3.5.1 \(\int (c+e x^2)^2 (a+c x^2+b x^4)^p \, dx\) [401]

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

[Out]

e^2*x*(b*x^4+c*x^2+a)^(1+p)/b/(5+4*p)-(a*e^2-b*c^2*(5+4*p))*x*(b*x^4+c*x^2+a)^p*AppellF1(1/2,-p,-p,3/2,-2*b*x^
2/(c-(-4*a*b+c^2)^(1/2)),-2*b*x^2/(c+(-4*a*b+c^2)^(1/2)))/b/(5+4*p)/((1+2*b*x^2/(c-(-4*a*b+c^2)^(1/2)))^p)/((1
+2*b*x^2/(c+(-4*a*b+c^2)^(1/2)))^p)+1/3*c*e*(8*b*p-2*e*p+10*b-3*e)*x^3*(b*x^4+c*x^2+a)^p*AppellF1(3/2,-p,-p,5/
2,-2*b*x^2/(c-(-4*a*b+c^2)^(1/2)),-2*b*x^2/(c+(-4*a*b+c^2)^(1/2)))/b/(5+4*p)/((1+2*b*x^2/(c-(-4*a*b+c^2)^(1/2)
))^p)/((1+2*b*x^2/(c+(-4*a*b+c^2)^(1/2)))^p)

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Rubi [A]
time = 0.22, antiderivative size = 345, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1220, 1217, 1119, 440, 1155, 524} \begin {gather*} 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)} \end {gather*}

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 440

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 524

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)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], 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])

Rule 1119

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 1155

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 1217

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 1220

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]

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*}

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Mathematica [A]
time = 0.47, size = 303, normalized size = 0.85 \begin {gather*} \frac {1}{15} x \left (\frac {c-\sqrt {-4 a b+c^2}+2 b x^2}{c-\sqrt {-4 a b+c^2}}\right )^{-p} \left (\frac {c+\sqrt {-4 a b+c^2}+2 b x^2}{c+\sqrt {-4 a b+c^2}}\right )^{-p} \left (a+c x^2+b x^4\right )^p \left (15 c^2 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 )+e x^2 \left (10 c 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 e x^2 F_1\left (\frac {5}{2};-p,-p;\frac {7}{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 )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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 = 0.03, size = 0, normalized size = 0.00 \[\int \left (e \,x^{2}+c \right )^{2} \left (b \,x^{4}+c \,x^{2}+a \right )^{p}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (e\,x^2+c\right )}^2\,{\left (b\,x^4+c\,x^2+a\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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