Integrand size = 19, antiderivative size = 81 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^6} \, dx=-\frac {3 b \sqrt {a x^2+b x^3}}{4 x^2}-\frac {\left (a x^2+b x^3\right )^{3/2}}{2 x^5}-\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{4 \sqrt {a}} \] Output:
-3/4*b*(b*x^3+a*x^2)^(1/2)/x^2-1/2*(b*x^3+a*x^2)^(3/2)/x^5-3/4*b^2*arctanh (a^(1/2)*x/(b*x^3+a*x^2)^(1/2))/a^(1/2)
Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^6} \, dx=-\frac {\sqrt {x^2 (a+b x)} \left (\sqrt {a} \sqrt {a+b x} (2 a+5 b x)+3 b^2 x^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{4 \sqrt {a} x^3 \sqrt {a+b x}} \] Input:
Integrate[(a*x^2 + b*x^3)^(3/2)/x^6,x]
Output:
-1/4*(Sqrt[x^2*(a + b*x)]*(Sqrt[a]*Sqrt[a + b*x]*(2*a + 5*b*x) + 3*b^2*x^2 *ArcTanh[Sqrt[a + b*x]/Sqrt[a]]))/(Sqrt[a]*x^3*Sqrt[a + b*x])
Time = 0.38 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1926, 1926, 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.
\(\displaystyle \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^6} \, dx\) |
\(\Big \downarrow \) 1926 |
\(\displaystyle \frac {3}{4} b \int \frac {\sqrt {b x^3+a x^2}}{x^3}dx-\frac {\left (a x^2+b x^3\right )^{3/2}}{2 x^5}\) |
\(\Big \downarrow \) 1926 |
\(\displaystyle \frac {3}{4} b \left (\frac {1}{2} b \int \frac {1}{\sqrt {b x^3+a x^2}}dx-\frac {\sqrt {a x^2+b x^3}}{x^2}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{2 x^5}\) |
\(\Big \downarrow \) 1914 |
\(\displaystyle \frac {3}{4} b \left (-b \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}-\frac {\sqrt {a x^2+b x^3}}{x^2}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{2 x^5}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {3}{4} b \left (-\frac {b \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{\sqrt {a}}-\frac {\sqrt {a x^2+b x^3}}{x^2}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{2 x^5}\) |
Input:
Int[(a*x^2 + b*x^3)^(3/2)/x^6,x]
Output:
-1/2*(a*x^2 + b*x^3)^(3/2)/x^5 + (3*b*(-(Sqrt[a*x^2 + b*x^3]/x^2) - (b*Arc Tanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/Sqrt[a]))/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]
Time = 0.55 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.83
method | result | size |
risch | \(-\frac {\left (5 b x +2 a \right ) \sqrt {x^{2} \left (b x +a \right )}}{4 x^{3}}-\frac {3 b^{2} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {x^{2} \left (b x +a \right )}}{4 \sqrt {a}\, x \sqrt {b x +a}}\) | \(67\) |
default | \(\frac {\left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (-3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{2} x^{2}+3 a^{\frac {3}{2}} \sqrt {b x +a}-5 \left (b x +a \right )^{\frac {3}{2}} \sqrt {a}\right )}{4 x^{5} \left (b x +a \right )^{\frac {3}{2}} \sqrt {a}}\) | \(74\) |
pseudoelliptic | \(-\frac {-\frac {15 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{5} x^{5}}{128}+\sqrt {b x +a}\, \left (\frac {15 \sqrt {a}\, b^{4} x^{4}}{128}-\frac {5 a^{\frac {3}{2}} b^{3} x^{3}}{64}+\frac {a^{\frac {5}{2}} b^{2} x^{2}}{16}+\frac {11 a^{\frac {7}{2}} b x}{8}+a^{\frac {9}{2}}\right )}{5 a^{\frac {7}{2}} x^{5}}\) | \(82\) |
Input:
int((b*x^3+a*x^2)^(3/2)/x^6,x,method=_RETURNVERBOSE)
Output:
-1/4*(5*b*x+2*a)/x^3*(x^2*(b*x+a))^(1/2)-3/4*b^2/a^(1/2)*arctanh((b*x+a)^( 1/2)/a^(1/2))*(x^2*(b*x+a))^(1/2)/x/(b*x+a)^(1/2)
Time = 0.09 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.96 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^6} \, dx=\left [\frac {3 \, \sqrt {a} b^{2} x^{3} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) - 2 \, \sqrt {b x^{3} + a x^{2}} {\left (5 \, a b x + 2 \, a^{2}\right )}}{8 \, a x^{3}}, \frac {3 \, \sqrt {-a} b^{2} x^{3} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) - \sqrt {b x^{3} + a x^{2}} {\left (5 \, a b x + 2 \, a^{2}\right )}}{4 \, a x^{3}}\right ] \] Input:
integrate((b*x^3+a*x^2)^(3/2)/x^6,x, algorithm="fricas")
Output:
[1/8*(3*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)*(5*a*b*x + 2*a^2))/(a*x^3), 1/4*(3*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)*(5*a*b*x + 2*a^2))/(a*x^3)]
\[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^6} \, dx=\int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}{x^{6}}\, dx \] Input:
integrate((b*x**3+a*x**2)**(3/2)/x**6,x)
Output:
Integral((x**2*(a + b*x))**(3/2)/x**6, x)
\[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^6} \, dx=\int { \frac {{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}}}{x^{6}} \,d x } \] Input:
integrate((b*x^3+a*x^2)^(3/2)/x^6,x, algorithm="maxima")
Output:
integrate((b*x^3 + a*x^2)^(3/2)/x^6, x)
Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^6} \, dx=\frac {\frac {3 \, b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-a}} - \frac {5 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{3} \mathrm {sgn}\left (x\right ) - 3 \, \sqrt {b x + a} a b^{3} \mathrm {sgn}\left (x\right )}{b^{2} x^{2}}}{4 \, b} \] Input:
integrate((b*x^3+a*x^2)^(3/2)/x^6,x, algorithm="giac")
Output:
1/4*(3*b^3*arctan(sqrt(b*x + a)/sqrt(-a))*sgn(x)/sqrt(-a) - (5*(b*x + a)^( 3/2)*b^3*sgn(x) - 3*sqrt(b*x + a)*a*b^3*sgn(x))/(b^2*x^2))/b
Timed out. \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^6} \, dx=\int \frac {{\left (b\,x^3+a\,x^2\right )}^{3/2}}{x^6} \,d x \] Input:
int((a*x^2 + b*x^3)^(3/2)/x^6,x)
Output:
int((a*x^2 + b*x^3)^(3/2)/x^6, x)
Time = 0.21 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^6} \, dx=\frac {-4 \sqrt {b x +a}\, a^{2}-10 \sqrt {b x +a}\, a b x +3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{2} x^{2}-3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{2} x^{2}}{8 a \,x^{2}} \] Input:
int((b*x^3+a*x^2)^(3/2)/x^6,x)
Output:
( - 4*sqrt(a + b*x)*a**2 - 10*sqrt(a + b*x)*a*b*x + 3*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*b**2*x**2 - 3*sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*b**2*x **2)/(8*a*x**2)