Integrand size = 23, antiderivative size = 116 \[ \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\frac {d^4 x \sqrt {d^2-e^2 x^2}}{16 e}+\frac {d^2 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e}-\frac {(6 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}+\frac {d^6 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^2} \]
1/24*d^2*x*(-e^2*x^2+d^2)^(3/2)/e-1/30*(5*e*x+6*d)*(-e^2*x^2+d^2)^(5/2)/e^ 2+1/16*d^6*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^2+1/16*d^4*x*(-e^2*x^2+d^2)^ (1/2)/e
Time = 0.00 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.98 \[ \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=-\frac {\sqrt {d^2-e^2 x^2} \left (48 d^5+15 d^4 e x-96 d^3 e^2 x^2-70 d^2 e^3 x^3+48 d e^4 x^4+40 e^5 x^5\right )+30 d^6 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{240 e^2} \]
-1/240*(Sqrt[d^2 - e^2*x^2]*(48*d^5 + 15*d^4*e*x - 96*d^3*e^2*x^2 - 70*d^2 *e^3*x^3 + 48*d*e^4*x^4 + 40*e^5*x^5) + 30*d^6*ArcTan[(e*x)/(Sqrt[d^2] - S qrt[d^2 - e^2*x^2])])/e^2
Time = 0.23 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.21, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {533, 27, 455, 211, 211, 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.
\(\displaystyle \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {\int d e (d+6 e x) \left (d^2-e^2 x^2\right )^{3/2}dx}{6 e^2}-\frac {x \left (d^2-e^2 x^2\right )^{5/2}}{6 e}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {d \int (d+6 e x) \left (d^2-e^2 x^2\right )^{3/2}dx}{6 e}-\frac {x \left (d^2-e^2 x^2\right )^{5/2}}{6 e}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {d \left (d \int \left (d^2-e^2 x^2\right )^{3/2}dx-\frac {6 \left (d^2-e^2 x^2\right )^{5/2}}{5 e}\right )}{6 e}-\frac {x \left (d^2-e^2 x^2\right )^{5/2}}{6 e}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {d \left (d \left (\frac {3}{4} d^2 \int \sqrt {d^2-e^2 x^2}dx+\frac {1}{4} x \left (d^2-e^2 x^2\right )^{3/2}\right )-\frac {6 \left (d^2-e^2 x^2\right )^{5/2}}{5 e}\right )}{6 e}-\frac {x \left (d^2-e^2 x^2\right )^{5/2}}{6 e}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {d \left (d \left (\frac {3}{4} d^2 \left (\frac {1}{2} d^2 \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )+\frac {1}{4} x \left (d^2-e^2 x^2\right )^{3/2}\right )-\frac {6 \left (d^2-e^2 x^2\right )^{5/2}}{5 e}\right )}{6 e}-\frac {x \left (d^2-e^2 x^2\right )^{5/2}}{6 e}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {d \left (d \left (\frac {3}{4} d^2 \left (\frac {1}{2} d^2 \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )+\frac {1}{4} x \left (d^2-e^2 x^2\right )^{3/2}\right )-\frac {6 \left (d^2-e^2 x^2\right )^{5/2}}{5 e}\right )}{6 e}-\frac {x \left (d^2-e^2 x^2\right )^{5/2}}{6 e}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {d \left (d \left (\frac {3}{4} d^2 \left (\frac {d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e}+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )+\frac {1}{4} x \left (d^2-e^2 x^2\right )^{3/2}\right )-\frac {6 \left (d^2-e^2 x^2\right )^{5/2}}{5 e}\right )}{6 e}-\frac {x \left (d^2-e^2 x^2\right )^{5/2}}{6 e}\) |
-1/6*(x*(d^2 - e^2*x^2)^(5/2))/e + (d*((-6*(d^2 - e^2*x^2)^(5/2))/(5*e) + d*((x*(d^2 - e^2*x^2)^(3/2))/4 + (3*d^2*((x*Sqrt[d^2 - e^2*x^2])/2 + (d^2* ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e)))/4)))/(6*e)
3.1.6.3.1 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)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[d*x^m*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 2))), x] - Simp[1/(b*(m + 2* p + 2)) Int[x^(m - 1)*(a + b*x^2)^p*Simp[a*d*m - b*c*(m + 2*p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 0] && GtQ[p, -1] && Integer Q[2*p]
Time = 0.30 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.93
method | result | size |
risch | \(-\frac {\left (40 e^{5} x^{5}+48 d \,e^{4} x^{4}-70 d^{2} e^{3} x^{3}-96 d^{3} e^{2} x^{2}+15 d^{4} e x +48 d^{5}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{240 e^{2}}+\frac {d^{6} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{16 e \sqrt {e^{2}}}\) | \(108\) |
default | \(e \left (-\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{6 e^{2}}+\frac {d^{2} \left (\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 d^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )}{6 e^{2}}\right )-\frac {d \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{5 e^{2}}\) | \(126\) |
-1/240*(40*e^5*x^5+48*d*e^4*x^4-70*d^2*e^3*x^3-96*d^3*e^2*x^2+15*d^4*e*x+4 8*d^5)/e^2*(-e^2*x^2+d^2)^(1/2)+1/16*d^6/e/(e^2)^(1/2)*arctan((e^2)^(1/2)* x/(-e^2*x^2+d^2)^(1/2))
Time = 0.35 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.91 \[ \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=-\frac {30 \, d^{6} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (40 \, e^{5} x^{5} + 48 \, d e^{4} x^{4} - 70 \, d^{2} e^{3} x^{3} - 96 \, d^{3} e^{2} x^{2} + 15 \, d^{4} e x + 48 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{240 \, e^{2}} \]
-1/240*(30*d^6*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (40*e^5*x^5 + 4 8*d*e^4*x^4 - 70*d^2*e^3*x^3 - 96*d^3*e^2*x^2 + 15*d^4*e*x + 48*d^5)*sqrt( -e^2*x^2 + d^2))/e^2
Time = 0.55 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.36 \[ \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\begin {cases} \frac {d^{6} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{16 e} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {d^{5}}{5 e^{2}} - \frac {d^{4} x}{16 e} + \frac {2 d^{3} x^{2}}{5} + \frac {7 d^{2} e x^{3}}{24} - \frac {d e^{2} x^{4}}{5} - \frac {e^{3} x^{5}}{6}\right ) & \text {for}\: e^{2} \neq 0 \\\left (\frac {d x^{2}}{2} + \frac {e x^{3}}{3}\right ) \left (d^{2}\right )^{\frac {3}{2}} & \text {otherwise} \end {cases} \]
Piecewise((d**6*Piecewise((log(-2*e**2*x + 2*sqrt(-e**2)*sqrt(d**2 - e**2* x**2))/sqrt(-e**2), Ne(d**2, 0)), (x*log(x)/sqrt(-e**2*x**2), True))/(16*e ) + sqrt(d**2 - e**2*x**2)*(-d**5/(5*e**2) - d**4*x/(16*e) + 2*d**3*x**2/5 + 7*d**2*e*x**3/24 - d*e**2*x**4/5 - e**3*x**5/6), Ne(e**2, 0)), ((d*x**2 /2 + e*x**3/3)*(d**2)**(3/2), True))
Time = 0.29 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.98 \[ \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\frac {d^{6} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{16 \, \sqrt {e^{2}} e} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{4} x}{16 \, e} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2} x}{24 \, e} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} x}{6 \, e} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{5 \, e^{2}} \]
1/16*d^6*arcsin(e^2*x/(d*sqrt(e^2)))/(sqrt(e^2)*e) + 1/16*sqrt(-e^2*x^2 + d^2)*d^4*x/e + 1/24*(-e^2*x^2 + d^2)^(3/2)*d^2*x/e - 1/6*(-e^2*x^2 + d^2)^ (5/2)*x/e - 1/5*(-e^2*x^2 + d^2)^(5/2)*d/e^2
Time = 0.31 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.81 \[ \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\frac {d^{6} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{16 \, e {\left | e \right |}} - \frac {1}{240} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (\frac {48 \, d^{5}}{e^{2}} + {\left (\frac {15 \, d^{4}}{e} - 2 \, {\left (48 \, d^{3} + {\left (35 \, d^{2} e - 4 \, {\left (5 \, e^{3} x + 6 \, d e^{2}\right )} x\right )} x\right )} x\right )} x\right )} \]
1/16*d^6*arcsin(e*x/d)*sgn(d)*sgn(e)/(e*abs(e)) - 1/240*sqrt(-e^2*x^2 + d^ 2)*(48*d^5/e^2 + (15*d^4/e - 2*(48*d^3 + (35*d^2*e - 4*(5*e^3*x + 6*d*e^2) *x)*x)*x)*x)
Timed out. \[ \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\int x\,{\left (d^2-e^2\,x^2\right )}^{3/2}\,\left (d+e\,x\right ) \,d x \]