Integrand size = 19, antiderivative size = 110 \[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x^8} \, dx=-\frac {a^2 \sqrt {a x^2+b x^3}}{2 x^3}-\frac {9 a b \sqrt {a x^2+b x^3}}{4 x^2}+\frac {2 b^2 \sqrt {a x^2+b x^3}}{x}-\frac {15}{4} \sqrt {a} b^2 \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right ) \] Output:
-1/2*a^2*(b*x^3+a*x^2)^(1/2)/x^3-9/4*a*b*(b*x^3+a*x^2)^(1/2)/x^2+2*b^2*(b* x^3+a*x^2)^(1/2)/x-15/4*a^(1/2)*b^2*arctanh((b*x^3+a*x^2)^(1/2)/a^(1/2)/x)
Time = 0.12 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.80 \[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x^8} \, dx=-\frac {\sqrt {x^2 (a+b x)} \left (\sqrt {a+b x} \left (2 a^2+9 a b x-8 b^2 x^2\right )+15 \sqrt {a} b^2 x^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{4 x^3 \sqrt {a+b x}} \] Input:
Integrate[(a*x^2 + b*x^3)^(5/2)/x^8,x]
Output:
-1/4*(Sqrt[x^2*(a + b*x)]*(Sqrt[a + b*x]*(2*a^2 + 9*a*b*x - 8*b^2*x^2) + 1 5*Sqrt[a]*b^2*x^2*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]))/(x^3*Sqrt[a + b*x])
Time = 0.46 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1926, 1926, 1927, 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.
\(\displaystyle \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x^8} \, dx\) |
\(\Big \downarrow \) 1926 |
\(\displaystyle \frac {5}{4} b \int \frac {\left (b x^3+a x^2\right )^{3/2}}{x^5}dx-\frac {\left (a x^2+b x^3\right )^{5/2}}{2 x^7}\) |
\(\Big \downarrow \) 1926 |
\(\displaystyle \frac {5}{4} b \left (\frac {3}{2} b \int \frac {\sqrt {b x^3+a x^2}}{x^2}dx-\frac {\left (a x^2+b x^3\right )^{3/2}}{x^4}\right )-\frac {\left (a x^2+b x^3\right )^{5/2}}{2 x^7}\) |
\(\Big \downarrow \) 1927 |
\(\displaystyle \frac {5}{4} b \left (\frac {3}{2} b \left (a \int \frac {1}{\sqrt {b x^3+a x^2}}dx+\frac {2 \sqrt {a x^2+b x^3}}{x}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{x^4}\right )-\frac {\left (a x^2+b x^3\right )^{5/2}}{2 x^7}\) |
\(\Big \downarrow \) 1914 |
\(\displaystyle \frac {5}{4} b \left (\frac {3}{2} b \left (\frac {2 \sqrt {a x^2+b x^3}}{x}-2 a \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{x^4}\right )-\frac {\left (a x^2+b x^3\right )^{5/2}}{2 x^7}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {5}{4} b \left (\frac {3}{2} b \left (\frac {2 \sqrt {a x^2+b x^3}}{x}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{x^4}\right )-\frac {\left (a x^2+b x^3\right )^{5/2}}{2 x^7}\) |
Input:
Int[(a*x^2 + b*x^3)^(5/2)/x^8,x]
Output:
-1/2*(a*x^2 + b*x^3)^(5/2)/x^7 + (5*b*(-((a*x^2 + b*x^3)^(3/2)/x^4) + (3*b *((2*Sqrt[a*x^2 + b*x^3])/x - 2*Sqrt[a]*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b *x^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*x)^(m + 1)*((a*x^j + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a* (n - j)*(p/(c^j*(m + n*p + 1))) Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p - 1) , x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[p] && LtQ[0, j, n] && (Int egersQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && NeQ[m + n*p + 1, 0]
Time = 0.58 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.73
method | result | size |
risch | \(-\frac {a \left (9 b x +2 a \right ) \sqrt {x^{2} \left (b x +a \right )}}{4 x^{3}}+\frac {b^{2} \left (16 \sqrt {b x +a}-30 \sqrt {a}\, \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )\right ) \sqrt {x^{2} \left (b x +a \right )}}{8 x \sqrt {b x +a}}\) | \(80\) |
default | \(-\frac {\left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}} \left (-8 b^{2} x^{2} \sqrt {a}\, \sqrt {b x +a}+15 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a \,b^{2} x^{2}+9 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {3}{2}}-7 \sqrt {b x +a}\, a^{\frac {5}{2}}\right )}{4 x^{7} \left (b x +a \right )^{\frac {5}{2}} \sqrt {a}}\) | \(93\) |
pseudoelliptic | \(-\frac {37 \left (\frac {105 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) x^{7} b^{7}}{4736}+\sqrt {b x +a}\, \left (-\frac {105 \sqrt {a}\, b^{6} x^{6}}{4736}+\frac {35 a^{\frac {3}{2}} b^{5} x^{5}}{2368}-\frac {7 a^{\frac {5}{2}} b^{4} x^{4}}{592}+\frac {3 a^{\frac {7}{2}} b^{3} x^{3}}{296}+a^{\frac {9}{2}} b^{2} x^{2}+\frac {58 a^{\frac {11}{2}} b x}{37}+\frac {24 a^{\frac {13}{2}}}{37}\right )\right )}{168 a^{\frac {9}{2}} x^{7}}\) | \(105\) |
Input:
int((b*x^3+a*x^2)^(5/2)/x^8,x,method=_RETURNVERBOSE)
Output:
-1/4*a*(9*b*x+2*a)/x^3*(x^2*(b*x+a))^(1/2)+1/8*b^2*(16*(b*x+a)^(1/2)-30*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 = 0.08 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.53 \[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x^8} \, dx=\left [\frac {15 \, \sqrt {a} b^{2} x^{3} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (8 \, b^{2} x^{2} - 9 \, a b x - 2 \, a^{2}\right )} \sqrt {b x^{3} + a x^{2}}}{8 \, x^{3}}, \frac {15 \, \sqrt {-a} b^{2} x^{3} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + {\left (8 \, b^{2} x^{2} - 9 \, a b x - 2 \, a^{2}\right )} \sqrt {b x^{3} + a x^{2}}}{4 \, x^{3}}\right ] \] Input:
integrate((b*x^3+a*x^2)^(5/2)/x^8,x, algorithm="fricas")
Output:
[1/8*(15*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*(8*b^2*x^2 - 9*a*b*x - 2*a^2)*sqrt(b*x^3 + a*x^2))/x^3, 1/4*(1 5*sqrt(-a)*b^2*x^3*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(b*x^2 + a*x)) + (8 *b^2*x^2 - 9*a*b*x - 2*a^2)*sqrt(b*x^3 + a*x^2))/x^3]
\[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x^8} \, dx=\int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {5}{2}}}{x^{8}}\, dx \] Input:
integrate((b*x**3+a*x**2)**(5/2)/x**8,x)
Output:
Integral((x**2*(a + b*x))**(5/2)/x**8, x)
\[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x^8} \, dx=\int { \frac {{\left (b x^{3} + a x^{2}\right )}^{\frac {5}{2}}}{x^{8}} \,d x } \] Input:
integrate((b*x^3+a*x^2)^(5/2)/x^8,x, algorithm="maxima")
Output:
integrate((b*x^3 + a*x^2)^(5/2)/x^8, x)
Time = 0.13 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.80 \[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x^8} \, dx=\frac {\frac {15 \, a b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-a}} + 8 \, \sqrt {b x + a} b^{3} \mathrm {sgn}\left (x\right ) - \frac {9 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{3} \mathrm {sgn}\left (x\right ) - 7 \, \sqrt {b x + a} a^{2} b^{3} \mathrm {sgn}\left (x\right )}{b^{2} x^{2}}}{4 \, b} \] Input:
integrate((b*x^3+a*x^2)^(5/2)/x^8,x, algorithm="giac")
Output:
1/4*(15*a*b^3*arctan(sqrt(b*x + a)/sqrt(-a))*sgn(x)/sqrt(-a) + 8*sqrt(b*x + a)*b^3*sgn(x) - (9*(b*x + a)^(3/2)*a*b^3*sgn(x) - 7*sqrt(b*x + a)*a^2*b^ 3*sgn(x))/(b^2*x^2))/b
Timed out. \[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x^8} \, dx=\int \frac {{\left (b\,x^3+a\,x^2\right )}^{5/2}}{x^8} \,d x \] Input:
int((a*x^2 + b*x^3)^(5/2)/x^8,x)
Output:
int((a*x^2 + b*x^3)^(5/2)/x^8, x)
Time = 0.32 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.76 \[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x^8} \, dx=\frac {-4 \sqrt {b x +a}\, a^{2}-18 \sqrt {b x +a}\, a b x +16 \sqrt {b x +a}\, b^{2} x^{2}+15 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{2} x^{2}-15 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{2} x^{2}}{8 x^{2}} \] Input:
int((b*x^3+a*x^2)^(5/2)/x^8,x)
Output:
( - 4*sqrt(a + b*x)*a**2 - 18*sqrt(a + b*x)*a*b*x + 16*sqrt(a + b*x)*b**2* x**2 + 15*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*b**2*x**2 - 15*sqrt(a)*log( sqrt(a + b*x) + sqrt(a))*b**2*x**2)/(8*x**2)