Integrand size = 27, antiderivative size = 192 \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {d^2 (d-e x)^4}{e^4 \sqrt {d^2-e^2 x^2}}+\frac {101 d^4 \sqrt {d^2-e^2 x^2}}{5 e^4}-\frac {19 d^3 x \sqrt {d^2-e^2 x^2}}{2 e^3}+\frac {18 d^2 x^2 \sqrt {d^2-e^2 x^2}}{5 e^2}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{e}+\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}+\frac {27 d^5 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^4} \]
27/2*d^5*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^4+d^2*(-e*x+d)^4/e^4/(-e^2*x^2 +d^2)^(1/2)+101/5*d^4*(-e^2*x^2+d^2)^(1/2)/e^4-19/2*d^3*x*(-e^2*x^2+d^2)^( 1/2)/e^3+18/5*d^2*x^2*(-e^2*x^2+d^2)^(1/2)/e^2-d*x^3*(-e^2*x^2+d^2)^(1/2)/ e+1/5*x^4*(-e^2*x^2+d^2)^(1/2)
Time = 0.32 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.67 \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {\frac {e \sqrt {d^2-e^2 x^2} \left (212 d^5+77 d^4 e x-29 d^3 e^2 x^2+16 d^2 e^3 x^3-8 d e^4 x^4+2 e^5 x^5\right )}{d+e x}+135 d^5 \sqrt {-e^2} \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{10 e^5} \]
((e*Sqrt[d^2 - e^2*x^2]*(212*d^5 + 77*d^4*e*x - 29*d^3*e^2*x^2 + 16*d^2*e^ 3*x^3 - 8*d*e^4*x^4 + 2*e^5*x^5))/(d + e*x) + 135*d^5*Sqrt[-e^2]*Log[-(Sqr t[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(10*e^5)
Time = 0.68 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.14, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {563, 2346, 25, 2346, 27, 2346, 27, 2346, 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.
| \(\displaystyle \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx\) |
| \(\Big \downarrow \) 563 |
| \(\displaystyle \frac {\int \frac {8 d^5-8 e x d^4+8 e^2 x^2 d^3-7 e^3 x^3 d^2+4 e^4 x^4 d-e^5 x^5}{\sqrt {d^2-e^2 x^2}}dx}{e^3}+\frac {8 d^5 \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}\) |
| \(\Big \downarrow \) 2346 |
| \(\displaystyle \frac {\frac {1}{5} e^3 x^4 \sqrt {d^2-e^2 x^2}-\frac {\int -\frac {20 d x^4 e^6-39 d^2 x^3 e^5+40 d^3 x^2 e^4-40 d^4 x e^3+40 d^5 e^2}{\sqrt {d^2-e^2 x^2}}dx}{5 e^2}}{e^3}+\frac {8 d^5 \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {20 d x^4 e^6-39 d^2 x^3 e^5+40 d^3 x^2 e^4-40 d^4 x e^3+40 d^5 e^2}{\sqrt {d^2-e^2 x^2}}dx}{5 e^2}+\frac {1}{5} e^3 x^4 \sqrt {d^2-e^2 x^2}}{e^3}+\frac {8 d^5 \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}\) |
| \(\Big \downarrow \) 2346 |
| \(\displaystyle \frac {\frac {-\frac {\int -\frac {4 \left (-39 d^2 x^3 e^7+55 d^3 x^2 e^6-40 d^4 x e^5+40 d^5 e^4\right )}{\sqrt {d^2-e^2 x^2}}dx}{4 e^2}-5 d e^4 x^3 \sqrt {d^2-e^2 x^2}}{5 e^2}+\frac {1}{5} e^3 x^4 \sqrt {d^2-e^2 x^2}}{e^3}+\frac {8 d^5 \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {-39 d^2 x^3 e^7+55 d^3 x^2 e^6-40 d^4 x e^5+40 d^5 e^4}{\sqrt {d^2-e^2 x^2}}dx}{e^2}-5 d e^4 x^3 \sqrt {d^2-e^2 x^2}}{5 e^2}+\frac {1}{5} e^3 x^4 \sqrt {d^2-e^2 x^2}}{e^3}+\frac {8 d^5 \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}\) |
| \(\Big \downarrow \) 2346 |
| \(\displaystyle \frac {\frac {\frac {13 d^2 e^5 x^2 \sqrt {d^2-e^2 x^2}-\frac {\int -\frac {3 \left (55 d^3 x^2 e^8-66 d^4 x e^7+40 d^5 e^6\right )}{\sqrt {d^2-e^2 x^2}}dx}{3 e^2}}{e^2}-5 d e^4 x^3 \sqrt {d^2-e^2 x^2}}{5 e^2}+\frac {1}{5} e^3 x^4 \sqrt {d^2-e^2 x^2}}{e^3}+\frac {8 d^5 \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\frac {\int \frac {55 d^3 x^2 e^8-66 d^4 x e^7+40 d^5 e^6}{\sqrt {d^2-e^2 x^2}}dx}{e^2}+13 d^2 e^5 x^2 \sqrt {d^2-e^2 x^2}}{e^2}-5 d e^4 x^3 \sqrt {d^2-e^2 x^2}}{5 e^2}+\frac {1}{5} e^3 x^4 \sqrt {d^2-e^2 x^2}}{e^3}+\frac {8 d^5 \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}\) |
| \(\Big \downarrow \) 2346 |
| \(\displaystyle \frac {\frac {\frac {\frac {-\frac {\int -\frac {3 d^4 e^8 (45 d-44 e x)}{\sqrt {d^2-e^2 x^2}}dx}{2 e^2}-\frac {55}{2} d^3 e^6 x \sqrt {d^2-e^2 x^2}}{e^2}+13 d^2 e^5 x^2 \sqrt {d^2-e^2 x^2}}{e^2}-5 d e^4 x^3 \sqrt {d^2-e^2 x^2}}{5 e^2}+\frac {1}{5} e^3 x^4 \sqrt {d^2-e^2 x^2}}{e^3}+\frac {8 d^5 \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\frac {\frac {3}{2} d^4 e^6 \int \frac {45 d-44 e x}{\sqrt {d^2-e^2 x^2}}dx-\frac {55}{2} d^3 e^6 x \sqrt {d^2-e^2 x^2}}{e^2}+13 d^2 e^5 x^2 \sqrt {d^2-e^2 x^2}}{e^2}-5 d e^4 x^3 \sqrt {d^2-e^2 x^2}}{5 e^2}+\frac {1}{5} e^3 x^4 \sqrt {d^2-e^2 x^2}}{e^3}+\frac {8 d^5 \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}\) |
| \(\Big \downarrow \) 455 |
| \(\displaystyle \frac {\frac {\frac {\frac {\frac {3}{2} d^4 e^6 \left (45 d \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx+\frac {44 \sqrt {d^2-e^2 x^2}}{e}\right )-\frac {55}{2} d^3 e^6 x \sqrt {d^2-e^2 x^2}}{e^2}+13 d^2 e^5 x^2 \sqrt {d^2-e^2 x^2}}{e^2}-5 d e^4 x^3 \sqrt {d^2-e^2 x^2}}{5 e^2}+\frac {1}{5} e^3 x^4 \sqrt {d^2-e^2 x^2}}{e^3}+\frac {8 d^5 \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {\frac {\frac {\frac {3}{2} d^4 e^6 \left (45 d \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}+\frac {44 \sqrt {d^2-e^2 x^2}}{e}\right )-\frac {55}{2} d^3 e^6 x \sqrt {d^2-e^2 x^2}}{e^2}+13 d^2 e^5 x^2 \sqrt {d^2-e^2 x^2}}{e^2}-5 d e^4 x^3 \sqrt {d^2-e^2 x^2}}{5 e^2}+\frac {1}{5} e^3 x^4 \sqrt {d^2-e^2 x^2}}{e^3}+\frac {8 d^5 \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {\frac {\frac {3}{2} d^4 e^6 \left (\frac {45 d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e}+\frac {44 \sqrt {d^2-e^2 x^2}}{e}\right )-\frac {55}{2} d^3 e^6 x \sqrt {d^2-e^2 x^2}}{e^2}+13 d^2 e^5 x^2 \sqrt {d^2-e^2 x^2}}{e^2}-5 d e^4 x^3 \sqrt {d^2-e^2 x^2}}{5 e^2}+\frac {1}{5} e^3 x^4 \sqrt {d^2-e^2 x^2}}{e^3}+\frac {8 d^5 \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}\) |
(8*d^5*Sqrt[d^2 - e^2*x^2])/(e^4*(d + e*x)) + ((e^3*x^4*Sqrt[d^2 - e^2*x^2 ])/5 + (-5*d*e^4*x^3*Sqrt[d^2 - e^2*x^2] + (13*d^2*e^5*x^2*Sqrt[d^2 - e^2* x^2] + ((-55*d^3*e^6*x*Sqrt[d^2 - e^2*x^2])/2 + (3*d^4*e^6*((44*Sqrt[d^2 - e^2*x^2])/e + (45*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e))/2)/e^2)/e^2)/( 5*e^2))/e^3
3.2.100.3.1 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_Symbo l] :> Simp[(-(-c)^(m - n - 2))*d^(2*n - m + 3)*(Sqrt[a + b*x^2]/(2^(n + 1)* b^(n + 2)*(c + d*x))), x] - Simp[d^(2*n - m + 2)/b^(n + 1) Int[(1/Sqrt[a + b*x^2])*ExpandToSum[(2^(-n - 1)*(-c)^(m - n - 1) - d^m*x^m*(-c + d*x)^(-n - 1))/(c + d*x), x], x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2 , 0] && IGtQ[m, 0] && ILtQ[n, 0] && EqQ[n + p, -3/2]
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 = 0.46 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.74
| method | result | size |
| risch | \(\frac {\left (2 e^{4} x^{4}-10 d \,e^{3} x^{3}+26 d^{2} e^{2} x^{2}-55 d^{3} e x +132 d^{4}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{10 e^{4}}+\frac {27 d^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{3} \sqrt {e^{2}}}+\frac {8 d^{5} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{5} \left (x +\frac {d}{e}\right )}\) | \(142\) |
| default | \(\text {Expression too large to display}\) | \(1086\) |
1/10*(2*e^4*x^4-10*d*e^3*x^3+26*d^2*e^2*x^2-55*d^3*e*x+132*d^4)/e^4*(-e^2* x^2+d^2)^(1/2)+27/2*d^5/e^3/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2 )^(1/2))+8*d^5/e^5/(x+d/e)*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)
Time = 0.26 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.70 \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {212 \, d^{5} e x + 212 \, d^{6} - 270 \, {\left (d^{5} e x + d^{6}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (2 \, e^{5} x^{5} - 8 \, d e^{4} x^{4} + 16 \, d^{2} e^{3} x^{3} - 29 \, d^{3} e^{2} x^{2} + 77 \, d^{4} e x + 212 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{10 \, {\left (e^{5} x + d e^{4}\right )}} \]
1/10*(212*d^5*e*x + 212*d^6 - 270*(d^5*e*x + d^6)*arctan(-(d - sqrt(-e^2*x ^2 + d^2))/(e*x)) + (2*e^5*x^5 - 8*d*e^4*x^4 + 16*d^2*e^3*x^3 - 29*d^3*e^2 *x^2 + 77*d^4*e*x + 212*d^5)*sqrt(-e^2*x^2 + d^2))/(e^5*x + d*e^4)
\[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\int \frac {x^{3} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}}}{\left (d + e x\right )^{4}}\, dx \]
Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 407, normalized size of antiderivative = 2.12 \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=-\frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{3}}{2 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} - \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4}}{2 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} + \frac {15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{5}}{e^{5} x + d e^{4}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}}{e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}} + \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3}}{2 \, {\left (e^{5} x + d e^{4}\right )}} - \frac {3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{4 \, {\left (e^{5} x + d e^{4}\right )}} + \frac {3 i \, d^{5} \arcsin \left (\frac {e x}{d} + 2\right )}{2 \, e^{4}} + \frac {15 \, d^{5} \arcsin \left (\frac {e x}{d}\right )}{e^{4}} - \frac {3 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{3} x}{2 \, e^{3}} - \frac {3 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{4}}{e^{4}} + \frac {15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4}}{2 \, e^{4}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d x}{4 \, e^{3}} - \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{4 \, e^{4}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{5 \, e^{4}} \]
-1/2*(-e^2*x^2 + d^2)^(5/2)*d^3/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3 *e^4) - 5/2*(-e^2*x^2 + d^2)^(3/2)*d^4/(e^6*x^2 + 2*d*e^5*x + d^2*e^4) + 1 5*sqrt(-e^2*x^2 + d^2)*d^5/(e^5*x + d*e^4) + (-e^2*x^2 + d^2)^(5/2)*d^2/(e ^6*x^2 + 2*d*e^5*x + d^2*e^4) + 5/2*(-e^2*x^2 + d^2)^(3/2)*d^3/(e^5*x + d* e^4) - 3/4*(-e^2*x^2 + d^2)^(5/2)*d/(e^5*x + d*e^4) + 3/2*I*d^5*arcsin(e*x /d + 2)/e^4 + 15*d^5*arcsin(e*x/d)/e^4 - 3/2*sqrt(e^2*x^2 + 4*d*e*x + 3*d^ 2)*d^3*x/e^3 - 3*sqrt(e^2*x^2 + 4*d*e*x + 3*d^2)*d^4/e^4 + 15/2*sqrt(-e^2* x^2 + d^2)*d^4/e^4 + 1/4*(-e^2*x^2 + d^2)^(3/2)*d*x/e^3 - 5/4*(-e^2*x^2 + d^2)^(3/2)*d^2/e^4 + 1/5*(-e^2*x^2 + d^2)^(5/2)/e^4
Time = 0.33 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.66 \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {27 \, d^{5} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{2 \, e^{3} {\left | e \right |}} + \frac {1}{10} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left ({\left (2 \, {\left ({\left (x - \frac {5 \, d}{e}\right )} x + \frac {13 \, d^{2}}{e^{2}}\right )} x - \frac {55 \, d^{3}}{e^{3}}\right )} x + \frac {132 \, d^{4}}{e^{4}}\right )} - \frac {16 \, d^{5}}{e^{3} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )} {\left | e \right |}} \]
27/2*d^5*arcsin(e*x/d)*sgn(d)*sgn(e)/(e^3*abs(e)) + 1/10*sqrt(-e^2*x^2 + d ^2)*((2*((x - 5*d/e)*x + 13*d^2/e^2)*x - 55*d^3/e^3)*x + 132*d^4/e^4) - 16 *d^5/(e^3*((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 1)*abs(e))
Timed out. \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\int \frac {x^3\,{\left (d^2-e^2\,x^2\right )}^{5/2}}{{\left (d+e\,x\right )}^4} \,d x \]