Integrand size = 19, antiderivative size = 84 \[ \int \frac {\sqrt {a x^2+b x^3}}{x^4} \, dx=-\frac {\sqrt {a x^2+b x^3}}{2 x^3}-\frac {b \sqrt {a x^2+b x^3}}{4 a x^2}+\frac {b^2 \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{4 a^{3/2}} \] Output:
-1/2*(b*x^3+a*x^2)^(1/2)/x^3-1/4*b*(b*x^3+a*x^2)^(1/2)/a/x^2+1/4*b^2*arcta nh(a^(1/2)*x/(b*x^3+a*x^2)^(1/2))/a^(3/2)
Time = 0.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt {a x^2+b x^3}}{x^4} \, dx=\frac {\sqrt {x^2 (a+b x)} \left (-\sqrt {a} \sqrt {a+b x} (2 a+b x)+b^2 x^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{4 a^{3/2} x^3 \sqrt {a+b x}} \] Input:
Integrate[Sqrt[a*x^2 + b*x^3]/x^4,x]
Output:
(Sqrt[x^2*(a + b*x)]*(-(Sqrt[a]*Sqrt[a + b*x]*(2*a + b*x)) + b^2*x^2*ArcTa nh[Sqrt[a + b*x]/Sqrt[a]]))/(4*a^(3/2)*x^3*Sqrt[a + b*x])
Time = 0.38 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1926, 1931, 1914, 219}
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.
\(\displaystyle \int \frac {\sqrt {a x^2+b x^3}}{x^4} \, dx\) |
\(\Big \downarrow \) 1926 |
\(\displaystyle \frac {1}{4} b \int \frac {1}{x \sqrt {b x^3+a x^2}}dx-\frac {\sqrt {a x^2+b x^3}}{2 x^3}\) |
\(\Big \downarrow \) 1931 |
\(\displaystyle \frac {1}{4} b \left (-\frac {b \int \frac {1}{\sqrt {b x^3+a x^2}}dx}{2 a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )-\frac {\sqrt {a x^2+b x^3}}{2 x^3}\) |
\(\Big \downarrow \) 1914 |
\(\displaystyle \frac {1}{4} b \left (\frac {b \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}}{a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )-\frac {\sqrt {a x^2+b x^3}}{2 x^3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{3/2}}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )-\frac {\sqrt {a x^2+b x^3}}{2 x^3}\) |
Input:
Int[Sqrt[a*x^2 + b*x^3]/x^4,x]
Output:
-1/2*Sqrt[a*x^2 + b*x^3]/x^3 + (b*(-(Sqrt[a*x^2 + b*x^3]/(a*x^2)) + (b*Arc Tanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/a^(3/2)))/4
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp[2/(2 - n) Subst[Int[1/(1 - a*x^2), x], x, x/Sqrt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]
Int[((c_.)*(x_))^(m_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a*x^j + b*x^n)^p/(c*(m + j*p + 1))), x] - Simp[b*p *((n - j)/(c^n*(m + j*p + 1))) Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && !IntegerQ[p] && LtQ[0, j, n] && (Integer sQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && LtQ[m + j*p + 1, 0]
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[c^(j - 1)*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Simp[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))) I nt[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[ m + j*p + 1, 0]
Time = 0.48 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.73
method | result | size |
pseudoelliptic | \(-\frac {\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{3} x^{3}-\left (\sqrt {a}\, b^{2} x^{2}-\frac {2 a^{\frac {3}{2}} b x}{3}-\frac {8 a^{\frac {5}{2}}}{3}\right ) \sqrt {b x +a}}{8 a^{\frac {5}{2}} x^{3}}\) | \(61\) |
risch | \(-\frac {\left (b x +2 a \right ) \sqrt {x^{2} \left (b x +a \right )}}{4 x^{3} a}+\frac {b^{2} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {x^{2} \left (b x +a \right )}}{4 a^{\frac {3}{2}} x \sqrt {b x +a}}\) | \(69\) |
default | \(-\frac {\sqrt {b \,x^{3}+a \,x^{2}}\, \left (a^{\frac {3}{2}} \left (b x +a \right )^{\frac {3}{2}}-\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a \,b^{2} x^{2}+\sqrt {b x +a}\, a^{\frac {5}{2}}\right )}{4 x^{3} \sqrt {b x +a}\, a^{\frac {5}{2}}}\) | \(73\) |
Input:
int((b*x^3+a*x^2)^(1/2)/x^4,x,method=_RETURNVERBOSE)
Output:
-1/8/a^(5/2)*(arctanh((b*x+a)^(1/2)/a^(1/2))*b^3*x^3-(a^(1/2)*b^2*x^2-2/3* a^(3/2)*b*x-8/3*a^(5/2))*(b*x+a)^(1/2))/x^3
Time = 0.09 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.83 \[ \int \frac {\sqrt {a x^2+b x^3}}{x^4} \, dx=\left [\frac {\sqrt {a} b^{2} x^{3} \log \left (\frac {b x^{2} + 2 \, a x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) - 2 \, \sqrt {b x^{3} + a x^{2}} {\left (a b x + 2 \, a^{2}\right )}}{8 \, a^{2} x^{3}}, -\frac {\sqrt {-a} b^{2} x^{3} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + \sqrt {b x^{3} + a x^{2}} {\left (a b x + 2 \, a^{2}\right )}}{4 \, a^{2} x^{3}}\right ] \] Input:
integrate((b*x^3+a*x^2)^(1/2)/x^4,x, algorithm="fricas")
Output:
[1/8*(sqrt(a)*b^2*x^3*log((b*x^2 + 2*a*x + 2*sqrt(b*x^3 + a*x^2)*sqrt(a))/ x^2) - 2*sqrt(b*x^3 + a*x^2)*(a*b*x + 2*a^2))/(a^2*x^3), -1/4*(sqrt(-a)*b^ 2*x^3*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(b*x^2 + a*x)) + sqrt(b*x^3 + a* x^2)*(a*b*x + 2*a^2))/(a^2*x^3)]
\[ \int \frac {\sqrt {a x^2+b x^3}}{x^4} \, dx=\int \frac {\sqrt {x^{2} \left (a + b x\right )}}{x^{4}}\, dx \] Input:
integrate((b*x**3+a*x**2)**(1/2)/x**4,x)
Output:
Integral(sqrt(x**2*(a + b*x))/x**4, x)
\[ \int \frac {\sqrt {a x^2+b x^3}}{x^4} \, dx=\int { \frac {\sqrt {b x^{3} + a x^{2}}}{x^{4}} \,d x } \] Input:
integrate((b*x^3+a*x^2)^(1/2)/x^4,x, algorithm="maxima")
Output:
integrate(sqrt(b*x^3 + a*x^2)/x^4, x)
Time = 0.25 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt {a x^2+b x^3}}{x^4} \, dx=-\frac {\frac {b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-a} a} + \frac {{\left (b x + a\right )}^{\frac {3}{2}} b^{3} \mathrm {sgn}\left (x\right ) + \sqrt {b x + a} a b^{3} \mathrm {sgn}\left (x\right )}{a b^{2} x^{2}}}{4 \, b} \] Input:
integrate((b*x^3+a*x^2)^(1/2)/x^4,x, algorithm="giac")
Output:
-1/4*(b^3*arctan(sqrt(b*x + a)/sqrt(-a))*sgn(x)/(sqrt(-a)*a) + ((b*x + a)^ (3/2)*b^3*sgn(x) + sqrt(b*x + a)*a*b^3*sgn(x))/(a*b^2*x^2))/b
Timed out. \[ \int \frac {\sqrt {a x^2+b x^3}}{x^4} \, dx=\int \frac {\sqrt {b\,x^3+a\,x^2}}{x^4} \,d x \] Input:
int((a*x^2 + b*x^3)^(1/2)/x^4,x)
Output:
int((a*x^2 + b*x^3)^(1/2)/x^4, x)
Time = 0.24 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt {a x^2+b x^3}}{x^4} \, dx=\frac {-4 \sqrt {b x +a}\, a^{2}-2 \sqrt {b x +a}\, a b x -\sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{2} x^{2}+\sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{2} x^{2}}{8 a^{2} x^{2}} \] Input:
int((b*x^3+a*x^2)^(1/2)/x^4,x)
Output:
( - 4*sqrt(a + b*x)*a**2 - 2*sqrt(a + b*x)*a*b*x - sqrt(a)*log(sqrt(a + b* x) - sqrt(a))*b**2*x**2 + sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*b**2*x**2)/ (8*a**2*x**2)