Integrand size = 27, antiderivative size = 120 \[ \int \frac {x^3}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {d^2 (d-e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d-e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {32 (d-e x)}{15 e^4 \sqrt {d^2-e^2 x^2}}+\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4} \]
1/5*d^2*(-e*x+d)^3/e^4/(-e^2*x^2+d^2)^(5/2)-13/15*d*(-e*x+d)^2/e^4/(-e^2*x ^2+d^2)^(3/2)+arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^4+32/15*(-e*x+d)/e^4/(-e^ 2*x^2+d^2)^(1/2)
Time = 0.34 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.72 \[ \int \frac {x^3}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (22 d^2+51 d e x+32 e^2 x^2\right )}{15 e^4 (d+e x)^3}-\frac {2 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{e^4} \]
(Sqrt[d^2 - e^2*x^2]*(22*d^2 + 51*d*e*x + 32*e^2*x^2))/(15*e^4*(d + e*x)^3 ) - (2*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/e^4
Time = 0.43 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.15, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {570, 529, 25, 2166, 27, 665, 27, 224, 216}
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.
| \(\displaystyle \int \frac {x^3}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx\) |
| \(\Big \downarrow \) 570 |
| \(\displaystyle \int \frac {x^3 (d-e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 529 |
| \(\displaystyle \frac {d^2 (d-e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int -\frac {(d-e x)^2 \left (\frac {3 d^3}{e^3}-\frac {5 x d^2}{e^2}+\frac {5 x^2 d}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {(d-e x)^2 \left (\frac {3 d^3}{e^3}-\frac {5 x d^2}{e^2}+\frac {5 x^2 d}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d}+\frac {d^2 (d-e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 2166 |
| \(\displaystyle \frac {-\frac {\int \frac {d^2 (17 d-15 e x) (d-e x)}{e^3 \left (d^2-e^2 x^2\right )^{3/2}}dx}{3 d}-\frac {13 d^2 (d-e x)^2}{3 e^4 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}+\frac {d^2 (d-e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {d \int \frac {(17 d-15 e x) (d-e x)}{\left (d^2-e^2 x^2\right )^{3/2}}dx}{3 e^3}-\frac {13 d^2 (d-e x)^2}{3 e^4 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}+\frac {d^2 (d-e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 665 |
| \(\displaystyle \frac {-\frac {d \left (\frac {\int -\frac {15 e}{\sqrt {d^2-e^2 x^2}}dx}{e}-\frac {32 (d-e x)}{e \sqrt {d^2-e^2 x^2}}\right )}{3 e^3}-\frac {13 d^2 (d-e x)^2}{3 e^4 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}+\frac {d^2 (d-e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {d \left (-15 \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx-\frac {32 (d-e x)}{e \sqrt {d^2-e^2 x^2}}\right )}{3 e^3}-\frac {13 d^2 (d-e x)^2}{3 e^4 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}+\frac {d^2 (d-e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {-\frac {d \left (-15 \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}-\frac {32 (d-e x)}{e \sqrt {d^2-e^2 x^2}}\right )}{3 e^3}-\frac {13 d^2 (d-e x)^2}{3 e^4 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}+\frac {d^2 (d-e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {-\frac {d \left (-\frac {15 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e}-\frac {32 (d-e x)}{e \sqrt {d^2-e^2 x^2}}\right )}{3 e^3}-\frac {13 d^2 (d-e x)^2}{3 e^4 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}+\frac {d^2 (d-e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}\) |
(d^2*(d - e*x)^3)/(5*e^4*(d^2 - e^2*x^2)^(5/2)) + ((-13*d^2*(d - e*x)^2)/( 3*e^4*(d^2 - e^2*x^2)^(3/2)) - (d*((-32*(d - e*x))/(e*Sqrt[d^2 - e^2*x^2]) - (15*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e))/(3*e^3))/(5*d)
3.2.81.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
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[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c^(2*n)/a^n Int[(e*x)^m*((a + b*x^2)^(n + p)/(c - d*x)^ n), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*c^2 + a*d^2, 0] && I LtQ[n, -1] && !(IGtQ[m, 0] && ILtQ[m + n, 0] && !GtQ[p, 1])
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( x_)^2)^(3/2), x_Symbol] :> Simp[(-2^(m - 1))*d^(m - 2)*(e*f + d*g)^n*((d + e*x)/(c*e^(n - 1)*Sqrt[a + c*x^2])), x] + Simp[1/(c*e^(n - 2)) Int[Expand ToSum[(2^(m - 1)*d^(m - 1)*(e*f + d*g)^n - e^n*(d + e*x)^(m - 1)*(f + g*x)^ n)/(d - e*x), x]/Sqrt[a + c*x^2], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && EqQ[c*d^2 + a*e^2, 0] && IGtQ[m, 0] && IGtQ[n, 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]
Leaf count of result is larger than twice the leaf count of optimal. \(318\) vs. \(2(106)=212\).
Time = 0.43 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.66
| method | result | size |
| default | \(\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{3} \sqrt {e^{2}}}+\frac {3 d^{2} \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}\right )}{e^{5}}+\frac {3 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{5} \left (x +\frac {d}{e}\right )}-\frac {d^{3} \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 d e \left (x +\frac {d}{e}\right )^{3}}+\frac {2 e \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}\right )}{5 d}\right )}{e^{6}}\) | \(319\) |
1/e^3/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))+3/e^5*d^2*(-1 /3/d/e/(x+d/e)^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)-1/3/d^2/(x+d/e)*(-(x +d/e)^2*e^2+2*d*e*(x+d/e))^(1/2))+3/e^5/(x+d/e)*(-(x+d/e)^2*e^2+2*d*e*(x+d /e))^(1/2)-d^3/e^6*(-1/5/d/e/(x+d/e)^3*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2 )+2/5*e/d*(-1/3/d/e/(x+d/e)^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)-1/3/d^2 /(x+d/e)*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)))
Time = 0.28 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.31 \[ \int \frac {x^3}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {22 \, e^{3} x^{3} + 66 \, d e^{2} x^{2} + 66 \, d^{2} e x + 22 \, d^{3} - 30 \, {\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (32 \, e^{2} x^{2} + 51 \, d e x + 22 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} \]
1/15*(22*e^3*x^3 + 66*d*e^2*x^2 + 66*d^2*e*x + 22*d^3 - 30*(e^3*x^3 + 3*d* e^2*x^2 + 3*d^2*e*x + d^3)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (32 *e^2*x^2 + 51*d*e*x + 22*d^2)*sqrt(-e^2*x^2 + d^2))/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4)
\[ \int \frac {x^3}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\int \frac {x^{3}}{\sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )^{3}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.13 \[ \int \frac {x^3}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{5 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} - \frac {13 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{15 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} + \frac {32 \, \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{5} x + d e^{4}\right )}} + \frac {\arcsin \left (\frac {e x}{d}\right )}{e^{4}} \]
1/5*sqrt(-e^2*x^2 + d^2)*d^2/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^ 4) - 13/15*sqrt(-e^2*x^2 + d^2)*d/(e^6*x^2 + 2*d*e^5*x + d^2*e^4) + 32/15* sqrt(-e^2*x^2 + d^2)/(e^5*x + d*e^4) + arcsin(e*x/d)/e^4
Time = 0.32 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.54 \[ \int \frac {x^3}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {\arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{e^{3} {\left | e \right |}} - \frac {2 \, {\left (\frac {95 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{e^{2} x} + \frac {145 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e^{4} x^{2}} + \frac {75 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3}}{e^{6} x^{3}} + \frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4}}{e^{8} x^{4}} + 22\right )}}{15 \, e^{3} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )}^{5} {\left | e \right |}} \]
arcsin(e*x/d)*sgn(d)*sgn(e)/(e^3*abs(e)) - 2/15*(95*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 145*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2/(e^4*x^2 ) + 75*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^3/(e^6*x^3) + 15*(d*e + sqrt(-e ^2*x^2 + d^2)*abs(e))^4/(e^8*x^4) + 22)/(e^3*((d*e + sqrt(-e^2*x^2 + d^2)* abs(e))/(e^2*x) + 1)^5*abs(e))
Timed out. \[ \int \frac {x^3}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\int \frac {x^3}{\sqrt {d^2-e^2\,x^2}\,{\left (d+e\,x\right )}^3} \,d x \]