Integrand size = 25, antiderivative size = 119 \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{d+e x} \, dx=\frac {d^2 \left (d^2-e^2 x^2\right )^p}{2 e^3 p}-\frac {\left (d^2-e^2 x^2\right )^{1+p}}{2 e^3 (1+p)}+\frac {x^3 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},1-p,\frac {5}{2},\frac {e^2 x^2}{d^2}\right )}{3 d} \]
1/2*d^2*(-e^2*x^2+d^2)^p/e^3/p-1/2*(-e^2*x^2+d^2)^(p+1)/e^3/(p+1)+1/3*x^3* (-e^2*x^2+d^2)^p*hypergeom([3/2, 1-p],[5/2],e^2*x^2/d^2)/d/((1-e^2*x^2/d^2 )^p)
Time = 0.34 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.66 \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{d+e x} \, dx=-\frac {\left (1+\frac {e x}{d}\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (\left (1+\frac {e x}{d}\right )^p \left (-e^2 x^2 \left (1-\frac {e^2 x^2}{d^2}\right )^p+d^2 \left (-1+\left (1-\frac {e^2 x^2}{d^2}\right )^p\right )\right )+2 d e (1+p) x \left (1+\frac {e x}{d}\right )^p \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )+d (d-e x) \left (2-\frac {2 e^2 x^2}{d^2}\right )^p \operatorname {Hypergeometric2F1}\left (1-p,1+p,2+p,\frac {d-e x}{2 d}\right )\right )}{2 e^3 (1+p)} \]
-1/2*((d^2 - e^2*x^2)^p*((1 + (e*x)/d)^p*(-(e^2*x^2*(1 - (e^2*x^2)/d^2)^p) + d^2*(-1 + (1 - (e^2*x^2)/d^2)^p)) + 2*d*e*(1 + p)*x*(1 + (e*x)/d)^p*Hyp ergeometric2F1[1/2, -p, 3/2, (e^2*x^2)/d^2] + d*(d - e*x)*(2 - (2*e^2*x^2) /d^2)^p*Hypergeometric2F1[1 - p, 1 + p, 2 + p, (d - e*x)/(2*d)]))/(e^3*(1 + p)*(1 + (e*x)/d)^p*(1 - (e^2*x^2)/d^2)^p)
Time = 0.28 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {583, 542, 243, 53, 279, 278, 2009}
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.
| \(\displaystyle \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{d+e x} \, dx\) |
| \(\Big \downarrow \) 583 |
| \(\displaystyle \int x^2 (d-e x) \left (d^2-e^2 x^2\right )^{p-1}dx\) |
| \(\Big \downarrow \) 542 |
| \(\displaystyle d \int x^2 \left (d^2-e^2 x^2\right )^{p-1}dx-e \int x^3 \left (d^2-e^2 x^2\right )^{p-1}dx\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle d \int x^2 \left (d^2-e^2 x^2\right )^{p-1}dx-\frac {1}{2} e \int x^2 \left (d^2-e^2 x^2\right )^{p-1}dx^2\) |
| \(\Big \downarrow \) 53 |
| \(\displaystyle d \int x^2 \left (d^2-e^2 x^2\right )^{p-1}dx-\frac {1}{2} e \int \left (\frac {d^2 \left (d^2-e^2 x^2\right )^{p-1}}{e^2}-\frac {\left (d^2-e^2 x^2\right )^p}{e^2}\right )dx^2\) |
| \(\Big \downarrow \) 279 |
| \(\displaystyle \frac {\left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \int x^2 \left (1-\frac {e^2 x^2}{d^2}\right )^{p-1}dx}{d}-\frac {1}{2} e \int \left (\frac {d^2 \left (d^2-e^2 x^2\right )^{p-1}}{e^2}-\frac {\left (d^2-e^2 x^2\right )^p}{e^2}\right )dx^2\) |
| \(\Big \downarrow \) 278 |
| \(\displaystyle \frac {x^3 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},1-p,\frac {5}{2},\frac {e^2 x^2}{d^2}\right )}{3 d}-\frac {1}{2} e \int \left (\frac {d^2 \left (d^2-e^2 x^2\right )^{p-1}}{e^2}-\frac {\left (d^2-e^2 x^2\right )^p}{e^2}\right )dx^2\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x^3 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},1-p,\frac {5}{2},\frac {e^2 x^2}{d^2}\right )}{3 d}-\frac {1}{2} e \left (\frac {\left (d^2-e^2 x^2\right )^{p+1}}{e^4 (p+1)}-\frac {d^2 \left (d^2-e^2 x^2\right )^p}{e^4 p}\right )\) |
-1/2*(e*(-((d^2*(d^2 - e^2*x^2)^p)/(e^4*p)) + (d^2 - e^2*x^2)^(1 + p)/(e^4 *(1 + p)))) + (x^3*(d^2 - e^2*x^2)^p*Hypergeometric2F1[3/2, 1 - p, 5/2, (e ^2*x^2)/d^2])/(3*d*(1 - (e^2*x^2)/d^2)^p)
3.3.69.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(c*x)^m* (1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && !(ILtQ[p, 0] || GtQ[a, 0])
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c Int[x^m*(a + b*x^2)^p, x], x] + Simp[d Int[x^(m + 1)*(a + b*x^2 )^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && IntegerQ[m] && !IntegerQ[2*p]
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c^(2*n)/a^n Int[(e*x)^m*((a + b*x^2)^(n + p)/(c - d*x)^ n), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*c^2 + a*d^2, 0] && I LtQ[n, 0]
\[\int \frac {x^{2} \left (-e^{2} x^{2}+d^{2}\right )^{p}}{e x +d}d x\]
\[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{d+e x} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{2}}{e x + d} \,d x } \]
Result contains complex when optimal does not.
Time = 12.29 (sec) , antiderivative size = 14895, normalized size of antiderivative = 125.17 \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{d+e x} \, dx=\text {Too large to display} \]
Piecewise((0**p*d**4*d**(2*p + 2)*p*log(d**2/(e**2*x**2))*gamma(1/2 - p)*g amma(p + 1)/(-2*d**4*e**3*p*gamma(1/2 - p)*gamma(p + 1) - 2*d**4*e**3*gamm a(1/2 - p)*gamma(p + 1) + 2*d**2*e**5*p*x**2*gamma(1/2 - p)*gamma(p + 1) + 2*d**2*e**5*x**2*gamma(1/2 - p)*gamma(p + 1)) - 0**p*d**4*d**(2*p + 2)*p* log(d**2/(e**2*x**2) - 1)*gamma(1/2 - p)*gamma(p + 1)/(-2*d**4*e**3*p*gamm a(1/2 - p)*gamma(p + 1) - 2*d**4*e**3*gamma(1/2 - p)*gamma(p + 1) + 2*d**2 *e**5*p*x**2*gamma(1/2 - p)*gamma(p + 1) + 2*d**2*e**5*x**2*gamma(1/2 - p) *gamma(p + 1)) - 2*0**p*d**4*d**(2*p + 2)*p*acoth(d/(e*x))*gamma(1/2 - p)* gamma(p + 1)/(-2*d**4*e**3*p*gamma(1/2 - p)*gamma(p + 1) - 2*d**4*e**3*gam ma(1/2 - p)*gamma(p + 1) + 2*d**2*e**5*p*x**2*gamma(1/2 - p)*gamma(p + 1) + 2*d**2*e**5*x**2*gamma(1/2 - p)*gamma(p + 1)) + 0**p*d**4*d**(2*p + 2)*l og(d**2/(e**2*x**2))*gamma(1/2 - p)*gamma(p + 1)/(-2*d**4*e**3*p*gamma(1/2 - p)*gamma(p + 1) - 2*d**4*e**3*gamma(1/2 - p)*gamma(p + 1) + 2*d**2*e**5 *p*x**2*gamma(1/2 - p)*gamma(p + 1) + 2*d**2*e**5*x**2*gamma(1/2 - p)*gamm a(p + 1)) - 0**p*d**4*d**(2*p + 2)*log(d**2/(e**2*x**2) - 1)*gamma(1/2 - p )*gamma(p + 1)/(-2*d**4*e**3*p*gamma(1/2 - p)*gamma(p + 1) - 2*d**4*e**3*g amma(1/2 - p)*gamma(p + 1) + 2*d**2*e**5*p*x**2*gamma(1/2 - p)*gamma(p + 1 ) + 2*d**2*e**5*x**2*gamma(1/2 - p)*gamma(p + 1)) - 2*0**p*d**4*d**(2*p + 2)*acoth(d/(e*x))*gamma(1/2 - p)*gamma(p + 1)/(-2*d**4*e**3*p*gamma(1/2 - p)*gamma(p + 1) - 2*d**4*e**3*gamma(1/2 - p)*gamma(p + 1) + 2*d**2*e**5...
\[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{d+e x} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{2}}{e x + d} \,d x } \]
\[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{d+e x} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{2}}{e x + d} \,d x } \]
Timed out. \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^p}{d+e x} \, dx=\int \frac {x^2\,{\left (d^2-e^2\,x^2\right )}^p}{d+e\,x} \,d x \]