3.400 \(\int (c+e x^2)^3 (a+c x^2+b x^4)^p \, dx\)

Optimal. Leaf size=498 \[ \frac{c x \left (-3 a b e^2 (4 p+7)+a e^3 (2 p+5)+b^2 c^2 \left (16 p^2+48 p+35\right )\right ) \left (\frac{2 b x^2}{c-\sqrt{c^2-4 a b}}+1\right )^{-p} \left (a+b x^4+c x^2\right )^p \left (\frac{2 b x^2}{\sqrt{c^2-4 a b}+c}+1\right )^{-p} F_1\left (\frac{1}{2};-p,-p;\frac{3}{2};-\frac{2 b x^2}{c-\sqrt{c^2-4 a b}},-\frac{2 b x^2}{c+\sqrt{c^2-4 a b}}\right )}{b^2 (4 p+5) (4 p+7)}+\frac{e x^3 \left (-3 b e \left (a e (4 p+5)+c^2 \left (8 p^2+26 p+21\right )\right )+3 b^2 c^2 \left (16 p^2+48 p+35\right )+c^2 e^2 \left (4 p^2+16 p+15\right )\right ) \left (\frac{2 b x^2}{c-\sqrt{c^2-4 a b}}+1\right )^{-p} \left (a+b x^4+c x^2\right )^p \left (\frac{2 b x^2}{\sqrt{c^2-4 a b}+c}+1\right )^{-p} F_1\left (\frac{3}{2};-p,-p;\frac{5}{2};-\frac{2 b x^2}{c-\sqrt{c^2-4 a b}},-\frac{2 b x^2}{c+\sqrt{c^2-4 a b}}\right )}{3 b^2 (4 p+5) (4 p+7)}+\frac{c e^2 x (12 b p+21 b-2 e p-5 e) \left (a+b x^4+c x^2\right )^{p+1}}{b^2 (4 p+5) (4 p+7)}+\frac{e^3 x^3 \left (a+b x^4+c x^2\right )^{p+1}}{b (4 p+7)} \]

[Out]

(c*e^2*(21*b - 5*e + 12*b*p - 2*e*p)*x*(a + c*x^2 + b*x^4)^(1 + p))/(b^2*(5 + 4*p)*(7 + 4*p)) + (e^3*x^3*(a +
c*x^2 + b*x^4)^(1 + p))/(b*(7 + 4*p)) + (c*(a*e^3*(5 + 2*p) - 3*a*b*e^2*(7 + 4*p) + b^2*c^2*(35 + 48*p + 16*p^
2))*x*(a + c*x^2 + b*x^4)^p*AppellF1[1/2, -p, -p, 3/2, (-2*b*x^2)/(c - Sqrt[-4*a*b + c^2]), (-2*b*x^2)/(c + Sq
rt[-4*a*b + c^2])])/(b^2*(5 + 4*p)*(7 + 4*p)*(1 + (2*b*x^2)/(c - Sqrt[-4*a*b + c^2]))^p*(1 + (2*b*x^2)/(c + Sq
rt[-4*a*b + c^2]))^p) + (e*(c^2*e^2*(15 + 16*p + 4*p^2) + 3*b^2*c^2*(35 + 48*p + 16*p^2) - 3*b*e*(a*e*(5 + 4*p
) + c^2*(21 + 26*p + 8*p^2)))*x^3*(a + c*x^2 + b*x^4)^p*AppellF1[3/2, -p, -p, 5/2, (-2*b*x^2)/(c - Sqrt[-4*a*b
+ c^2]), (-2*b*x^2)/(c + Sqrt[-4*a*b + c^2])])/(3*b^2*(5 + 4*p)*(7 + 4*p)*(1 + (2*b*x^2)/(c - Sqrt[-4*a*b + c
^2]))^p*(1 + (2*b*x^2)/(c + Sqrt[-4*a*b + c^2]))^p)

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Rubi [A]  time = 0.813798, antiderivative size = 498, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {1206, 1679, 1203, 1105, 429, 1141, 510} \[ \frac{c x \left (-3 a b e^2 (4 p+7)+a e^3 (2 p+5)+b^2 c^2 \left (16 p^2+48 p+35\right )\right ) \left (\frac{2 b x^2}{c-\sqrt{c^2-4 a b}}+1\right )^{-p} \left (a+b x^4+c x^2\right )^p \left (\frac{2 b x^2}{\sqrt{c^2-4 a b}+c}+1\right )^{-p} F_1\left (\frac{1}{2};-p,-p;\frac{3}{2};-\frac{2 b x^2}{c-\sqrt{c^2-4 a b}},-\frac{2 b x^2}{c+\sqrt{c^2-4 a b}}\right )}{b^2 (4 p+5) (4 p+7)}+\frac{e x^3 \left (-3 b e \left (a e (4 p+5)+c^2 \left (8 p^2+26 p+21\right )\right )+3 b^2 c^2 \left (16 p^2+48 p+35\right )+c^2 e^2 \left (4 p^2+16 p+15\right )\right ) \left (\frac{2 b x^2}{c-\sqrt{c^2-4 a b}}+1\right )^{-p} \left (a+b x^4+c x^2\right )^p \left (\frac{2 b x^2}{\sqrt{c^2-4 a b}+c}+1\right )^{-p} F_1\left (\frac{3}{2};-p,-p;\frac{5}{2};-\frac{2 b x^2}{c-\sqrt{c^2-4 a b}},-\frac{2 b x^2}{c+\sqrt{c^2-4 a b}}\right )}{3 b^2 (4 p+5) (4 p+7)}+\frac{c e^2 x (12 b p+21 b-2 e p-5 e) \left (a+b x^4+c x^2\right )^{p+1}}{b^2 (4 p+5) (4 p+7)}+\frac{e^3 x^3 \left (a+b x^4+c x^2\right )^{p+1}}{b (4 p+7)} \]

Antiderivative was successfully verified.

[In]

Int[(c + e*x^2)^3*(a + c*x^2 + b*x^4)^p,x]

[Out]

(c*e^2*(21*b - 5*e + 12*b*p - 2*e*p)*x*(a + c*x^2 + b*x^4)^(1 + p))/(b^2*(5 + 4*p)*(7 + 4*p)) + (e^3*x^3*(a +
c*x^2 + b*x^4)^(1 + p))/(b*(7 + 4*p)) + (c*(a*e^3*(5 + 2*p) - 3*a*b*e^2*(7 + 4*p) + b^2*c^2*(35 + 48*p + 16*p^
2))*x*(a + c*x^2 + b*x^4)^p*AppellF1[1/2, -p, -p, 3/2, (-2*b*x^2)/(c - Sqrt[-4*a*b + c^2]), (-2*b*x^2)/(c + Sq
rt[-4*a*b + c^2])])/(b^2*(5 + 4*p)*(7 + 4*p)*(1 + (2*b*x^2)/(c - Sqrt[-4*a*b + c^2]))^p*(1 + (2*b*x^2)/(c + Sq
rt[-4*a*b + c^2]))^p) + (e*(c^2*e^2*(15 + 16*p + 4*p^2) + 3*b^2*c^2*(35 + 48*p + 16*p^2) - 3*b*e*(a*e*(5 + 4*p
) + c^2*(21 + 26*p + 8*p^2)))*x^3*(a + c*x^2 + b*x^4)^p*AppellF1[3/2, -p, -p, 5/2, (-2*b*x^2)/(c - Sqrt[-4*a*b
+ c^2]), (-2*b*x^2)/(c + Sqrt[-4*a*b + c^2])])/(3*b^2*(5 + 4*p)*(7 + 4*p)*(1 + (2*b*x^2)/(c - Sqrt[-4*a*b + c
^2]))^p*(1 + (2*b*x^2)/(c + Sqrt[-4*a*b + c^2]))^p)

Rule 1206

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(e^q*x^(2*q - 3)*(
a + b*x^2 + c*x^4)^(p + 1))/(c*(4*p + 2*q + 1)), x] + Dist[1/(c*(4*p + 2*q + 1)), Int[(a + b*x^2 + c*x^4)^p*Ex
pandToSum[c*(4*p + 2*q + 1)*(d + e*x^2)^q - a*(2*q - 3)*e^q*x^(2*q - 4) - b*(2*p + 2*q - 1)*e^q*x^(2*q - 2) -
c*(4*p + 2*q + 1)*e^q*x^(2*q), x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && IGtQ[q, 1]

Rule 1679

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{q = Expon[Pq, x^2], e = Coeff[Pq, x^2,
Expon[Pq, x^2]]}, Simp[(e*x^(2*q - 3)*(a + b*x^2 + c*x^4)^(p + 1))/(c*(2*q + 4*p + 1)), x] + Dist[1/(c*(2*q +
4*p + 1)), Int[(a + b*x^2 + c*x^4)^p*ExpandToSum[c*(2*q + 4*p + 1)*Pq - a*e*(2*q - 3)*x^(2*q - 4) - b*e*(2*q
+ 2*p - 1)*x^(2*q - 2) - c*e*(2*q + 4*p + 1)*x^(2*q), x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2]
&& Expon[Pq, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] &&  !LtQ[p, -1]

Rule 1203

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x
^2)*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a
*e^2, 0]

Rule 1105

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[(a^IntPart[p]*
(a + b*x^2 + c*x^4)^FracPart[p])/((1 + (2*c*x^2)/(b + q))^FracPart[p]*(1 + (2*c*x^2)/(b - q))^FracPart[p]), In
t[(1 + (2*c*x^2)/(b + q))^p*(1 + (2*c*x^2)/(b - q))^p, x], x]] /; FreeQ[{a, b, c, p}, x] && NeQ[b^2 - 4*a*c, 0
]

Rule 429

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 1141

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^2 +
c*x^4)^FracPart[p])/((1 + (2*c*x^2)/(b + Rt[b^2 - 4*a*c, 2]))^FracPart[p]*(1 + (2*c*x^2)/(b - Rt[b^2 - 4*a*c,
2]))^FracPart[p]), Int[(d*x)^m*(1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]))^p*(1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c
]))^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x]

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \left (c+e x^2\right )^3 \left (a+c x^2+b x^4\right )^p \, dx &=\frac{e^3 x^3 \left (a+c x^2+b x^4\right )^{1+p}}{b (7+4 p)}+\frac{\int \left (a+c x^2+b x^4\right )^p \left (b c^3 (7+4 p)-3 e \left (a e^2-b c^2 (7+4 p)\right ) x^2+c e^2 (21 b-5 e+12 b p-2 e p) x^4\right ) \, dx}{b (7+4 p)}\\ &=\frac{c e^2 (21 b-5 e+12 b p-2 e p) x \left (a+c x^2+b x^4\right )^{1+p}}{b^2 (5+4 p) (7+4 p)}+\frac{e^3 x^3 \left (a+c x^2+b x^4\right )^{1+p}}{b (7+4 p)}+\frac{\int \left (c \left (a e^3 (5+2 p)-3 a b e^2 (7+4 p)+b^2 c^2 \left (35+48 p+16 p^2\right )\right )+e \left (c^2 e^2 \left (15+16 p+4 p^2\right )+3 b^2 c^2 \left (35+48 p+16 p^2\right )-3 b e \left (a e (5+4 p)+c^2 \left (21+26 p+8 p^2\right )\right )\right ) x^2\right ) \left (a+c x^2+b x^4\right )^p \, dx}{b^2 (5+4 p) (7+4 p)}\\ &=\frac{c e^2 (21 b-5 e+12 b p-2 e p) x \left (a+c x^2+b x^4\right )^{1+p}}{b^2 (5+4 p) (7+4 p)}+\frac{e^3 x^3 \left (a+c x^2+b x^4\right )^{1+p}}{b (7+4 p)}+\frac{\int \left (c \left (a e^3 (5+2 p)-3 a b e^2 (7+4 p)+b^2 c^2 \left (35+48 p+16 p^2\right )\right ) \left (a+c x^2+b x^4\right )^p+e \left (c^2 e^2 \left (15+16 p+4 p^2\right )+3 b^2 c^2 \left (35+48 p+16 p^2\right )-3 b e \left (a e (5+4 p)+c^2 \left (21+26 p+8 p^2\right )\right )\right ) x^2 \left (a+c x^2+b x^4\right )^p\right ) \, dx}{b^2 (5+4 p) (7+4 p)}\\ &=\frac{c e^2 (21 b-5 e+12 b p-2 e p) x \left (a+c x^2+b x^4\right )^{1+p}}{b^2 (5+4 p) (7+4 p)}+\frac{e^3 x^3 \left (a+c x^2+b x^4\right )^{1+p}}{b (7+4 p)}+\frac{\left (c \left (a e^3 (5+2 p)-3 a b e^2 (7+4 p)+b^2 c^2 \left (35+48 p+16 p^2\right )\right )\right ) \int \left (a+c x^2+b x^4\right )^p \, dx}{b^2 (5+4 p) (7+4 p)}+\frac{\left (e \left (c^2 e^2 \left (15+16 p+4 p^2\right )+3 b^2 c^2 \left (35+48 p+16 p^2\right )-3 b e \left (a e (5+4 p)+c^2 \left (21+26 p+8 p^2\right )\right )\right )\right ) \int x^2 \left (a+c x^2+b x^4\right )^p \, dx}{b^2 (5+4 p) (7+4 p)}\\ &=\frac{c e^2 (21 b-5 e+12 b p-2 e p) x \left (a+c x^2+b x^4\right )^{1+p}}{b^2 (5+4 p) (7+4 p)}+\frac{e^3 x^3 \left (a+c x^2+b x^4\right )^{1+p}}{b (7+4 p)}+\frac{\left (c \left (a e^3 (5+2 p)-3 a b e^2 (7+4 p)+b^2 c^2 \left (35+48 p+16 p^2\right )\right ) \left (1+\frac{2 b x^2}{c-\sqrt{-4 a b+c^2}}\right )^{-p} \left (1+\frac{2 b x^2}{c+\sqrt{-4 a b+c^2}}\right )^{-p} \left (a+c x^2+b x^4\right )^p\right ) \int \left (1+\frac{2 b x^2}{c-\sqrt{-4 a b+c^2}}\right )^p \left (1+\frac{2 b x^2}{c+\sqrt{-4 a b+c^2}}\right )^p \, dx}{b^2 (5+4 p) (7+4 p)}+\frac{\left (e \left (c^2 e^2 \left (15+16 p+4 p^2\right )+3 b^2 c^2 \left (35+48 p+16 p^2\right )-3 b e \left (a e (5+4 p)+c^2 \left (21+26 p+8 p^2\right )\right )\right ) \left (1+\frac{2 b x^2}{c-\sqrt{-4 a b+c^2}}\right )^{-p} \left (1+\frac{2 b x^2}{c+\sqrt{-4 a b+c^2}}\right )^{-p} \left (a+c x^2+b x^4\right )^p\right ) \int x^2 \left (1+\frac{2 b x^2}{c-\sqrt{-4 a b+c^2}}\right )^p \left (1+\frac{2 b x^2}{c+\sqrt{-4 a b+c^2}}\right )^p \, dx}{b^2 (5+4 p) (7+4 p)}\\ &=\frac{c e^2 (21 b-5 e+12 b p-2 e p) x \left (a+c x^2+b x^4\right )^{1+p}}{b^2 (5+4 p) (7+4 p)}+\frac{e^3 x^3 \left (a+c x^2+b x^4\right )^{1+p}}{b (7+4 p)}+\frac{c \left (a e^3 (5+2 p)-3 a b e^2 (7+4 p)+b^2 c^2 \left (35+48 p+16 p^2\right )\right ) x \left (1+\frac{2 b x^2}{c-\sqrt{-4 a b+c^2}}\right )^{-p} \left (1+\frac{2 b x^2}{c+\sqrt{-4 a b+c^2}}\right )^{-p} \left (a+c x^2+b x^4\right )^p F_1\left (\frac{1}{2};-p,-p;\frac{3}{2};-\frac{2 b x^2}{c-\sqrt{-4 a b+c^2}},-\frac{2 b x^2}{c+\sqrt{-4 a b+c^2}}\right )}{b^2 (5+4 p) (7+4 p)}+\frac{e \left (c^2 e^2 \left (15+16 p+4 p^2\right )+3 b^2 c^2 \left (35+48 p+16 p^2\right )-3 b e \left (a e (5+4 p)+c^2 \left (21+26 p+8 p^2\right )\right )\right ) x^3 \left (1+\frac{2 b x^2}{c-\sqrt{-4 a b+c^2}}\right )^{-p} \left (1+\frac{2 b x^2}{c+\sqrt{-4 a b+c^2}}\right )^{-p} \left (a+c x^2+b x^4\right )^p F_1\left (\frac{3}{2};-p,-p;\frac{5}{2};-\frac{2 b x^2}{c-\sqrt{-4 a b+c^2}},-\frac{2 b x^2}{c+\sqrt{-4 a b+c^2}}\right )}{3 b^2 (5+4 p) (7+4 p)}\\ \end{align*}

Mathematica [A]  time = 0.518336, size = 373, normalized size = 0.75 \[ \frac{1}{35} x \left (\frac{-\sqrt{c^2-4 a b}+2 b x^2+c}{c-\sqrt{c^2-4 a b}}\right )^{-p} \left (\frac{\sqrt{c^2-4 a b}+2 b x^2+c}{\sqrt{c^2-4 a b}+c}\right )^{-p} \left (a+b x^4+c x^2\right )^p \left (e x^2 \left (e x^2 \left (5 e x^2 F_1\left (\frac{7}{2};-p,-p;\frac{9}{2};-\frac{2 b x^2}{c+\sqrt{c^2-4 a b}},\frac{2 b x^2}{\sqrt{c^2-4 a b}-c}\right )+21 c F_1\left (\frac{5}{2};-p,-p;\frac{7}{2};-\frac{2 b x^2}{c+\sqrt{c^2-4 a b}},\frac{2 b x^2}{\sqrt{c^2-4 a b}-c}\right )\right )+35 c^2 F_1\left (\frac{3}{2};-p,-p;\frac{5}{2};-\frac{2 b x^2}{c+\sqrt{c^2-4 a b}},\frac{2 b x^2}{\sqrt{c^2-4 a b}-c}\right )\right )+35 c^3 F_1\left (\frac{1}{2};-p,-p;\frac{3}{2};-\frac{2 b x^2}{c+\sqrt{c^2-4 a b}},\frac{2 b x^2}{\sqrt{c^2-4 a b}-c}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + e*x^2)^3*(a + c*x^2 + b*x^4)^p,x]

[Out]

(x*(a + c*x^2 + b*x^4)^p*(35*c^3*AppellF1[1/2, -p, -p, 3/2, (-2*b*x^2)/(c + Sqrt[-4*a*b + c^2]), (2*b*x^2)/(-c
+ Sqrt[-4*a*b + c^2])] + e*x^2*(35*c^2*AppellF1[3/2, -p, -p, 5/2, (-2*b*x^2)/(c + Sqrt[-4*a*b + c^2]), (2*b*x
^2)/(-c + Sqrt[-4*a*b + c^2])] + e*x^2*(21*c*AppellF1[5/2, -p, -p, 7/2, (-2*b*x^2)/(c + Sqrt[-4*a*b + c^2]), (
2*b*x^2)/(-c + Sqrt[-4*a*b + c^2])] + 5*e*x^2*AppellF1[7/2, -p, -p, 9/2, (-2*b*x^2)/(c + Sqrt[-4*a*b + c^2]),
(2*b*x^2)/(-c + Sqrt[-4*a*b + c^2])]))))/(35*((c - Sqrt[-4*a*b + c^2] + 2*b*x^2)/(c - Sqrt[-4*a*b + c^2]))^p*(
(c + Sqrt[-4*a*b + c^2] + 2*b*x^2)/(c + Sqrt[-4*a*b + c^2]))^p)

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Maple [F]  time = 0.051, size = 0, normalized size = 0. \begin{align*} \int \left ( e{x}^{2}+c \right ) ^{3} \left ( b{x}^{4}+c{x}^{2}+a \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+c)^3*(b*x^4+c*x^2+a)^p,x)

[Out]

int((e*x^2+c)^3*(b*x^4+c*x^2+a)^p,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x^{2} + c\right )}^{3}{\left (b x^{4} + c x^{2} + a\right )}^{p}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+c)^3*(b*x^4+c*x^2+a)^p,x, algorithm="maxima")

[Out]

integrate((e*x^2 + c)^3*(b*x^4 + c*x^2 + a)^p, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (e^{3} x^{6} + 3 \, c e^{2} x^{4} + 3 \, c^{2} e x^{2} + c^{3}\right )}{\left (b x^{4} + c x^{2} + a\right )}^{p}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+c)^3*(b*x^4+c*x^2+a)^p,x, algorithm="fricas")

[Out]

integral((e^3*x^6 + 3*c*e^2*x^4 + 3*c^2*e*x^2 + c^3)*(b*x^4 + c*x^2 + a)^p, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+c)**3*(b*x**4+c*x**2+a)**p,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x^{2} + c\right )}^{3}{\left (b x^{4} + c x^{2} + a\right )}^{p}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+c)^3*(b*x^4+c*x^2+a)^p,x, algorithm="giac")

[Out]

integrate((e*x^2 + c)^3*(b*x^4 + c*x^2 + a)^p, x)