Integrand size = 13, antiderivative size = 107 \[ \int \frac {(a+b x)^{5/2}}{x^5} \, dx=-\frac {a^2 \sqrt {a+b x}}{4 x^4}-\frac {17 a b \sqrt {a+b x}}{24 x^3}-\frac {59 b^2 \sqrt {a+b x}}{96 x^2}-\frac {5 b^3 \sqrt {a+b x}}{64 a x}+\frac {5 b^4 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{3/2}} \] Output:
-1/4*a^2*(b*x+a)^(1/2)/x^4-17/24*a*b*(b*x+a)^(1/2)/x^3-59/96*b^2*(b*x+a)^( 1/2)/x^2-5/64*b^3*(b*x+a)^(1/2)/a/x+5/64*b^4*arctanh((b*x+a)^(1/2)/a^(1/2) )/a^(3/2)
Time = 0.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.73 \[ \int \frac {(a+b x)^{5/2}}{x^5} \, dx=-\frac {\sqrt {a+b x} \left (48 a^3+136 a^2 b x+118 a b^2 x^2+15 b^3 x^3\right )}{192 a x^4}+\frac {5 b^4 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{3/2}} \] Input:
Integrate[(a + b*x)^(5/2)/x^5,x]
Output:
-1/192*(Sqrt[a + b*x]*(48*a^3 + 136*a^2*b*x + 118*a*b^2*x^2 + 15*b^3*x^3)) /(a*x^4) + (5*b^4*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(64*a^(3/2))
Time = 0.17 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {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.
\(\displaystyle \int \frac {(a+b x)^{5/2}}{x^5} \, dx\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {5}{8} b \int \frac {(a+b x)^{3/2}}{x^4}dx-\frac {(a+b x)^{5/2}}{4 x^4}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {5}{8} b \left (\frac {1}{2} b \int \frac {\sqrt {a+b x}}{x^3}dx-\frac {(a+b x)^{3/2}}{3 x^3}\right )-\frac {(a+b x)^{5/2}}{4 x^4}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {5}{8} b \left (\frac {1}{2} b \left (\frac {1}{4} b \int \frac {1}{x^2 \sqrt {a+b x}}dx-\frac {\sqrt {a+b x}}{2 x^2}\right )-\frac {(a+b x)^{3/2}}{3 x^3}\right )-\frac {(a+b x)^{5/2}}{4 x^4}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {5}{8} b \left (\frac {1}{2} b \left (\frac {1}{4} b \left (-\frac {b \int \frac {1}{x \sqrt {a+b x}}dx}{2 a}-\frac {\sqrt {a+b x}}{a x}\right )-\frac {\sqrt {a+b x}}{2 x^2}\right )-\frac {(a+b x)^{3/2}}{3 x^3}\right )-\frac {(a+b x)^{5/2}}{4 x^4}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {5}{8} b \left (\frac {1}{2} b \left (\frac {1}{4} b \left (-\frac {\int \frac {1}{\frac {a+b x}{b}-\frac {a}{b}}d\sqrt {a+b x}}{a}-\frac {\sqrt {a+b x}}{a x}\right )-\frac {\sqrt {a+b x}}{2 x^2}\right )-\frac {(a+b x)^{3/2}}{3 x^3}\right )-\frac {(a+b x)^{5/2}}{4 x^4}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {5}{8} b \left (\frac {1}{2} b \left (\frac {1}{4} b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b x}}{a x}\right )-\frac {\sqrt {a+b x}}{2 x^2}\right )-\frac {(a+b x)^{3/2}}{3 x^3}\right )-\frac {(a+b x)^{5/2}}{4 x^4}\) |
Input:
Int[(a + b*x)^(5/2)/x^5,x]
Output:
-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
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_)^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 = 0.17 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.63
method | result | size |
risch | \(-\frac {\sqrt {b x +a}\, \left (15 b^{3} x^{3}+118 a \,b^{2} x^{2}+136 a^{2} b x +48 a^{3}\right )}{192 x^{4} a}+\frac {5 b^{4} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{64 a^{\frac {3}{2}}}\) | \(67\) |
pseudoelliptic | \(-\frac {5 \left (-\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{4} x^{4}+\sqrt {b x +a}\, \left (\sqrt {a}\, b^{3} x^{3}+\frac {118 a^{\frac {3}{2}} b^{2} x^{2}}{15}+\frac {136 a^{\frac {5}{2}} b x}{15}+\frac {16 a^{\frac {7}{2}}}{5}\right )\right )}{64 a^{\frac {3}{2}} x^{4}}\) | \(72\) |
derivativedivides | \(2 b^{4} \left (-\frac {\frac {5 \left (b x +a \right )^{\frac {7}{2}}}{128 a}+\frac {73 \left (b x +a \right )^{\frac {5}{2}}}{384}-\frac {55 a \left (b x +a \right )^{\frac {3}{2}}}{384}+\frac {5 a^{2} \sqrt {b x +a}}{128}}{b^{4} x^{4}}+\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{128 a^{\frac {3}{2}}}\right )\) | \(76\) |
default | \(2 b^{4} \left (-\frac {\frac {5 \left (b x +a \right )^{\frac {7}{2}}}{128 a}+\frac {73 \left (b x +a \right )^{\frac {5}{2}}}{384}-\frac {55 a \left (b x +a \right )^{\frac {3}{2}}}{384}+\frac {5 a^{2} \sqrt {b x +a}}{128}}{b^{4} x^{4}}+\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{128 a^{\frac {3}{2}}}\right )\) | \(76\) |
Input:
int((b*x+a)^(5/2)/x^5,x,method=_RETURNVERBOSE)
Output:
-1/192*(b*x+a)^(1/2)*(15*b^3*x^3+118*a*b^2*x^2+136*a^2*b*x+48*a^3)/x^4/a+5 /64*b^4*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(3/2)
Time = 0.08 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.53 \[ \int \frac {(a+b x)^{5/2}}{x^5} \, dx=\left [\frac {15 \, \sqrt {a} b^{4} x^{4} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (15 \, a b^{3} x^{3} + 118 \, a^{2} b^{2} x^{2} + 136 \, a^{3} b x + 48 \, a^{4}\right )} \sqrt {b x + a}}{384 \, a^{2} x^{4}}, -\frac {15 \, \sqrt {-a} b^{4} x^{4} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x + a}}\right ) + {\left (15 \, a b^{3} x^{3} + 118 \, a^{2} b^{2} x^{2} + 136 \, a^{3} b x + 48 \, a^{4}\right )} \sqrt {b x + a}}{192 \, a^{2} x^{4}}\right ] \] Input:
integrate((b*x+a)^(5/2)/x^5,x, algorithm="fricas")
Output:
[1/384*(15*sqrt(a)*b^4*x^4*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(15*a*b^3*x^3 + 118*a^2*b^2*x^2 + 136*a^3*b*x + 48*a^4)*sqrt(b*x + a))/( a^2*x^4), -1/192*(15*sqrt(-a)*b^4*x^4*arctan(sqrt(-a)/sqrt(b*x + a)) + (15 *a*b^3*x^3 + 118*a^2*b^2*x^2 + 136*a^3*b*x + 48*a^4)*sqrt(b*x + a))/(a^2*x ^4)]
Time = 6.28 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.45 \[ \int \frac {(a+b x)^{5/2}}{x^5} \, dx=- \frac {a^{3}}{4 \sqrt {b} x^{\frac {9}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {23 a^{2} \sqrt {b}}{24 x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {127 a b^{\frac {3}{2}}}{96 x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {133 b^{\frac {5}{2}}}{192 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {5 b^{\frac {7}{2}}}{64 a \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {5 b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{64 a^{\frac {3}{2}}} \] Input:
integrate((b*x+a)**(5/2)/x**5,x)
Output:
-a**3/(4*sqrt(b)*x**(9/2)*sqrt(a/(b*x) + 1)) - 23*a**2*sqrt(b)/(24*x**(7/2 )*sqrt(a/(b*x) + 1)) - 127*a*b**(3/2)/(96*x**(5/2)*sqrt(a/(b*x) + 1)) - 13 3*b**(5/2)/(192*x**(3/2)*sqrt(a/(b*x) + 1)) - 5*b**(7/2)/(64*a*sqrt(x)*sqr t(a/(b*x) + 1)) + 5*b**4*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(64*a**(3/2))
Time = 0.10 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.35 \[ \int \frac {(a+b x)^{5/2}}{x^5} \, dx=-\frac {5 \, b^{4} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{128 \, a^{\frac {3}{2}}} - \frac {15 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{4} + 73 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{4} - 55 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{4} + 15 \, \sqrt {b x + a} a^{3} b^{4}}{192 \, {\left ({\left (b x + a\right )}^{4} a - 4 \, {\left (b x + a\right )}^{3} a^{2} + 6 \, {\left (b x + a\right )}^{2} a^{3} - 4 \, {\left (b x + a\right )} a^{4} + a^{5}\right )}} \] Input:
integrate((b*x+a)^(5/2)/x^5,x, algorithm="maxima")
Output:
-5/128*b^4*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/a^(3/2 ) - 1/192*(15*(b*x + a)^(7/2)*b^4 + 73*(b*x + a)^(5/2)*a*b^4 - 55*(b*x + a )^(3/2)*a^2*b^4 + 15*sqrt(b*x + a)*a^3*b^4)/((b*x + a)^4*a - 4*(b*x + a)^3 *a^2 + 6*(b*x + a)^2*a^3 - 4*(b*x + a)*a^4 + a^5)
Time = 0.13 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.93 \[ \int \frac {(a+b x)^{5/2}}{x^5} \, dx=-\frac {\frac {15 \, b^{5} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {15 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{5} + 73 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{5} - 55 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{5} + 15 \, \sqrt {b x + a} a^{3} b^{5}}{a b^{4} x^{4}}}{192 \, b} \] Input:
integrate((b*x+a)^(5/2)/x^5,x, algorithm="giac")
Output:
-1/192*(15*b^5*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a) + (15*(b*x + a) ^(7/2)*b^5 + 73*(b*x + a)^(5/2)*a*b^5 - 55*(b*x + a)^(3/2)*a^2*b^5 + 15*sq rt(b*x + a)*a^3*b^5)/(a*b^4*x^4))/b
Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.74 \[ \int \frac {(a+b x)^{5/2}}{x^5} \, dx=\frac {55\,a\,{\left (a+b\,x\right )}^{3/2}}{192\,x^4}-\frac {5\,a^2\,\sqrt {a+b\,x}}{64\,x^4}-\frac {5\,{\left (a+b\,x\right )}^{7/2}}{64\,a\,x^4}-\frac {73\,{\left (a+b\,x\right )}^{5/2}}{192\,x^4}-\frac {b^4\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{64\,a^{3/2}} \] Input:
int((a + b*x)^(5/2)/x^5,x)
Output:
(55*a*(a + b*x)^(3/2))/(192*x^4) - (5*a^2*(a + b*x)^(1/2))/(64*x^4) - (5*( a + b*x)^(7/2))/(64*a*x^4) - (b^4*atan(((a + b*x)^(1/2)*1i)/a^(1/2))*5i)/( 64*a^(3/2)) - (73*(a + b*x)^(5/2))/(192*x^4)
Time = 0.16 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^{5/2}}{x^5} \, dx=\frac {-96 \sqrt {b x +a}\, a^{4}-272 \sqrt {b x +a}\, a^{3} b x -236 \sqrt {b x +a}\, a^{2} b^{2} x^{2}-30 \sqrt {b x +a}\, a \,b^{3} x^{3}-15 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{4} x^{4}+15 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{4} x^{4}}{384 a^{2} x^{4}} \] Input:
int((b*x+a)^(5/2)/x^5,x)
Output:
( - 96*sqrt(a + b*x)*a**4 - 272*sqrt(a + b*x)*a**3*b*x - 236*sqrt(a + b*x) *a**2*b**2*x**2 - 30*sqrt(a + b*x)*a*b**3*x**3 - 15*sqrt(a)*log(sqrt(a + b *x) - sqrt(a))*b**4*x**4 + 15*sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*b**4*x* *4)/(384*a**2*x**4)