Integrand size = 19, antiderivative size = 136 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{13}} \, dx=-\frac {\left (b x^2+c x^4\right )^{3/2}}{11 b x^{14}}+\frac {8 c \left (b x^2+c x^4\right )^{3/2}}{99 b^2 x^{12}}-\frac {16 c^2 \left (b x^2+c x^4\right )^{3/2}}{231 b^3 x^{10}}+\frac {64 c^3 \left (b x^2+c x^4\right )^{3/2}}{1155 b^4 x^8}-\frac {128 c^4 \left (b x^2+c x^4\right )^{3/2}}{3465 b^5 x^6} \] Output:
-1/11*(c*x^4+b*x^2)^(3/2)/b/x^14+8/99*c*(c*x^4+b*x^2)^(3/2)/b^2/x^12-16/23 1*c^2*(c*x^4+b*x^2)^(3/2)/b^3/x^10+64/1155*c^3*(c*x^4+b*x^2)^(3/2)/b^4/x^8 -128/3465*c^4*(c*x^4+b*x^2)^(3/2)/b^5/x^6
Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.58 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{13}} \, dx=\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (-315 b^5-35 b^4 c x^2+40 b^3 c^2 x^4-48 b^2 c^3 x^6+64 b c^4 x^8-128 c^5 x^{10}\right )}{3465 b^5 x^{12}} \] Input:
Integrate[Sqrt[b*x^2 + c*x^4]/x^13,x]
Output:
(Sqrt[x^2*(b + c*x^2)]*(-315*b^5 - 35*b^4*c*x^2 + 40*b^3*c^2*x^4 - 48*b^2* c^3*x^6 + 64*b*c^4*x^8 - 128*c^5*x^10))/(3465*b^5*x^12)
Time = 0.51 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1423, 1423, 1423, 1423, 1422}
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.
\(\displaystyle \int \frac {\sqrt {b x^2+c x^4}}{x^{13}} \, dx\) |
\(\Big \downarrow \) 1423 |
\(\displaystyle -\frac {8 c \int \frac {\sqrt {c x^4+b x^2}}{x^{11}}dx}{11 b}-\frac {\left (b x^2+c x^4\right )^{3/2}}{11 b x^{14}}\) |
\(\Big \downarrow \) 1423 |
\(\displaystyle -\frac {8 c \left (-\frac {2 c \int \frac {\sqrt {c x^4+b x^2}}{x^9}dx}{3 b}-\frac {\left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}}\right )}{11 b}-\frac {\left (b x^2+c x^4\right )^{3/2}}{11 b x^{14}}\) |
\(\Big \downarrow \) 1423 |
\(\displaystyle -\frac {8 c \left (-\frac {2 c \left (-\frac {4 c \int \frac {\sqrt {c x^4+b x^2}}{x^7}dx}{7 b}-\frac {\left (b x^2+c x^4\right )^{3/2}}{7 b x^{10}}\right )}{3 b}-\frac {\left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}}\right )}{11 b}-\frac {\left (b x^2+c x^4\right )^{3/2}}{11 b x^{14}}\) |
\(\Big \downarrow \) 1423 |
\(\displaystyle -\frac {8 c \left (-\frac {2 c \left (-\frac {4 c \left (-\frac {2 c \int \frac {\sqrt {c x^4+b x^2}}{x^5}dx}{5 b}-\frac {\left (b x^2+c x^4\right )^{3/2}}{5 b x^8}\right )}{7 b}-\frac {\left (b x^2+c x^4\right )^{3/2}}{7 b x^{10}}\right )}{3 b}-\frac {\left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}}\right )}{11 b}-\frac {\left (b x^2+c x^4\right )^{3/2}}{11 b x^{14}}\) |
\(\Big \downarrow \) 1422 |
\(\displaystyle -\frac {8 c \left (-\frac {2 c \left (-\frac {4 c \left (\frac {2 c \left (b x^2+c x^4\right )^{3/2}}{15 b^2 x^6}-\frac {\left (b x^2+c x^4\right )^{3/2}}{5 b x^8}\right )}{7 b}-\frac {\left (b x^2+c x^4\right )^{3/2}}{7 b x^{10}}\right )}{3 b}-\frac {\left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}}\right )}{11 b}-\frac {\left (b x^2+c x^4\right )^{3/2}}{11 b x^{14}}\) |
Input:
Int[Sqrt[b*x^2 + c*x^4]/x^13,x]
Output:
-1/11*(b*x^2 + c*x^4)^(3/2)/(b*x^14) - (8*c*(-1/9*(b*x^2 + c*x^4)^(3/2)/(b *x^12) - (2*c*(-1/7*(b*x^2 + c*x^4)^(3/2)/(b*x^10) - (4*c*(-1/5*(b*x^2 + c *x^4)^(3/2)/(b*x^8) + (2*c*(b*x^2 + c*x^4)^(3/2))/(15*b^2*x^6)))/(7*b)))/( 3*b)))/(11*b)
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [(-d)*(d*x)^(m - 1)*((b*x^2 + c*x^4)^(p + 1)/(2*b*(p + 1))), x] /; FreeQ[{b , c, d, m, p}, x] && !IntegerQ[p] && EqQ[m + 4*p + 3, 0]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [d*(d*x)^(m - 1)*((b*x^2 + c*x^4)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[c*( (m + 4*p + 3)/(b*d^2*(m + 2*p + 1))) Int[(d*x)^(m + 2)*(b*x^2 + c*x^4)^p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p] && ILtQ[Simplify[(m + 4*p + 3)/2], 0] && NeQ[m + 2*p + 1, 0]
Time = 0.23 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.53
method | result | size |
gosper | \(-\frac {\left (c \,x^{2}+b \right ) \left (128 c^{4} x^{8}-192 b \,c^{3} x^{6}+240 b^{2} c^{2} x^{4}-280 x^{2} b^{3} c +315 b^{4}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{3465 x^{12} b^{5}}\) | \(72\) |
default | \(-\frac {\left (c \,x^{2}+b \right ) \left (128 c^{4} x^{8}-192 b \,c^{3} x^{6}+240 b^{2} c^{2} x^{4}-280 x^{2} b^{3} c +315 b^{4}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{3465 x^{12} b^{5}}\) | \(72\) |
orering | \(-\frac {\left (c \,x^{2}+b \right ) \left (128 c^{4} x^{8}-192 b \,c^{3} x^{6}+240 b^{2} c^{2} x^{4}-280 x^{2} b^{3} c +315 b^{4}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{3465 x^{12} b^{5}}\) | \(72\) |
trager | \(-\frac {\left (128 c^{5} x^{10}-64 c^{4} x^{8} b +48 b^{2} c^{3} x^{6}-40 c^{2} x^{4} b^{3}+35 x^{2} c \,b^{4}+315 b^{5}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{3465 x^{12} b^{5}}\) | \(76\) |
risch | \(-\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (128 c^{5} x^{10}-64 c^{4} x^{8} b +48 b^{2} c^{3} x^{6}-40 c^{2} x^{4} b^{3}+35 x^{2} c \,b^{4}+315 b^{5}\right )}{3465 x^{12} b^{5}}\) | \(76\) |
pseudoelliptic | \(\frac {\left (-128 c^{5} x^{10}+64 c^{4} x^{8} b -48 b^{2} c^{3} x^{6}+40 c^{2} x^{4} b^{3}-35 x^{2} c \,b^{4}-315 b^{5}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{3465 x^{12} b^{5}}\) | \(76\) |
Input:
int((c*x^4+b*x^2)^(1/2)/x^13,x,method=_RETURNVERBOSE)
Output:
-1/3465*(c*x^2+b)*(128*c^4*x^8-192*b*c^3*x^6+240*b^2*c^2*x^4-280*b^3*c*x^2 +315*b^4)*(c*x^4+b*x^2)^(1/2)/x^12/b^5
Time = 0.25 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.55 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{13}} \, dx=-\frac {{\left (128 \, c^{5} x^{10} - 64 \, b c^{4} x^{8} + 48 \, b^{2} c^{3} x^{6} - 40 \, b^{3} c^{2} x^{4} + 35 \, b^{4} c x^{2} + 315 \, b^{5}\right )} \sqrt {c x^{4} + b x^{2}}}{3465 \, b^{5} x^{12}} \] Input:
integrate((c*x^4+b*x^2)^(1/2)/x^13,x, algorithm="fricas")
Output:
-1/3465*(128*c^5*x^10 - 64*b*c^4*x^8 + 48*b^2*c^3*x^6 - 40*b^3*c^2*x^4 + 3 5*b^4*c*x^2 + 315*b^5)*sqrt(c*x^4 + b*x^2)/(b^5*x^12)
\[ \int \frac {\sqrt {b x^2+c x^4}}{x^{13}} \, dx=\int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )}}{x^{13}}\, dx \] Input:
integrate((c*x**4+b*x**2)**(1/2)/x**13,x)
Output:
Integral(sqrt(x**2*(b + c*x**2))/x**13, x)
Time = 0.04 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.01 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{13}} \, dx=-\frac {128 \, \sqrt {c x^{4} + b x^{2}} c^{5}}{3465 \, b^{5} x^{2}} + \frac {64 \, \sqrt {c x^{4} + b x^{2}} c^{4}}{3465 \, b^{4} x^{4}} - \frac {16 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{1155 \, b^{3} x^{6}} + \frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{693 \, b^{2} x^{8}} - \frac {\sqrt {c x^{4} + b x^{2}} c}{99 \, b x^{10}} - \frac {\sqrt {c x^{4} + b x^{2}}}{11 \, x^{12}} \] Input:
integrate((c*x^4+b*x^2)^(1/2)/x^13,x, algorithm="maxima")
Output:
-128/3465*sqrt(c*x^4 + b*x^2)*c^5/(b^5*x^2) + 64/3465*sqrt(c*x^4 + b*x^2)* c^4/(b^4*x^4) - 16/1155*sqrt(c*x^4 + b*x^2)*c^3/(b^3*x^6) + 8/693*sqrt(c*x ^4 + b*x^2)*c^2/(b^2*x^8) - 1/99*sqrt(c*x^4 + b*x^2)*c/(b*x^10) - 1/11*sqr t(c*x^4 + b*x^2)/x^12
Time = 0.13 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.51 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{13}} \, dx=\frac {256 \, {\left (1386 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{12} c^{\frac {11}{2}} \mathrm {sgn}\left (x\right ) + 924 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} b c^{\frac {11}{2}} \mathrm {sgn}\left (x\right ) + 330 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} b^{2} c^{\frac {11}{2}} \mathrm {sgn}\left (x\right ) - 165 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} b^{3} c^{\frac {11}{2}} \mathrm {sgn}\left (x\right ) + 55 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} b^{4} c^{\frac {11}{2}} \mathrm {sgn}\left (x\right ) - 11 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} b^{5} c^{\frac {11}{2}} \mathrm {sgn}\left (x\right ) + b^{6} c^{\frac {11}{2}} \mathrm {sgn}\left (x\right )\right )}}{3465 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{11}} \] Input:
integrate((c*x^4+b*x^2)^(1/2)/x^13,x, algorithm="giac")
Output:
256/3465*(1386*(sqrt(c)*x - sqrt(c*x^2 + b))^12*c^(11/2)*sgn(x) + 924*(sqr t(c)*x - sqrt(c*x^2 + b))^10*b*c^(11/2)*sgn(x) + 330*(sqrt(c)*x - sqrt(c*x ^2 + b))^8*b^2*c^(11/2)*sgn(x) - 165*(sqrt(c)*x - sqrt(c*x^2 + b))^6*b^3*c ^(11/2)*sgn(x) + 55*(sqrt(c)*x - sqrt(c*x^2 + b))^4*b^4*c^(11/2)*sgn(x) - 11*(sqrt(c)*x - sqrt(c*x^2 + b))^2*b^5*c^(11/2)*sgn(x) + b^6*c^(11/2)*sgn( x))/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^11
Time = 19.06 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.01 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{13}} \, dx=\frac {8\,c^2\,\sqrt {c\,x^4+b\,x^2}}{693\,b^2\,x^8}-\frac {c\,\sqrt {c\,x^4+b\,x^2}}{99\,b\,x^{10}}-\frac {\sqrt {c\,x^4+b\,x^2}}{11\,x^{12}}-\frac {16\,c^3\,\sqrt {c\,x^4+b\,x^2}}{1155\,b^3\,x^6}+\frac {64\,c^4\,\sqrt {c\,x^4+b\,x^2}}{3465\,b^4\,x^4}-\frac {128\,c^5\,\sqrt {c\,x^4+b\,x^2}}{3465\,b^5\,x^2} \] Input:
int((b*x^2 + c*x^4)^(1/2)/x^13,x)
Output:
(8*c^2*(b*x^2 + c*x^4)^(1/2))/(693*b^2*x^8) - (c*(b*x^2 + c*x^4)^(1/2))/(9 9*b*x^10) - (b*x^2 + c*x^4)^(1/2)/(11*x^12) - (16*c^3*(b*x^2 + c*x^4)^(1/2 ))/(1155*b^3*x^6) + (64*c^4*(b*x^2 + c*x^4)^(1/2))/(3465*b^4*x^4) - (128*c ^5*(b*x^2 + c*x^4)^(1/2))/(3465*b^5*x^2)
Time = 0.25 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{13}} \, dx=\frac {-315 \sqrt {c \,x^{2}+b}\, b^{5}-35 \sqrt {c \,x^{2}+b}\, b^{4} c \,x^{2}+40 \sqrt {c \,x^{2}+b}\, b^{3} c^{2} x^{4}-48 \sqrt {c \,x^{2}+b}\, b^{2} c^{3} x^{6}+64 \sqrt {c \,x^{2}+b}\, b \,c^{4} x^{8}-128 \sqrt {c \,x^{2}+b}\, c^{5} x^{10}+128 \sqrt {c}\, c^{5} x^{11}}{3465 b^{5} x^{11}} \] Input:
int((c*x^4+b*x^2)^(1/2)/x^13,x)
Output:
( - 315*sqrt(b + c*x**2)*b**5 - 35*sqrt(b + c*x**2)*b**4*c*x**2 + 40*sqrt( b + c*x**2)*b**3*c**2*x**4 - 48*sqrt(b + c*x**2)*b**2*c**3*x**6 + 64*sqrt( b + c*x**2)*b*c**4*x**8 - 128*sqrt(b + c*x**2)*c**5*x**10 + 128*sqrt(c)*c* *5*x**11)/(3465*b**5*x**11)