Integrand size = 27, antiderivative size = 95 \[ \int \frac {x^2}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {d \sqrt {d^2-e^2 x^2}}{5 e^3 (d+e x)^3}+\frac {8 \sqrt {d^2-e^2 x^2}}{15 e^3 (d+e x)^2}-\frac {7 \sqrt {d^2-e^2 x^2}}{15 d e^3 (d+e x)} \]
-1/5*d*(-e^2*x^2+d^2)^(1/2)/e^3/(e*x+d)^3+8/15*(-e^2*x^2+d^2)^(1/2)/e^3/(e *x+d)^2-7/15*(-e^2*x^2+d^2)^(1/2)/d/e^3/(e*x+d)
Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.55 \[ \int \frac {x^2}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {\left (-2 d^2-6 d e x-7 e^2 x^2\right ) \sqrt {d^2-e^2 x^2}}{15 d e^3 (d+e x)^3} \]
Time = 0.26 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.42, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {581, 25, 27, 671, 461, 460}
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.
\(\displaystyle \int \frac {x^2}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx\) |
\(\Big \downarrow \) 581 |
\(\displaystyle \frac {\sqrt {d^2-e^2 x^2}}{e^3 (d+e x)^2}-\frac {\int -\frac {d (2 d+e x)}{(d+e x)^3 \sqrt {d^2-e^2 x^2}}dx}{e^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {d (2 d+e x)}{(d+e x)^3 \sqrt {d^2-e^2 x^2}}dx}{e^2}+\frac {\sqrt {d^2-e^2 x^2}}{e^3 (d+e x)^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {d \int \frac {2 d+e x}{(d+e x)^3 \sqrt {d^2-e^2 x^2}}dx}{e^2}+\frac {\sqrt {d^2-e^2 x^2}}{e^3 (d+e x)^2}\) |
\(\Big \downarrow \) 671 |
\(\displaystyle \frac {d \left (\frac {7}{5} \int \frac {1}{(d+e x)^2 \sqrt {d^2-e^2 x^2}}dx-\frac {\sqrt {d^2-e^2 x^2}}{5 e (d+e x)^3}\right )}{e^2}+\frac {\sqrt {d^2-e^2 x^2}}{e^3 (d+e x)^2}\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {d \left (\frac {7}{5} \left (\frac {\int \frac {1}{(d+e x) \sqrt {d^2-e^2 x^2}}dx}{3 d}-\frac {\sqrt {d^2-e^2 x^2}}{3 d e (d+e x)^2}\right )-\frac {\sqrt {d^2-e^2 x^2}}{5 e (d+e x)^3}\right )}{e^2}+\frac {\sqrt {d^2-e^2 x^2}}{e^3 (d+e x)^2}\) |
\(\Big \downarrow \) 460 |
\(\displaystyle \frac {d \left (\frac {7}{5} \left (-\frac {\sqrt {d^2-e^2 x^2}}{3 d^2 e (d+e x)}-\frac {\sqrt {d^2-e^2 x^2}}{3 d e (d+e x)^2}\right )-\frac {\sqrt {d^2-e^2 x^2}}{5 e (d+e x)^3}\right )}{e^2}+\frac {\sqrt {d^2-e^2 x^2}}{e^3 (d+e x)^2}\) |
Sqrt[d^2 - e^2*x^2]/(e^3*(d + e*x)^2) + (d*(-1/5*Sqrt[d^2 - e^2*x^2]/(e*(d + e*x)^3) + (7*(-1/3*Sqrt[d^2 - e^2*x^2]/(d*e*(d + e*x)^2) - Sqrt[d^2 - e ^2*x^2]/(3*d^2*e*(d + e*x))))/5))/e^2
3.2.82.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(b*c*n)), x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && EqQ[n + 2*p + 2, 0]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[Simpl ify[n + 2*p + 2]/(2*c*(n + p + 1)) Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x ], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && ILtQ[Simp lify[n + 2*p + 2], 0] && (LtQ[n, -1] || GtQ[n + p, 0])
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[(c + d*x)^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*d^(m - 1)*(m + n + 2*p + 1))), x] + Simp[1/(d^m*(m + n + 2*p + 1)) Int[(c + d*x)^n*(a + b*x^ 2)^p*ExpandToSum[d^m*(m + n + 2*p + 1)*x^m - (m + n + 2*p + 1)*(c + d*x)^m + c*(c + d*x)^(m - 2)*(c*(m + n - 1) + c*(m + n + 2*p + 1) + 2*d*(m + n + p )*x), x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] & & IGtQ[m, 1] && NeQ[m + n + 2*p + 1, 0] && (IntegerQ[2*p] || ILtQ[m + n, 0] )
Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ ), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m + p + 1))), x] + Simp[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d)*(m + p + 1)) Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && ((LtQ[m, -1] && !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]
Time = 0.40 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.52
method | result | size |
trager | \(-\frac {\left (7 e^{2} x^{2}+6 d e x +2 d^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{15 d \,e^{3} \left (e x +d \right )^{3}}\) | \(49\) |
gosper | \(-\frac {\left (-e x +d \right ) \left (7 e^{2} x^{2}+6 d e x +2 d^{2}\right )}{15 \left (e x +d \right )^{2} d \,e^{3} \sqrt {-e^{2} x^{2}+d^{2}}}\) | \(55\) |
default | \(-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{4} d \left (x +\frac {d}{e}\right )}+\frac {d^{2} \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 d e \left (x +\frac {d}{e}\right )^{3}}+\frac {2 e \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}\right )}{5 d}\right )}{e^{5}}-\frac {2 d \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}\right )}{e^{4}}\) | \(288\) |
Time = 0.26 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.09 \[ \int \frac {x^2}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {2 \, e^{3} x^{3} + 6 \, d e^{2} x^{2} + 6 \, d^{2} e x + 2 \, d^{3} + {\left (7 \, e^{2} x^{2} + 6 \, d e x + 2 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d e^{6} x^{3} + 3 \, d^{2} e^{5} x^{2} + 3 \, d^{3} e^{4} x + d^{4} e^{3}\right )}} \]
-1/15*(2*e^3*x^3 + 6*d*e^2*x^2 + 6*d^2*e*x + 2*d^3 + (7*e^2*x^2 + 6*d*e*x + 2*d^2)*sqrt(-e^2*x^2 + d^2))/(d*e^6*x^3 + 3*d^2*e^5*x^2 + 3*d^3*e^4*x + d^4*e^3)
\[ \int \frac {x^2}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\int \frac {x^{2}}{\sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )^{3}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.32 \[ \int \frac {x^2}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {-e^{2} x^{2} + d^{2}} d}{5 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} + \frac {8 \, \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} - \frac {7 \, \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d e^{4} x + d^{2} e^{3}\right )}} \]
-1/5*sqrt(-e^2*x^2 + d^2)*d/(e^6*x^3 + 3*d*e^5*x^2 + 3*d^2*e^4*x + d^3*e^3 ) + 8/15*sqrt(-e^2*x^2 + d^2)/(e^5*x^2 + 2*d*e^4*x + d^2*e^3) - 7/15*sqrt( -e^2*x^2 + d^2)/(d*e^4*x + d^2*e^3)
Time = 0.32 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.12 \[ \int \frac {x^2}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {4 \, {\left (\frac {5 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{e^{2} x} + \frac {10 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e^{4} x^{2}} + 1\right )}}{15 \, d e^{2} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )}^{5} {\left | e \right |}} \]
4/15*(5*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 10*(d*e + sqrt(-e^2* x^2 + d^2)*abs(e))^2/(e^4*x^2) + 1)/(d*e^2*((d*e + sqrt(-e^2*x^2 + d^2)*ab s(e))/(e^2*x) + 1)^5*abs(e))
Time = 11.69 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.51 \[ \int \frac {x^2}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {d^2-e^2\,x^2}\,\left (2\,d^2+6\,d\,e\,x+7\,e^2\,x^2\right )}{15\,d\,e^3\,{\left (d+e\,x\right )}^3} \]