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

Optimal. Leaf size=358 \[ -\frac{x \left (a e^2-b c^2 (4 p+5)\right ) \left (\frac{2 b x^2}{c-\sqrt{c^2-4 a b}}+1\right )^{-p} \left (a+b x^4+c x^2\right )^p \left (\frac{2 b x^2}{\sqrt{c^2-4 a b}+c}+1\right )^{-p} F_1\left (\frac{1}{2};-p,-p;\frac{3}{2};-\frac{2 b x^2}{c-\sqrt{c^2-4 a b}},-\frac{2 b x^2}{c+\sqrt{c^2-4 a b}}\right )}{b (4 p+5)}+\frac{c e x^3 (8 b p+10 b-2 e p-3 e) \left (\frac{2 b x^2}{c-\sqrt{c^2-4 a b}}+1\right )^{-p} \left (a+b x^4+c x^2\right )^p \left (\frac{2 b x^2}{\sqrt{c^2-4 a b}+c}+1\right )^{-p} F_1\left (\frac{3}{2};-p,-p;\frac{5}{2};-\frac{2 b x^2}{c-\sqrt{c^2-4 a b}},-\frac{2 b x^2}{c+\sqrt{c^2-4 a b}}\right )}{3 b (4 p+5)}+\frac{e^2 x \left (a+b x^4+c x^2\right )^{p+1}}{b (4 p+5)} \]

[Out]

(e^2*x*(a + c*x^2 + b*x^4)^(1 + p))/(b*(5 + 4*p)) - ((a*e^2 - b*c^2*(5 + 4*p))*x*(a + c*x^2 + b*x^4)^p*AppellF
1[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])])/(b*(5 + 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) + (c*e*(10*b - 3*e + 8*b*
p - 2*e*p)*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*(5 + 4*p)*(1 + (2*b*x^2)/(c - Sqrt[-4*a*b + c^2]))^p*(1 + (2*b*x^2)/(c + Sqr
t[-4*a*b + c^2]))^p)

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Rubi [A]  time = 0.356375, antiderivative size = 345, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1206, 1203, 1105, 429, 1141, 510} \[ x \left (c^2-\frac{a e^2}{4 b p+5 b}\right ) \left (\frac{2 b x^2}{c-\sqrt{c^2-4 a b}}+1\right )^{-p} \left (a+b x^4+c x^2\right )^p \left (\frac{2 b x^2}{\sqrt{c^2-4 a b}+c}+1\right )^{-p} F_1\left (\frac{1}{2};-p,-p;\frac{3}{2};-\frac{2 b x^2}{c-\sqrt{c^2-4 a b}},-\frac{2 b x^2}{c+\sqrt{c^2-4 a b}}\right )+\frac{1}{3} c e x^3 \left (2-\frac{e (2 p+3)}{b (4 p+5)}\right ) \left (\frac{2 b x^2}{c-\sqrt{c^2-4 a b}}+1\right )^{-p} \left (a+b x^4+c x^2\right )^p \left (\frac{2 b x^2}{\sqrt{c^2-4 a b}+c}+1\right )^{-p} F_1\left (\frac{3}{2};-p,-p;\frac{5}{2};-\frac{2 b x^2}{c-\sqrt{c^2-4 a b}},-\frac{2 b x^2}{c+\sqrt{c^2-4 a b}}\right )+\frac{e^2 x \left (a+b x^4+c x^2\right )^{p+1}}{b (4 p+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^2*x*(a + c*x^2 + b*x^4)^(1 + p))/(b*(5 + 4*p)) + ((c^2 - (a*e^2)/(5*b + 4*b*p))*x*(a + c*x^2 + b*x^4)^p*App
ellF1[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])])/((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) + (c*e*(2 - (e*(3 + 2*p))/(b*(5 +
4*p)))*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*(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 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 )^2 \left (a+c x^2+b x^4\right )^p \, dx &=\frac{e^2 x \left (a+c x^2+b x^4\right )^{1+p}}{b (5+4 p)}+\frac{\int \left (-a e^2+b c^2 (5+4 p)+c e (10 b-3 e+8 b p-2 e p) x^2\right ) \left (a+c x^2+b x^4\right )^p \, dx}{b (5+4 p)}\\ &=\frac{e^2 x \left (a+c x^2+b x^4\right )^{1+p}}{b (5+4 p)}+\frac{\int \left (-a e^2 \left (1-\frac{b c^2 (5+4 p)}{a e^2}\right ) \left (a+c x^2+b x^4\right )^p-c e (-10 b+3 e-8 b p+2 e p) x^2 \left (a+c x^2+b x^4\right )^p\right ) \, dx}{b (5+4 p)}\\ &=\frac{e^2 x \left (a+c x^2+b x^4\right )^{1+p}}{b (5+4 p)}+\left (c e \left (2-\frac{e (3+2 p)}{b (5+4 p)}\right )\right ) \int x^2 \left (a+c x^2+b x^4\right )^p \, dx-\left (-c^2+\frac{a e^2}{5 b+4 b p}\right ) \int \left (a+c x^2+b x^4\right )^p \, dx\\ &=\frac{e^2 x \left (a+c x^2+b x^4\right )^{1+p}}{b (5+4 p)}+\left (c e \left (2-\frac{e (3+2 p)}{b (5+4 p)}\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-\left (\left (-c^2+\frac{a e^2}{5 b+4 b p}\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\\ &=\frac{e^2 x \left (a+c x^2+b x^4\right )^{1+p}}{b (5+4 p)}+\left (c^2-\frac{a e^2}{5 b+4 b p}\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 )+\frac{1}{3} c e \left (2-\frac{e (3+2 p)}{b (5+4 p)}\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 )\\ \end{align*}

Mathematica [A]  time = 0.353722, size = 303, normalized size = 0.85 \[ \frac{1}{15} x \left (\frac{-\sqrt{c^2-4 a b}+2 b x^2+c}{c-\sqrt{c^2-4 a b}}\right )^{-p} \left (\frac{\sqrt{c^2-4 a b}+2 b x^2+c}{\sqrt{c^2-4 a b}+c}\right )^{-p} \left (a+b x^4+c x^2\right )^p \left (e x^2 \left (3 e x^2 F_1\left (\frac{5}{2};-p,-p;\frac{7}{2};-\frac{2 b x^2}{c+\sqrt{c^2-4 a b}},\frac{2 b x^2}{\sqrt{c^2-4 a b}-c}\right )+10 c F_1\left (\frac{3}{2};-p,-p;\frac{5}{2};-\frac{2 b x^2}{c+\sqrt{c^2-4 a b}},\frac{2 b x^2}{\sqrt{c^2-4 a b}-c}\right )\right )+15 c^2 F_1\left (\frac{1}{2};-p,-p;\frac{3}{2};-\frac{2 b x^2}{c+\sqrt{c^2-4 a b}},\frac{2 b x^2}{\sqrt{c^2-4 a b}-c}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(a + c*x^2 + b*x^4)^p*(15*c^2*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*(10*c*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*e*x^2*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])])))/(15*((c - Sqrt[-4*a*b + c^2] + 2*b*x^2)/(c - Sqrt[-4*a*b + c^2]))^p*((c + Sqr
t[-4*a*b + c^2] + 2*b*x^2)/(c + Sqrt[-4*a*b + c^2]))^p)

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Maple [F]  time = 0.041, size = 0, normalized size = 0. \begin{align*} \int \left ( e{x}^{2}+c \right ) ^{2} \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)^2*(b*x^4+c*x^2+a)^p,x)

[Out]

int((e*x^2+c)^2*(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 )}^{2}{\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)^2*(b*x^4+c*x^2+a)^p,x, algorithm="maxima")

[Out]

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

[Out]

integral((e^2*x^4 + 2*c*e*x^2 + c^2)*(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)**2*(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 )}^{2}{\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)^2*(b*x^4+c*x^2+a)^p,x, algorithm="giac")

[Out]

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