Integrand size = 20, antiderivative size = 207 \[ \int x^3 (d+e x)^3 \left (a+b x^2\right )^p \, dx=-\frac {a d \left (b d^2-3 a e^2\right ) \left (a+b x^2\right )^{1+p}}{2 b^3 (1+p)}+\frac {e^3 x^5 \left (a+b x^2\right )^{1+p}}{b (7+2 p)}+\frac {d \left (b d^2-6 a e^2\right ) \left (a+b x^2\right )^{2+p}}{2 b^3 (2+p)}+\frac {3 d e^2 \left (a+b x^2\right )^{3+p}}{2 b^3 (3+p)}-\frac {e \left (5 a e^2-3 b d^2 (7+2 p)\right ) x^5 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )}{5 b (7+2 p)} \]
-1/2*a*d*(-3*a*e^2+b*d^2)*(b*x^2+a)^(p+1)/b^3/(p+1)+e^3*x^5*(b*x^2+a)^(p+1 )/b/(7+2*p)+1/2*d*(-6*a*e^2+b*d^2)*(b*x^2+a)^(2+p)/b^3/(2+p)+3/2*d*e^2*(b* x^2+a)^(3+p)/b^3/(3+p)-1/5*e*(5*a*e^2-3*b*d^2*(7+2*p))*x^5*(b*x^2+a)^p*hyp ergeom([5/2, -p],[7/2],-b*x^2/a)/b/(7+2*p)/((1+b*x^2/a)^p)
Time = 0.26 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.95 \[ \int x^3 (d+e x)^3 \left (a+b x^2\right )^p \, dx=\frac {1}{70} \left (a+b x^2\right )^p \left (\frac {35 d^3 \left (a+b x^2\right ) \left (-a+b (1+p) x^2\right )}{b^2 (1+p) (2+p)}+\frac {105 d e^2 \left (a+b x^2\right ) \left (2 a^2-2 a b (1+p) x^2+b^2 \left (2+3 p+p^2\right ) x^4\right )}{b^3 (1+p) (2+p) (3+p)}+42 d^2 e x^5 \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )+10 e^3 x^7 \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-p,\frac {9}{2},-\frac {b x^2}{a}\right )\right ) \]
((a + b*x^2)^p*((35*d^3*(a + b*x^2)*(-a + b*(1 + p)*x^2))/(b^2*(1 + p)*(2 + p)) + (105*d*e^2*(a + b*x^2)*(2*a^2 - 2*a*b*(1 + p)*x^2 + b^2*(2 + 3*p + p^2)*x^4))/(b^3*(1 + p)*(2 + p)*(3 + p)) + (42*d^2*e*x^5*Hypergeometric2F 1[5/2, -p, 7/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p + (10*e^3*x^7*Hypergeomet ric2F1[7/2, -p, 9/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p))/70
Time = 0.37 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.95, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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.
\(\displaystyle \int x^3 (d+e x)^3 \left (a+b x^2\right )^p \, dx\) |
\(\Big \downarrow \) 543 |
\(\displaystyle \int x^3 \left (b x^2+a\right )^p \left (d^3+3 e^2 x^2 d\right )dx+\int x^4 \left (b x^2+a\right )^p \left (x^2 e^3+3 d^2 e\right )dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \int x^4 \left (b x^2+a\right )^p \left (x^2 e^3+3 d^2 e\right )dx+\frac {1}{2} \int d x^2 \left (b x^2+a\right )^p \left (d^2+3 e^2 x^2\right )dx^2\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int x^4 \left (b x^2+a\right )^p \left (x^2 e^3+3 d^2 e\right )dx+\frac {1}{2} d \int x^2 \left (b x^2+a\right )^p \left (d^2+3 e^2 x^2\right )dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} d \int \left (\frac {a \left (3 a e^2-b d^2\right ) \left (b x^2+a\right )^p}{b^2}+\frac {\left (b d^2-6 a e^2\right ) \left (b x^2+a\right )^{p+1}}{b^2}+\frac {3 e^2 \left (b x^2+a\right )^{p+2}}{b^2}\right )dx^2+\int x^4 \left (b x^2+a\right )^p \left (x^2 e^3+3 d^2 e\right )dx\) |
\(\Big \downarrow \) 363 |
\(\displaystyle \frac {1}{2} d \int \left (\frac {a \left (3 a e^2-b d^2\right ) \left (b x^2+a\right )^p}{b^2}+\frac {\left (b d^2-6 a e^2\right ) \left (b x^2+a\right )^{p+1}}{b^2}+\frac {3 e^2 \left (b x^2+a\right )^{p+2}}{b^2}\right )dx^2+e \left (3 d^2-\frac {5 a e^2}{2 b p+7 b}\right ) \int x^4 \left (b x^2+a\right )^pdx+\frac {e^3 x^5 \left (a+b x^2\right )^{p+1}}{b (2 p+7)}\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \frac {1}{2} d \int \left (\frac {a \left (3 a e^2-b d^2\right ) \left (b x^2+a\right )^p}{b^2}+\frac {\left (b d^2-6 a e^2\right ) \left (b x^2+a\right )^{p+1}}{b^2}+\frac {3 e^2 \left (b x^2+a\right )^{p+2}}{b^2}\right )dx^2+e \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 d^2-\frac {5 a e^2}{2 b p+7 b}\right ) \int x^4 \left (\frac {b x^2}{a}+1\right )^pdx+\frac {e^3 x^5 \left (a+b x^2\right )^{p+1}}{b (2 p+7)}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {1}{2} d \int \left (\frac {a \left (3 a e^2-b d^2\right ) \left (b x^2+a\right )^p}{b^2}+\frac {\left (b d^2-6 a e^2\right ) \left (b x^2+a\right )^{p+1}}{b^2}+\frac {3 e^2 \left (b x^2+a\right )^{p+2}}{b^2}\right )dx^2+\frac {1}{5} e x^5 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 d^2-\frac {5 a e^2}{2 b p+7 b}\right ) \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )+\frac {e^3 x^5 \left (a+b x^2\right )^{p+1}}{b (2 p+7)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} d \left (-\frac {a \left (b d^2-3 a e^2\right ) \left (a+b x^2\right )^{p+1}}{b^3 (p+1)}+\frac {\left (b d^2-6 a e^2\right ) \left (a+b x^2\right )^{p+2}}{b^3 (p+2)}+\frac {3 e^2 \left (a+b x^2\right )^{p+3}}{b^3 (p+3)}\right )+\frac {1}{5} e x^5 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 d^2-\frac {5 a e^2}{2 b p+7 b}\right ) \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )+\frac {e^3 x^5 \left (a+b x^2\right )^{p+1}}{b (2 p+7)}\) |
(e^3*x^5*(a + b*x^2)^(1 + p))/(b*(7 + 2*p)) + (d*(-((a*(b*d^2 - 3*a*e^2)*( a + b*x^2)^(1 + p))/(b^3*(1 + p))) + ((b*d^2 - 6*a*e^2)*(a + b*x^2)^(2 + p ))/(b^3*(2 + p)) + (3*e^2*(a + b*x^2)^(3 + p))/(b^3*(3 + p))))/2 + (e*(3*d ^2 - (5*a*e^2)/(7*b + 2*b*p))*x^5*(a + b*x^2)^p*Hypergeometric2F1[5/2, -p, 7/2, -((b*x^2)/a)])/(5*(1 + (b*x^2)/a)^p)
3.5.1.3.1 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])
\[\int x^{3} \left (e x +d \right )^{3} \left (b \,x^{2}+a \right )^{p}d x\]
\[ \int x^3 (d+e x)^3 \left (a+b x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{3} {\left (b x^{2} + a\right )}^{p} x^{3} \,d x } \]
Time = 17.32 (sec) , antiderivative size = 1329, normalized size of antiderivative = 6.42 \[ \int x^3 (d+e x)^3 \left (a+b x^2\right )^p \, dx=\text {Too large to display} \]
3*a**p*d**2*e*x**5*hyper((5/2, -p), (7/2,), b*x**2*exp_polar(I*pi)/a)/5 + a**p*e**3*x**7*hyper((7/2, -p), (9/2,), b*x**2*exp_polar(I*pi)/a)/7 + d**3 *Piecewise((a**p*x**4/4, Eq(b, 0)), (a*log(x - sqrt(-a/b))/(2*a*b**2 + 2*b **3*x**2) + a*log(x + sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2) + a/(2*a*b**2 + 2*b**3*x**2) + b*x**2*log(x - sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2) + b*x* *2*log(x + sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2), Eq(p, -2)), (-a*log(x - s qrt(-a/b))/(2*b**2) - a*log(x + sqrt(-a/b))/(2*b**2) + x**2/(2*b), Eq(p, - 1)), (-a**2*(a + b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + a*b*p*x**2 *(a + b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + b**2*p*x**4*(a + b*x* *2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + b**2*x**4*(a + b*x**2)**p/(2*b* *2*p**2 + 6*b**2*p + 4*b**2), True)) + 3*d*e**2*Piecewise((a**p*x**6/6, Eq (b, 0)), (2*a**2*log(x - sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5 *x**4) + 2*a**2*log(x + sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5* x**4) + 3*a**2/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 4*a*b*x**2*lo g(x - sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 4*a*b*x**2 *log(x + sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 4*a*b*x **2/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2*b**2*x**4*log(x - sqrt (-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2*b**2*x**4*log(x + sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4), Eq(p, -3)), (-2*a **2*log(x - sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) - 2*a**2*log(x + sqrt(...
\[ \int x^3 (d+e x)^3 \left (a+b x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{3} {\left (b x^{2} + a\right )}^{p} x^{3} \,d x } \]
1/2*(b^2*(p + 1)*x^4 + a*b*p*x^2 - a^2)*(b*x^2 + a)^p*d^3/((p^2 + 3*p + 2) *b^2) + integrate((e^3*x^6 + 3*d*e^2*x^5 + 3*d^2*e*x^4)*(b*x^2 + a)^p, x)
\[ \int x^3 (d+e x)^3 \left (a+b x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{3} {\left (b x^{2} + a\right )}^{p} x^{3} \,d x } \]
Timed out. \[ \int x^3 (d+e x)^3 \left (a+b x^2\right )^p \, dx=\int x^3\,{\left (b\,x^2+a\right )}^p\,{\left (d+e\,x\right )}^3 \,d x \]