Integrand size = 25, antiderivative size = 121 \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^p}{d+e x} \, dx=-\frac {d^3 \left (d^2-e^2 x^2\right )^p}{2 e^4 p}+\frac {d \left (d^2-e^2 x^2\right )^{1+p}}{2 e^4 (1+p)}-\frac {e x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},1-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )}{5 d^2} \]
-1/2*d^3*(-e^2*x^2+d^2)^p/e^4/p+1/2*d*(-e^2*x^2+d^2)^(p+1)/e^4/(p+1)-1/5*e *x^5*(-e^2*x^2+d^2)^p*hypergeom([5/2, 1-p],[7/2],e^2*x^2/d^2)/d^2/((1-e^2* x^2/d^2)^p)
Leaf count is larger than twice the leaf count of optimal. \(245\) vs. \(2(121)=242\).
Time = 0.41 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.02 \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^p}{d+e x} \, dx=\frac {\left (1+\frac {e x}{d}\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (6 d^2 e (1+p) x \left (1+\frac {e x}{d}\right )^p \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )+2 e^3 (1+p) x^3 \left (1+\frac {e x}{d}\right )^p \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},\frac {e^2 x^2}{d^2}\right )+3 d \left (\left (1+\frac {e x}{d}\right )^p \left (-e^2 x^2 \left (1-\frac {e^2 x^2}{d^2}\right )^p+d^2 \left (-1+\left (1-\frac {e^2 x^2}{d^2}\right )^p\right )\right )+d (d-e x) \left (2-\frac {2 e^2 x^2}{d^2}\right )^p \operatorname {Hypergeometric2F1}\left (1-p,1+p,2+p,\frac {d-e x}{2 d}\right )\right )\right )}{6 e^4 (1+p)} \]
((d^2 - e^2*x^2)^p*(6*d^2*e*(1 + p)*x*(1 + (e*x)/d)^p*Hypergeometric2F1[1/ 2, -p, 3/2, (e^2*x^2)/d^2] + 2*e^3*(1 + p)*x^3*(1 + (e*x)/d)^p*Hypergeomet ric2F1[3/2, -p, 5/2, (e^2*x^2)/d^2] + 3*d*((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)) + 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)] )))/(6*e^4*(1 + p)*(1 + (e*x)/d)^p*(1 - (e^2*x^2)/d^2)^p)
Time = 0.28 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, 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.
\(\displaystyle \int \frac {x^3 \left (d^2-e^2 x^2\right )^p}{d+e x} \, dx\) |
\(\Big \downarrow \) 583 |
\(\displaystyle \int x^3 (d-e x) \left (d^2-e^2 x^2\right )^{p-1}dx\) |
\(\Big \downarrow \) 542 |
\(\displaystyle d \int x^3 \left (d^2-e^2 x^2\right )^{p-1}dx-e \int x^4 \left (d^2-e^2 x^2\right )^{p-1}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} d \int x^2 \left (d^2-e^2 x^2\right )^{p-1}dx^2-e \int x^4 \left (d^2-e^2 x^2\right )^{p-1}dx\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {1}{2} d \int \left (\frac {d^2 \left (d^2-e^2 x^2\right )^{p-1}}{e^2}-\frac {\left (d^2-e^2 x^2\right )^p}{e^2}\right )dx^2-e \int x^4 \left (d^2-e^2 x^2\right )^{p-1}dx\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \frac {1}{2} d \int \left (\frac {d^2 \left (d^2-e^2 x^2\right )^{p-1}}{e^2}-\frac {\left (d^2-e^2 x^2\right )^p}{e^2}\right )dx^2-\frac {e \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \int x^4 \left (1-\frac {e^2 x^2}{d^2}\right )^{p-1}dx}{d^2}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {1}{2} d \int \left (\frac {d^2 \left (d^2-e^2 x^2\right )^{p-1}}{e^2}-\frac {\left (d^2-e^2 x^2\right )^p}{e^2}\right )dx^2-\frac {e x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},1-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )}{5 d^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} d \left (\frac {\left (d^2-e^2 x^2\right )^{p+1}}{e^4 (p+1)}-\frac {d^2 \left (d^2-e^2 x^2\right )^p}{e^4 p}\right )-\frac {e x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},1-p,\frac {7}{2},\frac {e^2 x^2}{d^2}\right )}{5 d^2}\) |
(d*(-((d^2*(d^2 - e^2*x^2)^p)/(e^4*p)) + (d^2 - e^2*x^2)^(1 + p)/(e^4*(1 + p))))/2 - (e*x^5*(d^2 - e^2*x^2)^p*Hypergeometric2F1[5/2, 1 - p, 7/2, (e^ 2*x^2)/d^2])/(5*d^2*(1 - (e^2*x^2)/d^2)^p)
3.3.68.3.1 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]
\[\int \frac {x^{3} \left (-e^{2} x^{2}+d^{2}\right )^{p}}{e x +d}d x\]
\[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^p}{d+e x} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{3}}{e x + d} \,d x } \]
Result contains complex when optimal does not.
Time = 62.78 (sec) , antiderivative size = 17065, normalized size of antiderivative = 141.03 \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^p}{d+e x} \, dx=\text {Too large to display} \]
Piecewise((-3*0**p*d**5*d**(2*p + 3)*p*log(d**2/(e**2*x**2))*gamma(-p - 1/ 2)*gamma(p + 1)/(-6*d**5*e**4*p*gamma(-p - 1/2)*gamma(p + 1) - 6*d**5*e**4 *gamma(-p - 1/2)*gamma(p + 1) + 6*d**3*e**6*p*x**2*gamma(-p - 1/2)*gamma(p + 1) + 6*d**3*e**6*x**2*gamma(-p - 1/2)*gamma(p + 1)) + 3*0**p*d**5*d**(2 *p + 3)*p*log(d**2/(e**2*x**2) - 1)*gamma(-p - 1/2)*gamma(p + 1)/(-6*d**5* e**4*p*gamma(-p - 1/2)*gamma(p + 1) - 6*d**5*e**4*gamma(-p - 1/2)*gamma(p + 1) + 6*d**3*e**6*p*x**2*gamma(-p - 1/2)*gamma(p + 1) + 6*d**3*e**6*x**2* gamma(-p - 1/2)*gamma(p + 1)) + 6*0**p*d**5*d**(2*p + 3)*p*acoth(d/(e*x))* gamma(-p - 1/2)*gamma(p + 1)/(-6*d**5*e**4*p*gamma(-p - 1/2)*gamma(p + 1) - 6*d**5*e**4*gamma(-p - 1/2)*gamma(p + 1) + 6*d**3*e**6*p*x**2*gamma(-p - 1/2)*gamma(p + 1) + 6*d**3*e**6*x**2*gamma(-p - 1/2)*gamma(p + 1)) - 3*0* *p*d**5*d**(2*p + 3)*log(d**2/(e**2*x**2))*gamma(-p - 1/2)*gamma(p + 1)/(- 6*d**5*e**4*p*gamma(-p - 1/2)*gamma(p + 1) - 6*d**5*e**4*gamma(-p - 1/2)*g amma(p + 1) + 6*d**3*e**6*p*x**2*gamma(-p - 1/2)*gamma(p + 1) + 6*d**3*e** 6*x**2*gamma(-p - 1/2)*gamma(p + 1)) + 3*0**p*d**5*d**(2*p + 3)*log(d**2/( e**2*x**2) - 1)*gamma(-p - 1/2)*gamma(p + 1)/(-6*d**5*e**4*p*gamma(-p - 1/ 2)*gamma(p + 1) - 6*d**5*e**4*gamma(-p - 1/2)*gamma(p + 1) + 6*d**3*e**6*p *x**2*gamma(-p - 1/2)*gamma(p + 1) + 6*d**3*e**6*x**2*gamma(-p - 1/2)*gamm a(p + 1)) + 6*0**p*d**5*d**(2*p + 3)*acoth(d/(e*x))*gamma(-p - 1/2)*gamma( p + 1)/(-6*d**5*e**4*p*gamma(-p - 1/2)*gamma(p + 1) - 6*d**5*e**4*gamma...
\[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^p}{d+e x} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{3}}{e x + d} \,d x } \]
\[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^p}{d+e x} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{3}}{e x + d} \,d x } \]
Timed out. \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^p}{d+e x} \, dx=\int \frac {x^3\,{\left (d^2-e^2\,x^2\right )}^p}{d+e\,x} \,d x \]