Integrand size = 24, antiderivative size = 113 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=-\frac {15 d \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 \left (d^2-e^2 x^2\right )^{3/2}}{2 e (d+e x)}-\frac {2 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^3}-\frac {15 d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e} \]
-5/2*(-e^2*x^2+d^2)^(3/2)/e/(e*x+d)-2*(-e^2*x^2+d^2)^(5/2)/e/(e*x+d)^3-15/ 2*d^2*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e-15/2*d*(-e^2*x^2+d^2)^(1/2)/e
Time = 0.33 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.79 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-24 d^2-7 d e x+e^2 x^2\right )}{2 e (d+e x)}+\frac {15 d^2 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{e} \]
(Sqrt[d^2 - e^2*x^2]*(-24*d^2 - 7*d*e*x + e^2*x^2))/(2*e*(d + e*x)) + (15* d^2*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/e
Time = 0.29 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.94, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {463, 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.
\(\displaystyle \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx\) |
\(\Big \downarrow \) 463 |
\(\displaystyle \int -\frac {7 d^2-4 e x d+e^2 x^2}{\sqrt {d^2-e^2 x^2}}dx-\frac {8 d^2 \sqrt {d^2-e^2 x^2}}{e (d+e x)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {7 d^2-4 e x d+e^2 x^2}{\sqrt {d^2-e^2 x^2}}dx-\frac {8 d^2 \sqrt {d^2-e^2 x^2}}{e (d+e x)}\) |
\(\Big \downarrow \) 2346 |
\(\displaystyle \frac {\int -\frac {d e^2 (15 d-8 e x)}{\sqrt {d^2-e^2 x^2}}dx}{2 e^2}-\frac {8 d^2 \sqrt {d^2-e^2 x^2}}{e (d+e x)}+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {d e^2 (15 d-8 e x)}{\sqrt {d^2-e^2 x^2}}dx}{2 e^2}-\frac {8 d^2 \sqrt {d^2-e^2 x^2}}{e (d+e x)}+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{2} d \int \frac {15 d-8 e x}{\sqrt {d^2-e^2 x^2}}dx-\frac {8 d^2 \sqrt {d^2-e^2 x^2}}{e (d+e x)}+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle -\frac {1}{2} d \left (15 d \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx+\frac {8 \sqrt {d^2-e^2 x^2}}{e}\right )-\frac {8 d^2 \sqrt {d^2-e^2 x^2}}{e (d+e x)}+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle -\frac {1}{2} d \left (15 d \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}+\frac {8 \sqrt {d^2-e^2 x^2}}{e}\right )-\frac {8 d^2 \sqrt {d^2-e^2 x^2}}{e (d+e x)}+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {1}{2} d \left (\frac {15 d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e}+\frac {8 \sqrt {d^2-e^2 x^2}}{e}\right )-\frac {8 d^2 \sqrt {d^2-e^2 x^2}}{e (d+e x)}+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\) |
(x*Sqrt[d^2 - e^2*x^2])/2 - (8*d^2*Sqrt[d^2 - e^2*x^2])/(e*(d + e*x)) - (d *((8*Sqrt[d^2 - e^2*x^2])/e + (15*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e)) /2
3.3.3.3.1 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[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-(-c)^(-n - 2))*d^(2*n + 3)*(Sqrt[a + b*x^2]/(2^(n + 1)*b^(n + 2)*(c + d*x ))), x] - Simp[d^(2*n + 2)/b^(n + 1) Int[(1/Sqrt[a + b*x^2])*ExpandToSum[ (2^(-n - 1)*(-c)^(-n - 1) - (-c + d*x)^(-n - 1))/(c + d*x), x], x], x] /; F reeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 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 = 0.44 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.94
method | result | size |
risch | \(-\frac {\left (-e x +8 d \right ) \sqrt {-e^{2} x^{2}+d^{2}}}{2 e}-\frac {15 d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}-\frac {8 d^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{2} \left (x +\frac {d}{e}\right )}\) | \(106\) |
default | \(\frac {-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{d e \left (x +\frac {d}{e}\right )^{4}}-\frac {3 e \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{d e \left (x +\frac {d}{e}\right )^{3}}+\frac {4 e \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{3 d e \left (x +\frac {d}{e}\right )^{2}}+\frac {5 e \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{5}+d e \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{8 e^{2}}+\frac {3 d^{2} \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{4 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )\right )}{3 d}\right )}{d}\right )}{d}}{e^{4}}\) | \(347\) |
-1/2*(-e*x+8*d)/e*(-e^2*x^2+d^2)^(1/2)-15/2*d^2/(e^2)^(1/2)*arctan((e^2)^( 1/2)*x/(-e^2*x^2+d^2)^(1/2))-8*d^2/e^2/(x+d/e)*(-(x+d/e)^2*e^2+2*d*e*(x+d/ e))^(1/2)
Time = 0.26 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.88 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=-\frac {24 \, d^{2} e x + 24 \, d^{3} - 30 \, {\left (d^{2} e x + d^{3}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (e^{2} x^{2} - 7 \, d e x - 24 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{2 \, {\left (e^{2} x + d e\right )}} \]
-1/2*(24*d^2*e*x + 24*d^3 - 30*(d^2*e*x + d^3)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (e^2*x^2 - 7*d*e*x - 24*d^2)*sqrt(-e^2*x^2 + d^2))/(e^2*x + d*e)
\[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}}}{\left (d + e x\right )^{4}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.19 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=-\frac {15 \, d^{2} \arcsin \left (\frac {e x}{d}\right )}{2 \, e} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{2 \, {\left (e^{4} x^{3} + 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x + d^{3} e\right )}} + \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d}{2 \, {\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} - \frac {15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{e^{2} x + d e} \]
-15/2*d^2*arcsin(e*x/d)/e + 1/2*(-e^2*x^2 + d^2)^(5/2)/(e^4*x^3 + 3*d*e^3* x^2 + 3*d^2*e^2*x + d^3*e) + 5/2*(-e^2*x^2 + d^2)^(3/2)*d/(e^3*x^2 + 2*d*e ^2*x + d^2*e) - 15*sqrt(-e^2*x^2 + d^2)*d^2/(e^2*x + d*e)
Time = 0.31 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.76 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=-\frac {15 \, d^{2} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{2 \, {\left | e \right |}} + \frac {1}{2} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (x - \frac {8 \, d}{e}\right )} + \frac {16 \, d^{2}}{{\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )} {\left | e \right |}} \]
-15/2*d^2*arcsin(e*x/d)*sgn(d)*sgn(e)/abs(e) + 1/2*sqrt(-e^2*x^2 + d^2)*(x - 8*d/e) + 16*d^2/(((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 1)*abs( e))
Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^{5/2}}{{\left (d+e\,x\right )}^4} \,d x \]