Integrand size = 23, antiderivative size = 83 \[ \int \frac {x (c+d x)}{\left (c^2-d^2 x^2\right )^{7/2}} \, dx=\frac {c+d x}{5 d^2 \left (c^2-d^2 x^2\right )^{5/2}}-\frac {x}{15 c^2 d \left (c^2-d^2 x^2\right )^{3/2}}-\frac {2 x}{15 c^4 d \sqrt {c^2-d^2 x^2}} \] Output:
1/5*(d*x+c)/d^2/(-d^2*x^2+c^2)^(5/2)-1/15*x/c^2/d/(-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 = 0.99 \[ \int \frac {x (c+d x)}{\left (c^2-d^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (3 c^4-3 c^3 d x+3 c^2 d^2 x^2+2 c d^3 x^3-2 d^4 x^4\right )}{15 c^4 d^2 (c-d x)^3 (c+d x)^2} \] Input:
Integrate[(x*(c + d*x))/(c^2 - d^2*x^2)^(7/2),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)^3*(c + d*x)^2)
Time = 0.30 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {530, 27, 209, 208}
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.
\(\displaystyle \int \frac {x (c+d x)}{\left (c^2-d^2 x^2\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 530 |
\(\displaystyle \frac {c+d x}{5 d^2 \left (c^2-d^2 x^2\right )^{5/2}}-\frac {\int \frac {c^2}{d \left (c^2-d^2 x^2\right )^{5/2}}dx}{5 c^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {c+d x}{5 d^2 \left (c^2-d^2 x^2\right )^{5/2}}-\frac {\int \frac {1}{\left (c^2-d^2 x^2\right )^{5/2}}dx}{5 d}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {c+d x}{5 d^2 \left (c^2-d^2 x^2\right )^{5/2}}-\frac {\frac {2 \int \frac {1}{\left (c^2-d^2 x^2\right )^{3/2}}dx}{3 c^2}+\frac {x}{3 c^2 \left (c^2-d^2 x^2\right )^{3/2}}}{5 d}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {c+d x}{5 d^2 \left (c^2-d^2 x^2\right )^{5/2}}-\frac {\frac {x}{3 c^2 \left (c^2-d^2 x^2\right )^{3/2}}+\frac {2 x}{3 c^4 \sqrt {c^2-d^2 x^2}}}{5 d}\) |
Input:
Int[(x*(c + d*x))/(c^2 - d^2*x^2)^(7/2),x]
Output:
(c + d*x)/(5*d^2*(c^2 - d^2*x^2)^(5/2)) - (x/(3*c^2*(c^2 - d^2*x^2)^(3/2)) + (2*x)/(3*c^4*Sqrt[c^2 - d^2*x^2]))/(5*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb ol] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Co eff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Po lynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x )*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Qx + e*(2*p + 3), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && EqQ[n, 1] && IntegerQ[2*p]
Time = 0.19 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.93
method | result | size |
gosper | \(\frac {\left (d x +c \right )^{2} \left (-d x +c \right ) \left (-2 d^{4} x^{4}+2 c \,d^{3} x^{3}+3 d^{2} c^{2} x^{2}-3 c^{3} d x +3 c^{4}\right )}{15 c^{4} d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {7}{2}}}\) | \(77\) |
orering | \(\frac {\left (d x +c \right )^{2} \left (-d x +c \right ) \left (-2 d^{4} x^{4}+2 c \,d^{3} x^{3}+3 d^{2} c^{2} x^{2}-3 c^{3} d x +3 c^{4}\right )}{15 c^{4} d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {7}{2}}}\) | \(77\) |
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 | \(\frac {c}{5 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}+d \left (\frac {x}{4 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}-\frac {c^{2} \left (\frac {x}{5 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 c^{4} \sqrt {-d^{2} x^{2}+c^{2}}}}{c^{2}}\right )}{4 d^{2}}\right )\) | \(120\) |
Input:
int(x*(d*x+c)/(-d^2*x^2+c^2)^(7/2),x,method=_RETURNVERBOSE)
Output:
1/15*(d*x+c)^2*(-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)^(7/2)
Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (71) = 142\).
Time = 0.14 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.07 \[ \int \frac {x (c+d x)}{\left (c^2-d^2 x^2\right )^{7/2}} \, dx=\frac {3 \, d^{5} x^{5} - 3 \, c d^{4} x^{4} - 6 \, c^{2} d^{3} x^{3} + 6 \, c^{3} d^{2} x^{2} + 3 \, c^{4} d x - 3 \, c^{5} + {\left (2 \, d^{4} x^{4} - 2 \, c d^{3} x^{3} - 3 \, c^{2} d^{2} x^{2} + 3 \, c^{3} d x - 3 \, c^{4}\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{15 \, {\left (c^{4} d^{7} x^{5} - c^{5} d^{6} x^{4} - 2 \, c^{6} d^{5} x^{3} + 2 \, c^{7} d^{4} x^{2} + c^{8} d^{3} x - c^{9} d^{2}\right )}} \] Input:
integrate(x*(d*x+c)/(-d^2*x^2+c^2)^(7/2),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)
Time = 6.37 (sec) , antiderivative size = 432, normalized size of antiderivative = 5.20 \[ \int \frac {x (c+d x)}{\left (c^2-d^2 x^2\right )^{7/2}} \, dx=c \left (\begin {cases} \frac {1}{5 c^{4} d^{2} \sqrt {c^{2} - d^{2} x^{2}} - 10 c^{2} d^{4} x^{2} \sqrt {c^{2} - d^{2} x^{2}} + 5 d^{6} x^{4} \sqrt {c^{2} - d^{2} x^{2}}} & \text {for}\: d \neq 0 \\\frac {x^{2}}{2 \left (c^{2}\right )^{\frac {7}{2}}} & \text {otherwise} \end {cases}\right ) + d \left (\begin {cases} - \frac {5 i c^{2} x^{3}}{15 c^{9} \sqrt {-1 + \frac {d^{2} x^{2}}{c^{2}}} - 30 c^{7} d^{2} x^{2} \sqrt {-1 + \frac {d^{2} x^{2}}{c^{2}}} + 15 c^{5} d^{4} x^{4} \sqrt {-1 + \frac {d^{2} x^{2}}{c^{2}}}} + \frac {2 i d^{2} x^{5}}{15 c^{9} \sqrt {-1 + \frac {d^{2} x^{2}}{c^{2}}} - 30 c^{7} d^{2} x^{2} \sqrt {-1 + \frac {d^{2} x^{2}}{c^{2}}} + 15 c^{5} d^{4} x^{4} \sqrt {-1 + \frac {d^{2} x^{2}}{c^{2}}}} & \text {for}\: \left |{\frac {d^{2} x^{2}}{c^{2}}}\right | > 1 \\\frac {5 c^{2} x^{3}}{15 c^{9} \sqrt {1 - \frac {d^{2} x^{2}}{c^{2}}} - 30 c^{7} d^{2} x^{2} \sqrt {1 - \frac {d^{2} x^{2}}{c^{2}}} + 15 c^{5} d^{4} x^{4} \sqrt {1 - \frac {d^{2} x^{2}}{c^{2}}}} - \frac {2 d^{2} x^{5}}{15 c^{9} \sqrt {1 - \frac {d^{2} x^{2}}{c^{2}}} - 30 c^{7} d^{2} x^{2} \sqrt {1 - \frac {d^{2} x^{2}}{c^{2}}} + 15 c^{5} d^{4} x^{4} \sqrt {1 - \frac {d^{2} x^{2}}{c^{2}}}} & \text {otherwise} \end {cases}\right ) \] Input:
integrate(x*(d*x+c)/(-d**2*x**2+c**2)**(7/2),x)
Output:
c*Piecewise((1/(5*c**4*d**2*sqrt(c**2 - d**2*x**2) - 10*c**2*d**4*x**2*sqr t(c**2 - d**2*x**2) + 5*d**6*x**4*sqrt(c**2 - d**2*x**2)), Ne(d, 0)), (x** 2/(2*(c**2)**(7/2)), True)) + d*Piecewise((-5*I*c**2*x**3/(15*c**9*sqrt(-1 + d**2*x**2/c**2) - 30*c**7*d**2*x**2*sqrt(-1 + d**2*x**2/c**2) + 15*c**5 *d**4*x**4*sqrt(-1 + d**2*x**2/c**2)) + 2*I*d**2*x**5/(15*c**9*sqrt(-1 + d **2*x**2/c**2) - 30*c**7*d**2*x**2*sqrt(-1 + d**2*x**2/c**2) + 15*c**5*d** 4*x**4*sqrt(-1 + d**2*x**2/c**2)), Abs(d**2*x**2/c**2) > 1), (5*c**2*x**3/ (15*c**9*sqrt(1 - d**2*x**2/c**2) - 30*c**7*d**2*x**2*sqrt(1 - d**2*x**2/c **2) + 15*c**5*d**4*x**4*sqrt(1 - d**2*x**2/c**2)) - 2*d**2*x**5/(15*c**9* sqrt(1 - d**2*x**2/c**2) - 30*c**7*d**2*x**2*sqrt(1 - d**2*x**2/c**2) + 15 *c**5*d**4*x**4*sqrt(1 - d**2*x**2/c**2)), True))
Time = 0.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.05 \[ \int \frac {x (c+d x)}{\left (c^2-d^2 x^2\right )^{7/2}} \, dx=\frac {x}{5 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} d} + \frac {c}{5 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} d^{2}} - \frac {x}{15 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{2} d} - \frac {2 \, x}{15 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{4} d} \] Input:
integrate(x*(d*x+c)/(-d^2*x^2+c^2)^(7/2),x, algorithm="maxima")
Output:
1/5*x/((-d^2*x^2 + c^2)^(5/2)*d) + 1/5*c/((-d^2*x^2 + c^2)^(5/2)*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)
\[ \int \frac {x (c+d x)}{\left (c^2-d^2 x^2\right )^{7/2}} \, dx=\int { \frac {{\left (d x + c\right )} x}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate(x*(d*x+c)/(-d^2*x^2+c^2)^(7/2),x, algorithm="giac")
Output:
integrate((d*x + c)*x/(-d^2*x^2 + c^2)^(7/2), x)
Time = 7.01 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.94 \[ \int \frac {x (c+d x)}{\left (c^2-d^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {c^2-d^2\,x^2}\,\left (3\,c^4-3\,c^3\,d\,x+3\,c^2\,d^2\,x^2+2\,c\,d^3\,x^3-2\,d^4\,x^4\right )}{15\,c^4\,d^2\,{\left (c+d\,x\right )}^2\,{\left (c-d\,x\right )}^3} \] Input:
int((x*(c + d*x))/(c^2 - d^2*x^2)^(7/2),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)^2*(c - d*x)^3)
Time = 0.23 (sec) , antiderivative size = 174, normalized size of antiderivative = 2.10 \[ \int \frac {x (c+d x)}{\left (c^2-d^2 x^2\right )^{7/2}} \, dx=\frac {3 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3}-3 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d x -3 \sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{2} x^{2}+3 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{3} x^{3}+3 c^{4}-3 c^{3} d x +3 c^{2} d^{2} x^{2}+2 c \,d^{3} x^{3}-2 d^{4} x^{4}}{15 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{4} d^{2} \left (d^{3} x^{3}-c \,d^{2} x^{2}-c^{2} d x +c^{3}\right )} \] Input:
int(x*(d*x+c)/(-d^2*x^2+c^2)^(7/2),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))