Integrand size = 17, antiderivative size = 120 \[ \int \frac {1}{x \left (a x+b x^2\right )^{5/2}} \, dx=\frac {2}{3 a x \left (a x+b x^2\right )^{3/2}}+\frac {16}{3 a^2 x^2 \sqrt {a x+b x^2}}-\frac {32 \sqrt {a x+b x^2}}{5 a^3 x^3}+\frac {128 b \sqrt {a x+b x^2}}{15 a^4 x^2}-\frac {256 b^2 \sqrt {a x+b x^2}}{15 a^5 x} \] Output:
2/3/a/x/(b*x^2+a*x)^(3/2)+16/3/a^2/x^2/(b*x^2+a*x)^(1/2)-32/5*(b*x^2+a*x)^ (1/2)/a^3/x^3+128/15*b*(b*x^2+a*x)^(1/2)/a^4/x^2-256/15*b^2*(b*x^2+a*x)^(1 /2)/a^5/x
Time = 0.13 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.53 \[ \int \frac {1}{x \left (a x+b x^2\right )^{5/2}} \, dx=-\frac {2 (a+b x) \left (3 a^4-8 a^3 b x+48 a^2 b^2 x^2+192 a b^3 x^3+128 b^4 x^4\right )}{15 a^5 (x (a+b x))^{5/2}} \] Input:
Integrate[1/(x*(a*x + b*x^2)^(5/2)),x]
Output:
(-2*(a + b*x)*(3*a^4 - 8*a^3*b*x + 48*a^2*b^2*x^2 + 192*a*b^3*x^3 + 128*b^ 4*x^4))/(15*a^5*(x*(a + b*x))^(5/2))
Time = 0.33 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.72, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {1129, 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.
\(\displaystyle \int \frac {1}{x \left (a x+b x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1129 |
\(\displaystyle -\frac {8 b \int \frac {1}{\left (b x^2+a x\right )^{5/2}}dx}{5 a}-\frac {2}{5 a x \left (a x+b x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 1089 |
\(\displaystyle -\frac {8 b \left (-\frac {8 b \int \frac {1}{\left (b x^2+a x\right )^{3/2}}dx}{3 a^2}-\frac {2 (a+2 b x)}{3 a^2 \left (a x+b x^2\right )^{3/2}}\right )}{5 a}-\frac {2}{5 a x \left (a x+b x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 1088 |
\(\displaystyle -\frac {8 b \left (\frac {16 b (a+2 b x)}{3 a^4 \sqrt {a x+b x^2}}-\frac {2 (a+2 b x)}{3 a^2 \left (a x+b x^2\right )^{3/2}}\right )}{5 a}-\frac {2}{5 a x \left (a x+b x^2\right )^{3/2}}\) |
Input:
Int[1/(x*(a*x + b*x^2)^(5/2)),x]
Output:
-2/(5*a*x*(a*x + b*x^2)^(3/2)) - (8*b*((-2*(a + 2*b*x))/(3*a^2*(a*x + b*x^ 2)^(3/2)) + (16*b*(a + 2*b*x))/(3*a^4*Sqrt[a*x + b*x^2])))/(5*a)
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])
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(-e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((m + p + 1)*(2* c*d - b*e))), x] + Simp[c*(Simplify[m + 2*p + 2]/((m + p + 1)*(2*c*d - b*e) )) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d , e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ILtQ[Simplify[m + 2*p + 2], 0]
Time = 0.37 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.52
method | result | size |
gosper | \(-\frac {2 \left (b x +a \right ) \left (128 b^{4} x^{4}+192 a \,b^{3} x^{3}+48 a^{2} b^{2} x^{2}-8 a^{3} b x +3 a^{4}\right )}{15 a^{5} \left (b \,x^{2}+a x \right )^{\frac {5}{2}}}\) | \(63\) |
orering | \(-\frac {2 \left (b x +a \right ) \left (128 b^{4} x^{4}+192 a \,b^{3} x^{3}+48 a^{2} b^{2} x^{2}-8 a^{3} b x +3 a^{4}\right )}{15 a^{5} \left (b \,x^{2}+a x \right )^{\frac {5}{2}}}\) | \(63\) |
pseudoelliptic | \(-\frac {2 \left (\frac {128}{3} b^{4} x^{4}+64 a \,b^{3} x^{3}+16 a^{2} b^{2} x^{2}-\frac {8}{3} a^{3} b x +a^{4}\right )}{5 \sqrt {x \left (b x +a \right )}\, x^{2} \left (b x +a \right ) a^{5}}\) | \(64\) |
trager | \(-\frac {2 \left (128 b^{4} x^{4}+192 a \,b^{3} x^{3}+48 a^{2} b^{2} x^{2}-8 a^{3} b x +3 a^{4}\right ) \sqrt {b \,x^{2}+a x}}{15 a^{5} x^{3} \left (b x +a \right )^{2}}\) | \(68\) |
default | \(-\frac {2}{5 a x \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {8 b \left (-\frac {2 \left (2 b x +a \right )}{3 a^{2} \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}+\frac {16 b \left (2 b x +a \right )}{3 a^{4} \sqrt {b \,x^{2}+a x}}\right )}{5 a}\) | \(73\) |
risch | \(-\frac {2 \left (b x +a \right ) \left (73 b^{2} x^{2}-14 a b x +3 a^{2}\right )}{15 a^{5} x^{2} \sqrt {x \left (b x +a \right )}}-\frac {2 b^{3} \left (11 b x +12 a \right ) x}{3 \sqrt {x \left (b x +a \right )}\, \left (b x +a \right ) a^{5}}\) | \(76\) |
Input:
int(1/x/(b*x^2+a*x)^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/15*(b*x+a)*(128*b^4*x^4+192*a*b^3*x^3+48*a^2*b^2*x^2-8*a^3*b*x+3*a^4)/a ^5/(b*x^2+a*x)^(5/2)
Time = 0.09 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.69 \[ \int \frac {1}{x \left (a x+b x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (128 \, b^{4} x^{4} + 192 \, a b^{3} x^{3} + 48 \, a^{2} b^{2} x^{2} - 8 \, a^{3} b x + 3 \, a^{4}\right )} \sqrt {b x^{2} + a x}}{15 \, {\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}} \] Input:
integrate(1/x/(b*x^2+a*x)^(5/2),x, algorithm="fricas")
Output:
-2/15*(128*b^4*x^4 + 192*a*b^3*x^3 + 48*a^2*b^2*x^2 - 8*a^3*b*x + 3*a^4)*s qrt(b*x^2 + a*x)/(a^5*b^2*x^5 + 2*a^6*b*x^4 + a^7*x^3)
\[ \int \frac {1}{x \left (a x+b x^2\right )^{5/2}} \, dx=\int \frac {1}{x \left (x \left (a + b x\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/x/(b*x**2+a*x)**(5/2),x)
Output:
Integral(1/(x*(x*(a + b*x))**(5/2)), x)
Time = 0.03 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x \left (a x+b x^2\right )^{5/2}} \, dx=\frac {32 \, b^{2} x}{15 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{3}} - \frac {256 \, b^{3} x}{15 \, \sqrt {b x^{2} + a x} a^{5}} + \frac {16 \, b}{15 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{2}} - \frac {128 \, b^{2}}{15 \, \sqrt {b x^{2} + a x} a^{4}} - \frac {2}{5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a x} \] Input:
integrate(1/x/(b*x^2+a*x)^(5/2),x, algorithm="maxima")
Output:
32/15*b^2*x/((b*x^2 + a*x)^(3/2)*a^3) - 256/15*b^3*x/(sqrt(b*x^2 + a*x)*a^ 5) + 16/15*b/((b*x^2 + a*x)^(3/2)*a^2) - 128/15*b^2/(sqrt(b*x^2 + a*x)*a^4 ) - 2/5/((b*x^2 + a*x)^(3/2)*a*x)
\[ \int \frac {1}{x \left (a x+b x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a x\right )}^{\frac {5}{2}} x} \,d x } \] Input:
integrate(1/x/(b*x^2+a*x)^(5/2),x, algorithm="giac")
Output:
integrate(1/((b*x^2 + a*x)^(5/2)*x), x)
Time = 10.32 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.56 \[ \int \frac {1}{x \left (a x+b x^2\right )^{5/2}} \, dx=-\frac {2\,\sqrt {b\,x^2+a\,x}\,\left (3\,a^4-8\,a^3\,b\,x+48\,a^2\,b^2\,x^2+192\,a\,b^3\,x^3+128\,b^4\,x^4\right )}{15\,a^5\,x^3\,{\left (a+b\,x\right )}^2} \] Input:
int(1/(x*(a*x + b*x^2)^(5/2)),x)
Output:
-(2*(a*x + b*x^2)^(1/2)*(3*a^4 + 128*b^4*x^4 + 192*a*b^3*x^3 + 48*a^2*b^2* x^2 - 8*a^3*b*x))/(15*a^5*x^3*(a + b*x)^2)
Time = 0.24 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x \left (a x+b x^2\right )^{5/2}} \, dx=\frac {\frac {256 \sqrt {b}\, \sqrt {b x +a}\, a \,b^{2} x^{3}}{15}+\frac {256 \sqrt {b}\, \sqrt {b x +a}\, b^{3} x^{4}}{15}-\frac {2 \sqrt {x}\, a^{4}}{5}+\frac {16 \sqrt {x}\, a^{3} b x}{15}-\frac {32 \sqrt {x}\, a^{2} b^{2} x^{2}}{5}-\frac {128 \sqrt {x}\, a \,b^{3} x^{3}}{5}-\frac {256 \sqrt {x}\, b^{4} x^{4}}{15}}{\sqrt {b x +a}\, a^{5} x^{3} \left (b x +a \right )} \] Input:
int(1/x/(b*x^2+a*x)^(5/2),x)
Output:
(2*(128*sqrt(b)*sqrt(a + b*x)*a*b**2*x**3 + 128*sqrt(b)*sqrt(a + b*x)*b**3 *x**4 - 3*sqrt(x)*a**4 + 8*sqrt(x)*a**3*b*x - 48*sqrt(x)*a**2*b**2*x**2 - 192*sqrt(x)*a*b**3*x**3 - 128*sqrt(x)*b**4*x**4))/(15*sqrt(a + b*x)*a**5*x **3*(a + b*x))