Integrand size = 25, antiderivative size = 85 \[ \int \frac {x}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {x}{15 c^2 d \left (c^2-d^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 (c+d x) \left (c^2-d^2 x^2\right )^{3/2}}+\frac {2 x}{15 c^4 d \sqrt {c^2-d^2 x^2}} \] Output:
1/15*x/c^2/d/(-d^2*x^2+c^2)^(3/2)+1/5/d^2/(d*x+c)/(-d^2*x^2+c^2)^(3/2)+2/1 5*x/c^4/d/(-d^2*x^2+c^2)^(1/2)
Time = 0.32 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.96 \[ \int \frac {x}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (3 c^4+3 c^3 d x+3 c^2 d^2 x^2-2 c d^3 x^3-2 d^4 x^4\right )}{15 c^4 d^2 (c-d x)^2 (c+d x)^3} \] Input:
Integrate[x/((c + d*x)*(c^2 - d^2*x^2)^(5/2)),x]
Output:
(Sqrt[c^2 - d^2*x^2]*(3*c^4 + 3*c^3*d*x + 3*c^2*d^2*x^2 - 2*c*d^3*x^3 - 2* d^4*x^4))/(15*c^4*d^2*(c - d*x)^2*(c + d*x)^3)
Time = 0.33 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {565, 245, 242}
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.
\(\displaystyle \int \frac {x}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 565 |
\(\displaystyle \frac {c}{5 d^2 \left (c^2-d^2 x^2\right )^{5/2}}-d \int \frac {x^2}{\left (c^2-d^2 x^2\right )^{7/2}}dx\) |
\(\Big \downarrow \) 245 |
\(\displaystyle \frac {c}{5 d^2 \left (c^2-d^2 x^2\right )^{5/2}}-d \left (\frac {x^3}{3 c^2 \left (c^2-d^2 x^2\right )^{5/2}}-\frac {2 d^2 \int \frac {x^4}{\left (c^2-d^2 x^2\right )^{7/2}}dx}{3 c^2}\right )\) |
\(\Big \downarrow \) 242 |
\(\displaystyle \frac {c}{5 d^2 \left (c^2-d^2 x^2\right )^{5/2}}-d \left (\frac {x^3}{3 c^2 \left (c^2-d^2 x^2\right )^{5/2}}-\frac {2 d^2 x^5}{15 c^4 \left (c^2-d^2 x^2\right )^{5/2}}\right )\) |
Input:
Int[x/((c + d*x)*(c^2 - d^2*x^2)^(5/2)),x]
Output:
c/(5*d^2*(c^2 - d^2*x^2)^(5/2)) - d*(x^3/(3*c^2*(c^2 - d^2*x^2)^(5/2)) - ( 2*d^2*x^5)/(15*c^4*(c^2 - d^2*x^2)^(5/2)))
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, p}, x ] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + 2*(p + 1) + 1)/(a*(m + 1))) Int[x^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, m, p}, x] && ILtQ[Si mplify[(m + 1)/2 + p + 1], 0] && NeQ[m, -1]
Int[((x_)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)), x_Symbol] :> Sim p[a*((a + b*x^2)^p/(2*b*c*p)), x] + Simp[b/d Int[x^2*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0]
Time = 0.21 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.82
method | result | size |
gosper | \(\frac {\left (-d x +c \right ) \left (-2 d^{4} x^{4}-2 c \,d^{3} x^{3}+3 d^{2} c^{2} x^{2}+3 c^{3} d x +3 c^{4}\right )}{15 c^{4} d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\) | \(70\) |
orering | \(\frac {\left (-d x +c \right ) \left (-2 d^{4} x^{4}-2 c \,d^{3} x^{3}+3 d^{2} c^{2} x^{2}+3 c^{3} d x +3 c^{4}\right )}{15 c^{4} d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\) | \(70\) |
trager | \(\frac {\left (-2 d^{4} x^{4}-2 c \,d^{3} x^{3}+3 d^{2} c^{2} x^{2}+3 c^{3} d x +3 c^{4}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{15 c^{4} \left (d x +c \right )^{3} \left (-d x +c \right )^{2} d^{2}}\) | \(79\) |
default | \(\frac {\frac {x}{3 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{4} \sqrt {-d^{2} x^{2}+c^{2}}}}{d}-\frac {c \left (-\frac {1}{5 c d \left (x +\frac {c}{d}\right ) \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}+\frac {4 d \left (-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{6 c^{2} d^{2} \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{3 d^{2} c^{4} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{5 c}\right )}{d^{2}}\) | \(212\) |
Input:
int(x/(d*x+c)/(-d^2*x^2+c^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/15*(-d*x+c)*(-2*d^4*x^4-2*c*d^3*x^3+3*c^2*d^2*x^2+3*c^3*d*x+3*c^4)/c^4/d ^2/(-d^2*x^2+c^2)^(5/2)
Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (73) = 146\).
Time = 0.09 (sec) , antiderivative size = 171, normalized size of antiderivative = 2.01 \[ \int \frac {x}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {3 \, d^{5} x^{5} + 3 \, c d^{4} x^{4} - 6 \, c^{2} d^{3} x^{3} - 6 \, c^{3} d^{2} x^{2} + 3 \, c^{4} d x + 3 \, c^{5} - {\left (2 \, d^{4} x^{4} + 2 \, c d^{3} x^{3} - 3 \, c^{2} d^{2} x^{2} - 3 \, c^{3} d x - 3 \, c^{4}\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{15 \, {\left (c^{4} d^{7} x^{5} + c^{5} d^{6} x^{4} - 2 \, c^{6} d^{5} x^{3} - 2 \, c^{7} d^{4} x^{2} + c^{8} d^{3} x + c^{9} d^{2}\right )}} \] Input:
integrate(x/(d*x+c)/(-d^2*x^2+c^2)^(5/2),x, algorithm="fricas")
Output:
1/15*(3*d^5*x^5 + 3*c*d^4*x^4 - 6*c^2*d^3*x^3 - 6*c^3*d^2*x^2 + 3*c^4*d*x + 3*c^5 - (2*d^4*x^4 + 2*c*d^3*x^3 - 3*c^2*d^2*x^2 - 3*c^3*d*x - 3*c^4)*sq rt(-d^2*x^2 + c^2))/(c^4*d^7*x^5 + c^5*d^6*x^4 - 2*c^6*d^5*x^3 - 2*c^7*d^4 *x^2 + c^8*d^3*x + c^9*d^2)
\[ \int \frac {x}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int \frac {x}{\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {5}{2}} \left (c + d x\right )}\, dx \] Input:
integrate(x/(d*x+c)/(-d**2*x**2+c**2)**(5/2),x)
Output:
Integral(x/((-(-c + d*x)*(c + d*x))**(5/2)*(c + d*x)), x)
Time = 0.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.06 \[ \int \frac {x}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {1}{5 \, {\left ({\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{3} x + {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c d^{2}\right )}} + \frac {x}{15 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{2} d} + \frac {2 \, x}{15 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{4} d} \] Input:
integrate(x/(d*x+c)/(-d^2*x^2+c^2)^(5/2),x, algorithm="maxima")
Output:
1/5/((-d^2*x^2 + c^2)^(3/2)*d^3*x + (-d^2*x^2 + c^2)^(3/2)*c*d^2) + 1/15*x /((-d^2*x^2 + c^2)^(3/2)*c^2*d) + 2/15*x/(sqrt(-d^2*x^2 + c^2)*c^4*d)
\[ \int \frac {x}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int { \frac {x}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} {\left (d x + c\right )}} \,d x } \] Input:
integrate(x/(d*x+c)/(-d^2*x^2+c^2)^(5/2),x, algorithm="giac")
Output:
integrate(x/((-d^2*x^2 + c^2)^(5/2)*(d*x + c)), x)
Time = 6.83 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92 \[ \int \frac {x}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {c^2-d^2\,x^2}\,\left (3\,c^4+3\,c^3\,d\,x+3\,c^2\,d^2\,x^2-2\,c\,d^3\,x^3-2\,d^4\,x^4\right )}{15\,c^4\,d^2\,{\left (c+d\,x\right )}^3\,{\left (c-d\,x\right )}^2} \] Input:
int(x/((c^2 - d^2*x^2)^(5/2)*(c + d*x)),x)
Output:
((c^2 - d^2*x^2)^(1/2)*(3*c^4 - 2*d^4*x^4 - 2*c*d^3*x^3 + 3*c^2*d^2*x^2 + 3*c^3*d*x))/(15*c^4*d^2*(c + d*x)^3*(c - d*x)^2)
Time = 0.18 (sec) , antiderivative size = 174, normalized size of antiderivative = 2.05 \[ \int \frac {x}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {3 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3}+3 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d x -3 \sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{2} x^{2}-3 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{3} x^{3}+3 c^{4}+3 c^{3} d x +3 c^{2} d^{2} x^{2}-2 c \,d^{3} x^{3}-2 d^{4} x^{4}}{15 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{4} d^{2} \left (-d^{3} x^{3}-c \,d^{2} x^{2}+c^{2} d x +c^{3}\right )} \] Input:
int(x/(d*x+c)/(-d^2*x^2+c^2)^(5/2),x)
Output:
(3*sqrt(c**2 - d**2*x**2)*c**3 + 3*sqrt(c**2 - d**2*x**2)*c**2*d*x - 3*sqr t(c**2 - d**2*x**2)*c*d**2*x**2 - 3*sqrt(c**2 - d**2*x**2)*d**3*x**3 + 3*c **4 + 3*c**3*d*x + 3*c**2*d**2*x**2 - 2*c*d**3*x**3 - 2*d**4*x**4)/(15*sqr t(c**2 - d**2*x**2)*c**4*d**2*(c**3 + c**2*d*x - c*d**2*x**2 - d**3*x**3))