Integrand size = 19, antiderivative size = 109 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^7} \, dx=-\frac {b \sqrt {a x^2+b x^3}}{4 x^3}-\frac {b^2 \sqrt {a x^2+b x^3}}{8 a x^2}-\frac {\left (a x^2+b x^3\right )^{3/2}}{3 x^6}+\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{8 a^{3/2}} \] Output:
-1/4*b*(b*x^3+a*x^2)^(1/2)/x^3-1/8*b^2*(b*x^3+a*x^2)^(1/2)/a/x^2-1/3*(b*x^ 3+a*x^2)^(3/2)/x^6+1/8*b^3*arctanh(a^(1/2)*x/(b*x^3+a*x^2)^(1/2))/a^(3/2)
Time = 0.03 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^7} \, dx=\frac {\sqrt {x^2 (a+b x)} \left (-\sqrt {a} \sqrt {a+b x} \left (8 a^2+14 a b x+3 b^2 x^2\right )+3 b^3 x^3 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{24 a^{3/2} x^4 \sqrt {a+b x}} \] Input:
Integrate[(a*x^2 + b*x^3)^(3/2)/x^7,x]
Output:
(Sqrt[x^2*(a + b*x)]*(-(Sqrt[a]*Sqrt[a + b*x]*(8*a^2 + 14*a*b*x + 3*b^2*x^ 2)) + 3*b^3*x^3*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]))/(24*a^(3/2)*x^4*Sqrt[a + b*x])
Time = 0.45 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1926, 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.
\(\displaystyle \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^7} \, dx\) |
\(\Big \downarrow \) 1926 |
\(\displaystyle \frac {1}{2} b \int \frac {\sqrt {b x^3+a x^2}}{x^4}dx-\frac {\left (a x^2+b x^3\right )^{3/2}}{3 x^6}\) |
\(\Big \downarrow \) 1926 |
\(\displaystyle \frac {1}{2} b \left (\frac {1}{4} b \int \frac {1}{x \sqrt {b x^3+a x^2}}dx-\frac {\sqrt {a x^2+b x^3}}{2 x^3}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{3 x^6}\) |
\(\Big \downarrow \) 1931 |
\(\displaystyle \frac {1}{2} b \left (\frac {1}{4} b \left (-\frac {b \int \frac {1}{\sqrt {b x^3+a x^2}}dx}{2 a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )-\frac {\sqrt {a x^2+b x^3}}{2 x^3}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{3 x^6}\) |
\(\Big \downarrow \) 1914 |
\(\displaystyle \frac {1}{2} b \left (\frac {1}{4} b \left (\frac {b \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}}{a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )-\frac {\sqrt {a x^2+b x^3}}{2 x^3}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{3 x^6}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} b \left (\frac {1}{4} b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{3/2}}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )-\frac {\sqrt {a x^2+b x^3}}{2 x^3}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{3 x^6}\) |
Input:
Int[(a*x^2 + b*x^3)^(3/2)/x^7,x]
Output:
-1/3*(a*x^2 + b*x^3)^(3/2)/x^6 + (b*(-1/2*Sqrt[a*x^2 + b*x^3]/x^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))/2
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 = 0.56 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.74
method | result | size |
risch | \(-\frac {\left (3 b^{2} x^{2}+14 a b x +8 a^{2}\right ) \sqrt {x^{2} \left (b x +a \right )}}{24 x^{4} a}+\frac {b^{3} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {x^{2} \left (b x +a \right )}}{8 a^{\frac {3}{2}} x \sqrt {b x +a}}\) | \(81\) |
default | \(-\frac {\left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (3 \left (b x +a \right )^{\frac {5}{2}} a^{\frac {3}{2}}-3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a \,b^{3} x^{3}+8 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {5}{2}}-3 a^{\frac {7}{2}} \sqrt {b x +a}\right )}{24 x^{6} \left (b x +a \right )^{\frac {3}{2}} a^{\frac {5}{2}}}\) | \(87\) |
pseudoelliptic | \(-\frac {13 \left (\frac {105 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{6} x^{6}}{1664}+\sqrt {b x +a}\, \left (-\frac {105 \sqrt {a}\, b^{5} x^{5}}{1664}+\frac {35 a^{\frac {3}{2}} b^{4} x^{4}}{832}-\frac {7 a^{\frac {5}{2}} b^{3} x^{3}}{208}+\frac {3 a^{\frac {7}{2}} b^{2} x^{2}}{104}+a^{\frac {9}{2}} b x +\frac {10 a^{\frac {11}{2}}}{13}\right )\right )}{60 a^{\frac {9}{2}} x^{6}}\) | \(94\) |
Input:
int((b*x^3+a*x^2)^(3/2)/x^7,x,method=_RETURNVERBOSE)
Output:
-1/24*(3*b^2*x^2+14*a*b*x+8*a^2)/x^4/a*(x^2*(b*x+a))^(1/2)+1/8/a^(3/2)*b^3 *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 = 180, normalized size of antiderivative = 1.65 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^7} \, dx=\left [\frac {3 \, \sqrt {a} b^{3} x^{4} \log \left (\frac {b x^{2} + 2 \, a x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) - 2 \, {\left (3 \, a b^{2} x^{2} + 14 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{48 \, a^{2} x^{4}}, -\frac {3 \, \sqrt {-a} b^{3} x^{4} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + {\left (3 \, a b^{2} x^{2} + 14 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{24 \, a^{2} x^{4}}\right ] \] Input:
integrate((b*x^3+a*x^2)^(3/2)/x^7,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 + 14*a^2*b*x + 8*a^3)*sqrt(b*x^3 + a*x^2))/(a^2*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 + 14*a^2*b*x + 8*a^3)*sqrt(b*x^3 + a*x^2))/(a^2*x^4 )]
\[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^7} \, dx=\int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}{x^{7}}\, dx \] Input:
integrate((b*x**3+a*x**2)**(3/2)/x**7,x)
Output:
Integral((x**2*(a + b*x))**(3/2)/x**7, x)
\[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^7} \, dx=\int { \frac {{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}}}{x^{7}} \,d x } \] Input:
integrate((b*x^3+a*x^2)^(3/2)/x^7,x, algorithm="maxima")
Output:
integrate((b*x^3 + a*x^2)^(3/2)/x^7, x)
Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.73 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^7} \, dx=-\frac {1}{24} \, b^{3} {\left (\frac {3 \, \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-a} a} + \frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) + 8 \, {\left (b x + a\right )}^{\frac {3}{2}} a \mathrm {sgn}\left (x\right ) - 3 \, \sqrt {b x + a} a^{2} \mathrm {sgn}\left (x\right )}{a b^{3} x^{3}}\right )} \] Input:
integrate((b*x^3+a*x^2)^(3/2)/x^7,x, algorithm="giac")
Output:
-1/24*b^3*(3*arctan(sqrt(b*x + a)/sqrt(-a))*sgn(x)/(sqrt(-a)*a) + (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*b^3*x^3))
Timed out. \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^7} \, dx=\int \frac {{\left (b\,x^3+a\,x^2\right )}^{3/2}}{x^7} \,d x \] Input:
int((a*x^2 + b*x^3)^(3/2)/x^7,x)
Output:
int((a*x^2 + b*x^3)^(3/2)/x^7, x)
Time = 0.22 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^7} \, dx=\frac {-16 \sqrt {b x +a}\, a^{3}-28 \sqrt {b x +a}\, a^{2} b x -6 \sqrt {b x +a}\, a \,b^{2} x^{2}-3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{3} x^{3}+3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{3} x^{3}}{48 a^{2} x^{3}} \] Input:
int((b*x^3+a*x^2)^(3/2)/x^7,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)