Integrand size = 18, antiderivative size = 146 \[ \int \frac {A+B x}{x^5 \sqrt {a+b x}} \, dx=-\frac {A \sqrt {a+b x}}{4 a x^4}+\frac {(7 A b-8 a B) \sqrt {a+b x}}{24 a^2 x^3}-\frac {5 b (7 A b-8 a B) \sqrt {a+b x}}{96 a^3 x^2}+\frac {5 b^2 (7 A b-8 a B) \sqrt {a+b x}}{64 a^4 x}-\frac {5 b^3 (7 A b-8 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{9/2}} \]
-5/64*b^3*(7*A*b-8*B*a)*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(9/2)-1/4*A*(b*x+ a)^(1/2)/a/x^4+1/24*(7*A*b-8*B*a)*(b*x+a)^(1/2)/a^2/x^3-5/96*b*(7*A*b-8*B* a)*(b*x+a)^(1/2)/a^3/x^2+5/64*b^2*(7*A*b-8*B*a)*(b*x+a)^(1/2)/a^4/x
Time = 0.22 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.77 \[ \int \frac {A+B x}{x^5 \sqrt {a+b x}} \, dx=\frac {\sqrt {a+b x} \left (105 A b^3 x^3-16 a^3 (3 A+4 B x)+8 a^2 b x (7 A+10 B x)-10 a b^2 x^2 (7 A+12 B x)\right )}{192 a^4 x^4}+\frac {5 b^3 (-7 A b+8 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{9/2}} \]
(Sqrt[a + b*x]*(105*A*b^3*x^3 - 16*a^3*(3*A + 4*B*x) + 8*a^2*b*x*(7*A + 10 *B*x) - 10*a*b^2*x^2*(7*A + 12*B*x)))/(192*a^4*x^4) + (5*b^3*(-7*A*b + 8*a *B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(64*a^(9/2))
Time = 0.22 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {87, 52, 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.
| \(\displaystyle \int \frac {A+B x}{x^5 \sqrt {a+b x}} \, dx\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle -\frac {(7 A b-8 a B) \int \frac {1}{x^4 \sqrt {a+b x}}dx}{8 a}-\frac {A \sqrt {a+b x}}{4 a x^4}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {(7 A b-8 a B) \left (-\frac {5 b \int \frac {1}{x^3 \sqrt {a+b x}}dx}{6 a}-\frac {\sqrt {a+b x}}{3 a x^3}\right )}{8 a}-\frac {A \sqrt {a+b x}}{4 a x^4}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {(7 A b-8 a B) \left (-\frac {5 b \left (-\frac {3 b \int \frac {1}{x^2 \sqrt {a+b x}}dx}{4 a}-\frac {\sqrt {a+b x}}{2 a x^2}\right )}{6 a}-\frac {\sqrt {a+b x}}{3 a x^3}\right )}{8 a}-\frac {A \sqrt {a+b x}}{4 a x^4}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {(7 A b-8 a B) \left (-\frac {5 b \left (-\frac {3 b \left (-\frac {b \int \frac {1}{x \sqrt {a+b x}}dx}{2 a}-\frac {\sqrt {a+b x}}{a x}\right )}{4 a}-\frac {\sqrt {a+b x}}{2 a x^2}\right )}{6 a}-\frac {\sqrt {a+b x}}{3 a x^3}\right )}{8 a}-\frac {A \sqrt {a+b x}}{4 a x^4}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {(7 A b-8 a B) \left (-\frac {5 b \left (-\frac {3 b \left (-\frac {\int \frac {1}{\frac {a+b x}{b}-\frac {a}{b}}d\sqrt {a+b x}}{a}-\frac {\sqrt {a+b x}}{a x}\right )}{4 a}-\frac {\sqrt {a+b x}}{2 a x^2}\right )}{6 a}-\frac {\sqrt {a+b x}}{3 a x^3}\right )}{8 a}-\frac {A \sqrt {a+b x}}{4 a x^4}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {(7 A b-8 a B) \left (-\frac {5 b \left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b x}}{a x}\right )}{4 a}-\frac {\sqrt {a+b x}}{2 a x^2}\right )}{6 a}-\frac {\sqrt {a+b x}}{3 a x^3}\right )}{8 a}-\frac {A \sqrt {a+b x}}{4 a x^4}\) |
-1/4*(A*Sqrt[a + b*x])/(a*x^4) - ((7*A*b - 8*a*B)*(-1/3*Sqrt[a + b*x]/(a*x ^3) - (5*b*(-1/2*Sqrt[a + b*x]/(a*x^2) - (3*b*(-(Sqrt[a + b*x]/(a*x)) + (b *ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/a^(3/2)))/(4*a)))/(6*a)))/(8*a)
3.5.32.3.1 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] - 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 = 101, normalized size of antiderivative = 0.69
| method | result | size |
| pseudoelliptic | \(\frac {-\frac {35 x^{4} b^{3} \left (A b -\frac {8 B a}{7}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{64}+\frac {7 \sqrt {b x +a}\, \left (-\frac {5 x^{2} \left (\frac {12 B x}{7}+A \right ) b^{2} a^{\frac {3}{2}}}{4}+b x \left (\frac {10 B x}{7}+A \right ) a^{\frac {5}{2}}+\frac {2 \left (-4 B x -3 A \right ) a^{\frac {7}{2}}}{7}+\frac {15 A \sqrt {a}\, b^{3} x^{3}}{8}\right )}{24}}{a^{\frac {9}{2}} x^{4}}\) | \(101\) |
| risch | \(-\frac {\sqrt {b x +a}\, \left (-105 A \,b^{3} x^{3}+120 B a \,b^{2} x^{3}+70 a A \,b^{2} x^{2}-80 B \,a^{2} b \,x^{2}-56 a^{2} A b x +64 a^{3} B x +48 a^{3} A \right )}{192 a^{4} x^{4}}-\frac {5 b^{3} \left (7 A b -8 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{64 a^{\frac {9}{2}}}\) | \(107\) |
| derivativedivides | \(2 b^{3} \left (-\frac {-\frac {5 \left (7 A b -8 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{128 a^{4}}+\frac {55 \left (7 A b -8 B a \right ) \left (b x +a \right )^{\frac {5}{2}}}{384 a^{3}}-\frac {73 \left (7 A b -8 B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{384 a^{2}}+\frac {\left (93 A b -88 B a \right ) \sqrt {b x +a}}{128 a}}{b^{4} x^{4}}-\frac {5 \left (7 A b -8 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{128 a^{\frac {9}{2}}}\right )\) | \(126\) |
| default | \(2 b^{3} \left (-\frac {-\frac {5 \left (7 A b -8 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{128 a^{4}}+\frac {55 \left (7 A b -8 B a \right ) \left (b x +a \right )^{\frac {5}{2}}}{384 a^{3}}-\frac {73 \left (7 A b -8 B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{384 a^{2}}+\frac {\left (93 A b -88 B a \right ) \sqrt {b x +a}}{128 a}}{b^{4} x^{4}}-\frac {5 \left (7 A b -8 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{128 a^{\frac {9}{2}}}\right )\) | \(126\) |
7/24*(-15/8*x^4*b^3*(A*b-8/7*B*a)*arctanh((b*x+a)^(1/2)/a^(1/2))+(b*x+a)^( 1/2)*(-5/4*x^2*(12/7*B*x+A)*b^2*a^(3/2)+b*x*(10/7*B*x+A)*a^(5/2)+2/7*(-4*B *x-3*A)*a^(7/2)+15/8*A*a^(1/2)*b^3*x^3))/a^(9/2)/x^4
Time = 0.23 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.77 \[ \int \frac {A+B x}{x^5 \sqrt {a+b x}} \, dx=\left [-\frac {15 \, {\left (8 \, B a b^{3} - 7 \, A b^{4}\right )} \sqrt {a} x^{4} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (48 \, A a^{4} + 15 \, {\left (8 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3} - 10 \, {\left (8 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{2} + 8 \, {\left (8 \, B a^{4} - 7 \, A a^{3} b\right )} x\right )} \sqrt {b x + a}}{384 \, a^{5} x^{4}}, -\frac {15 \, {\left (8 \, B a b^{3} - 7 \, A b^{4}\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (48 \, A a^{4} + 15 \, {\left (8 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3} - 10 \, {\left (8 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{2} + 8 \, {\left (8 \, B a^{4} - 7 \, A a^{3} b\right )} x\right )} \sqrt {b x + a}}{192 \, a^{5} x^{4}}\right ] \]
[-1/384*(15*(8*B*a*b^3 - 7*A*b^4)*sqrt(a)*x^4*log((b*x - 2*sqrt(b*x + a)*s qrt(a) + 2*a)/x) + 2*(48*A*a^4 + 15*(8*B*a^2*b^2 - 7*A*a*b^3)*x^3 - 10*(8* B*a^3*b - 7*A*a^2*b^2)*x^2 + 8*(8*B*a^4 - 7*A*a^3*b)*x)*sqrt(b*x + a))/(a^ 5*x^4), -1/192*(15*(8*B*a*b^3 - 7*A*b^4)*sqrt(-a)*x^4*arctan(sqrt(b*x + a) *sqrt(-a)/a) + (48*A*a^4 + 15*(8*B*a^2*b^2 - 7*A*a*b^3)*x^3 - 10*(8*B*a^3* b - 7*A*a^2*b^2)*x^2 + 8*(8*B*a^4 - 7*A*a^3*b)*x)*sqrt(b*x + a))/(a^5*x^4) ]
Leaf count of result is larger than twice the leaf count of optimal. 303 vs. \(2 (141) = 282\).
Time = 56.87 (sec) , antiderivative size = 303, normalized size of antiderivative = 2.08 \[ \int \frac {A+B x}{x^5 \sqrt {a+b x}} \, dx=- \frac {A}{4 \sqrt {b} x^{\frac {9}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {A \sqrt {b}}{24 a x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {7 A b^{\frac {3}{2}}}{96 a^{2} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {35 A b^{\frac {5}{2}}}{192 a^{3} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {35 A b^{\frac {7}{2}}}{64 a^{4} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} - \frac {35 A b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{64 a^{\frac {9}{2}}} - \frac {B}{3 \sqrt {b} x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {B \sqrt {b}}{12 a x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {5 B b^{\frac {3}{2}}}{24 a^{2} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {5 B b^{\frac {5}{2}}}{8 a^{3} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {5 B b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{8 a^{\frac {7}{2}}} \]
-A/(4*sqrt(b)*x**(9/2)*sqrt(a/(b*x) + 1)) + A*sqrt(b)/(24*a*x**(7/2)*sqrt( a/(b*x) + 1)) - 7*A*b**(3/2)/(96*a**2*x**(5/2)*sqrt(a/(b*x) + 1)) + 35*A*b **(5/2)/(192*a**3*x**(3/2)*sqrt(a/(b*x) + 1)) + 35*A*b**(7/2)/(64*a**4*sqr t(x)*sqrt(a/(b*x) + 1)) - 35*A*b**4*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(64*a **(9/2)) - B/(3*sqrt(b)*x**(7/2)*sqrt(a/(b*x) + 1)) + B*sqrt(b)/(12*a*x**( 5/2)*sqrt(a/(b*x) + 1)) - 5*B*b**(3/2)/(24*a**2*x**(3/2)*sqrt(a/(b*x) + 1) ) - 5*B*b**(5/2)/(8*a**3*sqrt(x)*sqrt(a/(b*x) + 1)) + 5*B*b**3*asinh(sqrt( a)/(sqrt(b)*sqrt(x)))/(8*a**(7/2))
Time = 0.29 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.34 \[ \int \frac {A+B x}{x^5 \sqrt {a+b x}} \, dx=-\frac {1}{384} \, b^{4} {\left (\frac {2 \, {\left (15 \, {\left (8 \, B a - 7 \, A b\right )} {\left (b x + a\right )}^{\frac {7}{2}} - 55 \, {\left (8 \, B a^{2} - 7 \, A a b\right )} {\left (b x + a\right )}^{\frac {5}{2}} + 73 \, {\left (8 \, B a^{3} - 7 \, A a^{2} b\right )} {\left (b x + a\right )}^{\frac {3}{2}} - 3 \, {\left (88 \, B a^{4} - 93 \, A a^{3} b\right )} \sqrt {b x + a}\right )}}{{\left (b x + a\right )}^{4} a^{4} b - 4 \, {\left (b x + a\right )}^{3} a^{5} b + 6 \, {\left (b x + a\right )}^{2} a^{6} b - 4 \, {\left (b x + a\right )} a^{7} b + a^{8} b} + \frac {15 \, {\left (8 \, B a - 7 \, A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {9}{2}} b}\right )} \]
-1/384*b^4*(2*(15*(8*B*a - 7*A*b)*(b*x + a)^(7/2) - 55*(8*B*a^2 - 7*A*a*b) *(b*x + a)^(5/2) + 73*(8*B*a^3 - 7*A*a^2*b)*(b*x + a)^(3/2) - 3*(88*B*a^4 - 93*A*a^3*b)*sqrt(b*x + a))/((b*x + a)^4*a^4*b - 4*(b*x + a)^3*a^5*b + 6* (b*x + a)^2*a^6*b - 4*(b*x + a)*a^7*b + a^8*b) + 15*(8*B*a - 7*A*b)*log((s qrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/(a^(9/2)*b))
Time = 0.28 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.21 \[ \int \frac {A+B x}{x^5 \sqrt {a+b x}} \, dx=-\frac {\frac {15 \, {\left (8 \, B a b^{4} - 7 \, A b^{5}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{4}} + \frac {120 \, {\left (b x + a\right )}^{\frac {7}{2}} B a b^{4} - 440 \, {\left (b x + a\right )}^{\frac {5}{2}} B a^{2} b^{4} + 584 \, {\left (b x + a\right )}^{\frac {3}{2}} B a^{3} b^{4} - 264 \, \sqrt {b x + a} B a^{4} b^{4} - 105 \, {\left (b x + a\right )}^{\frac {7}{2}} A b^{5} + 385 \, {\left (b x + a\right )}^{\frac {5}{2}} A a b^{5} - 511 \, {\left (b x + a\right )}^{\frac {3}{2}} A a^{2} b^{5} + 279 \, \sqrt {b x + a} A a^{3} b^{5}}{a^{4} b^{4} x^{4}}}{192 \, b} \]
-1/192*(15*(8*B*a*b^4 - 7*A*b^5)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)* a^4) + (120*(b*x + a)^(7/2)*B*a*b^4 - 440*(b*x + a)^(5/2)*B*a^2*b^4 + 584* (b*x + a)^(3/2)*B*a^3*b^4 - 264*sqrt(b*x + a)*B*a^4*b^4 - 105*(b*x + a)^(7 /2)*A*b^5 + 385*(b*x + a)^(5/2)*A*a*b^5 - 511*(b*x + a)^(3/2)*A*a^2*b^5 + 279*sqrt(b*x + a)*A*a^3*b^5)/(a^4*b^4*x^4))/b
Time = 0.48 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.24 \[ \int \frac {A+B x}{x^5 \sqrt {a+b x}} \, dx=\frac {\frac {73\,\left (7\,A\,b^4-8\,B\,a\,b^3\right )\,{\left (a+b\,x\right )}^{3/2}}{192\,a^2}-\frac {55\,\left (7\,A\,b^4-8\,B\,a\,b^3\right )\,{\left (a+b\,x\right )}^{5/2}}{192\,a^3}+\frac {5\,\left (7\,A\,b^4-8\,B\,a\,b^3\right )\,{\left (a+b\,x\right )}^{7/2}}{64\,a^4}-\frac {\left (93\,A\,b^4-88\,B\,a\,b^3\right )\,\sqrt {a+b\,x}}{64\,a}}{{\left (a+b\,x\right )}^4-4\,a^3\,\left (a+b\,x\right )-4\,a\,{\left (a+b\,x\right )}^3+6\,a^2\,{\left (a+b\,x\right )}^2+a^4}-\frac {5\,b^3\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (7\,A\,b-8\,B\,a\right )}{64\,a^{9/2}} \]
((73*(7*A*b^4 - 8*B*a*b^3)*(a + b*x)^(3/2))/(192*a^2) - (55*(7*A*b^4 - 8*B *a*b^3)*(a + b*x)^(5/2))/(192*a^3) + (5*(7*A*b^4 - 8*B*a*b^3)*(a + b*x)^(7 /2))/(64*a^4) - ((93*A*b^4 - 88*B*a*b^3)*(a + b*x)^(1/2))/(64*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) - (5*b ^3*atanh((a + b*x)^(1/2)/a^(1/2))*(7*A*b - 8*B*a))/(64*a^(9/2))