Integrand size = 26, antiderivative size = 140 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx=-\frac {c (6 b B-A c) \sqrt {b x^2+c x^4}}{16 b x^3}-\frac {(6 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{24 b x^7}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{6 b x^{11}}-\frac {c^2 (6 b B-A c) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{16 b^{3/2}} \] Output:
-1/16*c*(-A*c+6*B*b)*(c*x^4+b*x^2)^(1/2)/b/x^3-1/24*(-A*c+6*B*b)*(c*x^4+b* x^2)^(3/2)/b/x^7-1/6*A*(c*x^4+b*x^2)^(5/2)/b/x^11-1/16*c^2*(-A*c+6*B*b)*ar ctanh(b^(1/2)*x/(c*x^4+b*x^2)^(1/2))/b^(3/2)
Time = 0.22 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.94 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx=\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (-\sqrt {b} \sqrt {b+c x^2} \left (6 b B x^2 \left (2 b+5 c x^2\right )+A \left (8 b^2+14 b c x^2+3 c^2 x^4\right )\right )+3 c^2 (-6 b B+A c) x^6 \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )\right )}{48 b^{3/2} x^7 \sqrt {b+c x^2}} \] Input:
Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^10,x]
Output:
(Sqrt[x^2*(b + c*x^2)]*(-(Sqrt[b]*Sqrt[b + c*x^2]*(6*b*B*x^2*(2*b + 5*c*x^ 2) + A*(8*b^2 + 14*b*c*x^2 + 3*c^2*x^4))) + 3*c^2*(-6*b*B + A*c)*x^6*ArcTa nh[Sqrt[b + c*x^2]/Sqrt[b]]))/(48*b^(3/2)*x^7*Sqrt[b + c*x^2])
Time = 0.49 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1944, 1425, 1425, 1400, 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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx\) |
\(\Big \downarrow \) 1944 |
\(\displaystyle \frac {(6 b B-A c) \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^8}dx}{6 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{6 b x^{11}}\) |
\(\Big \downarrow \) 1425 |
\(\displaystyle \frac {(6 b B-A c) \left (\frac {3}{4} c \int \frac {\sqrt {c x^4+b x^2}}{x^4}dx-\frac {\left (b x^2+c x^4\right )^{3/2}}{4 x^7}\right )}{6 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{6 b x^{11}}\) |
\(\Big \downarrow \) 1425 |
\(\displaystyle \frac {(6 b B-A c) \left (\frac {3}{4} c \left (\frac {1}{2} c \int \frac {1}{\sqrt {c x^4+b x^2}}dx-\frac {\sqrt {b x^2+c x^4}}{2 x^3}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{4 x^7}\right )}{6 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{6 b x^{11}}\) |
\(\Big \downarrow \) 1400 |
\(\displaystyle \frac {(6 b B-A c) \left (\frac {3}{4} c \left (-\frac {1}{2} c \int \frac {1}{1-\frac {b x^2}{c x^4+b x^2}}d\frac {x}{\sqrt {c x^4+b x^2}}-\frac {\sqrt {b x^2+c x^4}}{2 x^3}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{4 x^7}\right )}{6 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{6 b x^{11}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {(6 b B-A c) \left (\frac {3}{4} c \left (-\frac {c \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 \sqrt {b}}-\frac {\sqrt {b x^2+c x^4}}{2 x^3}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{4 x^7}\right )}{6 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{6 b x^{11}}\) |
Input:
Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^10,x]
Output:
-1/6*(A*(b*x^2 + c*x^4)^(5/2))/(b*x^11) + ((6*b*B - A*c)*(-1/4*(b*x^2 + c* x^4)^(3/2)/x^7 + (3*c*(-1/2*Sqrt[b*x^2 + c*x^4]/x^3 - (c*ArcTanh[(Sqrt[b]* x)/Sqrt[b*x^2 + c*x^4]])/(2*Sqrt[b])))/4))/(6*b)
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[(b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> -Subst[Int[1/(1 - b*x ^2), x], x, x/Sqrt[b*x^2 + c*x^4]] /; FreeQ[{b, c}, x]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [(d*x)^(m + 1)*((b*x^2 + c*x^4)^p/(d*(m + 2*p + 1))), x] - Simp[2*c*(p/(d^4 *(m + 2*p + 1))) Int[(d*x)^(m + 4)*(b*x^2 + c*x^4)^(p - 1), x], x] /; Fre eQ[{b, c, d, m, p}, x] && !IntegerQ[p] && GtQ[p, 0] && LtQ[m + 2*p + 1, 0]
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Simp[(a*d*(m + j*p + 1) - b *c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)) Int[(e*x)^(m + n)*(a*x^ j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m + j*p, -1 ] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && ( GtQ[e, 0] || IntegersQ[j, n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1 , 0]
Time = 1.21 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.91
method | result | size |
risch | \(-\frac {\left (3 x^{4} A \,c^{2}+30 x^{4} B b c +14 A b c \,x^{2}+12 x^{2} B \,b^{2}+8 b^{2} A \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{48 x^{7} b}+\frac {\left (A c -6 B b \right ) c^{2} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{16 b^{\frac {3}{2}} x \sqrt {c \,x^{2}+b}}\) | \(128\) |
default | \(\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (3 A \,b^{\frac {3}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) c^{3} x^{6}-A \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{3} x^{6}-18 B \,b^{\frac {5}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) c^{2} x^{6}+6 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} b \,c^{2} x^{6}+A \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{2} x^{4}-3 A \sqrt {c \,x^{2}+b}\, b \,c^{3} x^{6}-6 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} b c \,x^{4}+18 B \sqrt {c \,x^{2}+b}\, b^{2} c^{2} x^{6}+2 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} b c \,x^{2}-12 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{2} x^{2}-8 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{2}\right )}{48 x^{9} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{3}}\) | \(259\) |
Input:
int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^10,x,method=_RETURNVERBOSE)
Output:
-1/48*(3*A*c^2*x^4+30*B*b*c*x^4+14*A*b*c*x^2+12*B*b^2*x^2+8*A*b^2)/x^7/b*( x^2*(c*x^2+b))^(1/2)+1/16*(A*c-6*B*b)*c^2/b^(3/2)*ln((2*b+2*b^(1/2)*(c*x^2 +b)^(1/2))/x)*(x^2*(c*x^2+b))^(1/2)/x/(c*x^2+b)^(1/2)
Time = 0.12 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.75 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx=\left [-\frac {3 \, {\left (6 \, B b c^{2} - A c^{3}\right )} \sqrt {b} x^{7} \log \left (-\frac {c x^{3} + 2 \, b x + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, {\left (3 \, {\left (10 \, B b^{2} c + A b c^{2}\right )} x^{4} + 8 \, A b^{3} + 2 \, {\left (6 \, B b^{3} + 7 \, A b^{2} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{96 \, b^{2} x^{7}}, \frac {3 \, {\left (6 \, B b c^{2} - A c^{3}\right )} \sqrt {-b} x^{7} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{b x}\right ) - {\left (3 \, {\left (10 \, B b^{2} c + A b c^{2}\right )} x^{4} + 8 \, A b^{3} + 2 \, {\left (6 \, B b^{3} + 7 \, A b^{2} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{48 \, b^{2} x^{7}}\right ] \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^10,x, algorithm="fricas")
Output:
[-1/96*(3*(6*B*b*c^2 - A*c^3)*sqrt(b)*x^7*log(-(c*x^3 + 2*b*x + 2*sqrt(c*x ^4 + b*x^2)*sqrt(b))/x^3) + 2*(3*(10*B*b^2*c + A*b*c^2)*x^4 + 8*A*b^3 + 2* (6*B*b^3 + 7*A*b^2*c)*x^2)*sqrt(c*x^4 + b*x^2))/(b^2*x^7), 1/48*(3*(6*B*b* c^2 - A*c^3)*sqrt(-b)*x^7*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(b*x)) - (3* (10*B*b^2*c + A*b*c^2)*x^4 + 8*A*b^3 + 2*(6*B*b^3 + 7*A*b^2*c)*x^2)*sqrt(c *x^4 + b*x^2))/(b^2*x^7)]
\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{10}}\, dx \] Input:
integrate((B*x**2+A)*(c*x**4+b*x**2)**(3/2)/x**10,x)
Output:
Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**10, x)
\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{10}} \,d x } \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^10,x, algorithm="maxima")
Output:
integrate((c*x^4 + b*x^2)^(3/2)*(B*x^2 + A)/x^10, x)
Time = 0.29 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.13 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx=\frac {1}{48} \, c^{3} {\left (\frac {3 \, {\left (6 \, B b \mathrm {sgn}\left (x\right ) - A c \mathrm {sgn}\left (x\right )\right )} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b c} - \frac {30 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} B b \mathrm {sgn}\left (x\right ) - 48 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} B b^{2} \mathrm {sgn}\left (x\right ) + 18 \, \sqrt {c x^{2} + b} B b^{3} \mathrm {sgn}\left (x\right ) + 3 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} A c \mathrm {sgn}\left (x\right ) + 8 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} A b c \mathrm {sgn}\left (x\right ) - 3 \, \sqrt {c x^{2} + b} A b^{2} c \mathrm {sgn}\left (x\right )}{b c^{4} x^{6}}\right )} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^10,x, algorithm="giac")
Output:
1/48*c^3*(3*(6*B*b*sgn(x) - A*c*sgn(x))*arctan(sqrt(c*x^2 + b)/sqrt(-b))/( sqrt(-b)*b*c) - (30*(c*x^2 + b)^(5/2)*B*b*sgn(x) - 48*(c*x^2 + b)^(3/2)*B* b^2*sgn(x) + 18*sqrt(c*x^2 + b)*B*b^3*sgn(x) + 3*(c*x^2 + b)^(5/2)*A*c*sgn (x) + 8*(c*x^2 + b)^(3/2)*A*b*c*sgn(x) - 3*sqrt(c*x^2 + b)*A*b^2*c*sgn(x)) /(b*c^4*x^6))
Timed out. \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{10}} \,d x \] Input:
int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^10,x)
Output:
int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^10, x)
Time = 0.23 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.63 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx=\frac {-8 \sqrt {c \,x^{2}+b}\, a \,b^{3}-14 \sqrt {c \,x^{2}+b}\, a \,b^{2} c \,x^{2}-3 \sqrt {c \,x^{2}+b}\, a b \,c^{2} x^{4}-12 \sqrt {c \,x^{2}+b}\, b^{4} x^{2}-30 \sqrt {c \,x^{2}+b}\, b^{3} c \,x^{4}-3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,c^{3} x^{6}+18 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} c^{2} x^{6}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,c^{3} x^{6}-18 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} c^{2} x^{6}}{48 b^{2} x^{6}} \] Input:
int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^10,x)
Output:
( - 8*sqrt(b + c*x**2)*a*b**3 - 14*sqrt(b + c*x**2)*a*b**2*c*x**2 - 3*sqrt (b + c*x**2)*a*b*c**2*x**4 - 12*sqrt(b + c*x**2)*b**4*x**2 - 30*sqrt(b + c *x**2)*b**3*c*x**4 - 3*sqrt(b)*log((sqrt(b + c*x**2) - sqrt(b) + sqrt(c)*x )/sqrt(b))*a*c**3*x**6 + 18*sqrt(b)*log((sqrt(b + c*x**2) - sqrt(b) + sqrt (c)*x)/sqrt(b))*b**2*c**2*x**6 + 3*sqrt(b)*log((sqrt(b + c*x**2) + sqrt(b) + sqrt(c)*x)/sqrt(b))*a*c**3*x**6 - 18*sqrt(b)*log((sqrt(b + c*x**2) + sq rt(b) + sqrt(c)*x)/sqrt(b))*b**2*c**2*x**6)/(48*b**2*x**6)