Integrand size = 13, antiderivative size = 114 \[ \int \frac {(a+b x)^{9/2}}{x^4} \, dx=\frac {105}{8} a b^3 \sqrt {a+b x}+\frac {35}{8} b^3 (a+b x)^{3/2}-\frac {21 b^2 (a+b x)^{5/2}}{8 x}-\frac {3 b (a+b x)^{7/2}}{4 x^2}-\frac {(a+b x)^{9/2}}{3 x^3}-\frac {105}{8} a^{3/2} b^3 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]
35/8*b^3*(b*x+a)^(3/2)-21/8*b^2*(b*x+a)^(5/2)/x-3/4*b*(b*x+a)^(7/2)/x^2-1/ 3*(b*x+a)^(9/2)/x^3-105/8*a^(3/2)*b^3*arctanh((b*x+a)^(1/2)/a^(1/2))+105/8 *a*b^3*(b*x+a)^(1/2)
Time = 0.13 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.75 \[ \int \frac {(a+b x)^{9/2}}{x^4} \, dx=\frac {1}{24} \left (\frac {\sqrt {a+b x} \left (-8 a^4-50 a^3 b x-165 a^2 b^2 x^2+208 a b^3 x^3+16 b^4 x^4\right )}{x^3}-315 a^{3/2} b^3 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right ) \]
((Sqrt[a + b*x]*(-8*a^4 - 50*a^3*b*x - 165*a^2*b^2*x^2 + 208*a*b^3*x^3 + 1 6*b^4*x^4))/x^3 - 315*a^(3/2)*b^3*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/24
Time = 0.19 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {51, 51, 51, 60, 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.
\(\displaystyle \int \frac {(a+b x)^{9/2}}{x^4} \, dx\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {3}{2} b \int \frac {(a+b x)^{7/2}}{x^3}dx-\frac {(a+b x)^{9/2}}{3 x^3}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {3}{2} b \left (\frac {7}{4} b \int \frac {(a+b x)^{5/2}}{x^2}dx-\frac {(a+b x)^{7/2}}{2 x^2}\right )-\frac {(a+b x)^{9/2}}{3 x^3}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {3}{2} b \left (\frac {7}{4} b \left (\frac {5}{2} b \int \frac {(a+b x)^{3/2}}{x}dx-\frac {(a+b x)^{5/2}}{x}\right )-\frac {(a+b x)^{7/2}}{2 x^2}\right )-\frac {(a+b x)^{9/2}}{3 x^3}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {3}{2} b \left (\frac {7}{4} b \left (\frac {5}{2} b \left (a \int \frac {\sqrt {a+b x}}{x}dx+\frac {2}{3} (a+b x)^{3/2}\right )-\frac {(a+b x)^{5/2}}{x}\right )-\frac {(a+b x)^{7/2}}{2 x^2}\right )-\frac {(a+b x)^{9/2}}{3 x^3}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {3}{2} b \left (\frac {7}{4} b \left (\frac {5}{2} b \left (a \left (a \int \frac {1}{x \sqrt {a+b x}}dx+2 \sqrt {a+b x}\right )+\frac {2}{3} (a+b x)^{3/2}\right )-\frac {(a+b x)^{5/2}}{x}\right )-\frac {(a+b x)^{7/2}}{2 x^2}\right )-\frac {(a+b x)^{9/2}}{3 x^3}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {3}{2} b \left (\frac {7}{4} b \left (\frac {5}{2} b \left (a \left (\frac {2 a \int \frac {1}{\frac {a+b x}{b}-\frac {a}{b}}d\sqrt {a+b x}}{b}+2 \sqrt {a+b x}\right )+\frac {2}{3} (a+b x)^{3/2}\right )-\frac {(a+b x)^{5/2}}{x}\right )-\frac {(a+b x)^{7/2}}{2 x^2}\right )-\frac {(a+b x)^{9/2}}{3 x^3}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {3}{2} b \left (\frac {7}{4} b \left (\frac {5}{2} b \left (a \left (2 \sqrt {a+b x}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )+\frac {2}{3} (a+b x)^{3/2}\right )-\frac {(a+b x)^{5/2}}{x}\right )-\frac {(a+b x)^{7/2}}{2 x^2}\right )-\frac {(a+b x)^{9/2}}{3 x^3}\) |
-1/3*(a + b*x)^(9/2)/x^3 + (3*b*(-1/2*(a + b*x)^(7/2)/x^2 + (7*b*(-((a + b *x)^(5/2)/x) + (5*b*((2*(a + b*x)^(3/2))/3 + a*(2*Sqrt[a + b*x] - 2*Sqrt[a ]*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])))/2))/4))/2
3.4.20.3.1 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/(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_)^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 = 0.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.68
method | result | size |
risch | \(-\frac {a^{2} \sqrt {b x +a}\, \left (165 b^{2} x^{2}+50 a b x +8 a^{2}\right )}{24 x^{3}}+\frac {b^{3} \left (\frac {32 \left (b x +a \right )^{\frac {3}{2}}}{3}+128 a \sqrt {b x +a}-210 a^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )\right )}{16}\) | \(78\) |
pseudoelliptic | \(-\frac {105 \left (\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{2} b^{3} x^{3}-\frac {16 \left (\sqrt {a}\, b^{4} x^{4}+13 a^{\frac {3}{2}} b^{3} x^{3}-\frac {165 a^{\frac {5}{2}} b^{2} x^{2}}{16}-\frac {25 a^{\frac {7}{2}} b x}{8}-\frac {a^{\frac {9}{2}}}{2}\right ) \sqrt {b x +a}}{315}\right )}{8 \sqrt {a}\, x^{3}}\) | \(86\) |
derivativedivides | \(2 b^{3} \left (\frac {\left (b x +a \right )^{\frac {3}{2}}}{3}+4 a \sqrt {b x +a}-a^{2} \left (-\frac {-\frac {55 \left (b x +a \right )^{\frac {5}{2}}}{16}+\frac {35 a \left (b x +a \right )^{\frac {3}{2}}}{6}-\frac {41 a^{2} \sqrt {b x +a}}{16}}{b^{3} x^{3}}+\frac {105 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 \sqrt {a}}\right )\right )\) | \(89\) |
default | \(2 b^{3} \left (\frac {\left (b x +a \right )^{\frac {3}{2}}}{3}+4 a \sqrt {b x +a}-a^{2} \left (-\frac {-\frac {55 \left (b x +a \right )^{\frac {5}{2}}}{16}+\frac {35 a \left (b x +a \right )^{\frac {3}{2}}}{6}-\frac {41 a^{2} \sqrt {b x +a}}{16}}{b^{3} x^{3}}+\frac {105 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 \sqrt {a}}\right )\right )\) | \(89\) |
-1/24*a^2*(b*x+a)^(1/2)*(165*b^2*x^2+50*a*b*x+8*a^2)/x^3+1/16*b^3*(32/3*(b *x+a)^(3/2)+128*a*(b*x+a)^(1/2)-210*a^(3/2)*arctanh((b*x+a)^(1/2)/a^(1/2)) )
Time = 0.23 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.56 \[ \int \frac {(a+b x)^{9/2}}{x^4} \, dx=\left [\frac {315 \, a^{\frac {3}{2}} b^{3} x^{3} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (16 \, b^{4} x^{4} + 208 \, a b^{3} x^{3} - 165 \, a^{2} b^{2} x^{2} - 50 \, a^{3} b x - 8 \, a^{4}\right )} \sqrt {b x + a}}{48 \, x^{3}}, \frac {315 \, \sqrt {-a} a b^{3} x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (16 \, b^{4} x^{4} + 208 \, a b^{3} x^{3} - 165 \, a^{2} b^{2} x^{2} - 50 \, a^{3} b x - 8 \, a^{4}\right )} \sqrt {b x + a}}{24 \, x^{3}}\right ] \]
[1/48*(315*a^(3/2)*b^3*x^3*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(16*b^4*x^4 + 208*a*b^3*x^3 - 165*a^2*b^2*x^2 - 50*a^3*b*x - 8*a^4)*sqrt (b*x + a))/x^3, 1/24*(315*sqrt(-a)*a*b^3*x^3*arctan(sqrt(b*x + a)*sqrt(-a) /a) + (16*b^4*x^4 + 208*a*b^3*x^3 - 165*a^2*b^2*x^2 - 50*a^3*b*x - 8*a^4)* sqrt(b*x + a))/x^3]
Time = 13.53 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.61 \[ \int \frac {(a+b x)^{9/2}}{x^4} \, dx=- \frac {105 a^{\frac {3}{2}} b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{8} - \frac {a^{5}}{3 \sqrt {b} x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {29 a^{4} \sqrt {b}}{12 x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {215 a^{3} b^{\frac {3}{2}}}{24 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {43 a^{2} b^{\frac {5}{2}}}{24 \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {28 a b^{\frac {7}{2}} \sqrt {x}}{3 \sqrt {\frac {a}{b x} + 1}} + \frac {2 b^{\frac {9}{2}} x^{\frac {3}{2}}}{3 \sqrt {\frac {a}{b x} + 1}} \]
-105*a**(3/2)*b**3*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/8 - a**5/(3*sqrt(b)*x* *(7/2)*sqrt(a/(b*x) + 1)) - 29*a**4*sqrt(b)/(12*x**(5/2)*sqrt(a/(b*x) + 1) ) - 215*a**3*b**(3/2)/(24*x**(3/2)*sqrt(a/(b*x) + 1)) + 43*a**2*b**(5/2)/( 24*sqrt(x)*sqrt(a/(b*x) + 1)) + 28*a*b**(7/2)*sqrt(x)/(3*sqrt(a/(b*x) + 1) ) + 2*b**(9/2)*x**(3/2)/(3*sqrt(a/(b*x) + 1))
Time = 0.44 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.27 \[ \int \frac {(a+b x)^{9/2}}{x^4} \, dx=\frac {105}{16} \, a^{\frac {3}{2}} b^{3} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + \frac {2}{3} \, {\left (b x + a\right )}^{\frac {3}{2}} b^{3} + 8 \, \sqrt {b x + a} a b^{3} - \frac {165 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} b^{3} - 280 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} b^{3} + 123 \, \sqrt {b x + a} a^{4} b^{3}}{24 \, {\left ({\left (b x + a\right )}^{3} - 3 \, {\left (b x + a\right )}^{2} a + 3 \, {\left (b x + a\right )} a^{2} - a^{3}\right )}} \]
105/16*a^(3/2)*b^3*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)) ) + 2/3*(b*x + a)^(3/2)*b^3 + 8*sqrt(b*x + a)*a*b^3 - 1/24*(165*(b*x + a)^ (5/2)*a^2*b^3 - 280*(b*x + a)^(3/2)*a^3*b^3 + 123*sqrt(b*x + a)*a^4*b^3)/( (b*x + a)^3 - 3*(b*x + a)^2*a + 3*(b*x + a)*a^2 - a^3)
Time = 0.30 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.98 \[ \int \frac {(a+b x)^{9/2}}{x^4} \, dx=\frac {\frac {315 \, a^{2} b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 16 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{4} + 192 \, \sqrt {b x + a} a b^{4} - \frac {165 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} b^{4} - 280 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} b^{4} + 123 \, \sqrt {b x + a} a^{4} b^{4}}{b^{3} x^{3}}}{24 \, b} \]
1/24*(315*a^2*b^4*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) + 16*(b*x + a)^( 3/2)*b^4 + 192*sqrt(b*x + a)*a*b^4 - (165*(b*x + a)^(5/2)*a^2*b^4 - 280*(b *x + a)^(3/2)*a^3*b^4 + 123*sqrt(b*x + a)*a^4*b^4)/(b^3*x^3))/b
Time = 0.17 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.15 \[ \int \frac {(a+b x)^{9/2}}{x^4} \, dx=\frac {2\,b^3\,{\left (a+b\,x\right )}^{3/2}}{3}+\frac {\frac {41\,a^4\,b^3\,\sqrt {a+b\,x}}{8}-\frac {35\,a^3\,b^3\,{\left (a+b\,x\right )}^{3/2}}{3}+\frac {55\,a^2\,b^3\,{\left (a+b\,x\right )}^{5/2}}{8}}{3\,a\,{\left (a+b\,x\right )}^2-3\,a^2\,\left (a+b\,x\right )-{\left (a+b\,x\right )}^3+a^3}+8\,a\,b^3\,\sqrt {a+b\,x}+\frac {a^{3/2}\,b^3\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,105{}\mathrm {i}}{8} \]