Integrand size = 27, antiderivative size = 148 \[ \int \frac {x^6}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {x^5 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^3 (5 d-6 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {x (5 d-8 e x)}{5 e^6 \sqrt {d^2-e^2 x^2}}-\frac {16 \sqrt {d^2-e^2 x^2}}{5 e^7}-\frac {d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^7} \]
1/5*x^5*(-e*x+d)/e^2/(-e^2*x^2+d^2)^(5/2)-1/15*x^3*(-6*e*x+5*d)/e^4/(-e^2* x^2+d^2)^(3/2)-d*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^7+1/5*x*(-8*e*x+5*d)/e ^6/(-e^2*x^2+d^2)^(1/2)-16/5*(-e^2*x^2+d^2)^(1/2)/e^7
Time = 0.47 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.88 \[ \int \frac {x^6}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-48 d^5-33 d^4 e x+87 d^3 e^2 x^2+52 d^2 e^3 x^3-38 d e^4 x^4-15 e^5 x^5\right )}{15 e^7 (-d+e x)^2 (d+e x)^3}+\frac {2 d \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{e^7} \]
(Sqrt[d^2 - e^2*x^2]*(-48*d^5 - 33*d^4*e*x + 87*d^3*e^2*x^2 + 52*d^2*e^3*x ^3 - 38*d*e^4*x^4 - 15*e^5*x^5))/(15*e^7*(-d + e*x)^2*(d + e*x)^3) + (2*d* ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/e^7
Time = 0.39 (sec) , antiderivative size = 172, 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 = {568, 530, 2345, 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.
| \(\displaystyle \int \frac {x^6}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 568 |
| \(\displaystyle \frac {x^5}{5 e^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int \frac {x^4 (5 d-6 e x)}{\left (d^2-e^2 x^2\right )^{5/2}}dx}{5 e^2}\) |
| \(\Big \downarrow \) 530 |
| \(\displaystyle \frac {x^5}{5 e^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}-\frac {-\frac {\int \frac {\frac {5 d^5}{e^4}-\frac {18 x d^4}{e^3}+\frac {15 x^2 d^3}{e^2}-\frac {18 x^3 d^2}{e}}{\left (d^2-e^2 x^2\right )^{3/2}}dx}{3 d^2}-\frac {d^3 (6 d-5 e x)}{3 e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 e^2}\) |
| \(\Big \downarrow \) 2345 |
| \(\displaystyle \frac {x^5}{5 e^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}-\frac {-\frac {-\frac {\int \frac {3 d^4 (5 d-6 e x)}{e^4 \sqrt {d^2-e^2 x^2}}dx}{d^2}-\frac {4 d^3 (9 d-5 e x)}{e^5 \sqrt {d^2-e^2 x^2}}}{3 d^2}-\frac {d^3 (6 d-5 e x)}{3 e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 e^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {x^5}{5 e^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}-\frac {-\frac {-\frac {3 d^2 \int \frac {5 d-6 e x}{\sqrt {d^2-e^2 x^2}}dx}{e^4}-\frac {4 d^3 (9 d-5 e x)}{e^5 \sqrt {d^2-e^2 x^2}}}{3 d^2}-\frac {d^3 (6 d-5 e x)}{3 e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 e^2}\) |
| \(\Big \downarrow \) 455 |
| \(\displaystyle \frac {x^5}{5 e^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}-\frac {-\frac {-\frac {3 d^2 \left (5 d \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx+\frac {6 \sqrt {d^2-e^2 x^2}}{e}\right )}{e^4}-\frac {4 d^3 (9 d-5 e x)}{e^5 \sqrt {d^2-e^2 x^2}}}{3 d^2}-\frac {d^3 (6 d-5 e x)}{3 e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 e^2}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {x^5}{5 e^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}-\frac {-\frac {-\frac {3 d^2 \left (5 d \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}+\frac {6 \sqrt {d^2-e^2 x^2}}{e}\right )}{e^4}-\frac {4 d^3 (9 d-5 e x)}{e^5 \sqrt {d^2-e^2 x^2}}}{3 d^2}-\frac {d^3 (6 d-5 e x)}{3 e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 e^2}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {x^5}{5 e^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}-\frac {-\frac {-\frac {3 d^2 \left (\frac {5 d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e}+\frac {6 \sqrt {d^2-e^2 x^2}}{e}\right )}{e^4}-\frac {4 d^3 (9 d-5 e x)}{e^5 \sqrt {d^2-e^2 x^2}}}{3 d^2}-\frac {d^3 (6 d-5 e x)}{3 e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 e^2}\) |
x^5/(5*e^2*(d + e*x)*(d^2 - e^2*x^2)^(3/2)) - (-1/3*(d^3*(6*d - 5*e*x))/(e ^5*(d^2 - e^2*x^2)^(3/2)) - ((-4*d^3*(9*d - 5*e*x))/(e^5*Sqrt[d^2 - e^2*x^ 2]) - (3*d^2*((6*Sqrt[d^2 - e^2*x^2])/e + (5*d*ArcTan[(e*x)/Sqrt[d^2 - e^2 *x^2]])/e))/e^4)/(3*d^2))/(5*e^2)
3.2.37.3.1 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_Symb ol] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Co eff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Po lynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x )*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Qx + e*(2*p + 3), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && EqQ[n, 1] && IntegerQ[2*p]
Int[((x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)), x_Symbol] : > Simp[x^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*p*(c + d*x))), x] + Simp[1/(2*d^ 2*p) Int[x^(m - 2)*(a + b*x^2)^p*(c*(m - 1) - d*m*x), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && IGtQ[m, 1] && LtQ[p, -1]
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. \(280\) vs. \(2(130)=260\).
Time = 0.43 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.90
| method | result | size |
| risch | \(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{e^{7}}-\frac {d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{6} \sqrt {e^{2}}}+\frac {23 d^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{60 e^{9} \left (x +\frac {d}{e}\right )^{2}}-\frac {493 d \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{240 e^{8} \left (x +\frac {d}{e}\right )}+\frac {d^{2} \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{24 e^{9} \left (x -\frac {d}{e}\right )^{2}}+\frac {25 d \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{48 e^{8} \left (x -\frac {d}{e}\right )}-\frac {d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{20 e^{10} \left (x +\frac {d}{e}\right )^{3}}\) | \(281\) |
| default | \(\frac {-\frac {x^{4}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {4 d^{2} \left (\frac {x^{2}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {2 d^{2}}{3 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}\right )}{e^{2}}}{e}+\frac {d^{2} \left (\frac {x^{2}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {2 d^{2}}{3 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}\right )}{e^{3}}+\frac {d^{4}}{3 e^{7} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {d^{5} \left (\frac {x}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{6}}-\frac {d \left (\frac {x^{3}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}}{e^{2}}\right )}{e^{2}}-\frac {d^{3} \left (\frac {x}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {d^{2} \left (\frac {x}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2}}\right )}{e^{4}}+\frac {d^{6} \left (-\frac {1}{5 d e \left (x +\frac {d}{e}\right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}+\frac {4 e \left (-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{6 d^{2} e^{2} \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{3 e^{2} d^{4} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{5 d}\right )}{e^{7}}\) | \(533\) |
-(-e^2*x^2+d^2)^(1/2)/e^7-d/e^6/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2 +d^2)^(1/2))+23/60*d^2/e^9/(x+d/e)^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)- 493/240*d/e^8/(x+d/e)*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)+1/24*d^2/e^9/(x -d/e)^2*(-(x-d/e)^2*e^2-2*d*e*(x-d/e))^(1/2)+25/48*d/e^8/(x-d/e)*(-(x-d/e) ^2*e^2-2*d*e*(x-d/e))^(1/2)-1/20*d^3/e^10/(x+d/e)^3*(-(x+d/e)^2*e^2+2*d*e* (x+d/e))^(1/2)
Time = 0.28 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.74 \[ \int \frac {x^6}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=-\frac {48 \, d e^{5} x^{5} + 48 \, d^{2} e^{4} x^{4} - 96 \, d^{3} e^{3} x^{3} - 96 \, d^{4} e^{2} x^{2} + 48 \, d^{5} e x + 48 \, d^{6} - 30 \, {\left (d e^{5} x^{5} + d^{2} e^{4} x^{4} - 2 \, d^{3} e^{3} x^{3} - 2 \, d^{4} e^{2} x^{2} + d^{5} e x + d^{6}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (15 \, e^{5} x^{5} + 38 \, d e^{4} x^{4} - 52 \, d^{2} e^{3} x^{3} - 87 \, d^{3} e^{2} x^{2} + 33 \, d^{4} e x + 48 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{12} x^{5} + d e^{11} x^{4} - 2 \, d^{2} e^{10} x^{3} - 2 \, d^{3} e^{9} x^{2} + d^{4} e^{8} x + d^{5} e^{7}\right )}} \]
-1/15*(48*d*e^5*x^5 + 48*d^2*e^4*x^4 - 96*d^3*e^3*x^3 - 96*d^4*e^2*x^2 + 4 8*d^5*e*x + 48*d^6 - 30*(d*e^5*x^5 + d^2*e^4*x^4 - 2*d^3*e^3*x^3 - 2*d^4*e ^2*x^2 + d^5*e*x + d^6)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (15*e^ 5*x^5 + 38*d*e^4*x^4 - 52*d^2*e^3*x^3 - 87*d^3*e^2*x^2 + 33*d^4*e*x + 48*d ^5)*sqrt(-e^2*x^2 + d^2))/(e^12*x^5 + d*e^11*x^4 - 2*d^2*e^10*x^3 - 2*d^3* e^9*x^2 + d^4*e^8*x + d^5*e^7)
\[ \int \frac {x^6}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\int \frac {x^{6}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}} \left (d + e x\right )}\, dx \]
Time = 0.31 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.75 \[ \int \frac {x^6}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=-\frac {d^{5}}{5 \, {\left ({\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{8} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d e^{7}\right )}} - \frac {x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3}} - \frac {5 \, d x^{3}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}} + \frac {20 \, d^{2} x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{5}} + \frac {64 \, d^{3} x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{6}} + \frac {x^{2}}{3 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{5}} - \frac {14 \, d^{4}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{7}} - \frac {52 \, d x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{6}} - \frac {d \arcsin \left (\frac {e x}{d}\right )}{e^{7}} + \frac {4 \, d^{2}}{3 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{7}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{3 \, e^{7}} \]
-1/5*d^5/((-e^2*x^2 + d^2)^(3/2)*e^8*x + (-e^2*x^2 + d^2)^(3/2)*d*e^7) - x ^4/((-e^2*x^2 + d^2)^(3/2)*e^3) - 5*d*x^3/((-e^2*x^2 + d^2)^(3/2)*e^4) + 2 0/3*d^2*x^2/((-e^2*x^2 + d^2)^(3/2)*e^5) + 64/15*d^3*x/((-e^2*x^2 + d^2)^( 3/2)*e^6) + 1/3*x^2/(sqrt(-e^2*x^2 + d^2)*e^5) - 14/3*d^4/((-e^2*x^2 + d^2 )^(3/2)*e^7) - 52/15*d*x/(sqrt(-e^2*x^2 + d^2)*e^6) - d*arcsin(e*x/d)/e^7 + 4/3*d^2/(sqrt(-e^2*x^2 + d^2)*e^7) + 1/3*sqrt(-e^2*x^2 + d^2)/e^7
\[ \int \frac {x^6}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\int { \frac {x^{6}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} {\left (e x + d\right )}} \,d x } \]
Timed out. \[ \int \frac {x^6}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\int \frac {x^6}{{\left (d^2-e^2\,x^2\right )}^{5/2}\,\left (d+e\,x\right )} \,d x \]