Integrand size = 27, antiderivative size = 146 \[ \int \frac {x^4}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d^2 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {24 d (d-e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {3 d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5} \]
-1/5*d^3*(-e*x+d)^3/e^5/(-e^2*x^2+d^2)^(5/2)+6/5*d^2*(-e*x+d)^2/e^5/(-e^2* x^2+d^2)^(3/2)-3*d*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^5-24/5*d*(-e*x+d)/e^ 5/(-e^2*x^2+d^2)^(1/2)-(-e^2*x^2+d^2)^(1/2)/e^5
Time = 0.39 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.68 \[ \int \frac {x^4}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-24 d^3-57 d^2 e x-39 d e^2 x^2-5 e^3 x^3\right )}{5 e^5 (d+e x)^3}+\frac {6 d \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{e^5} \]
(Sqrt[d^2 - e^2*x^2]*(-24*d^3 - 57*d^2*e*x - 39*d*e^2*x^2 - 5*e^3*x^3))/(5 *e^5*(d + e*x)^3) + (6*d*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/ e^5
Time = 0.59 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.14, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {570, 529, 2166, 27, 2166, 27, 455, 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.
\(\displaystyle \int \frac {x^4}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx\) |
\(\Big \downarrow \) 570 |
\(\displaystyle \int \frac {x^4 (d-e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}}dx\) |
\(\Big \downarrow \) 529 |
\(\displaystyle -\frac {\int \frac {(d-e x)^2 \left (\frac {3 d^4}{e^4}-\frac {5 x d^3}{e^3}+\frac {5 x^2 d^2}{e^2}-\frac {5 x^3 d}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d}-\frac {d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 2166 |
\(\displaystyle -\frac {-\frac {\int \frac {3 (d-e x) \left (\frac {9 d^4}{e^4}-\frac {10 x d^3}{e^3}+\frac {5 x^2 d^2}{e^2}\right )}{\left (d^2-e^2 x^2\right )^{3/2}}dx}{3 d}-\frac {6 d^3 (d-e x)^2}{e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}-\frac {d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {-\frac {\int \frac {(d-e x) \left (\frac {9 d^4}{e^4}-\frac {10 x d^3}{e^3}+\frac {5 x^2 d^2}{e^2}\right )}{\left (d^2-e^2 x^2\right )^{3/2}}dx}{d}-\frac {6 d^3 (d-e x)^2}{e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}-\frac {d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 2166 |
\(\displaystyle -\frac {-\frac {-\frac {\int \frac {5 d^3 (3 d-e x)}{e^4 \sqrt {d^2-e^2 x^2}}dx}{d}-\frac {24 d^3 (d-e x)}{e^5 \sqrt {d^2-e^2 x^2}}}{d}-\frac {6 d^3 (d-e x)^2}{e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}-\frac {d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {-\frac {-\frac {5 d^2 \int \frac {3 d-e x}{\sqrt {d^2-e^2 x^2}}dx}{e^4}-\frac {24 d^3 (d-e x)}{e^5 \sqrt {d^2-e^2 x^2}}}{d}-\frac {6 d^3 (d-e x)^2}{e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}-\frac {d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle -\frac {-\frac {-\frac {5 d^2 \left (3 d \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx+\frac {\sqrt {d^2-e^2 x^2}}{e}\right )}{e^4}-\frac {24 d^3 (d-e x)}{e^5 \sqrt {d^2-e^2 x^2}}}{d}-\frac {6 d^3 (d-e x)^2}{e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}-\frac {d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle -\frac {-\frac {-\frac {5 d^2 \left (3 d \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}+\frac {\sqrt {d^2-e^2 x^2}}{e}\right )}{e^4}-\frac {24 d^3 (d-e x)}{e^5 \sqrt {d^2-e^2 x^2}}}{d}-\frac {6 d^3 (d-e x)^2}{e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}-\frac {d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {-\frac {-\frac {5 d^2 \left (\frac {3 d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e}+\frac {\sqrt {d^2-e^2 x^2}}{e}\right )}{e^4}-\frac {24 d^3 (d-e x)}{e^5 \sqrt {d^2-e^2 x^2}}}{d}-\frac {6 d^3 (d-e x)^2}{e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}-\frac {d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}\) |
-1/5*(d^3*(d - e*x)^3)/(e^5*(d^2 - e^2*x^2)^(5/2)) - ((-6*d^3*(d - e*x)^2) /(e^5*(d^2 - e^2*x^2)^(3/2)) - ((-24*d^3*(d - e*x))/(e^5*Sqrt[d^2 - e^2*x^ 2]) - (5*d^2*(Sqrt[d^2 - e^2*x^2]/e + (3*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2 ]])/e))/e^4)/d)/(5*d)
3.2.80.3.1 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[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
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[(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 = 0.46 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.28
method | result | size |
risch | \(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{e^{5}}-\frac {3 d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{4} \sqrt {e^{2}}}-\frac {24 d \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 e^{6} \left (x +\frac {d}{e}\right )}+\frac {6 d^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 e^{7} \left (x +\frac {d}{e}\right )^{2}}-\frac {d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 e^{8} \left (x +\frac {d}{e}\right )^{3}}\) | \(187\) |
default | \(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{e^{5}}-\frac {3 d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{4} \sqrt {e^{2}}}+\frac {d^{4} \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 d e \left (x +\frac {d}{e}\right )^{3}}+\frac {2 e \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}\right )}{5 d}\right )}{e^{7}}-\frac {4 d^{3} \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}\right )}{e^{6}}-\frac {6 d \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{6} \left (x +\frac {d}{e}\right )}\) | \(340\) |
-(-e^2*x^2+d^2)^(1/2)/e^5-3*d/e^4/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x ^2+d^2)^(1/2))-24/5*d/e^6/(x+d/e)*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)+6/5 *d^2/e^7/(x+d/e)^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)-1/5*d^3/e^8/(x+d/e )^3*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)
Time = 0.30 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.19 \[ \int \frac {x^4}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {24 \, d e^{3} x^{3} + 72 \, d^{2} e^{2} x^{2} + 72 \, d^{3} e x + 24 \, d^{4} - 30 \, {\left (d e^{3} x^{3} + 3 \, d^{2} e^{2} x^{2} + 3 \, d^{3} e x + d^{4}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (5 \, e^{3} x^{3} + 39 \, d e^{2} x^{2} + 57 \, d^{2} e x + 24 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{5 \, {\left (e^{8} x^{3} + 3 \, d e^{7} x^{2} + 3 \, d^{2} e^{6} x + d^{3} e^{5}\right )}} \]
-1/5*(24*d*e^3*x^3 + 72*d^2*e^2*x^2 + 72*d^3*e*x + 24*d^4 - 30*(d*e^3*x^3 + 3*d^2*e^2*x^2 + 3*d^3*e*x + d^4)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x )) + (5*e^3*x^3 + 39*d*e^2*x^2 + 57*d^2*e*x + 24*d^3)*sqrt(-e^2*x^2 + d^2) )/(e^8*x^3 + 3*d*e^7*x^2 + 3*d^2*e^6*x + d^3*e^5)
\[ \int \frac {x^4}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\int \frac {x^{4}}{\sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )^{3}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.10 \[ \int \frac {x^4}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{3}}{5 \, {\left (e^{8} x^{3} + 3 \, d e^{7} x^{2} + 3 \, d^{2} e^{6} x + d^{3} e^{5}\right )}} + \frac {6 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{5 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} - \frac {24 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{5 \, {\left (e^{6} x + d e^{5}\right )}} - \frac {3 \, d \arcsin \left (\frac {e x}{d}\right )}{e^{5}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{e^{5}} \]
-1/5*sqrt(-e^2*x^2 + d^2)*d^3/(e^8*x^3 + 3*d*e^7*x^2 + 3*d^2*e^6*x + d^3*e ^5) + 6/5*sqrt(-e^2*x^2 + d^2)*d^2/(e^7*x^2 + 2*d*e^6*x + d^2*e^5) - 24/5* sqrt(-e^2*x^2 + d^2)*d/(e^6*x + d*e^5) - 3*d*arcsin(e*x/d)/e^5 - sqrt(-e^2 *x^2 + d^2)/e^5
Time = 0.31 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.45 \[ \int \frac {x^4}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {3 \, d \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{e^{4} {\left | e \right |}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{e^{5}} + \frac {2 \, {\left (19 \, d + \frac {80 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d}{e^{2} x} + \frac {120 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d}{e^{4} x^{2}} + \frac {70 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3} d}{e^{6} x^{3}} + \frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4} d}{e^{8} x^{4}}\right )}}{5 \, e^{4} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )}^{5} {\left | e \right |}} \]
-3*d*arcsin(e*x/d)*sgn(d)*sgn(e)/(e^4*abs(e)) - sqrt(-e^2*x^2 + d^2)/e^5 + 2/5*(19*d + 80*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))*d/(e^2*x) + 120*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2*d/(e^4*x^2) + 70*(d*e + sqrt(-e^2*x^2 + d^ 2)*abs(e))^3*d/(e^6*x^3) + 15*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^4*d/(e^8 *x^4))/(e^4*((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 1)^5*abs(e))
Timed out. \[ \int \frac {x^4}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\int \frac {x^4}{\sqrt {d^2-e^2\,x^2}\,{\left (d+e\,x\right )}^3} \,d x \]