Integrand size = 27, antiderivative size = 118 \[ \int \frac {x^3 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d+e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {32 (d+e x)}{15 e^4 \sqrt {d^2-e^2 x^2}}-\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4} \]
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 = 0.37 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.73 \[ \int \frac {x^3 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\frac {\sqrt {d^2-e^2 x^2} \left (22 d^2-51 d e x+32 e^2 x^2\right )}{(d-e x)^3}+30 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{15 e^4} \]
((Sqrt[d^2 - e^2*x^2]*(22*d^2 - 51*d*e*x + 32*e^2*x^2))/(d - e*x)^3 + 30*A rcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/(15*e^4)
Time = 0.39 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {529, 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.
| \(\displaystyle \int \frac {x^3 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx\) |
| \(\Big \downarrow \) 529 |
| \(\displaystyle \frac {d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d+e x)^2 \left (\frac {3 d^3}{e^3}+\frac {5 x d^2}{e^2}+\frac {5 x^2 d}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d}\) |
| \(\Big \downarrow \) 2166 |
| \(\displaystyle \frac {d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\frac {13 d^2 (d+e x)^2}{3 e^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int \frac {d^2 (d+e x) (17 d+15 e x)}{e^3 \left (d^2-e^2 x^2\right )^{3/2}}dx}{3 d}}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\frac {13 d^2 (d+e x)^2}{3 e^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {d \int \frac {(d+e x) (17 d+15 e x)}{\left (d^2-e^2 x^2\right )^{3/2}}dx}{3 e^3}}{5 d}\) |
| \(\Big \downarrow \) 665 |
| \(\displaystyle \frac {d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\frac {13 d^2 (d+e x)^2}{3 e^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {d \left (\frac {32 (d+e x)}{e \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {15 e}{\sqrt {d^2-e^2 x^2}}dx}{e}\right )}{3 e^3}}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\frac {13 d^2 (d+e x)^2}{3 e^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {d \left (\frac {32 (d+e x)}{e \sqrt {d^2-e^2 x^2}}-15 \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx\right )}{3 e^3}}{5 d}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\frac {13 d^2 (d+e x)^2}{3 e^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {d \left (\frac {32 (d+e x)}{e \sqrt {d^2-e^2 x^2}}-15 \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{3 e^3}}{5 d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\frac {13 d^2 (d+e x)^2}{3 e^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {d \left (\frac {32 (d+e x)}{e \sqrt {d^2-e^2 x^2}}-\frac {15 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e}\right )}{3 e^3}}{5 d}\) |
(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)
3.1.85.3.1 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[(((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. \(376\) vs. \(2(104)=208\).
Time = 0.40 (sec) , antiderivative size = 377, normalized size of antiderivative = 3.19
| method | result | size |
| default | \(e^{3} \left (\frac {x^{5}}{5 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {\frac {x^{3}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}}{e^{2}}}{e^{2}}\right )+d^{3} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )+3 d \,e^{2} \left (\frac {x^{4}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {4 d^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )}{e^{2}}\right )+3 d^{2} e \left (\frac {x^{3}}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {3 d^{2} \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )}{2 e^{2}}\right )\) | \(377\) |
e^3*(1/5*x^5/e^2/(-e^2*x^2+d^2)^(5/2)-1/e^2*(1/3*x^3/e^2/(-e^2*x^2+d^2)^(3 /2)-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)))))+d^3*(1/3*x^2/e^2/(-e^2*x^2+d^2)^(5/2)-2/15*d^2 /e^4/(-e^2*x^2+d^2)^(5/2))+3*d*e^2*(x^4/e^2/(-e^2*x^2+d^2)^(5/2)-4*d^2/e^2 *(1/3*x^2/e^2/(-e^2*x^2+d^2)^(5/2)-2/15*d^2/e^4/(-e^2*x^2+d^2)^(5/2)))+3*d ^2*e*(1/2*x^3/e^2/(-e^2*x^2+d^2)^(5/2)-3/2*d^2/e^2*(1/4*x/e^2/(-e^2*x^2+d^ 2)^(5/2)-1/4*d^2/e^2*(1/5*x/d^2/(-e^2*x^2+d^2)^(5/2)+4/5/d^2*(1/3*x/d^2/(- e^2*x^2+d^2)^(3/2)+2/3*x/d^4/(-e^2*x^2+d^2)^(1/2)))))
Time = 0.28 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.36 \[ \int \frac {x^3 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {22 \, e^{3} x^{3} - 66 \, d e^{2} x^{2} + 66 \, d^{2} e x - 22 \, d^{3} + 30 \, {\left (e^{3} x^{3} - 3 \, d e^{2} x^{2} + 3 \, d^{2} e x - d^{3}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (32 \, e^{2} x^{2} - 51 \, d e x + 22 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{7} x^{3} - 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x - d^{3} e^{4}\right )}} \]
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)
\[ \int \frac {x^3 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {x^{3} \left (d + e x\right )^{3}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 308 vs. \(2 (104) = 208\).
Time = 0.30 (sec) , antiderivative size = 308, normalized size of antiderivative = 2.61 \[ \int \frac {x^3 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {1}{15} \, e^{3} x {\left (\frac {15 \, x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {20 \, d^{2} x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {8 \, d^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{6}}\right )} - \frac {1}{3} \, e x {\left (\frac {3 \, x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}} - \frac {2 \, d^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}}\right )} + \frac {3 \, d x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {3 \, d^{2} x^{3}}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} - \frac {11 \, d^{3} x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {9 \, d^{4} x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}} + \frac {22 \, d^{5}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {17 \, d^{2} x}{30 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3}} + \frac {2 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{3}} - \frac {\arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{\sqrt {e^{2}} e^{3}} \]
1/15*e^3*x*(15*x^4/((-e^2*x^2 + d^2)^(5/2)*e^2) - 20*d^2*x^2/((-e^2*x^2 + d^2)^(5/2)*e^4) + 8*d^4/((-e^2*x^2 + d^2)^(5/2)*e^6)) - 1/3*e*x*(3*x^2/((- e^2*x^2 + d^2)^(3/2)*e^2) - 2*d^2/((-e^2*x^2 + d^2)^(3/2)*e^4)) + 3*d*x^4/ (-e^2*x^2 + d^2)^(5/2) + 3/2*d^2*x^3/((-e^2*x^2 + d^2)^(5/2)*e) - 11/3*d^3 *x^2/((-e^2*x^2 + d^2)^(5/2)*e^2) - 9/10*d^4*x/((-e^2*x^2 + d^2)^(5/2)*e^3 ) + 22/15*d^5/((-e^2*x^2 + d^2)^(5/2)*e^4) + 17/30*d^2*x/((-e^2*x^2 + d^2) ^(3/2)*e^3) + 2/15*x/(sqrt(-e^2*x^2 + d^2)*e^3) - arcsin(e^2*x/(d*sqrt(e^2 )))/(sqrt(e^2)*e^3)
Time = 0.30 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.58 \[ \int \frac {x^3 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {\arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{e^{3} {\left | e \right |}} - \frac {2 \, {\left (\frac {95 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{e^{2} x} - \frac {145 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e^{4} x^{2}} + \frac {75 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3}}{e^{6} x^{3}} - \frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4}}{e^{8} x^{4}} - 22\right )}}{15 \, e^{3} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} - 1\right )}^{5} {\left | e \right |}} \]
-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. \[ \int \frac {x^3 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {x^3\,{\left (d+e\,x\right )}^3}{{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \]