Integrand size = 19, antiderivative size = 112 \[ \int \frac {\sqrt {a x^2+b x^3}}{x^5} \, dx=-\frac {\sqrt {a x^2+b x^3}}{3 x^4}-\frac {b \sqrt {a x^2+b x^3}}{12 a x^3}+\frac {b^2 \sqrt {a x^2+b x^3}}{8 a^2 x^2}-\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{8 a^{5/2}} \] Output:
-1/3*(b*x^3+a*x^2)^(1/2)/x^4-1/12*b*(b*x^3+a*x^2)^(1/2)/a/x^3+1/8*b^2*(b*x ^3+a*x^2)^(1/2)/a^2/x^2-1/8*b^3*arctanh(a^(1/2)*x/(b*x^3+a*x^2)^(1/2))/a^( 5/2)
Time = 0.03 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt {a x^2+b x^3}}{x^5} \, dx=-\frac {\sqrt {x^2 (a+b x)} \left (\sqrt {a} \sqrt {a+b x} \left (8 a^2+2 a b x-3 b^2 x^2\right )+3 b^3 x^3 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{24 a^{5/2} x^4 \sqrt {a+b x}} \] Input:
Integrate[Sqrt[a*x^2 + b*x^3]/x^5,x]
Output:
-1/24*(Sqrt[x^2*(a + b*x)]*(Sqrt[a]*Sqrt[a + b*x]*(8*a^2 + 2*a*b*x - 3*b^2 *x^2) + 3*b^3*x^3*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]))/(a^(5/2)*x^4*Sqrt[a + b *x])
Time = 0.45 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1926, 1931, 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.
\(\displaystyle \int \frac {\sqrt {a x^2+b x^3}}{x^5} \, dx\) |
\(\Big \downarrow \) 1926 |
\(\displaystyle \frac {1}{6} b \int \frac {1}{x^2 \sqrt {b x^3+a x^2}}dx-\frac {\sqrt {a x^2+b x^3}}{3 x^4}\) |
\(\Big \downarrow \) 1931 |
\(\displaystyle \frac {1}{6} b \left (-\frac {3 b \int \frac {1}{x \sqrt {b x^3+a x^2}}dx}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )-\frac {\sqrt {a x^2+b x^3}}{3 x^4}\) |
\(\Big \downarrow \) 1931 |
\(\displaystyle \frac {1}{6} b \left (-\frac {3 b \left (-\frac {b \int \frac {1}{\sqrt {b x^3+a x^2}}dx}{2 a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )-\frac {\sqrt {a x^2+b x^3}}{3 x^4}\) |
\(\Big \downarrow \) 1914 |
\(\displaystyle \frac {1}{6} b \left (-\frac {3 b \left (\frac {b \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}}{a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )-\frac {\sqrt {a x^2+b x^3}}{3 x^4}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{6} b \left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{3/2}}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )-\frac {\sqrt {a x^2+b x^3}}{3 x^4}\) |
Input:
Int[Sqrt[a*x^2 + b*x^3]/x^5,x]
Output:
-1/3*Sqrt[a*x^2 + b*x^3]/x^4 + (b*(-1/2*Sqrt[a*x^2 + b*x^3]/(a*x^3) - (3*b *(-(Sqrt[a*x^2 + b*x^3]/(a*x^2)) + (b*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x ^3]])/a^(3/2)))/(4*a)))/6
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 = 0.50 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.64
method | result | size |
pseudoelliptic | \(-\frac {5 \left (-\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{4} x^{4}+\sqrt {b x +a}\, \left (\sqrt {a}\, b^{3} x^{3}-\frac {2 a^{\frac {3}{2}} b^{2} x^{2}}{3}+\frac {8 a^{\frac {5}{2}} b x}{15}+\frac {16 a^{\frac {7}{2}}}{5}\right )\right )}{64 a^{\frac {7}{2}} x^{4}}\) | \(72\) |
risch | \(-\frac {\left (-3 b^{2} x^{2}+2 a b x +8 a^{2}\right ) \sqrt {x^{2} \left (b x +a \right )}}{24 x^{4} a^{2}}-\frac {b^{3} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {x^{2} \left (b x +a \right )}}{8 a^{\frac {5}{2}} x \sqrt {b x +a}}\) | \(81\) |
default | \(-\frac {\sqrt {b \,x^{3}+a \,x^{2}}\, \left (3 a^{\frac {9}{2}} \sqrt {b x +a}+8 a^{\frac {7}{2}} \left (b x +a \right )^{\frac {3}{2}}-3 a^{\frac {5}{2}} \left (b x +a \right )^{\frac {5}{2}}+3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{2} b^{3} x^{3}\right )}{24 x^{4} \sqrt {b x +a}\, a^{\frac {9}{2}}}\) | \(89\) |
Input:
int((b*x^3+a*x^2)^(1/2)/x^5,x,method=_RETURNVERBOSE)
Output:
-5/64*(-arctanh((b*x+a)^(1/2)/a^(1/2))*b^4*x^4+(b*x+a)^(1/2)*(a^(1/2)*b^3* x^3-2/3*a^(3/2)*b^2*x^2+8/15*a^(5/2)*b*x+16/5*a^(7/2)))/a^(7/2)/x^4
Time = 0.08 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.61 \[ \int \frac {\sqrt {a x^2+b x^3}}{x^5} \, dx=\left [\frac {3 \, \sqrt {a} b^{3} x^{4} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (3 \, a b^{2} x^{2} - 2 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{48 \, a^{3} x^{4}}, \frac {3 \, \sqrt {-a} b^{3} x^{4} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + {\left (3 \, a b^{2} x^{2} - 2 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{24 \, a^{3} x^{4}}\right ] \] Input:
integrate((b*x^3+a*x^2)^(1/2)/x^5,x, algorithm="fricas")
Output:
[1/48*(3*sqrt(a)*b^3*x^4*log((b*x^2 + 2*a*x - 2*sqrt(b*x^3 + a*x^2)*sqrt(a ))/x^2) + 2*(3*a*b^2*x^2 - 2*a^2*b*x - 8*a^3)*sqrt(b*x^3 + a*x^2))/(a^3*x^ 4), 1/24*(3*sqrt(-a)*b^3*x^4*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(b*x^2 + a*x)) + (3*a*b^2*x^2 - 2*a^2*b*x - 8*a^3)*sqrt(b*x^3 + a*x^2))/(a^3*x^4)]
\[ \int \frac {\sqrt {a x^2+b x^3}}{x^5} \, dx=\int \frac {\sqrt {x^{2} \left (a + b x\right )}}{x^{5}}\, dx \] Input:
integrate((b*x**3+a*x**2)**(1/2)/x**5,x)
Output:
Integral(sqrt(x**2*(a + b*x))/x**5, x)
\[ \int \frac {\sqrt {a x^2+b x^3}}{x^5} \, dx=\int { \frac {\sqrt {b x^{3} + a x^{2}}}{x^{5}} \,d x } \] Input:
integrate((b*x^3+a*x^2)^(1/2)/x^5,x, algorithm="maxima")
Output:
integrate(sqrt(b*x^3 + a*x^2)/x^5, x)
Time = 0.24 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.71 \[ \int \frac {\sqrt {a x^2+b x^3}}{x^5} \, dx=\frac {1}{24} \, b^{3} {\left (\frac {3 \, \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-a} a^{2}} + \frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) - 8 \, {\left (b x + a\right )}^{\frac {3}{2}} a \mathrm {sgn}\left (x\right ) - 3 \, \sqrt {b x + a} a^{2} \mathrm {sgn}\left (x\right )}{a^{2} b^{3} x^{3}}\right )} \] Input:
integrate((b*x^3+a*x^2)^(1/2)/x^5,x, algorithm="giac")
Output:
1/24*b^3*(3*arctan(sqrt(b*x + a)/sqrt(-a))*sgn(x)/(sqrt(-a)*a^2) + (3*(b*x + a)^(5/2)*sgn(x) - 8*(b*x + a)^(3/2)*a*sgn(x) - 3*sqrt(b*x + a)*a^2*sgn( x))/(a^2*b^3*x^3))
Timed out. \[ \int \frac {\sqrt {a x^2+b x^3}}{x^5} \, dx=\int \frac {\sqrt {b\,x^3+a\,x^2}}{x^5} \,d x \] Input:
int((a*x^2 + b*x^3)^(1/2)/x^5,x)
Output:
int((a*x^2 + b*x^3)^(1/2)/x^5, x)
Time = 0.22 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.80 \[ \int \frac {\sqrt {a x^2+b x^3}}{x^5} \, dx=\frac {-16 \sqrt {b x +a}\, a^{3}-4 \sqrt {b x +a}\, a^{2} b x +6 \sqrt {b x +a}\, a \,b^{2} x^{2}+3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{3} x^{3}-3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{3} x^{3}}{48 a^{3} x^{3}} \] Input:
int((b*x^3+a*x^2)^(1/2)/x^5,x)
Output:
( - 16*sqrt(a + b*x)*a**3 - 4*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)*lo g(sqrt(a + b*x) + sqrt(a))*b**3*x**3)/(48*a**3*x**3)