Integrand size = 13, antiderivative size = 85 \[ \int \frac {(a+b x)^{3/2}}{x^4} \, dx=-\frac {a \sqrt {a+b x}}{3 x^3}-\frac {7 b \sqrt {a+b x}}{12 x^2}-\frac {b^2 \sqrt {a+b x}}{8 a x}+\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{3/2}} \] Output:
-1/3*a*(b*x+a)^(1/2)/x^3-7/12*b*(b*x+a)^(1/2)/x^2-1/8*b^2*(b*x+a)^(1/2)/a/ x+1/8*b^3*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(3/2)
Time = 0.11 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.79 \[ \int \frac {(a+b x)^{3/2}}{x^4} \, dx=-\frac {\sqrt {a+b x} \left (8 a^2+14 a b x+3 b^2 x^2\right )}{24 a x^3}+\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{3/2}} \] Input:
Integrate[(a + b*x)^(3/2)/x^4,x]
Output:
-1/24*(Sqrt[a + b*x]*(8*a^2 + 14*a*b*x + 3*b^2*x^2))/(a*x^3) + (b^3*ArcTan h[Sqrt[a + b*x]/Sqrt[a]])/(8*a^(3/2))
Time = 0.16 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {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.
\(\displaystyle \int \frac {(a+b x)^{3/2}}{x^4} \, dx\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} b \int \frac {\sqrt {a+b x}}{x^3}dx-\frac {(a+b x)^{3/2}}{3 x^3}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} b \left (\frac {1}{4} b \int \frac {1}{x^2 \sqrt {a+b x}}dx-\frac {\sqrt {a+b x}}{2 x^2}\right )-\frac {(a+b x)^{3/2}}{3 x^3}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{2} b \left (\frac {1}{4} b \left (-\frac {b \int \frac {1}{x \sqrt {a+b x}}dx}{2 a}-\frac {\sqrt {a+b x}}{a x}\right )-\frac {\sqrt {a+b x}}{2 x^2}\right )-\frac {(a+b x)^{3/2}}{3 x^3}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} b \left (\frac {1}{4} b \left (-\frac {\int \frac {1}{\frac {a+b x}{b}-\frac {a}{b}}d\sqrt {a+b x}}{a}-\frac {\sqrt {a+b x}}{a x}\right )-\frac {\sqrt {a+b x}}{2 x^2}\right )-\frac {(a+b x)^{3/2}}{3 x^3}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} b \left (\frac {1}{4} b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b x}}{a x}\right )-\frac {\sqrt {a+b x}}{2 x^2}\right )-\frac {(a+b x)^{3/2}}{3 x^3}\) |
Input:
Int[(a + b*x)^(3/2)/x^4,x]
Output:
-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
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 = 0.15 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.66
method | result | size |
risch | \(-\frac {\sqrt {b x +a}\, \left (3 b^{2} x^{2}+14 a b x +8 a^{2}\right )}{24 x^{3} a}+\frac {b^{3} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\) | \(56\) |
pseudoelliptic | \(\frac {\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{3} x^{3}-\left (\sqrt {a}\, b^{2} x^{2}+\frac {14 a^{\frac {3}{2}} b x}{3}+\frac {8 a^{\frac {5}{2}}}{3}\right ) \sqrt {b x +a}}{8 a^{\frac {3}{2}} x^{3}}\) | \(61\) |
derivativedivides | \(2 b^{3} \left (-\frac {\frac {\left (b x +a \right )^{\frac {5}{2}}}{16 a}+\frac {\left (b x +a \right )^{\frac {3}{2}}}{6}-\frac {a \sqrt {b x +a}}{16}}{b^{3} x^{3}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 a^{\frac {3}{2}}}\right )\) | \(64\) |
default | \(2 b^{3} \left (-\frac {\frac {\left (b x +a \right )^{\frac {5}{2}}}{16 a}+\frac {\left (b x +a \right )^{\frac {3}{2}}}{6}-\frac {a \sqrt {b x +a}}{16}}{b^{3} x^{3}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 a^{\frac {3}{2}}}\right )\) | \(64\) |
Input:
int((b*x+a)^(3/2)/x^4,x,method=_RETURNVERBOSE)
Output:
-1/24*(b*x+a)^(1/2)*(3*b^2*x^2+14*a*b*x+8*a^2)/x^3/a+1/8*b^3*arctanh((b*x+ a)^(1/2)/a^(1/2))/a^(3/2)
Time = 0.08 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.67 \[ \int \frac {(a+b x)^{3/2}}{x^4} \, dx=\left [\frac {3 \, \sqrt {a} b^{3} x^{3} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (3 \, a b^{2} x^{2} + 14 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{48 \, a^{2} x^{3}}, -\frac {3 \, \sqrt {-a} b^{3} x^{3} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x + a}}\right ) + {\left (3 \, a b^{2} x^{2} + 14 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{24 \, a^{2} x^{3}}\right ] \] Input:
integrate((b*x+a)^(3/2)/x^4,x, algorithm="fricas")
Output:
[1/48*(3*sqrt(a)*b^3*x^3*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2* (3*a*b^2*x^2 + 14*a^2*b*x + 8*a^3)*sqrt(b*x + a))/(a^2*x^3), -1/24*(3*sqrt (-a)*b^3*x^3*arctan(sqrt(-a)/sqrt(b*x + a)) + (3*a*b^2*x^2 + 14*a^2*b*x + 8*a^3)*sqrt(b*x + a))/(a^2*x^3)]
Time = 3.06 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.46 \[ \int \frac {(a+b x)^{3/2}}{x^4} \, dx=- \frac {a^{2}}{3 \sqrt {b} x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {11 a \sqrt {b}}{12 x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {17 b^{\frac {3}{2}}}{24 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {b^{\frac {5}{2}}}{8 a \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{8 a^{\frac {3}{2}}} \] Input:
integrate((b*x+a)**(3/2)/x**4,x)
Output:
-a**2/(3*sqrt(b)*x**(7/2)*sqrt(a/(b*x) + 1)) - 11*a*sqrt(b)/(12*x**(5/2)*s qrt(a/(b*x) + 1)) - 17*b**(3/2)/(24*x**(3/2)*sqrt(a/(b*x) + 1)) - b**(5/2) /(8*a*sqrt(x)*sqrt(a/(b*x) + 1)) + b**3*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/( 8*a**(3/2))
Time = 0.11 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.40 \[ \int \frac {(a+b x)^{3/2}}{x^4} \, dx=-\frac {b^{3} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{16 \, a^{\frac {3}{2}}} - \frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{3} + 8 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{3} - 3 \, \sqrt {b x + a} a^{2} b^{3}}{24 \, {\left ({\left (b x + a\right )}^{3} a - 3 \, {\left (b x + a\right )}^{2} a^{2} + 3 \, {\left (b x + a\right )} a^{3} - a^{4}\right )}} \] Input:
integrate((b*x+a)^(3/2)/x^4,x, algorithm="maxima")
Output:
-1/16*b^3*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/a^(3/2) - 1/24*(3*(b*x + a)^(5/2)*b^3 + 8*(b*x + a)^(3/2)*a*b^3 - 3*sqrt(b*x + a) *a^2*b^3)/((b*x + a)^3*a - 3*(b*x + a)^2*a^2 + 3*(b*x + a)*a^3 - a^4)
Time = 0.12 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.85 \[ \int \frac {(a+b x)^{3/2}}{x^4} \, dx=-\frac {1}{24} \, b^{3} {\left (\frac {3 \, \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} + 8 \, {\left (b x + a\right )}^{\frac {3}{2}} a - 3 \, \sqrt {b x + a} a^{2}}{a b^{3} x^{3}}\right )} \] Input:
integrate((b*x+a)^(3/2)/x^4,x, algorithm="giac")
Output:
-1/24*b^3*(3*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a) + (3*(b*x + a)^(5 /2) + 8*(b*x + a)^(3/2)*a - 3*sqrt(b*x + a)*a^2)/(a*b^3*x^3))
Time = 0.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.75 \[ \int \frac {(a+b x)^{3/2}}{x^4} \, dx=\frac {a\,\sqrt {a+b\,x}}{8\,x^3}-\frac {{\left (a+b\,x\right )}^{5/2}}{8\,a\,x^3}-\frac {{\left (a+b\,x\right )}^{3/2}}{3\,x^3}-\frac {b^3\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i}}{8\,a^{3/2}} \] Input:
int((a + b*x)^(3/2)/x^4,x)
Output:
(a*(a + b*x)^(1/2))/(8*x^3) - (a + b*x)^(5/2)/(8*a*x^3) - (b^3*atan(((a + b*x)^(1/2)*1i)/a^(1/2))*1i)/(8*a^(3/2)) - (a + b*x)^(3/2)/(3*x^3)
Time = 0.16 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.06 \[ \int \frac {(a+b x)^{3/2}}{x^4} \, dx=\frac {-16 \sqrt {b x +a}\, a^{3}-28 \sqrt {b x +a}\, a^{2} b x -6 \sqrt {b x +a}\, a \,b^{2} x^{2}-3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{3} x^{3}+3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{3} x^{3}}{48 a^{2} x^{3}} \] Input:
int((b*x+a)^(3/2)/x^4,x)
Output:
( - 16*sqrt(a + b*x)*a**3 - 28*sqrt(a + b*x)*a**2*b*x - 6*sqrt(a + b*x)*a* b**2*x**2 - 3*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*b**3*x**3 + 3*sqrt(a)*l og(sqrt(a + b*x) + sqrt(a))*b**3*x**3)/(48*a**2*x**3)