Integrand size = 18, antiderivative size = 146 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^5} \, dx=\frac {(5 A b-8 a B) \sqrt {a+b x}}{24 a x^3}+\frac {b (5 A b-8 a B) \sqrt {a+b x}}{96 a^2 x^2}-\frac {b^2 (5 A b-8 a B) \sqrt {a+b x}}{64 a^3 x}-\frac {A (a+b x)^{3/2}}{4 a x^4}+\frac {b^3 (5 A b-8 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{7/2}} \]
-1/4*A*(b*x+a)^(3/2)/a/x^4+1/64*b^3*(5*A*b-8*B*a)*arctanh((b*x+a)^(1/2)/a^ (1/2))/a^(7/2)+1/24*(5*A*b-8*B*a)*(b*x+a)^(1/2)/a/x^3+1/96*b*(5*A*b-8*B*a) *(b*x+a)^(1/2)/a^2/x^2-1/64*b^2*(5*A*b-8*B*a)*(b*x+a)^(1/2)/a^3/x
Time = 0.20 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.75 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^5} \, dx=-\frac {\sqrt {a+b x} \left (15 A b^3 x^3+8 a^2 b x (A+2 B x)+16 a^3 (3 A+4 B x)-2 a b^2 x^2 (5 A+12 B x)\right )}{192 a^3 x^4}+\frac {b^3 (5 A b-8 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{7/2}} \]
-1/192*(Sqrt[a + b*x]*(15*A*b^3*x^3 + 8*a^2*b*x*(A + 2*B*x) + 16*a^3*(3*A + 4*B*x) - 2*a*b^2*x^2*(5*A + 12*B*x)))/(a^3*x^4) + (b^3*(5*A*b - 8*a*B)*A rcTanh[Sqrt[a + b*x]/Sqrt[a]])/(64*a^(7/2))
Time = 0.21 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.88, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {87, 51, 52, 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.
| \(\displaystyle \int \frac {\sqrt {a+b x} (A+B x)}{x^5} \, dx\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle -\frac {(5 A b-8 a B) \int \frac {\sqrt {a+b x}}{x^4}dx}{8 a}-\frac {A (a+b x)^{3/2}}{4 a x^4}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle -\frac {(5 A b-8 a B) \left (\frac {1}{6} b \int \frac {1}{x^3 \sqrt {a+b x}}dx-\frac {\sqrt {a+b x}}{3 x^3}\right )}{8 a}-\frac {A (a+b x)^{3/2}}{4 a x^4}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {(5 A b-8 a B) \left (\frac {1}{6} b \left (-\frac {3 b \int \frac {1}{x^2 \sqrt {a+b x}}dx}{4 a}-\frac {\sqrt {a+b x}}{2 a x^2}\right )-\frac {\sqrt {a+b x}}{3 x^3}\right )}{8 a}-\frac {A (a+b x)^{3/2}}{4 a x^4}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {(5 A b-8 a B) \left (\frac {1}{6} b \left (-\frac {3 b \left (-\frac {b \int \frac {1}{x \sqrt {a+b x}}dx}{2 a}-\frac {\sqrt {a+b x}}{a x}\right )}{4 a}-\frac {\sqrt {a+b x}}{2 a x^2}\right )-\frac {\sqrt {a+b x}}{3 x^3}\right )}{8 a}-\frac {A (a+b x)^{3/2}}{4 a x^4}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {(5 A b-8 a B) \left (\frac {1}{6} b \left (-\frac {3 b \left (-\frac {\int \frac {1}{\frac {a+b x}{b}-\frac {a}{b}}d\sqrt {a+b x}}{a}-\frac {\sqrt {a+b x}}{a x}\right )}{4 a}-\frac {\sqrt {a+b x}}{2 a x^2}\right )-\frac {\sqrt {a+b x}}{3 x^3}\right )}{8 a}-\frac {A (a+b x)^{3/2}}{4 a x^4}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {(5 A b-8 a B) \left (\frac {1}{6} b \left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b x}}{a x}\right )}{4 a}-\frac {\sqrt {a+b x}}{2 a x^2}\right )-\frac {\sqrt {a+b x}}{3 x^3}\right )}{8 a}-\frac {A (a+b x)^{3/2}}{4 a x^4}\) |
-1/4*(A*(a + b*x)^(3/2))/(a*x^4) - ((5*A*b - 8*a*B)*(-1/3*Sqrt[a + b*x]/x^ 3 + (b*(-1/2*Sqrt[a + b*x]/(a*x^2) - (3*b*(-(Sqrt[a + b*x]/(a*x)) + (b*Arc Tanh[Sqrt[a + b*x]/Sqrt[a]])/a^(3/2)))/(4*a)))/6))/(8*a)
3.4.96.3.1 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 = 0.53 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.70
| method | result | size |
| pseudoelliptic | \(-\frac {\left (-\frac {15}{8} A \,b^{4}+3 B a \,b^{3}\right ) x^{4} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\sqrt {b x +a}\, \left (-\frac {5 x^{2} b^{2} \left (\frac {12 B x}{5}+A \right ) a^{\frac {3}{2}}}{4}+b x \left (2 B x +A \right ) a^{\frac {5}{2}}+\left (8 B x +6 A \right ) a^{\frac {7}{2}}+\frac {15 A \sqrt {a}\, b^{3} x^{3}}{8}\right )}{24 a^{\frac {7}{2}} x^{4}}\) | \(102\) |
| risch | \(-\frac {\sqrt {b x +a}\, \left (15 A \,b^{3} x^{3}-24 B a \,b^{2} x^{3}-10 a A \,b^{2} x^{2}+16 B \,a^{2} b \,x^{2}+8 a^{2} A b x +64 a^{3} B x +48 a^{3} A \right )}{192 x^{4} a^{3}}+\frac {b^{3} \left (5 A b -8 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{64 a^{\frac {7}{2}}}\) | \(107\) |
| derivativedivides | \(2 b^{3} \left (-\frac {\frac {\left (5 A b -8 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{128 a^{3}}-\frac {11 \left (5 A b -8 B a \right ) \left (b x +a \right )^{\frac {5}{2}}}{384 a^{2}}+\frac {\left (73 A b -40 B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{384 a}+\left (\frac {5 A b}{128}-\frac {B a}{16}\right ) \sqrt {b x +a}}{b^{4} x^{4}}+\frac {\left (5 A b -8 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{128 a^{\frac {7}{2}}}\right )\) | \(122\) |
| default | \(2 b^{3} \left (-\frac {\frac {\left (5 A b -8 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{128 a^{3}}-\frac {11 \left (5 A b -8 B a \right ) \left (b x +a \right )^{\frac {5}{2}}}{384 a^{2}}+\frac {\left (73 A b -40 B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{384 a}+\left (\frac {5 A b}{128}-\frac {B a}{16}\right ) \sqrt {b x +a}}{b^{4} x^{4}}+\frac {\left (5 A b -8 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{128 a^{\frac {7}{2}}}\right )\) | \(122\) |
-1/24*((-15/8*A*b^4+3*B*a*b^3)*x^4*arctanh((b*x+a)^(1/2)/a^(1/2))+(b*x+a)^ (1/2)*(-5/4*x^2*b^2*(12/5*B*x+A)*a^(3/2)+b*x*(2*B*x+A)*a^(5/2)+(8*B*x+6*A) *a^(7/2)+15/8*A*a^(1/2)*b^3*x^3))/a^(7/2)/x^4
Time = 0.23 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.77 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^5} \, dx=\left [-\frac {3 \, {\left (8 \, B a b^{3} - 5 \, A b^{4}\right )} \sqrt {a} x^{4} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (48 \, A a^{4} - 3 \, {\left (8 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{3} + 2 \, {\left (8 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2} + 8 \, {\left (8 \, B a^{4} + A a^{3} b\right )} x\right )} \sqrt {b x + a}}{384 \, a^{4} x^{4}}, \frac {3 \, {\left (8 \, B a b^{3} - 5 \, A b^{4}\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - {\left (48 \, A a^{4} - 3 \, {\left (8 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{3} + 2 \, {\left (8 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2} + 8 \, {\left (8 \, B a^{4} + A a^{3} b\right )} x\right )} \sqrt {b x + a}}{192 \, a^{4} x^{4}}\right ] \]
[-1/384*(3*(8*B*a*b^3 - 5*A*b^4)*sqrt(a)*x^4*log((b*x + 2*sqrt(b*x + a)*sq rt(a) + 2*a)/x) + 2*(48*A*a^4 - 3*(8*B*a^2*b^2 - 5*A*a*b^3)*x^3 + 2*(8*B*a ^3*b - 5*A*a^2*b^2)*x^2 + 8*(8*B*a^4 + A*a^3*b)*x)*sqrt(b*x + a))/(a^4*x^4 ), 1/192*(3*(8*B*a*b^3 - 5*A*b^4)*sqrt(-a)*x^4*arctan(sqrt(b*x + a)*sqrt(- a)/a) - (48*A*a^4 - 3*(8*B*a^2*b^2 - 5*A*a*b^3)*x^3 + 2*(8*B*a^3*b - 5*A*a ^2*b^2)*x^2 + 8*(8*B*a^4 + A*a^3*b)*x)*sqrt(b*x + a))/(a^4*x^4)]
Leaf count of result is larger than twice the leaf count of optimal. 296 vs. \(2 (134) = 268\).
Time = 80.36 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.03 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^5} \, dx=- \frac {A a}{4 \sqrt {b} x^{\frac {9}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {7 A \sqrt {b}}{24 x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {A b^{\frac {3}{2}}}{96 a x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {5 A b^{\frac {5}{2}}}{192 a^{2} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {5 A b^{\frac {7}{2}}}{64 a^{3} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {5 A b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{64 a^{\frac {7}{2}}} - \frac {B a}{3 \sqrt {b} x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {5 B \sqrt {b}}{12 x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {B b^{\frac {3}{2}}}{24 a x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {B b^{\frac {5}{2}}}{8 a^{2} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} - \frac {B b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{8 a^{\frac {5}{2}}} \]
-A*a/(4*sqrt(b)*x**(9/2)*sqrt(a/(b*x) + 1)) - 7*A*sqrt(b)/(24*x**(7/2)*sqr t(a/(b*x) + 1)) + A*b**(3/2)/(96*a*x**(5/2)*sqrt(a/(b*x) + 1)) - 5*A*b**(5 /2)/(192*a**2*x**(3/2)*sqrt(a/(b*x) + 1)) - 5*A*b**(7/2)/(64*a**3*sqrt(x)* sqrt(a/(b*x) + 1)) + 5*A*b**4*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(64*a**(7/2 )) - B*a/(3*sqrt(b)*x**(7/2)*sqrt(a/(b*x) + 1)) - 5*B*sqrt(b)/(12*x**(5/2) *sqrt(a/(b*x) + 1)) + B*b**(3/2)/(24*a*x**(3/2)*sqrt(a/(b*x) + 1)) + B*b** (5/2)/(8*a**2*sqrt(x)*sqrt(a/(b*x) + 1)) - B*b**3*asinh(sqrt(a)/(sqrt(b)*s qrt(x)))/(8*a**(5/2))
Time = 0.29 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.34 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^5} \, dx=\frac {1}{384} \, b^{4} {\left (\frac {2 \, {\left (3 \, {\left (8 \, B a - 5 \, A b\right )} {\left (b x + a\right )}^{\frac {7}{2}} - 11 \, {\left (8 \, B a^{2} - 5 \, A a b\right )} {\left (b x + a\right )}^{\frac {5}{2}} + {\left (40 \, B a^{3} - 73 \, A a^{2} b\right )} {\left (b x + a\right )}^{\frac {3}{2}} + 3 \, {\left (8 \, B a^{4} - 5 \, A a^{3} b\right )} \sqrt {b x + a}\right )}}{{\left (b x + a\right )}^{4} a^{3} b - 4 \, {\left (b x + a\right )}^{3} a^{4} b + 6 \, {\left (b x + a\right )}^{2} a^{5} b - 4 \, {\left (b x + a\right )} a^{6} b + a^{7} b} + \frac {3 \, {\left (8 \, B a - 5 \, A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {7}{2}} b}\right )} \]
1/384*b^4*(2*(3*(8*B*a - 5*A*b)*(b*x + a)^(7/2) - 11*(8*B*a^2 - 5*A*a*b)*( b*x + a)^(5/2) + (40*B*a^3 - 73*A*a^2*b)*(b*x + a)^(3/2) + 3*(8*B*a^4 - 5* A*a^3*b)*sqrt(b*x + a))/((b*x + a)^4*a^3*b - 4*(b*x + a)^3*a^4*b + 6*(b*x + a)^2*a^5*b - 4*(b*x + a)*a^6*b + a^7*b) + 3*(8*B*a - 5*A*b)*log((sqrt(b* x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/(a^(7/2)*b))
Time = 0.29 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^5} \, dx=\frac {\frac {3 \, {\left (8 \, B a b^{4} - 5 \, A b^{5}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} + \frac {24 \, {\left (b x + a\right )}^{\frac {7}{2}} B a b^{4} - 88 \, {\left (b x + a\right )}^{\frac {5}{2}} B a^{2} b^{4} + 40 \, {\left (b x + a\right )}^{\frac {3}{2}} B a^{3} b^{4} + 24 \, \sqrt {b x + a} B a^{4} b^{4} - 15 \, {\left (b x + a\right )}^{\frac {7}{2}} A b^{5} + 55 \, {\left (b x + a\right )}^{\frac {5}{2}} A a b^{5} - 73 \, {\left (b x + a\right )}^{\frac {3}{2}} A a^{2} b^{5} - 15 \, \sqrt {b x + a} A a^{3} b^{5}}{a^{3} b^{4} x^{4}}}{192 \, b} \]
1/192*(3*(8*B*a*b^4 - 5*A*b^5)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^ 3) + (24*(b*x + a)^(7/2)*B*a*b^4 - 88*(b*x + a)^(5/2)*B*a^2*b^4 + 40*(b*x + a)^(3/2)*B*a^3*b^4 + 24*sqrt(b*x + a)*B*a^4*b^4 - 15*(b*x + a)^(7/2)*A*b ^5 + 55*(b*x + a)^(5/2)*A*a*b^5 - 73*(b*x + a)^(3/2)*A*a^2*b^5 - 15*sqrt(b *x + a)*A*a^3*b^5)/(a^3*b^4*x^4))/b
Time = 0.57 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.22 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^5} \, dx=\frac {b^3\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (5\,A\,b-8\,B\,a\right )}{64\,a^{7/2}}-\frac {\left (\frac {5\,A\,b^4}{64}-\frac {B\,a\,b^3}{8}\right )\,\sqrt {a+b\,x}-\frac {11\,\left (5\,A\,b^4-8\,B\,a\,b^3\right )\,{\left (a+b\,x\right )}^{5/2}}{192\,a^2}+\frac {\left (5\,A\,b^4-8\,B\,a\,b^3\right )\,{\left (a+b\,x\right )}^{7/2}}{64\,a^3}+\frac {\left (73\,A\,b^4-40\,B\,a\,b^3\right )\,{\left (a+b\,x\right )}^{3/2}}{192\,a}}{{\left (a+b\,x\right )}^4-4\,a^3\,\left (a+b\,x\right )-4\,a\,{\left (a+b\,x\right )}^3+6\,a^2\,{\left (a+b\,x\right )}^2+a^4} \]
(b^3*atanh((a + b*x)^(1/2)/a^(1/2))*(5*A*b - 8*B*a))/(64*a^(7/2)) - (((5*A *b^4)/64 - (B*a*b^3)/8)*(a + b*x)^(1/2) - (11*(5*A*b^4 - 8*B*a*b^3)*(a + b *x)^(5/2))/(192*a^2) + ((5*A*b^4 - 8*B*a*b^3)*(a + b*x)^(7/2))/(64*a^3) + ((73*A*b^4 - 40*B*a*b^3)*(a + b*x)^(3/2))/(192*a))/((a + b*x)^4 - 4*a^3*(a + b*x) - 4*a*(a + b*x)^3 + 6*a^2*(a + b*x)^2 + a^4)