Integrand size = 25, antiderivative size = 218 \[ \int x^4 (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx=-\frac {2 d^6 \left (d^2-e^2 x^2\right )^{1+p}}{e^5 (1+p)}-\frac {3 d x^5 \left (d^2-e^2 x^2\right )^{1+p}}{7+2 p}+\frac {9 d^4 \left (d^2-e^2 x^2\right )^{2+p}}{2 e^5 (2+p)}-\frac {3 d^2 \left (d^2-e^2 x^2\right )^{3+p}}{e^5 (3+p)}+\frac {\left (d^2-e^2 x^2\right )^{4+p}}{2 e^5 (4+p)}+\frac {2 d^3 (11+p) x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )}{5 (7+2 p)} \]
-2*d^6*(-e^2*x^2+d^2)^(p+1)/e^5/(p+1)-3*d*x^5*(-e^2*x^2+d^2)^(p+1)/(7+2*p) +9/2*d^4*(-e^2*x^2+d^2)^(2+p)/e^5/(2+p)-3*d^2*(-e^2*x^2+d^2)^(3+p)/e^5/(3+ p)+1/2*(-e^2*x^2+d^2)^(4+p)/e^5/(4+p)+2/5*d^3*(11+p)*x^5*(-e^2*x^2+d^2)^p* hypergeom([5/2, -p],[7/2],e^2*x^2/d^2)/(7+2*p)/((1-e^2*x^2/d^2)^p)
Time = 0.41 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.00 \[ \int x^4 (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx=\frac {1}{70} \left (d^2-e^2 x^2\right )^p \left (-\frac {35 \left (d^2-e^2 x^2\right ) \left (6 d^6 (5+p)+6 d^4 e^2 \left (5+6 p+p^2\right ) x^2+3 d^2 e^4 \left (10+17 p+8 p^2+p^3\right ) x^4+e^6 \left (6+11 p+6 p^2+p^3\right ) x^6\right )}{e^5 (1+p) (2+p) (3+p) (4+p)}+14 d^3 x^5 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )+30 d e^2 x^7 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-p,\frac {9}{2},\frac {e^2 x^2}{d^2}\right )\right ) \]
((d^2 - e^2*x^2)^p*((-35*(d^2 - e^2*x^2)*(6*d^6*(5 + p) + 6*d^4*e^2*(5 + 6 *p + p^2)*x^2 + 3*d^2*e^4*(10 + 17*p + 8*p^2 + p^3)*x^4 + e^6*(6 + 11*p + 6*p^2 + p^3)*x^6))/(e^5*(1 + p)*(2 + p)*(3 + p)*(4 + p)) + (14*d^3*x^5*Hyp ergeometric2F1[5/2, -p, 7/2, (e^2*x^2)/d^2])/(1 - (e^2*x^2)/d^2)^p + (30*d *e^2*x^7*Hypergeometric2F1[7/2, -p, 9/2, (e^2*x^2)/d^2])/(1 - (e^2*x^2)/d^ 2)^p))/70
Time = 0.38 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {543, 354, 27, 86, 363, 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.
| \(\displaystyle \int x^4 (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx\) |
| \(\Big \downarrow \) 543 |
| \(\displaystyle \int x^5 \left (d^2-e^2 x^2\right )^p \left (x^2 e^3+3 d^2 e\right )dx+\int x^4 \left (d^2-e^2 x^2\right )^p \left (d^3+3 e^2 x^2 d\right )dx\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {1}{2} \int e x^4 \left (d^2-e^2 x^2\right )^p \left (3 d^2+e^2 x^2\right )dx^2+\int x^4 \left (d^2-e^2 x^2\right )^p \left (d^3+3 e^2 x^2 d\right )dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} e \int x^4 \left (d^2-e^2 x^2\right )^p \left (3 d^2+e^2 x^2\right )dx^2+\int x^4 \left (d^2-e^2 x^2\right )^p \left (d^3+3 e^2 x^2 d\right )dx\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \int x^4 \left (d^2-e^2 x^2\right )^p \left (d^3+3 e^2 x^2 d\right )dx+\frac {1}{2} e \int \left (\frac {4 d^6 \left (d^2-e^2 x^2\right )^p}{e^4}-\frac {9 d^4 \left (d^2-e^2 x^2\right )^{p+1}}{e^4}+\frac {6 d^2 \left (d^2-e^2 x^2\right )^{p+2}}{e^4}-\frac {\left (d^2-e^2 x^2\right )^{p+3}}{e^4}\right )dx^2\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle \frac {2 d^3 (p+11) \int x^4 \left (d^2-e^2 x^2\right )^pdx}{2 p+7}+\frac {1}{2} e \int \left (\frac {4 d^6 \left (d^2-e^2 x^2\right )^p}{e^4}-\frac {9 d^4 \left (d^2-e^2 x^2\right )^{p+1}}{e^4}+\frac {6 d^2 \left (d^2-e^2 x^2\right )^{p+2}}{e^4}-\frac {\left (d^2-e^2 x^2\right )^{p+3}}{e^4}\right )dx^2-\frac {3 d x^5 \left (d^2-e^2 x^2\right )^{p+1}}{2 p+7}\) |
| \(\Big \downarrow \) 279 |
| \(\displaystyle \frac {2 d^3 (p+11) \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \int x^4 \left (1-\frac {e^2 x^2}{d^2}\right )^pdx}{2 p+7}+\frac {1}{2} e \int \left (\frac {4 d^6 \left (d^2-e^2 x^2\right )^p}{e^4}-\frac {9 d^4 \left (d^2-e^2 x^2\right )^{p+1}}{e^4}+\frac {6 d^2 \left (d^2-e^2 x^2\right )^{p+2}}{e^4}-\frac {\left (d^2-e^2 x^2\right )^{p+3}}{e^4}\right )dx^2-\frac {3 d x^5 \left (d^2-e^2 x^2\right )^{p+1}}{2 p+7}\) |
| \(\Big \downarrow \) 278 |
| \(\displaystyle \frac {1}{2} e \int \left (\frac {4 d^6 \left (d^2-e^2 x^2\right )^p}{e^4}-\frac {9 d^4 \left (d^2-e^2 x^2\right )^{p+1}}{e^4}+\frac {6 d^2 \left (d^2-e^2 x^2\right )^{p+2}}{e^4}-\frac {\left (d^2-e^2 x^2\right )^{p+3}}{e^4}\right )dx^2-\frac {3 d x^5 \left (d^2-e^2 x^2\right )^{p+1}}{2 p+7}+\frac {2 d^3 (p+11) x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )}{5 (2 p+7)}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 d x^5 \left (d^2-e^2 x^2\right )^{p+1}}{2 p+7}+\frac {2 d^3 (p+11) x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )}{5 (2 p+7)}+\frac {1}{2} e \left (-\frac {6 d^2 \left (d^2-e^2 x^2\right )^{p+3}}{e^6 (p+3)}+\frac {\left (d^2-e^2 x^2\right )^{p+4}}{e^6 (p+4)}-\frac {4 d^6 \left (d^2-e^2 x^2\right )^{p+1}}{e^6 (p+1)}+\frac {9 d^4 \left (d^2-e^2 x^2\right )^{p+2}}{e^6 (p+2)}\right )\) |
(-3*d*x^5*(d^2 - e^2*x^2)^(1 + p))/(7 + 2*p) + (e*((-4*d^6*(d^2 - e^2*x^2) ^(1 + p))/(e^6*(1 + p)) + (9*d^4*(d^2 - e^2*x^2)^(2 + p))/(e^6*(2 + p)) - (6*d^2*(d^2 - e^2*x^2)^(3 + p))/(e^6*(3 + p)) + (d^2 - e^2*x^2)^(4 + p)/(e ^6*(4 + p))))/2 + (2*d^3*(11 + p)*x^5*(d^2 - e^2*x^2)^p*Hypergeometric2F1[ 5/2, -p, 7/2, (e^2*x^2)/d^2])/(5*(7 + 2*p)*(1 - (e^2*x^2)/d^2)^p)
3.3.59.3.1 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_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> Module[{k}, Int[x^m*Sum[Binomial[n, 2*k]*c^(n - 2*k)*d^(2*k)*x^(2*k), {k, 0, n/2}]*(a + b*x^2)^p, x] + Int[x^(m + 1)*Sum[Binomial[n, 2*k + 1]*c^ (n - 2*k - 1)*d^(2*k + 1)*x^(2*k), {k, 0, (n - 1)/2}]*(a + b*x^2)^p, x]] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[n, 1] && IntegerQ[m] && !IntegerQ[2*p] && !(EqQ[m, 1] && EqQ[b*c^2 + a*d^2, 0])
\[\int x^{4} \left (e x +d \right )^{3} \left (-e^{2} x^{2}+d^{2}\right )^{p}d x\]
\[ \int x^4 (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{3} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{4} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 1945 vs. \(2 (185) = 370\).
Time = 3.36 (sec) , antiderivative size = 2966, normalized size of antiderivative = 13.61 \[ \int x^4 (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx=\text {Too large to display} \]
d**3*d**(2*p)*x**5*hyper((5/2, -p), (7/2,), e**2*x**2*exp_polar(2*I*pi)/d* *2)/5 + 3*d**2*e*Piecewise((x**6*(d**2)**p/6, Eq(e, 0)), (-2*d**4*log(-d/e + x)/(4*d**4*e**6 - 8*d**2*e**8*x**2 + 4*e**10*x**4) - 2*d**4*log(d/e + x )/(4*d**4*e**6 - 8*d**2*e**8*x**2 + 4*e**10*x**4) - 3*d**4/(4*d**4*e**6 - 8*d**2*e**8*x**2 + 4*e**10*x**4) + 4*d**2*e**2*x**2*log(-d/e + x)/(4*d**4* e**6 - 8*d**2*e**8*x**2 + 4*e**10*x**4) + 4*d**2*e**2*x**2*log(d/e + x)/(4 *d**4*e**6 - 8*d**2*e**8*x**2 + 4*e**10*x**4) + 4*d**2*e**2*x**2/(4*d**4*e **6 - 8*d**2*e**8*x**2 + 4*e**10*x**4) - 2*e**4*x**4*log(-d/e + x)/(4*d**4 *e**6 - 8*d**2*e**8*x**2 + 4*e**10*x**4) - 2*e**4*x**4*log(d/e + x)/(4*d** 4*e**6 - 8*d**2*e**8*x**2 + 4*e**10*x**4), Eq(p, -3)), (-2*d**4*log(-d/e + x)/(-2*d**2*e**6 + 2*e**8*x**2) - 2*d**4*log(d/e + x)/(-2*d**2*e**6 + 2*e **8*x**2) - 2*d**4/(-2*d**2*e**6 + 2*e**8*x**2) + 2*d**2*e**2*x**2*log(-d/ e + x)/(-2*d**2*e**6 + 2*e**8*x**2) + 2*d**2*e**2*x**2*log(d/e + x)/(-2*d* *2*e**6 + 2*e**8*x**2) + e**4*x**4/(-2*d**2*e**6 + 2*e**8*x**2), Eq(p, -2) ), (-d**4*log(-d/e + x)/(2*e**6) - d**4*log(d/e + x)/(2*e**6) - d**2*x**2/ (2*e**4) - x**4/(4*e**2), Eq(p, -1)), (-2*d**6*(d**2 - e**2*x**2)**p/(2*e* *6*p**3 + 12*e**6*p**2 + 22*e**6*p + 12*e**6) - 2*d**4*e**2*p*x**2*(d**2 - e**2*x**2)**p/(2*e**6*p**3 + 12*e**6*p**2 + 22*e**6*p + 12*e**6) - d**2*e **4*p**2*x**4*(d**2 - e**2*x**2)**p/(2*e**6*p**3 + 12*e**6*p**2 + 22*e**6* p + 12*e**6) - d**2*e**4*p*x**4*(d**2 - e**2*x**2)**p/(2*e**6*p**3 + 12...
\[ \int x^4 (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{3} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{4} \,d x } \]
\[ \int x^4 (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{3} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{4} \,d x } \]
Timed out. \[ \int x^4 (d+e x)^3 \left (d^2-e^2 x^2\right )^p \, dx=\int x^4\,{\left (d^2-e^2\,x^2\right )}^p\,{\left (d+e\,x\right )}^3 \,d x \]