Integrand size = 27, antiderivative size = 128 \[ \int \frac {x^5}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {x^4 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {x^2 (4 d-5 e x)}{3 e^4 \sqrt {d^2-e^2 x^2}}-\frac {(16 d-15 e x) \sqrt {d^2-e^2 x^2}}{6 e^6}-\frac {5 d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^6} \]
1/3*x^4*(-e*x+d)/e^2/(-e^2*x^2+d^2)^(3/2)-5/2*d^2*arctan(e*x/(-e^2*x^2+d^2 )^(1/2))/e^6-1/3*x^2*(-5*e*x+4*d)/e^4/(-e^2*x^2+d^2)^(1/2)-1/6*(-15*e*x+16 *d)*(-e^2*x^2+d^2)^(1/2)/e^6
Time = 0.35 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.94 \[ \int \frac {x^5}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (16 d^4+d^3 e x-23 d^2 e^2 x^2-3 d e^3 x^3+3 e^4 x^4\right )}{6 e^6 (-d+e x) (d+e x)^2}+\frac {5 d^2 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{e^6} \]
(Sqrt[d^2 - e^2*x^2]*(16*d^4 + d^3*e*x - 23*d^2*e^2*x^2 - 3*d*e^3*x^3 + 3* e^4*x^4))/(6*e^6*(-d + e*x)*(d + e*x)^2) + (5*d^2*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/e^6
Time = 0.37 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.27, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {568, 530, 25, 2346, 25, 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.
| \(\displaystyle \int \frac {x^5}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 568 |
| \(\displaystyle \frac {x^4}{3 e^2 (d+e x) \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {x^3 (4 d-5 e x)}{\left (d^2-e^2 x^2\right )^{3/2}}dx}{3 e^2}\) |
| \(\Big \downarrow \) 530 |
| \(\displaystyle \frac {x^4}{3 e^2 (d+e x) \sqrt {d^2-e^2 x^2}}-\frac {\frac {d^2 (4 d-5 e x)}{e^4 \sqrt {d^2-e^2 x^2}}-\frac {\int -\frac {\frac {5 d^4}{e^3}-\frac {4 x d^3}{e^2}+\frac {5 x^2 d^2}{e}}{\sqrt {d^2-e^2 x^2}}dx}{d^2}}{3 e^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {x^4}{3 e^2 (d+e x) \sqrt {d^2-e^2 x^2}}-\frac {\frac {\int \frac {\frac {5 d^4}{e^3}-\frac {4 x d^3}{e^2}+\frac {5 x^2 d^2}{e}}{\sqrt {d^2-e^2 x^2}}dx}{d^2}+\frac {d^2 (4 d-5 e x)}{e^4 \sqrt {d^2-e^2 x^2}}}{3 e^2}\) |
| \(\Big \downarrow \) 2346 |
| \(\displaystyle \frac {x^4}{3 e^2 (d+e x) \sqrt {d^2-e^2 x^2}}-\frac {\frac {-\frac {\int -\frac {d^3 (15 d-8 e x)}{e \sqrt {d^2-e^2 x^2}}dx}{2 e^2}-\frac {5 d^2 x \sqrt {d^2-e^2 x^2}}{2 e^3}}{d^2}+\frac {d^2 (4 d-5 e x)}{e^4 \sqrt {d^2-e^2 x^2}}}{3 e^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {x^4}{3 e^2 (d+e x) \sqrt {d^2-e^2 x^2}}-\frac {\frac {\frac {\int \frac {d^3 (15 d-8 e x)}{e \sqrt {d^2-e^2 x^2}}dx}{2 e^2}-\frac {5 d^2 x \sqrt {d^2-e^2 x^2}}{2 e^3}}{d^2}+\frac {d^2 (4 d-5 e x)}{e^4 \sqrt {d^2-e^2 x^2}}}{3 e^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {x^4}{3 e^2 (d+e x) \sqrt {d^2-e^2 x^2}}-\frac {\frac {\frac {d^3 \int \frac {15 d-8 e x}{\sqrt {d^2-e^2 x^2}}dx}{2 e^3}-\frac {5 d^2 x \sqrt {d^2-e^2 x^2}}{2 e^3}}{d^2}+\frac {d^2 (4 d-5 e x)}{e^4 \sqrt {d^2-e^2 x^2}}}{3 e^2}\) |
| \(\Big \downarrow \) 455 |
| \(\displaystyle \frac {x^4}{3 e^2 (d+e x) \sqrt {d^2-e^2 x^2}}-\frac {\frac {\frac {d^3 \left (15 d \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx+\frac {8 \sqrt {d^2-e^2 x^2}}{e}\right )}{2 e^3}-\frac {5 d^2 x \sqrt {d^2-e^2 x^2}}{2 e^3}}{d^2}+\frac {d^2 (4 d-5 e x)}{e^4 \sqrt {d^2-e^2 x^2}}}{3 e^2}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {x^4}{3 e^2 (d+e x) \sqrt {d^2-e^2 x^2}}-\frac {\frac {\frac {d^3 \left (15 d \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}+\frac {8 \sqrt {d^2-e^2 x^2}}{e}\right )}{2 e^3}-\frac {5 d^2 x \sqrt {d^2-e^2 x^2}}{2 e^3}}{d^2}+\frac {d^2 (4 d-5 e x)}{e^4 \sqrt {d^2-e^2 x^2}}}{3 e^2}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {x^4}{3 e^2 (d+e x) \sqrt {d^2-e^2 x^2}}-\frac {\frac {\frac {d^3 \left (\frac {15 d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e}+\frac {8 \sqrt {d^2-e^2 x^2}}{e}\right )}{2 e^3}-\frac {5 d^2 x \sqrt {d^2-e^2 x^2}}{2 e^3}}{d^2}+\frac {d^2 (4 d-5 e x)}{e^4 \sqrt {d^2-e^2 x^2}}}{3 e^2}\) |
x^4/(3*e^2*(d + e*x)*Sqrt[d^2 - e^2*x^2]) - ((d^2*(4*d - 5*e*x))/(e^4*Sqrt [d^2 - e^2*x^2]) + ((-5*d^2*x*Sqrt[d^2 - e^2*x^2])/(2*e^3) + (d^3*((8*Sqrt [d^2 - e^2*x^2])/e + (15*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e))/(2*e^3)) /d^2)/(3*e^2)
3.2.27.3.1 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 = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1)) Int[(a + b*x^2)^p*ExpandToS um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && !LeQ[p, -1]
Time = 0.41 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.58
| method | result | size |
| risch | \(-\frac {\left (-e x +2 d \right ) \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{6}}-\frac {5 d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{5} \sqrt {e^{2}}}+\frac {d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{6 e^{8} \left (x +\frac {d}{e}\right )^{2}}-\frac {25 d^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{12 e^{7} \left (x +\frac {d}{e}\right )}-\frac {d^{2} \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{4 e^{7} \left (x -\frac {d}{e}\right )}\) | \(202\) |
| default | \(\frac {-\frac {x^{3}}{2 e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {3 d^{2} \left (\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}\right )}{2 e^{2}}}{e}+\frac {d^{2} x}{e^{5} \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {d^{2} \left (\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}\right )}{e^{3}}-\frac {d \left (-\frac {x^{2}}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {2 d^{2}}{e^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2}}-\frac {d^{3}}{e^{6} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {d^{5} \left (-\frac {1}{3 d e \left (x +\frac {d}{e}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{3 e \,d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{e^{6}}\) | \(350\) |
-1/2*(-e*x+2*d)/e^6*(-e^2*x^2+d^2)^(1/2)-5/2*d^2/e^5/(e^2)^(1/2)*arctan((e ^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))+1/6*d^3/e^8/(x+d/e)^2*(-(x+d/e)^2*e^2+2* d*e*(x+d/e))^(1/2)-25/12*d^2/e^7/(x+d/e)*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1 /2)-1/4*d^2/e^7/(x-d/e)*(-(x-d/e)^2*e^2-2*d*e*(x-d/e))^(1/2)
Time = 0.27 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.48 \[ \int \frac {x^5}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=-\frac {16 \, d^{2} e^{3} x^{3} + 16 \, d^{3} e^{2} x^{2} - 16 \, d^{4} e x - 16 \, d^{5} - 30 \, {\left (d^{2} e^{3} x^{3} + d^{3} e^{2} x^{2} - d^{4} e x - d^{5}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (3 \, e^{4} x^{4} - 3 \, d e^{3} x^{3} - 23 \, d^{2} e^{2} x^{2} + d^{3} e x + 16 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{6 \, {\left (e^{9} x^{3} + d e^{8} x^{2} - d^{2} e^{7} x - d^{3} e^{6}\right )}} \]
-1/6*(16*d^2*e^3*x^3 + 16*d^3*e^2*x^2 - 16*d^4*e*x - 16*d^5 - 30*(d^2*e^3* x^3 + d^3*e^2*x^2 - d^4*e*x - d^5)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x )) - (3*e^4*x^4 - 3*d*e^3*x^3 - 23*d^2*e^2*x^2 + d^3*e*x + 16*d^4)*sqrt(-e ^2*x^2 + d^2))/(e^9*x^3 + d*e^8*x^2 - d^2*e^7*x - d^3*e^6)
\[ \int \frac {x^5}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\int \frac {x^{5}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}} \left (d + e x\right )}\, dx \]
Time = 0.29 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.18 \[ \int \frac {x^5}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {d^{4}}{3 \, {\left (\sqrt {-e^{2} x^{2} + d^{2}} e^{7} x + \sqrt {-e^{2} x^{2} + d^{2}} d e^{6}\right )}} - \frac {x^{3}}{2 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{3}} + \frac {d x^{2}}{\sqrt {-e^{2} x^{2} + d^{2}} e^{4}} + \frac {17 \, d^{2} x}{6 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{5}} - \frac {5 \, d^{2} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{6}} - \frac {3 \, d^{3}}{\sqrt {-e^{2} x^{2} + d^{2}} e^{6}} \]
1/3*d^4/(sqrt(-e^2*x^2 + d^2)*e^7*x + sqrt(-e^2*x^2 + d^2)*d*e^6) - 1/2*x^ 3/(sqrt(-e^2*x^2 + d^2)*e^3) + d*x^2/(sqrt(-e^2*x^2 + d^2)*e^4) + 17/6*d^2 *x/(sqrt(-e^2*x^2 + d^2)*e^5) - 5/2*d^2*arcsin(e*x/d)/e^6 - 3*d^3/(sqrt(-e ^2*x^2 + d^2)*e^6)
\[ \int \frac {x^5}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\int { \frac {x^{5}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} {\left (e x + d\right )}} \,d x } \]
Timed out. \[ \int \frac {x^5}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\int \frac {x^5}{{\left (d^2-e^2\,x^2\right )}^{3/2}\,\left (d+e\,x\right )} \,d x \]