Integrand size = 27, antiderivative size = 123 \[ \int \frac {x^4}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=-\frac {d^3 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {17 d^2 (d-e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 (15 d-13 e x)}{15 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5} \]
-1/5*d^3*(-e*x+d)^2/e^5/(-e^2*x^2+d^2)^(5/2)+17/15*d^2*(-e*x+d)/e^5/(-e^2* x^2+d^2)^(3/2)-arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^5-2/15*(-13*e*x+15*d)/e^ 5/(-e^2*x^2+d^2)^(1/2)
Time = 0.42 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.85 \[ \int \frac {x^4}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {\frac {\sqrt {d^2-e^2 x^2} \left (-16 d^3-17 d^2 e x+22 d e^2 x^2+26 e^3 x^3\right )}{(d-e x) (d+e x)^3}+30 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{15 e^5} \]
((Sqrt[d^2 - e^2*x^2]*(-16*d^3 - 17*d^2*e*x + 22*d*e^2*x^2 + 26*e^3*x^3))/ ((d - e*x)*(d + e*x)^3) + 30*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2] )])/(15*e^5)
Time = 0.45 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.16, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {570, 529, 2166, 2345, 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.
\(\displaystyle \int \frac {x^4}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 570 |
\(\displaystyle \int \frac {x^4 (d-e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}}dx\) |
\(\Big \downarrow \) 529 |
\(\displaystyle -\frac {\int \frac {(d-e x) \left (\frac {2 d^4}{e^4}-\frac {5 x d^3}{e^3}+\frac {5 x^2 d^2}{e^2}-\frac {5 x^3 d}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d}-\frac {d^3 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 2166 |
\(\displaystyle -\frac {-\frac {\int \frac {\frac {11 d^4}{e^4}-\frac {30 x d^3}{e^3}+\frac {15 x^2 d^2}{e^2}}{\left (d^2-e^2 x^2\right )^{3/2}}dx}{3 d}-\frac {17 d^3 (d-e x)}{3 e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}-\frac {d^3 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle -\frac {-\frac {-\frac {\int \frac {15 d^4}{e^4 \sqrt {d^2-e^2 x^2}}dx}{d^2}-\frac {2 d^2 (15 d-13 e x)}{e^5 \sqrt {d^2-e^2 x^2}}}{3 d}-\frac {17 d^3 (d-e x)}{3 e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}-\frac {d^3 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {-\frac {-\frac {15 d^2 \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx}{e^4}-\frac {2 d^2 (15 d-13 e x)}{e^5 \sqrt {d^2-e^2 x^2}}}{3 d}-\frac {17 d^3 (d-e x)}{3 e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}-\frac {d^3 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle -\frac {-\frac {-\frac {15 d^2 \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}}{e^4}-\frac {2 d^2 (15 d-13 e x)}{e^5 \sqrt {d^2-e^2 x^2}}}{3 d}-\frac {17 d^3 (d-e x)}{3 e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}-\frac {d^3 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {-\frac {-\frac {15 d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5}-\frac {2 d^2 (15 d-13 e x)}{e^5 \sqrt {d^2-e^2 x^2}}}{3 d}-\frac {17 d^3 (d-e x)}{3 e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}-\frac {d^3 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}\) |
-1/5*(d^3*(d - e*x)^2)/(e^5*(d^2 - e^2*x^2)^(5/2)) - ((-17*d^3*(d - e*x))/ (3*e^5*(d^2 - e^2*x^2)^(3/2)) - ((-2*d^2*(15*d - 13*e*x))/(e^5*Sqrt[d^2 - e^2*x^2]) - (15*d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e^5)/(3*d))/(5*d)
3.2.71.3.1 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[(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]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(362\) vs. \(2(109)=218\).
Time = 0.40 (sec) , antiderivative size = 363, normalized size of antiderivative = 2.95
method | result | size |
default | \(\frac {\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}}{e^{2}}+\frac {3 x}{\sqrt {-e^{2} x^{2}+d^{2}}\, e^{4}}-\frac {2 d}{e^{5} \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {d^{4} \left (-\frac {1}{5 d e \left (x +\frac {d}{e}\right )^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}+\frac {3 e \left (-\frac {1}{3 d e \left (x +\frac {d}{e}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{3 e \,d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{5 d}\right )}{e^{6}}-\frac {4 d^{3} \left (-\frac {1}{3 d e \left (x +\frac {d}{e}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{3 e \,d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{e^{5}}\) | \(363\) |
1/e^2*(x/e^2/(-e^2*x^2+d^2)^(1/2)-1/e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/( -e^2*x^2+d^2)^(1/2)))+3/(-e^2*x^2+d^2)^(1/2)/e^4*x-2*d/e^5/(-e^2*x^2+d^2)^ (1/2)+d^4/e^6*(-1/5/d/e/(x+d/e)^2/(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)+3/5 *e/d*(-1/3/d/e/(x+d/e)/(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)-1/3/e/d^3*(-2* (x+d/e)*e^2+2*d*e)/(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)))-4/e^5*d^3*(-1/3/ d/e/(x+d/e)/(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)-1/3/e/d^3*(-2*(x+d/e)*e^2 +2*d*e)/(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2))
Time = 0.29 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.39 \[ \int \frac {x^4}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=-\frac {16 \, e^{4} x^{4} + 32 \, d e^{3} x^{3} - 32 \, d^{3} e x - 16 \, d^{4} - 30 \, {\left (e^{4} x^{4} + 2 \, d e^{3} x^{3} - 2 \, d^{3} e x - d^{4}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (26 \, e^{3} x^{3} + 22 \, d e^{2} x^{2} - 17 \, d^{2} e x - 16 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{9} x^{4} + 2 \, d e^{8} x^{3} - 2 \, d^{3} e^{6} x - d^{4} e^{5}\right )}} \]
-1/15*(16*e^4*x^4 + 32*d*e^3*x^3 - 32*d^3*e*x - 16*d^4 - 30*(e^4*x^4 + 2*d *e^3*x^3 - 2*d^3*e*x - d^4)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (2 6*e^3*x^3 + 22*d*e^2*x^2 - 17*d^2*e*x - 16*d^3)*sqrt(-e^2*x^2 + d^2))/(e^9 *x^4 + 2*d*e^8*x^3 - 2*d^3*e^6*x - d^4*e^5)
\[ \int \frac {x^4}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\int \frac {x^{4}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}} \left (d + e x\right )^{2}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.38 \[ \int \frac {x^4}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=-\frac {d^{3}}{5 \, {\left (\sqrt {-e^{2} x^{2} + d^{2}} e^{7} x^{2} + 2 \, \sqrt {-e^{2} x^{2} + d^{2}} d e^{6} x + \sqrt {-e^{2} x^{2} + d^{2}} d^{2} e^{5}\right )}} + \frac {17 \, d^{2}}{15 \, {\left (\sqrt {-e^{2} x^{2} + d^{2}} e^{6} x + \sqrt {-e^{2} x^{2} + d^{2}} d e^{5}\right )}} + \frac {26 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{4}} - \frac {\arcsin \left (\frac {e x}{d}\right )}{e^{5}} - \frac {2 \, d}{\sqrt {-e^{2} x^{2} + d^{2}} e^{5}} \]
-1/5*d^3/(sqrt(-e^2*x^2 + d^2)*e^7*x^2 + 2*sqrt(-e^2*x^2 + d^2)*d*e^6*x + sqrt(-e^2*x^2 + d^2)*d^2*e^5) + 17/15*d^2/(sqrt(-e^2*x^2 + d^2)*e^6*x + sq rt(-e^2*x^2 + d^2)*d*e^5) + 26/15*x/(sqrt(-e^2*x^2 + d^2)*e^4) - arcsin(e* x/d)/e^5 - 2*d/(sqrt(-e^2*x^2 + d^2)*e^5)
Exception generated. \[ \int \frac {x^4}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\text {Exception raised: NotImplementedError} \]
Exception raised: NotImplementedError >> unable to parse Giac output: 1/ab s(sageVARe)*(1/32768*(-20480/3*sageVARe^16*sqrt(2*sageVARd*sageVARe*(sageV ARe*sageVARx+sageVARd)^-1/sageVARe-1)*(2*sageVARd*sageVARe*(sageVARe*sageV ARx+sageVARd)^-1/sa
Timed out. \[ \int \frac {x^4}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\int \frac {x^4}{{\left (d^2-e^2\,x^2\right )}^{3/2}\,{\left (d+e\,x\right )}^2} \,d x \]