseudoelliptic | \(-\frac {31 \left (\frac {15 x^{5} \left (A b -\frac {10 B a}{3}\right ) b^{4} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{248}+\sqrt {b x +a}\, \left (\frac {5 b^{3} x^{3} \left (5 B x +A \right ) a^{\frac {3}{2}}}{124}+b^{2} x^{2} \left (\frac {295 B x}{186}+A \right ) a^{\frac {5}{2}}+\frac {42 \left (\frac {85 B x}{63}+A \right ) x b \,a^{\frac {7}{2}}}{31}+\frac {4 \left (5 B x +4 A \right ) a^{\frac {9}{2}}}{31}-\frac {15 A \sqrt {a}\, b^{4} x^{4}}{248}\right )\right )}{80 a^{\frac {5}{2}} x^{5}}\) | \(118\) |
| risch | \(-\frac {\sqrt {b x +a}\, \left (-45 A \,b^{4} x^{4}+150 B a \,b^{3} x^{4}+30 A a \,b^{3} x^{3}+1180 B \,a^{2} b^{2} x^{3}+744 A \,a^{2} b^{2} x^{2}+1360 B \,a^{3} b \,x^{2}+1008 A \,a^{3} b x +480 B \,a^{4} x +384 A \,a^{4}\right )}{1920 x^{5} a^{2}}-\frac {b^{4} \left (3 A b -10 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{128 a^{\frac {5}{2}}}\) | \(131\) |
| derivativedivides | \(2 b^{4} \left (-\frac {-\frac {\left (3 A b -10 B a \right ) \left (b x +a \right )^{\frac {9}{2}}}{256 a^{2}}+\frac {\left (21 A b +58 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{384 a}+\left (\frac {A b}{10}-\frac {B a}{3}\right ) \left (b x +a \right )^{\frac {5}{2}}-\frac {7 a \left (3 A b -10 B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{384}+\frac {a^{2} \left (3 A b -10 B a \right ) \sqrt {b x +a}}{256}}{b^{5} x^{5}}-\frac {\left (3 A b -10 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{256 a^{\frac {5}{2}}}\right )\) | \(141\) |
| default | \(2 b^{4} \left (-\frac {-\frac {\left (3 A b -10 B a \right ) \left (b x +a \right )^{\frac {9}{2}}}{256 a^{2}}+\frac {\left (21 A b +58 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{384 a}+\left (\frac {A b}{10}-\frac {B a}{3}\right ) \left (b x +a \right )^{\frac {5}{2}}-\frac {7 a \left (3 A b -10 B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{384}+\frac {a^{2} \left (3 A b -10 B a \right ) \sqrt {b x +a}}{256}}{b^{5} x^{5}}-\frac {\left (3 A b -10 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{256 a^{\frac {5}{2}}}\right )\) | \(141\) |
-31/80*(15/248*x^5*(A*b-10/3*B*a)*b^4*arctanh((b*x+a)^(1/2)/a^(1/2))+(b*x+ a)^(1/2)*(5/124*b^3*x^3*(5*B*x+A)*a^(3/2)+b^2*x^2*(295/186*B*x+A)*a^(5/2)+ 42/31*(85/63*B*x+A)*x*b*a^(7/2)+4/31*(5*B*x+4*A)*a^(9/2)-15/248*A*a^(1/2)* b^4*x^4))/a^(5/2)/x^5
Time = 0.24 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.73 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^6} \, dx=\left [-\frac {15 \, {\left (10 \, B a b^{4} - 3 \, A b^{5}\right )} \sqrt {a} x^{5} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (384 \, A a^{5} + 15 \, {\left (10 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} + 10 \, {\left (118 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3}\right )} x^{3} + 8 \, {\left (170 \, B a^{4} b + 93 \, A a^{3} b^{2}\right )} x^{2} + 48 \, {\left (10 \, B a^{5} + 21 \, A a^{4} b\right )} x\right )} \sqrt {b x + a}}{3840 \, a^{3} x^{5}}, -\frac {15 \, {\left (10 \, B a b^{4} - 3 \, A b^{5}\right )} \sqrt {-a} x^{5} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (384 \, A a^{5} + 15 \, {\left (10 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} + 10 \, {\left (118 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3}\right )} x^{3} + 8 \, {\left (170 \, B a^{4} b + 93 \, A a^{3} b^{2}\right )} x^{2} + 48 \, {\left (10 \, B a^{5} + 21 \, A a^{4} b\right )} x\right )} \sqrt {b x + a}}{1920 \, a^{3} x^{5}}\right ] \]
[-1/3840*(15*(10*B*a*b^4 - 3*A*b^5)*sqrt(a)*x^5*log((b*x - 2*sqrt(b*x + a) *sqrt(a) + 2*a)/x) + 2*(384*A*a^5 + 15*(10*B*a^2*b^3 - 3*A*a*b^4)*x^4 + 10 *(118*B*a^3*b^2 + 3*A*a^2*b^3)*x^3 + 8*(170*B*a^4*b + 93*A*a^3*b^2)*x^2 + 48*(10*B*a^5 + 21*A*a^4*b)*x)*sqrt(b*x + a))/(a^3*x^5), -1/1920*(15*(10*B* a*b^4 - 3*A*b^5)*sqrt(-a)*x^5*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (384*A*a^ 5 + 15*(10*B*a^2*b^3 - 3*A*a*b^4)*x^4 + 10*(118*B*a^3*b^2 + 3*A*a^2*b^3)*x ^3 + 8*(170*B*a^4*b + 93*A*a^3*b^2)*x^2 + 48*(10*B*a^5 + 21*A*a^4*b)*x)*sq rt(b*x + a))/(a^3*x^5)]
Timed out. \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^6} \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.32 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^6} \, dx=-\frac {1}{3840} \, b^{5} {\left (\frac {2 \, {\left (15 \, {\left (10 \, B a - 3 \, A b\right )} {\left (b x + a\right )}^{\frac {9}{2}} + 10 \, {\left (58 \, B a^{2} + 21 \, A a b\right )} {\left (b x + a\right )}^{\frac {7}{2}} - 128 \, {\left (10 \, B a^{3} - 3 \, A a^{2} b\right )} {\left (b x + a\right )}^{\frac {5}{2}} + 70 \, {\left (10 \, B a^{4} - 3 \, A a^{3} b\right )} {\left (b x + a\right )}^{\frac {3}{2}} - 15 \, {\left (10 \, B a^{5} - 3 \, A a^{4} b\right )} \sqrt {b x + a}\right )}}{{\left (b x + a\right )}^{5} a^{2} b - 5 \, {\left (b x + a\right )}^{4} a^{3} b + 10 \, {\left (b x + a\right )}^{3} a^{4} b - 10 \, {\left (b x + a\right )}^{2} a^{5} b + 5 \, {\left (b x + a\right )} a^{6} b - a^{7} b} + \frac {15 \, {\left (10 \, B a - 3 \, A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}} b}\right )} \]
-1/3840*b^5*(2*(15*(10*B*a - 3*A*b)*(b*x + a)^(9/2) + 10*(58*B*a^2 + 21*A* a*b)*(b*x + a)^(7/2) - 128*(10*B*a^3 - 3*A*a^2*b)*(b*x + a)^(5/2) + 70*(10 *B*a^4 - 3*A*a^3*b)*(b*x + a)^(3/2) - 15*(10*B*a^5 - 3*A*a^4*b)*sqrt(b*x + a))/((b*x + a)^5*a^2*b - 5*(b*x + a)^4*a^3*b + 10*(b*x + a)^3*a^4*b - 10* (b*x + a)^2*a^5*b + 5*(b*x + a)*a^6*b - a^7*b) + 15*(10*B*a - 3*A*b)*log(( sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/(a^(5/2)*b))
Time = 0.30 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.18 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^6} \, dx=-\frac {\frac {15 \, {\left (10 \, B a b^{5} - 3 \, A b^{6}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {150 \, {\left (b x + a\right )}^{\frac {9}{2}} B a b^{5} + 580 \, {\left (b x + a\right )}^{\frac {7}{2}} B a^{2} b^{5} - 1280 \, {\left (b x + a\right )}^{\frac {5}{2}} B a^{3} b^{5} + 700 \, {\left (b x + a\right )}^{\frac {3}{2}} B a^{4} b^{5} - 150 \, \sqrt {b x + a} B a^{5} b^{5} - 45 \, {\left (b x + a\right )}^{\frac {9}{2}} A b^{6} + 210 \, {\left (b x + a\right )}^{\frac {7}{2}} A a b^{6} + 384 \, {\left (b x + a\right )}^{\frac {5}{2}} A a^{2} b^{6} - 210 \, {\left (b x + a\right )}^{\frac {3}{2}} A a^{3} b^{6} + 45 \, \sqrt {b x + a} A a^{4} b^{6}}{a^{2} b^{5} x^{5}}}{1920 \, b} \]
-1/1920*(15*(10*B*a*b^5 - 3*A*b^6)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a )*a^2) + (150*(b*x + a)^(9/2)*B*a*b^5 + 580*(b*x + a)^(7/2)*B*a^2*b^5 - 12 80*(b*x + a)^(5/2)*B*a^3*b^5 + 700*(b*x + a)^(3/2)*B*a^4*b^5 - 150*sqrt(b* x + a)*B*a^5*b^5 - 45*(b*x + a)^(9/2)*A*b^6 + 210*(b*x + a)^(7/2)*A*a*b^6 + 384*(b*x + a)^(5/2)*A*a^2*b^6 - 210*(b*x + a)^(3/2)*A*a^3*b^6 + 45*sqrt( b*x + a)*A*a^4*b^6)/(a^2*b^5*x^5))/b
Time = 0.21 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.23 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^6} \, dx=\frac {\left (\frac {A\,b^5}{5}-\frac {2\,B\,a\,b^4}{3}\right )\,{\left (a+b\,x\right )}^{5/2}+\left (\frac {3\,A\,a^2\,b^5}{128}-\frac {5\,B\,a^3\,b^4}{64}\right )\,\sqrt {a+b\,x}+\left (\frac {35\,B\,a^2\,b^4}{96}-\frac {7\,A\,a\,b^5}{64}\right )\,{\left (a+b\,x\right )}^{3/2}-\frac {\left (3\,A\,b^5-10\,B\,a\,b^4\right )\,{\left (a+b\,x\right )}^{9/2}}{128\,a^2}+\frac {\left (21\,A\,b^5+58\,B\,a\,b^4\right )\,{\left (a+b\,x\right )}^{7/2}}{192\,a}}{5\,a\,{\left (a+b\,x\right )}^4-5\,a^4\,\left (a+b\,x\right )-{\left (a+b\,x\right )}^5-10\,a^2\,{\left (a+b\,x\right )}^3+10\,a^3\,{\left (a+b\,x\right )}^2+a^5}-\frac {b^4\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (3\,A\,b-10\,B\,a\right )}{128\,a^{5/2}} \]
(((A*b^5)/5 - (2*B*a*b^4)/3)*(a + b*x)^(5/2) + ((3*A*a^2*b^5)/128 - (5*B*a ^3*b^4)/64)*(a + b*x)^(1/2) + ((35*B*a^2*b^4)/96 - (7*A*a*b^5)/64)*(a + b* x)^(3/2) - ((3*A*b^5 - 10*B*a*b^4)*(a + b*x)^(9/2))/(128*a^2) + ((21*A*b^5 + 58*B*a*b^4)*(a + b*x)^(7/2))/(192*a))/(5*a*(a + b*x)^4 - 5*a^4*(a + b*x ) - (a + b*x)^5 - 10*a^2*(a + b*x)^3 + 10*a^3*(a + b*x)^2 + a^5) - (b^4*at anh((a + b*x)^(1/2)/a^(1/2))*(3*A*b - 10*B*a))/(128*a^(5/2))