Integrand size = 26, antiderivative size = 163 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^7} \, dx=-\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 x^6 \left (a+b x^2\right )}-\frac {3 a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^4 \left (a+b x^2\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}+\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4} \log (x)}{a+b x^2} \] Output:
-1/6*a^3*((b*x^2+a)^2)^(1/2)/x^6/(b*x^2+a)-3/4*a^2*b*((b*x^2+a)^2)^(1/2)/x ^4/(b*x^2+a)-3/2*a*b^2*((b*x^2+a)^2)^(1/2)/x^2/(b*x^2+a)+b^3*((b*x^2+a)^2) ^(1/2)*ln(x)/(b*x^2+a)
Time = 0.32 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.63 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^7} \, dx=\frac {2 a^3 \sqrt {a^2}+9 \left (a^2\right )^{3/2} b x^2+18 a \sqrt {a^2} b^2 x^4-2 a^3 \sqrt {\left (a+b x^2\right )^2}-7 a^2 b x^2 \sqrt {\left (a+b x^2\right )^2}-11 a b^2 x^4 \sqrt {\left (a+b x^2\right )^2}-12 a b^3 x^6 \text {arctanh}\left (\frac {b x^2}{\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}}\right )-12 \sqrt {a^2} b^3 x^6 \log \left (x^2\right )+6 \sqrt {a^2} b^3 x^6 \log \left (a \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )\right )+6 \sqrt {a^2} b^3 x^6 \log \left (a \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{24 a x^6} \] Input:
Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^7,x]
Output:
(2*a^3*Sqrt[a^2] + 9*(a^2)^(3/2)*b*x^2 + 18*a*Sqrt[a^2]*b^2*x^4 - 2*a^3*Sq rt[(a + b*x^2)^2] - 7*a^2*b*x^2*Sqrt[(a + b*x^2)^2] - 11*a*b^2*x^4*Sqrt[(a + b*x^2)^2] - 12*a*b^3*x^6*ArcTanh[(b*x^2)/(Sqrt[a^2] - Sqrt[(a + b*x^2)^ 2])] - 12*Sqrt[a^2]*b^3*x^6*Log[x^2] + 6*Sqrt[a^2]*b^3*x^6*Log[a*(Sqrt[a^2 ] - b*x^2 - Sqrt[(a + b*x^2)^2])] + 6*Sqrt[a^2]*b^3*x^6*Log[a*(Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2])])/(24*a*x^6)
Time = 0.37 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.45, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1384, 27, 243, 49, 2009}
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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^7} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b^3 \left (b x^2+a\right )^3}{x^7}dx}{b^3 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^3}{x^7}dx}{a+b x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^3}{x^8}dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {a^3}{x^8}+\frac {3 b a^2}{x^6}+\frac {3 b^2 a}{x^4}+\frac {b^3}{x^2}\right )dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {a^3}{3 x^6}-\frac {3 a^2 b}{2 x^4}-\frac {3 a b^2}{x^2}+b^3 \log \left (x^2\right )\right )}{2 \left (a+b x^2\right )}\) |
Input:
Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^7,x]
Output:
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-1/3*a^3/x^6 - (3*a^2*b)/(2*x^4) - (3*a* b^2)/x^2 + b^3*Log[x^2]))/(2*(a + b*x^2))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S imp[(a + b*x^n + c*x^(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*Frac Part[p])) Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && NeQ[u, x^(n - 1)] && NeQ[u, x^(2*n - 1)] && !(EqQ[p, 1/2] && EqQ[u, x^(-2*n - 1)])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.13 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.31
method | result | size |
pseudoelliptic | \(\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (6 b^{3} \ln \left (x^{2}\right ) x^{6}-18 b^{2} x^{4} a -9 a^{2} b \,x^{2}-2 a^{3}\right )}{12 x^{6}}\) | \(50\) |
default | \(\frac {{\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}} \left (12 \ln \left (x \right ) x^{6} b^{3}-18 b^{2} x^{4} a -9 a^{2} b \,x^{2}-2 a^{3}\right )}{12 x^{6} \left (b \,x^{2}+a \right )^{3}}\) | \(60\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {3}{2} b^{2} x^{4} a -\frac {3}{4} a^{2} b \,x^{2}-\frac {1}{6} a^{3}\right )}{\left (b \,x^{2}+a \right ) x^{6}}+\frac {b^{3} \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (x \right )}{b \,x^{2}+a}\) | \(76\) |
Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^7,x,method=_RETURNVERBOSE)
Output:
1/12*csgn(b*x^2+a)*(6*b^3*ln(x^2)*x^6-18*b^2*x^4*a-9*a^2*b*x^2-2*a^3)/x^6
Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^7} \, dx=\frac {12 \, b^{3} x^{6} \log \left (x\right ) - 18 \, a b^{2} x^{4} - 9 \, a^{2} b x^{2} - 2 \, a^{3}}{12 \, x^{6}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^7,x, algorithm="fricas")
Output:
1/12*(12*b^3*x^6*log(x) - 18*a*b^2*x^4 - 9*a^2*b*x^2 - 2*a^3)/x^6
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^7} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}{x^{7}}\, dx \] Input:
integrate((b**2*x**4+2*a*b*x**2+a**2)**(3/2)/x**7,x)
Output:
Integral(((a + b*x**2)**2)**(3/2)/x**7, x)
Time = 0.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.20 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^7} \, dx=b^{3} \log \left (x\right ) - \frac {3 \, a b^{2}}{2 \, x^{2}} - \frac {3 \, a^{2} b}{4 \, x^{4}} - \frac {a^{3}}{6 \, x^{6}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^7,x, algorithm="maxima")
Output:
b^3*log(x) - 3/2*a*b^2/x^2 - 3/4*a^2*b/x^4 - 1/6*a^3/x^6
Time = 0.12 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.53 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^7} \, dx=\frac {1}{2} \, b^{3} \log \left (x^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {11 \, b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 18 \, a b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 9 \, a^{2} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 2 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{12 \, x^{6}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^7,x, algorithm="giac")
Output:
1/2*b^3*log(x^2)*sgn(b*x^2 + a) - 1/12*(11*b^3*x^6*sgn(b*x^2 + a) + 18*a*b ^2*x^4*sgn(b*x^2 + a) + 9*a^2*b*x^2*sgn(b*x^2 + a) + 2*a^3*sgn(b*x^2 + a)) /x^6
Timed out. \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^7} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}}{x^7} \,d x \] Input:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)/x^7,x)
Output:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)/x^7, x)
Time = 0.17 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^7} \, dx=\frac {12 \,\mathrm {log}\left (x \right ) b^{3} x^{6}-2 a^{3}-9 a^{2} b \,x^{2}-18 a \,b^{2} x^{4}}{12 x^{6}} \] Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^7,x)
Output:
(12*log(x)*b**3*x**6 - 2*a**3 - 9*a**2*b*x**2 - 18*a*b**2*x**4)/(12*x**6)