Integrand size = 27, antiderivative size = 121 \[ \int \frac {x^4 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {d^3 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {17 d^2 (d+e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (15 d+13 e x)}{15 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5} \]
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 = 0.40 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.86 \[ \int \frac {x^4 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\frac {\sqrt {d^2-e^2 x^2} \left (16 d^3-17 d^2 e x-22 d e^2 x^2+26 e^3 x^3\right )}{(d-e x)^3 (d+e x)}+30 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{15 e^5} \]
((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)^3*(d + e*x)) + 30*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2]) ])/(15*e^5)
Time = 0.42 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.17, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {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.
\(\displaystyle \int \frac {x^4 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 529 |
\(\displaystyle \frac {d^3 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d+e x) \left (\frac {2 d^4}{e^4}+\frac {5 x d^3}{e^3}+\frac {5 x^2 d^2}{e^2}+\frac {5 x^3 d}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d}\) |
\(\Big \downarrow \) 2166 |
\(\displaystyle \frac {d^3 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\frac {17 d^3 (d+e x)}{3 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int \frac {\frac {11 d^4}{e^4}+\frac {30 x d^3}{e^3}+\frac {15 x^2 d^2}{e^2}}{\left (d^2-e^2 x^2\right )^{3/2}}dx}{3 d}}{5 d}\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle \frac {d^3 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\frac {17 d^3 (d+e x)}{3 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\frac {2 d^2 (15 d+13 e x)}{e^5 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {15 d^4}{e^4 \sqrt {d^2-e^2 x^2}}dx}{d^2}}{3 d}}{5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {d^3 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\frac {17 d^3 (d+e x)}{3 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\frac {2 d^2 (15 d+13 e x)}{e^5 \sqrt {d^2-e^2 x^2}}-\frac {15 d^2 \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx}{e^4}}{3 d}}{5 d}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {d^3 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\frac {17 d^3 (d+e x)}{3 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\frac {2 d^2 (15 d+13 e x)}{e^5 \sqrt {d^2-e^2 x^2}}-\frac {15 d^2 \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}}{e^4}}{3 d}}{5 d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {d^3 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\frac {17 d^3 (d+e x)}{3 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\frac {2 d^2 (15 d+13 e x)}{e^5 \sqrt {d^2-e^2 x^2}}-\frac {15 d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5}}{3 d}}{5 d}\) |
(d^3*(d + e*x)^2)/(5*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)
3.1.45.3.1 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[(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. \(323\) vs. \(2(107)=214\).
Time = 0.41 (sec) , antiderivative size = 324, normalized size of antiderivative = 2.68
method | result | size |
default | \(e^{2} \left (\frac {x^{5}}{5 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {\frac {x^{3}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}}{e^{2}}}{e^{2}}\right )+d^{2} \left (\frac {x^{3}}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {3 d^{2} \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )}{2 e^{2}}\right )+2 d e \left (\frac {x^{4}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {4 d^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )}{e^{2}}\right )\) | \(324\) |
e^2*(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^2*(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)))))+2*d*e*(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)))
Time = 0.34 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.42 \[ \int \frac {x^4 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {16 \, e^{4} x^{4} - 32 \, d e^{3} x^{3} + 32 \, d^{3} e x - 16 \, d^{4} + 30 \, {\left (e^{4} x^{4} - 2 \, d e^{3} x^{3} + 2 \, d^{3} e x - d^{4}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (26 \, e^{3} x^{3} - 22 \, d e^{2} x^{2} - 17 \, d^{2} e x + 16 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{9} x^{4} - 2 \, d e^{8} x^{3} + 2 \, d^{3} e^{6} x - d^{4} e^{5}\right )}} \]
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)) - (26 *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)
\[ \int \frac {x^4 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {x^{4} \left (d + e x\right )^{2}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 310 vs. \(2 (107) = 214\).
Time = 0.43 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.56 \[ \int \frac {x^4 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {1}{15} \, e^{2} x {\left (\frac {15 \, x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {20 \, d^{2} x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {8 \, d^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{6}}\right )} - \frac {1}{3} \, x {\left (\frac {3 \, x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}} - \frac {2 \, d^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}}\right )} + \frac {2 \, d x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} + \frac {d^{2} x^{3}}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {8 \, d^{3} x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}} - \frac {3 \, d^{4} x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {16 \, d^{5}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{5}} + \frac {11 \, d^{2} x}{30 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}} - \frac {4 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{4}} - \frac {\arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{\sqrt {e^{2}} e^{4}} \]
1/15*e^2*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*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)) + 2*d*x^4/(( -e^2*x^2 + d^2)^(5/2)*e) + 1/2*d^2*x^3/((-e^2*x^2 + d^2)^(5/2)*e^2) - 8/3* d^3*x^2/((-e^2*x^2 + d^2)^(5/2)*e^3) - 3/10*d^4*x/((-e^2*x^2 + d^2)^(5/2)* e^4) + 16/15*d^5/((-e^2*x^2 + d^2)^(5/2)*e^5) + 11/30*d^2*x/((-e^2*x^2 + d ^2)^(3/2)*e^4) - 4/15*x/(sqrt(-e^2*x^2 + d^2)*e^4) - arcsin(e^2*x/(d*sqrt( e^2)))/(sqrt(e^2)*e^4)
\[ \int \frac {x^4 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{2} x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^4 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {x^4\,{\left (d+e\,x\right )}^2}{{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \]