Integrand size = 27, antiderivative size = 118 \[ \int \frac {x^3 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=-\frac {d x^2 \sqrt {d^2-e^2 x^2}}{3 e^2}+\frac {x^3 \sqrt {d^2-e^2 x^2}}{4 e}-\frac {d^2 (16 d-9 e x) \sqrt {d^2-e^2 x^2}}{24 e^4}-\frac {3 d^4 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^4} \]
-3/8*d^4*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^4-1/3*d*x^2*(-e^2*x^2+d^2)^(1/ 2)/e^2+1/4*x^3*(-e^2*x^2+d^2)^(1/2)/e-1/24*d^2*(-9*e*x+16*d)*(-e^2*x^2+d^2 )^(1/2)/e^4
Time = 0.22 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.78 \[ \int \frac {x^3 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-16 d^3+9 d^2 e x-8 d e^2 x^2+6 e^3 x^3\right )+18 d^4 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{24 e^4} \]
(Sqrt[d^2 - e^2*x^2]*(-16*d^3 + 9*d^2*e*x - 8*d*e^2*x^2 + 6*e^3*x^3) + 18* d^4*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/(24*e^4)
Time = 0.31 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.28, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.481, Rules used = {566, 533, 25, 27, 533, 25, 27, 533, 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.
\(\displaystyle \int \frac {x^3 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx\) |
\(\Big \downarrow \) 566 |
\(\displaystyle \int \frac {x^3 (d-e x)}{\sqrt {d^2-e^2 x^2}}dx\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {\int -\frac {d e x^2 (3 d-4 e x)}{\sqrt {d^2-e^2 x^2}}dx}{4 e^2}+\frac {x^3 \sqrt {d^2-e^2 x^2}}{4 e}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {x^3 \sqrt {d^2-e^2 x^2}}{4 e}-\frac {\int \frac {d e x^2 (3 d-4 e x)}{\sqrt {d^2-e^2 x^2}}dx}{4 e^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x^3 \sqrt {d^2-e^2 x^2}}{4 e}-\frac {d \int \frac {x^2 (3 d-4 e x)}{\sqrt {d^2-e^2 x^2}}dx}{4 e}\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {x^3 \sqrt {d^2-e^2 x^2}}{4 e}-\frac {d \left (\frac {\int -\frac {d e x (8 d-9 e x)}{\sqrt {d^2-e^2 x^2}}dx}{3 e^2}+\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e}\right )}{4 e}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {x^3 \sqrt {d^2-e^2 x^2}}{4 e}-\frac {d \left (\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e}-\frac {\int \frac {d e x (8 d-9 e x)}{\sqrt {d^2-e^2 x^2}}dx}{3 e^2}\right )}{4 e}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x^3 \sqrt {d^2-e^2 x^2}}{4 e}-\frac {d \left (\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e}-\frac {d \int \frac {x (8 d-9 e x)}{\sqrt {d^2-e^2 x^2}}dx}{3 e}\right )}{4 e}\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {x^3 \sqrt {d^2-e^2 x^2}}{4 e}-\frac {d \left (\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e}-\frac {d \left (\frac {\int -\frac {d e (9 d-16 e x)}{\sqrt {d^2-e^2 x^2}}dx}{2 e^2}+\frac {9 x \sqrt {d^2-e^2 x^2}}{2 e}\right )}{3 e}\right )}{4 e}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {x^3 \sqrt {d^2-e^2 x^2}}{4 e}-\frac {d \left (\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e}-\frac {d \left (\frac {9 x \sqrt {d^2-e^2 x^2}}{2 e}-\frac {\int \frac {d e (9 d-16 e x)}{\sqrt {d^2-e^2 x^2}}dx}{2 e^2}\right )}{3 e}\right )}{4 e}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x^3 \sqrt {d^2-e^2 x^2}}{4 e}-\frac {d \left (\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e}-\frac {d \left (\frac {9 x \sqrt {d^2-e^2 x^2}}{2 e}-\frac {d \int \frac {9 d-16 e x}{\sqrt {d^2-e^2 x^2}}dx}{2 e}\right )}{3 e}\right )}{4 e}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {x^3 \sqrt {d^2-e^2 x^2}}{4 e}-\frac {d \left (\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e}-\frac {d \left (\frac {9 x \sqrt {d^2-e^2 x^2}}{2 e}-\frac {d \left (9 d \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx+\frac {16 \sqrt {d^2-e^2 x^2}}{e}\right )}{2 e}\right )}{3 e}\right )}{4 e}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {x^3 \sqrt {d^2-e^2 x^2}}{4 e}-\frac {d \left (\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e}-\frac {d \left (\frac {9 x \sqrt {d^2-e^2 x^2}}{2 e}-\frac {d \left (9 d \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}+\frac {16 \sqrt {d^2-e^2 x^2}}{e}\right )}{2 e}\right )}{3 e}\right )}{4 e}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {x^3 \sqrt {d^2-e^2 x^2}}{4 e}-\frac {d \left (\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e}-\frac {d \left (\frac {9 x \sqrt {d^2-e^2 x^2}}{2 e}-\frac {d \left (\frac {9 d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e}+\frac {16 \sqrt {d^2-e^2 x^2}}{e}\right )}{2 e}\right )}{3 e}\right )}{4 e}\) |
(x^3*Sqrt[d^2 - e^2*x^2])/(4*e) - (d*((4*x^2*Sqrt[d^2 - e^2*x^2])/(3*e) - (d*((9*x*Sqrt[d^2 - e^2*x^2])/(2*e) - (d*((16*Sqrt[d^2 - e^2*x^2])/e + (9* d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e))/(2*e)))/(3*e)))/(4*e)
3.1.93.3.1 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_))*((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]
Int[((x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)), x_Symbol] : > Int[x^m*(a/c + b*(x/d))*(a + b*x^2)^(p - 1), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0]
Time = 0.39 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.73
method | result | size |
risch | \(-\frac {\left (-6 e^{3} x^{3}+8 d \,e^{2} x^{2}-9 d^{2} e x +16 d^{3}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{24 e^{4}}-\frac {3 d^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 e^{3} \sqrt {e^{2}}}\) | \(86\) |
default | \(\frac {-\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{4 e^{2}}+\frac {d^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{4 e^{2}}}{e}+\frac {d^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{e^{3}}+\frac {d \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{3 e^{4}}-\frac {d^{3} \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{e^{4}}\) | \(243\) |
-1/24*(-6*e^3*x^3+8*d*e^2*x^2-9*d^2*e*x+16*d^3)/e^4*(-e^2*x^2+d^2)^(1/2)-3 /8*d^4/e^3/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))
Time = 0.28 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.70 \[ \int \frac {x^3 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=\frac {18 \, d^{4} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (6 \, e^{3} x^{3} - 8 \, d e^{2} x^{2} + 9 \, d^{2} e x - 16 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{24 \, e^{4}} \]
1/24*(18*d^4*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (6*e^3*x^3 - 8*d* e^2*x^2 + 9*d^2*e*x - 16*d^3)*sqrt(-e^2*x^2 + d^2))/e^4
\[ \int \frac {x^3 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=\int \frac {x^{3} \sqrt {- \left (- d + e x\right ) \left (d + e x\right )}}{d + e x}\, dx \]
Time = 0.28 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.86 \[ \int \frac {x^3 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=-\frac {3 \, d^{4} \arcsin \left (\frac {e x}{d}\right )}{8 \, e^{4}} + \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} x}{8 \, e^{3}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{3}}{e^{4}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} x}{4 \, e^{3}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d}{3 \, e^{4}} \]
-3/8*d^4*arcsin(e*x/d)/e^4 + 5/8*sqrt(-e^2*x^2 + d^2)*d^2*x/e^3 - sqrt(-e^ 2*x^2 + d^2)*d^3/e^4 - 1/4*(-e^2*x^2 + d^2)^(3/2)*x/e^3 + 1/3*(-e^2*x^2 + d^2)^(3/2)*d/e^4
Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.64 \[ \int \frac {x^3 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=-\frac {3 \, d^{4} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{8 \, e^{3} {\left | e \right |}} + \frac {1}{24} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left ({\left (2 \, x {\left (\frac {3 \, x}{e} - \frac {4 \, d}{e^{2}}\right )} + \frac {9 \, d^{2}}{e^{3}}\right )} x - \frac {16 \, d^{3}}{e^{4}}\right )} \]
-3/8*d^4*arcsin(e*x/d)*sgn(d)*sgn(e)/(e^3*abs(e)) + 1/24*sqrt(-e^2*x^2 + d ^2)*((2*x*(3*x/e - 4*d/e^2) + 9*d^2/e^3)*x - 16*d^3/e^4)
Timed out. \[ \int \frac {x^3 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx=\int \frac {x^3\,\sqrt {d^2-e^2\,x^2}}{d+e\,x} \,d x \]