Integrand size = 27, antiderivative size = 97 \[ \int \frac {x^3 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {d^2 (d+e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 d (d+e x)}{5 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {5 d+2 e x}{5 d e^4 \sqrt {d^2-e^2 x^2}} \]
1/5*d^2*(e*x+d)^2/e^4/(-e^2*x^2+d^2)^(5/2)-4/5*d*(e*x+d)/e^4/(-e^2*x^2+d^2 )^(3/2)+1/5*(2*e*x+5*d)/d/e^4/(-e^2*x^2+d^2)^(1/2)
Time = 0.30 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.72 \[ \int \frac {x^3 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (2 d^3-4 d^2 e x+d e^2 x^2+2 e^3 x^3\right )}{5 d e^4 (d-e x)^3 (d+e x)} \]
(Sqrt[d^2 - e^2*x^2]*(2*d^3 - 4*d^2*e*x + d*e^2*x^2 + 2*e^3*x^3))/(5*d*e^4 *(d - e*x)^3*(d + e*x))
Time = 0.32 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {529, 2166, 27, 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.
\(\displaystyle \int \frac {x^3 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 529 |
\(\displaystyle \frac {d^2 (d+e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d+e x) \left (\frac {2 d^3}{e^3}+\frac {5 x d^2}{e^2}+\frac {5 x^2 d}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d}\) |
\(\Big \downarrow \) 2166 |
\(\displaystyle \frac {d^2 (d+e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\frac {4 d^2 (d+e x)}{e^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int \frac {3 d^2 (2 d+5 e x)}{e^3 \left (d^2-e^2 x^2\right )^{3/2}}dx}{3 d}}{5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {d^2 (d+e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\frac {4 d^2 (d+e x)}{e^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {d \int \frac {2 d+5 e x}{\left (d^2-e^2 x^2\right )^{3/2}}dx}{e^3}}{5 d}\) |
\(\Big \downarrow \) 453 |
\(\displaystyle \frac {d^2 (d+e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\frac {4 d^2 (d+e x)}{e^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {5 d+2 e x}{e^4 \sqrt {d^2-e^2 x^2}}}{5 d}\) |
(d^2*(d + e*x)^2)/(5*e^4*(d^2 - e^2*x^2)^(5/2)) - ((4*d^2*(d + e*x))/(e^4* (d^2 - e^2*x^2)^(3/2)) - (5*d + 2*e*x)/(e^4*Sqrt[d^2 - e^2*x^2]))/(5*d)
3.1.46.3.1 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[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : > With[{Qx = PolynomialQuotient[Pq, a*e + b*d*x, x], R = PolynomialRemainde r[Pq, a*e + b*d*x, x]}, Simp[(-d)*R*(d + e*x)^m*((a + b*x^2)^(p + 1)/(2*a*e *(p + 1))), x] + Simp[d/(2*a*(p + 1)) Int[(d + e*x)^(m - 1)*(a + b*x^2)^( p + 1)*ExpandToSum[2*a*e*(p + 1)*Qx + R*(m + 2*p + 2), x], x], x]] /; FreeQ [{a, b, d, e}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && ILtQ[p + 1/2, 0] && GtQ[m, 0]
Time = 0.38 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.67
method | result | size |
gosper | \(\frac {\left (-e x +d \right ) \left (e x +d \right )^{3} \left (2 e^{3} x^{3}+d \,e^{2} x^{2}-4 d^{2} e x +2 d^{3}\right )}{5 d \,e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}\) | \(65\) |
trager | \(\frac {\left (2 e^{3} x^{3}+d \,e^{2} x^{2}-4 d^{2} e x +2 d^{3}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{5 d \,e^{4} \left (-e x +d \right )^{3} \left (e x +d \right )}\) | \(67\) |
default | \(e^{2} \left (\frac {x^{4}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {4 d^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )}{e^{2}}\right )+d^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )+2 d e \left (\frac {x^{3}}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {3 d^{2} \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )}{2 e^{2}}\right )\) | \(261\) |
Time = 0.40 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.20 \[ \int \frac {x^3 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {2 \, e^{4} x^{4} - 4 \, d e^{3} x^{3} + 4 \, d^{3} e x - 2 \, d^{4} - {\left (2 \, e^{3} x^{3} + d e^{2} x^{2} - 4 \, d^{2} e x + 2 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{5 \, {\left (d e^{8} x^{4} - 2 \, d^{2} e^{7} x^{3} + 2 \, d^{4} e^{5} x - d^{5} e^{4}\right )}} \]
1/5*(2*e^4*x^4 - 4*d*e^3*x^3 + 4*d^3*e*x - 2*d^4 - (2*e^3*x^3 + d*e^2*x^2 - 4*d^2*e*x + 2*d^3)*sqrt(-e^2*x^2 + d^2))/(d*e^8*x^4 - 2*d^2*e^7*x^3 + 2* d^4*e^5*x - d^5*e^4)
\[ \int \frac {x^3 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {x^{3} \left (d + e x\right )^{2}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.60 \[ \int \frac {x^3 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {d x^{3}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} - \frac {d^{2} x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {3 \, d^{3} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}} + \frac {2 \, d^{4}}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {d x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3}} + \frac {2 \, x}{5 \, \sqrt {-e^{2} x^{2} + d^{2}} d e^{3}} \]
x^4/(-e^2*x^2 + d^2)^(5/2) + d*x^3/((-e^2*x^2 + d^2)^(5/2)*e) - d^2*x^2/(( -e^2*x^2 + d^2)^(5/2)*e^2) - 3/5*d^3*x/((-e^2*x^2 + d^2)^(5/2)*e^3) + 2/5* d^4/((-e^2*x^2 + d^2)^(5/2)*e^4) + 1/5*d*x/((-e^2*x^2 + d^2)^(3/2)*e^3) + 2/5*x/(sqrt(-e^2*x^2 + d^2)*d*e^3)
\[ \int \frac {x^3 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{2} x^{3}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}}} \,d x } \]
Time = 11.60 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.68 \[ \int \frac {x^3 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2\,x^2}\,\left (2\,d^3-4\,d^2\,e\,x+d\,e^2\,x^2+2\,e^3\,x^3\right )}{5\,d\,e^4\,\left (d+e\,x\right )\,{\left (d-e\,x\right )}^3} \]