Integrand size = 13, antiderivative size = 140 \[ \int \frac {1}{\left (b x+c x^2\right )^{7/2}} \, dx=\frac {2}{5 b \left (b x+c x^2\right )^{5/2}}+\frac {4}{3 b^2 x \left (b x+c x^2\right )^{3/2}}+\frac {32}{3 b^3 x^2 \sqrt {b x+c x^2}}-\frac {64 \sqrt {b x+c x^2}}{5 b^4 x^3}+\frac {256 c \sqrt {b x+c x^2}}{15 b^5 x^2}-\frac {512 c^2 \sqrt {b x+c x^2}}{15 b^6 x} \] Output:
2/5/b/(c*x^2+b*x)^(5/2)+4/3/b^2/x/(c*x^2+b*x)^(3/2)+32/3/b^3/x^2/(c*x^2+b* x)^(1/2)-64/5*(c*x^2+b*x)^(1/2)/b^4/x^3+256/15*c*(c*x^2+b*x)^(1/2)/b^5/x^2 -512/15*c^2*(c*x^2+b*x)^(1/2)/b^6/x
Time = 0.01 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.50 \[ \int \frac {1}{\left (b x+c x^2\right )^{7/2}} \, dx=-\frac {2 \left (3 b^5-10 b^4 c x+80 b^3 c^2 x^2+480 b^2 c^3 x^3+640 b c^4 x^4+256 c^5 x^5\right )}{15 b^6 (x (b+c x))^{5/2}} \] Input:
Integrate[(b*x + c*x^2)^(-7/2),x]
Output:
(-2*(3*b^5 - 10*b^4*c*x + 80*b^3*c^2*x^2 + 480*b^2*c^3*x^3 + 640*b*c^4*x^4 + 256*c^5*x^5))/(15*b^6*(x*(b + c*x))^(5/2))
Time = 0.34 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.64, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1089, 1089, 1088}
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.
\(\displaystyle \int \frac {1}{\left (b x+c x^2\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 1089 |
\(\displaystyle -\frac {16 c \int \frac {1}{\left (c x^2+b x\right )^{5/2}}dx}{5 b^2}-\frac {2 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 1089 |
\(\displaystyle -\frac {16 c \left (-\frac {8 c \int \frac {1}{\left (c x^2+b x\right )^{3/2}}dx}{3 b^2}-\frac {2 (b+2 c x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}\right )}{5 b^2}-\frac {2 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 1088 |
\(\displaystyle -\frac {2 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/2}}-\frac {16 c \left (\frac {16 c (b+2 c x)}{3 b^4 \sqrt {b x+c x^2}}-\frac {2 (b+2 c x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}\right )}{5 b^2}\) |
Input:
Int[(b*x + c*x^2)^(-7/2),x]
Output:
(-2*(b + 2*c*x))/(5*b^2*(b*x + c*x^2)^(5/2)) - (16*c*((-2*(b + 2*c*x))/(3* b^2*(b*x + c*x^2)^(3/2)) + (16*c*(b + 2*c*x))/(3*b^4*Sqrt[b*x + c*x^2])))/ (5*b^2)
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[-2*((b + 2*c*x)/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2])), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] - Simp[2*c*((2*p + 3)/((p + 1)*(b^2 - 4*a*c))) Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fre eQ[{a, b, c}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[3*p])
Time = 0.67 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.54
method | result | size |
gosper | \(-\frac {2 x \left (c x +b \right ) \left (256 c^{5} x^{5}+640 b \,x^{4} c^{4}+480 b^{2} c^{3} x^{3}+80 c^{2} x^{2} b^{3}-10 b^{4} c x +3 b^{5}\right )}{15 b^{6} \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}\) | \(75\) |
orering | \(-\frac {2 x \left (c x +b \right ) \left (256 c^{5} x^{5}+640 b \,x^{4} c^{4}+480 b^{2} c^{3} x^{3}+80 c^{2} x^{2} b^{3}-10 b^{4} c x +3 b^{5}\right )}{15 b^{6} \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}\) | \(75\) |
default | \(-\frac {2 \left (2 c x +b \right )}{5 b^{2} \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}-\frac {16 c \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )}{5 b^{2}}\) | \(76\) |
pseudoelliptic | \(\frac {-\frac {512}{15} c^{5} x^{5}-\frac {256}{3} b \,x^{4} c^{4}-64 b^{2} c^{3} x^{3}-\frac {32}{3} c^{2} x^{2} b^{3}+\frac {4}{3} b^{4} c x -\frac {2}{5} b^{5}}{x^{2} \left (c x +b \right )^{2} \sqrt {x \left (c x +b \right )}\, b^{6}}\) | \(77\) |
trager | \(-\frac {2 \left (256 c^{5} x^{5}+640 b \,x^{4} c^{4}+480 b^{2} c^{3} x^{3}+80 c^{2} x^{2} b^{3}-10 b^{4} c x +3 b^{5}\right ) \sqrt {c \,x^{2}+b x}}{15 b^{6} \left (c x +b \right )^{3} x^{3}}\) | \(79\) |
risch | \(-\frac {2 \left (c x +b \right ) \left (128 c^{2} x^{2}-19 c b x +3 b^{2}\right )}{15 b^{6} x^{2} \sqrt {x \left (c x +b \right )}}-\frac {2 c^{3} \left (128 c^{2} x^{2}+275 c b x +150 b^{2}\right ) x}{15 \sqrt {x \left (c x +b \right )}\, \left (c^{2} x^{2}+2 c b x +b^{2}\right ) b^{6}}\) | \(98\) |
Input:
int(1/(c*x^2+b*x)^(7/2),x,method=_RETURNVERBOSE)
Output:
-2/15*x*(c*x+b)*(256*c^5*x^5+640*b*c^4*x^4+480*b^2*c^3*x^3+80*b^3*c^2*x^2- 10*b^4*c*x+3*b^5)/b^6/(c*x^2+b*x)^(7/2)
Time = 0.09 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\left (b x+c x^2\right )^{7/2}} \, dx=-\frac {2 \, {\left (256 \, c^{5} x^{5} + 640 \, b c^{4} x^{4} + 480 \, b^{2} c^{3} x^{3} + 80 \, b^{3} c^{2} x^{2} - 10 \, b^{4} c x + 3 \, b^{5}\right )} \sqrt {c x^{2} + b x}}{15 \, {\left (b^{6} c^{3} x^{6} + 3 \, b^{7} c^{2} x^{5} + 3 \, b^{8} c x^{4} + b^{9} x^{3}\right )}} \] Input:
integrate(1/(c*x^2+b*x)^(7/2),x, algorithm="fricas")
Output:
-2/15*(256*c^5*x^5 + 640*b*c^4*x^4 + 480*b^2*c^3*x^3 + 80*b^3*c^2*x^2 - 10 *b^4*c*x + 3*b^5)*sqrt(c*x^2 + b*x)/(b^6*c^3*x^6 + 3*b^7*c^2*x^5 + 3*b^8*c *x^4 + b^9*x^3)
\[ \int \frac {1}{\left (b x+c x^2\right )^{7/2}} \, dx=\int \frac {1}{\left (b x + c x^{2}\right )^{\frac {7}{2}}}\, dx \] Input:
integrate(1/(c*x**2+b*x)**(7/2),x)
Output:
Integral((b*x + c*x**2)**(-7/2), x)
Time = 0.03 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\left (b x+c x^2\right )^{7/2}} \, dx=-\frac {4 \, c x}{5 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} b^{2}} + \frac {64 \, c^{2} x}{15 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{4}} - \frac {512 \, c^{3} x}{15 \, \sqrt {c x^{2} + b x} b^{6}} - \frac {2}{5 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} b} + \frac {32 \, c}{15 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{3}} - \frac {256 \, c^{2}}{15 \, \sqrt {c x^{2} + b x} b^{5}} \] Input:
integrate(1/(c*x^2+b*x)^(7/2),x, algorithm="maxima")
Output:
-4/5*c*x/((c*x^2 + b*x)^(5/2)*b^2) + 64/15*c^2*x/((c*x^2 + b*x)^(3/2)*b^4) - 512/15*c^3*x/(sqrt(c*x^2 + b*x)*b^6) - 2/5/((c*x^2 + b*x)^(5/2)*b) + 32 /15*c/((c*x^2 + b*x)^(3/2)*b^3) - 256/15*c^2/(sqrt(c*x^2 + b*x)*b^5)
Time = 0.14 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.53 \[ \int \frac {1}{\left (b x+c x^2\right )^{7/2}} \, dx=-\frac {2 \, {\left (2 \, {\left (8 \, {\left (2 \, {\left (4 \, x {\left (\frac {2 \, c^{5} x}{b^{6}} + \frac {5 \, c^{4}}{b^{5}}\right )} + \frac {15 \, c^{3}}{b^{4}}\right )} x + \frac {5 \, c^{2}}{b^{3}}\right )} x - \frac {5 \, c}{b^{2}}\right )} x + \frac {3}{b}\right )}}{15 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}}} \] Input:
integrate(1/(c*x^2+b*x)^(7/2),x, algorithm="giac")
Output:
-2/15*(2*(8*(2*(4*x*(2*c^5*x/b^6 + 5*c^4/b^5) + 15*c^3/b^4)*x + 5*c^2/b^3) *x - 5*c/b^2)*x + 3/b)/(c*x^2 + b*x)^(5/2)
Time = 5.03 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.69 \[ \int \frac {1}{\left (b x+c x^2\right )^{7/2}} \, dx=-\frac {6\,b^5+256\,b\,c^2\,{\left (c\,x^2+b\,x\right )}^2+512\,c^3\,x\,{\left (c\,x^2+b\,x\right )}^2-32\,b^3\,c\,\left (c\,x^2+b\,x\right )+12\,b^4\,c\,x-64\,b^2\,c^2\,x\,\left (c\,x^2+b\,x\right )}{15\,b^6\,{\left (c\,x^2+b\,x\right )}^{5/2}} \] Input:
int(1/(b*x + c*x^2)^(7/2),x)
Output:
-(6*b^5 + 256*b*c^2*(b*x + c*x^2)^2 + 512*c^3*x*(b*x + c*x^2)^2 - 32*b^3*c *(b*x + c*x^2) + 12*b^4*c*x - 64*b^2*c^2*x*(b*x + c*x^2))/(15*b^6*(b*x + c *x^2)^(5/2))
Time = 0.22 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\left (b x+c x^2\right )^{7/2}} \, dx=\frac {\frac {512 \sqrt {c}\, \sqrt {c x +b}\, b^{2} c^{2} x^{3}}{15}+\frac {1024 \sqrt {c}\, \sqrt {c x +b}\, b \,c^{3} x^{4}}{15}+\frac {512 \sqrt {c}\, \sqrt {c x +b}\, c^{4} x^{5}}{15}-\frac {2 \sqrt {x}\, b^{5}}{5}+\frac {4 \sqrt {x}\, b^{4} c x}{3}-\frac {32 \sqrt {x}\, b^{3} c^{2} x^{2}}{3}-64 \sqrt {x}\, b^{2} c^{3} x^{3}-\frac {256 \sqrt {x}\, b \,c^{4} x^{4}}{3}-\frac {512 \sqrt {x}\, c^{5} x^{5}}{15}}{\sqrt {c x +b}\, b^{6} x^{3} \left (c^{2} x^{2}+2 b c x +b^{2}\right )} \] Input:
int(1/(c*x^2+b*x)^(7/2),x)
Output:
(2*(256*sqrt(c)*sqrt(b + c*x)*b**2*c**2*x**3 + 512*sqrt(c)*sqrt(b + c*x)*b *c**3*x**4 + 256*sqrt(c)*sqrt(b + c*x)*c**4*x**5 - 3*sqrt(x)*b**5 + 10*sqr t(x)*b**4*c*x - 80*sqrt(x)*b**3*c**2*x**2 - 480*sqrt(x)*b**2*c**3*x**3 - 6 40*sqrt(x)*b*c**4*x**4 - 256*sqrt(x)*c**5*x**5))/(15*sqrt(b + c*x)*b**6*x* *3*(b**2 + 2*b*c*x + c**2*x**2))