Integrand size = 27, antiderivative size = 93 \[ \int \frac {x^2 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {d (d+e x)^3}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {8 (d+e x)^2}{15 e^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {7 (d+e x)}{15 d e^3 \sqrt {d^2-e^2 x^2}} \]
1/5*d*(e*x+d)^3/e^3/(-e^2*x^2+d^2)^(5/2)-8/15*(e*x+d)^2/e^3/(-e^2*x^2+d^2) ^(3/2)+7/15*(e*x+d)/d/e^3/(-e^2*x^2+d^2)^(1/2)
Time = 0.31 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.57 \[ \int \frac {x^2 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (2 d^2-6 d e x+7 e^2 x^2\right )}{15 d e^3 (d-e x)^3} \]
Time = 0.23 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.09, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {529, 27, 669, 453}
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.
\(\displaystyle \int \frac {x^2 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 529 |
\(\displaystyle \frac {d (d+e x)^3}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {d (d+e x)^2 (3 d+5 e x)}{e^2 \left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {d (d+e x)^3}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d+e x)^2 (3 d+5 e x)}{\left (d^2-e^2 x^2\right )^{5/2}}dx}{5 e^2}\) |
\(\Big \downarrow \) 669 |
\(\displaystyle \frac {d (d+e x)^3}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\frac {8 (d+e x)^2}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {7}{3} \int \frac {d+e x}{\left (d^2-e^2 x^2\right )^{3/2}}dx}{5 e^2}\) |
\(\Big \downarrow \) 453 |
\(\displaystyle \frac {d (d+e x)^3}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\frac {8 (d+e x)^2}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {7 (d+e x)}{3 d e \sqrt {d^2-e^2 x^2}}}{5 e^2}\) |
(d*(d + e*x)^3)/(5*e^3*(d^2 - e^2*x^2)^(5/2)) - ((8*(d + e*x)^2)/(3*e*(d^2 - e^2*x^2)^(3/2)) - (7*(d + e*x))/(3*d*e*Sqrt[d^2 - e^2*x^2]))/(5*e^2)
3.1.86.3.1 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_))/((a_) + (b_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[-(a* d - b*c*x)/(a*b*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b, c, d}, x]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m, a*d + b*c*x, x], R = PolynomialRem ainder[x^m, a*d + b*c*x, x]}, Simp[(-c)*R*(c + d*x)^n*((a + b*x^2)^(p + 1)/ (2*a*d*(p + 1))), x] + Simp[c/(2*a*(p + 1)) Int[(c + d*x)^(n - 1)*(a + b* x^2)^(p + 1)*ExpandToSum[2*a*d*(p + 1)*Qx + R*(n + 2*p + 2), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 1] && LtQ[p, -1] && EqQ[b* c^2 + a*d^2, 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* (p + 1))), x] - Simp[e*((m*(d*g + e*f) + 2*e*f*(p + 1))/(2*c*d*(p + 1))) Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && EqQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0]
Time = 0.42 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.54
method | result | size |
trager | \(\frac {\left (7 e^{2} x^{2}-6 d e x +2 d^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{15 d \,e^{3} \left (-e x +d \right )^{3}}\) | \(50\) |
gosper | \(\frac {\left (-e x +d \right ) \left (e x +d \right )^{4} \left (7 e^{2} x^{2}-6 d e x +2 d^{2}\right )}{15 d \,e^{3} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}\) | \(55\) |
default | \(e^{3} \left (\frac {x^{4}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {4 d^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )}{e^{2}}\right )+d^{3} \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )+3 d \,e^{2} \left (\frac {x^{3}}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {3 d^{2} \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )}{2 e^{2}}\right )+3 d^{2} e \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )\) | \(365\) |
Time = 0.26 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.14 \[ \int \frac {x^2 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {2 \, e^{3} x^{3} - 6 \, d e^{2} x^{2} + 6 \, d^{2} e x - 2 \, d^{3} - {\left (7 \, e^{2} x^{2} - 6 \, d e x + 2 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d e^{6} x^{3} - 3 \, d^{2} e^{5} x^{2} + 3 \, d^{3} e^{4} x - d^{4} e^{3}\right )}} \]
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)
\[ \int \frac {x^2 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {x^{2} \left (d + e x\right )^{3}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.66 \[ \int \frac {x^2 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {e x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {3 \, d x^{3}}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {d^{2} x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} - \frac {7 \, d^{3} x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} + \frac {2 \, d^{4}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}} + \frac {7 \, d x}{30 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}} + \frac {7 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d e^{2}} \]
e*x^4/(-e^2*x^2 + d^2)^(5/2) + 3/2*d*x^3/(-e^2*x^2 + d^2)^(5/2) - 1/3*d^2* x^2/((-e^2*x^2 + d^2)^(5/2)*e) - 7/10*d^3*x/((-e^2*x^2 + d^2)^(5/2)*e^2) + 2/15*d^4/((-e^2*x^2 + d^2)^(5/2)*e^3) + 7/30*d*x/((-e^2*x^2 + d^2)^(3/2)* e^2) + 7/15*x/(sqrt(-e^2*x^2 + d^2)*d*e^2)
Time = 0.31 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.14 \[ \int \frac {x^2 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {4 \, {\left (\frac {5 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{e^{2} x} - \frac {10 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e^{4} x^{2}} - 1\right )}}{15 \, d e^{2} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} - 1\right )}^{5} {\left | e \right |}} \]
-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)*a bs(e))/(e^2*x) - 1)^5*abs(e))
Time = 11.57 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.53 \[ \int \frac {x^2 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2\,x^2}\,\left (2\,d^2-6\,d\,e\,x+7\,e^2\,x^2\right )}{15\,d\,e^3\,{\left (d-e\,x\right )}^3} \]