Integrand size = 18, antiderivative size = 139 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^4} \, dx=\frac {5 b^2 (A b+6 a B) \sqrt {a+b x}}{8 a}-\frac {5 b (A b+6 a B) (a+b x)^{3/2}}{24 a x}-\frac {(A b+6 a B) (a+b x)^{5/2}}{12 a x^2}-\frac {A (a+b x)^{7/2}}{3 a x^3}-\frac {5 b^2 (A b+6 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 \sqrt {a}} \]
-5/24*b*(A*b+6*B*a)*(b*x+a)^(3/2)/a/x-1/12*(A*b+6*B*a)*(b*x+a)^(5/2)/a/x^2 -1/3*A*(b*x+a)^(7/2)/a/x^3-5/8*b^2*(A*b+6*B*a)*arctanh((b*x+a)^(1/2)/a^(1/ 2))/a^(1/2)+5/8*b^2*(A*b+6*B*a)*(b*x+a)^(1/2)/a
Time = 0.20 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.69 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^4} \, dx=-\frac {\sqrt {a+b x} \left (3 b^2 x^2 (11 A-16 B x)+4 a^2 (2 A+3 B x)+2 a b x (13 A+27 B x)\right )}{24 x^3}-\frac {5 b^2 (A b+6 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 \sqrt {a}} \]
-1/24*(Sqrt[a + b*x]*(3*b^2*x^2*(11*A - 16*B*x) + 4*a^2*(2*A + 3*B*x) + 2* a*b*x*(13*A + 27*B*x)))/x^3 - (5*b^2*(A*b + 6*a*B)*ArcTanh[Sqrt[a + b*x]/S qrt[a]])/(8*Sqrt[a])
Time = 0.21 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.81, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {87, 51, 51, 60, 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.
| \(\displaystyle \int \frac {(a+b x)^{5/2} (A+B x)}{x^4} \, dx\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {(6 a B+A b) \int \frac {(a+b x)^{5/2}}{x^3}dx}{6 a}-\frac {A (a+b x)^{7/2}}{3 a x^3}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {(6 a B+A b) \left (\frac {5}{4} b \int \frac {(a+b x)^{3/2}}{x^2}dx-\frac {(a+b x)^{5/2}}{2 x^2}\right )}{6 a}-\frac {A (a+b x)^{7/2}}{3 a x^3}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {(6 a B+A b) \left (\frac {5}{4} b \left (\frac {3}{2} b \int \frac {\sqrt {a+b x}}{x}dx-\frac {(a+b x)^{3/2}}{x}\right )-\frac {(a+b x)^{5/2}}{2 x^2}\right )}{6 a}-\frac {A (a+b x)^{7/2}}{3 a x^3}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(6 a B+A b) \left (\frac {5}{4} b \left (\frac {3}{2} b \left (a \int \frac {1}{x \sqrt {a+b x}}dx+2 \sqrt {a+b x}\right )-\frac {(a+b x)^{3/2}}{x}\right )-\frac {(a+b x)^{5/2}}{2 x^2}\right )}{6 a}-\frac {A (a+b x)^{7/2}}{3 a x^3}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(6 a B+A b) \left (\frac {5}{4} b \left (\frac {3}{2} b \left (\frac {2 a \int \frac {1}{\frac {a+b x}{b}-\frac {a}{b}}d\sqrt {a+b x}}{b}+2 \sqrt {a+b x}\right )-\frac {(a+b x)^{3/2}}{x}\right )-\frac {(a+b x)^{5/2}}{2 x^2}\right )}{6 a}-\frac {A (a+b x)^{7/2}}{3 a x^3}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(6 a B+A b) \left (\frac {5}{4} b \left (\frac {3}{2} b \left (2 \sqrt {a+b x}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )-\frac {(a+b x)^{3/2}}{x}\right )-\frac {(a+b x)^{5/2}}{2 x^2}\right )}{6 a}-\frac {A (a+b x)^{7/2}}{3 a x^3}\) |
-1/3*(A*(a + b*x)^(7/2))/(a*x^3) + ((A*b + 6*a*B)*(-1/2*(a + b*x)^(5/2)/x^ 2 + (5*b*(-((a + b*x)^(3/2)/x) + (3*b*(2*Sqrt[a + b*x] - 2*Sqrt[a]*ArcTanh [Sqrt[a + b*x]/Sqrt[a]]))/2))/4))/(6*a)
3.5.18.3.1 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/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
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 = 0.54 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.63
| method | result | size |
| pseudoelliptic | \(-\frac {11 \left (\frac {5 b^{2} x^{3} \left (A b +6 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{11}+\left (\frac {26 x b \left (\frac {27 B x}{13}+A \right ) a^{\frac {3}{2}}}{33}+\frac {4 \left (B x +\frac {2 A}{3}\right ) a^{\frac {5}{2}}}{11}+b^{2} x^{2} \sqrt {a}\, \left (-\frac {16 B x}{11}+A \right )\right ) \sqrt {b x +a}\right )}{8 \sqrt {a}\, x^{3}}\) | \(88\) |
| risch | \(-\frac {\sqrt {b x +a}\, \left (33 A \,b^{2} x^{2}+54 B a b \,x^{2}+26 a A b x +12 a^{2} B x +8 a^{2} A \right )}{24 x^{3}}+\frac {b^{2} \left (32 B \sqrt {b x +a}-\frac {2 \left (5 A b +30 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}\right )}{16}\) | \(93\) |
| derivativedivides | \(2 b^{2} \left (B \sqrt {b x +a}-\frac {\left (\frac {11 A b}{16}+\frac {9 B a}{8}\right ) \left (b x +a \right )^{\frac {5}{2}}+\left (-\frac {5}{6} a b A -2 a^{2} B \right ) \left (b x +a \right )^{\frac {3}{2}}+\left (\frac {7}{8} a^{3} B +\frac {5}{16} a^{2} b A \right ) \sqrt {b x +a}}{b^{3} x^{3}}-\frac {5 \left (A b +6 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 \sqrt {a}}\right )\) | \(109\) |
| default | \(2 b^{2} \left (B \sqrt {b x +a}-\frac {\left (\frac {11 A b}{16}+\frac {9 B a}{8}\right ) \left (b x +a \right )^{\frac {5}{2}}+\left (-\frac {5}{6} a b A -2 a^{2} B \right ) \left (b x +a \right )^{\frac {3}{2}}+\left (\frac {7}{8} a^{3} B +\frac {5}{16} a^{2} b A \right ) \sqrt {b x +a}}{b^{3} x^{3}}-\frac {5 \left (A b +6 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 \sqrt {a}}\right )\) | \(109\) |
-11/8/a^(1/2)*(5/11*b^2*x^3*(A*b+6*B*a)*arctanh((b*x+a)^(1/2)/a^(1/2))+(26 /33*x*b*(27/13*B*x+A)*a^(3/2)+4/11*(B*x+2/3*A)*a^(5/2)+b^2*x^2*a^(1/2)*(-1 6/11*B*x+A))*(b*x+a)^(1/2))/x^3
Time = 0.24 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.65 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^4} \, dx=\left [\frac {15 \, {\left (6 \, B a b^{2} + A b^{3}\right )} \sqrt {a} x^{3} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (48 \, B a b^{2} x^{3} - 8 \, A a^{3} - 3 \, {\left (18 \, B a^{2} b + 11 \, A a b^{2}\right )} x^{2} - 2 \, {\left (6 \, B a^{3} + 13 \, A a^{2} b\right )} x\right )} \sqrt {b x + a}}{48 \, a x^{3}}, \frac {15 \, {\left (6 \, B a b^{2} + A b^{3}\right )} \sqrt {-a} x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (48 \, B a b^{2} x^{3} - 8 \, A a^{3} - 3 \, {\left (18 \, B a^{2} b + 11 \, A a b^{2}\right )} x^{2} - 2 \, {\left (6 \, B a^{3} + 13 \, A a^{2} b\right )} x\right )} \sqrt {b x + a}}{24 \, a x^{3}}\right ] \]
[1/48*(15*(6*B*a*b^2 + A*b^3)*sqrt(a)*x^3*log((b*x - 2*sqrt(b*x + a)*sqrt( a) + 2*a)/x) + 2*(48*B*a*b^2*x^3 - 8*A*a^3 - 3*(18*B*a^2*b + 11*A*a*b^2)*x ^2 - 2*(6*B*a^3 + 13*A*a^2*b)*x)*sqrt(b*x + a))/(a*x^3), 1/24*(15*(6*B*a*b ^2 + A*b^3)*sqrt(-a)*x^3*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (48*B*a*b^2*x^ 3 - 8*A*a^3 - 3*(18*B*a^2*b + 11*A*a*b^2)*x^2 - 2*(6*B*a^3 + 13*A*a^2*b)*x )*sqrt(b*x + a))/(a*x^3)]
Time = 59.84 (sec) , antiderivative size = 348, normalized size of antiderivative = 2.50 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^4} \, dx=- \frac {A a^{3}}{3 \sqrt {b} x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {17 A a^{2} \sqrt {b}}{12 x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {35 A a b^{\frac {3}{2}}}{24 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {A b^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}}{\sqrt {x}} - \frac {3 A b^{\frac {5}{2}}}{8 \sqrt {x} \sqrt {\frac {a}{b x} + 1}} - \frac {5 A b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{8 \sqrt {a}} - \frac {7 B \sqrt {a} b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4} - \frac {B a^{3}}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {3 B a^{2} \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {2 B a b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{\sqrt {x}} - \frac {B a b^{\frac {3}{2}}}{4 \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + B b^{2} \left (\begin {cases} \frac {2 a \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 \sqrt {a + b x} & \text {for}\: b \neq 0 \\\sqrt {a} \log {\left (x \right )} & \text {otherwise} \end {cases}\right ) \]
-A*a**3/(3*sqrt(b)*x**(7/2)*sqrt(a/(b*x) + 1)) - 17*A*a**2*sqrt(b)/(12*x** (5/2)*sqrt(a/(b*x) + 1)) - 35*A*a*b**(3/2)/(24*x**(3/2)*sqrt(a/(b*x) + 1)) - A*b**(5/2)*sqrt(a/(b*x) + 1)/sqrt(x) - 3*A*b**(5/2)/(8*sqrt(x)*sqrt(a/( b*x) + 1)) - 5*A*b**3*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(8*sqrt(a)) - 7*B*s qrt(a)*b**2*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/4 - B*a**3/(2*sqrt(b)*x**(5/2 )*sqrt(a/(b*x) + 1)) - 3*B*a**2*sqrt(b)/(4*x**(3/2)*sqrt(a/(b*x) + 1)) - 2 *B*a*b**(3/2)*sqrt(a/(b*x) + 1)/sqrt(x) - B*a*b**(3/2)/(4*sqrt(x)*sqrt(a/( b*x) + 1)) + B*b**2*Piecewise((2*a*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) + 2*sqrt(a + b*x), Ne(b, 0)), (sqrt(a)*log(x), True))
Time = 0.28 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.21 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^4} \, dx=\frac {1}{48} \, b^{3} {\left (\frac {96 \, \sqrt {b x + a} B}{b} + \frac {15 \, {\left (6 \, B a + A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{\sqrt {a} b} - \frac {2 \, {\left (3 \, {\left (18 \, B a + 11 \, A b\right )} {\left (b x + a\right )}^{\frac {5}{2}} - 8 \, {\left (12 \, B a^{2} + 5 \, A a b\right )} {\left (b x + a\right )}^{\frac {3}{2}} + 3 \, {\left (14 \, B a^{3} + 5 \, A a^{2} b\right )} \sqrt {b x + a}\right )}}{{\left (b x + a\right )}^{3} b - 3 \, {\left (b x + a\right )}^{2} a b + 3 \, {\left (b x + a\right )} a^{2} b - a^{3} b}\right )} \]
1/48*b^3*(96*sqrt(b*x + a)*B/b + 15*(6*B*a + A*b)*log((sqrt(b*x + a) - sqr t(a))/(sqrt(b*x + a) + sqrt(a)))/(sqrt(a)*b) - 2*(3*(18*B*a + 11*A*b)*(b*x + a)^(5/2) - 8*(12*B*a^2 + 5*A*a*b)*(b*x + a)^(3/2) + 3*(14*B*a^3 + 5*A*a ^2*b)*sqrt(b*x + a))/((b*x + a)^3*b - 3*(b*x + a)^2*a*b + 3*(b*x + a)*a^2* b - a^3*b))
Time = 0.28 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.09 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^4} \, dx=\frac {48 \, \sqrt {b x + a} B b^{3} + \frac {15 \, {\left (6 \, B a b^{3} + A b^{4}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {54 \, {\left (b x + a\right )}^{\frac {5}{2}} B a b^{3} - 96 \, {\left (b x + a\right )}^{\frac {3}{2}} B a^{2} b^{3} + 42 \, \sqrt {b x + a} B a^{3} b^{3} + 33 \, {\left (b x + a\right )}^{\frac {5}{2}} A b^{4} - 40 \, {\left (b x + a\right )}^{\frac {3}{2}} A a b^{4} + 15 \, \sqrt {b x + a} A a^{2} b^{4}}{b^{3} x^{3}}}{24 \, b} \]
1/24*(48*sqrt(b*x + a)*B*b^3 + 15*(6*B*a*b^3 + A*b^4)*arctan(sqrt(b*x + a) /sqrt(-a))/sqrt(-a) - (54*(b*x + a)^(5/2)*B*a*b^3 - 96*(b*x + a)^(3/2)*B*a ^2*b^3 + 42*sqrt(b*x + a)*B*a^3*b^3 + 33*(b*x + a)^(5/2)*A*b^4 - 40*(b*x + a)^(3/2)*A*a*b^4 + 15*sqrt(b*x + a)*A*a^2*b^4)/(b^3*x^3))/b
Time = 0.45 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.31 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^4} \, dx=\frac {\left (\frac {11\,A\,b^3}{8}+\frac {9\,B\,a\,b^2}{4}\right )\,{\left (a+b\,x\right )}^{5/2}+\left (\frac {7\,B\,a^3\,b^2}{4}+\frac {5\,A\,a^2\,b^3}{8}\right )\,\sqrt {a+b\,x}-\left (4\,B\,a^2\,b^2+\frac {5\,A\,a\,b^3}{3}\right )\,{\left (a+b\,x\right )}^{3/2}}{3\,a\,{\left (a+b\,x\right )}^2-3\,a^2\,\left (a+b\,x\right )-{\left (a+b\,x\right )}^3+a^3}+2\,B\,b^2\,\sqrt {a+b\,x}-\frac {5\,b^2\,\mathrm {atanh}\left (\frac {5\,b^2\,\left (A\,b+6\,B\,a\right )\,\sqrt {a+b\,x}}{4\,\sqrt {a}\,\left (\frac {5\,A\,b^3}{4}+\frac {15\,B\,a\,b^2}{2}\right )}\right )\,\left (A\,b+6\,B\,a\right )}{8\,\sqrt {a}} \]
(((11*A*b^3)/8 + (9*B*a*b^2)/4)*(a + b*x)^(5/2) + ((5*A*a^2*b^3)/8 + (7*B* a^3*b^2)/4)*(a + b*x)^(1/2) - (4*B*a^2*b^2 + (5*A*a*b^3)/3)*(a + b*x)^(3/2 ))/(3*a*(a + b*x)^2 - 3*a^2*(a + b*x) - (a + b*x)^3 + a^3) + 2*B*b^2*(a + b*x)^(1/2) - (5*b^2*atanh((5*b^2*(A*b + 6*B*a)*(a + b*x)^(1/2))/(4*a^(1/2) *((5*A*b^3)/4 + (15*B*a*b^2)/2)))*(A*b + 6*B*a))/(8*a^(1/2))