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

Optimal. Leaf size=150 \[ -\frac{x \left (a+b x^4\right )^p \left (\frac{b x^4}{a}+1\right )^{-p} \left (a e^2-b c^2 (4 p+5)\right ) \, _2F_1\left (\frac{1}{4},-p;\frac{5}{4};-\frac{b x^4}{a}\right )}{b (4 p+5)}+\frac{2}{3} c e x^3 \left (a+b x^4\right )^p \left (\frac{b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac{3}{4},-p;\frac{7}{4};-\frac{b x^4}{a}\right )+\frac{e^2 x \left (a+b x^4\right )^{p+1}}{b (4 p+5)} \]

[Out]

(e^2*x*(a + b*x^4)^(1 + p))/(b*(5 + 4*p)) - ((a*e^2 - b*c^2*(5 + 4*p))*x*(a + b*x^4)^p*Hypergeometric2F1[1/4,
-p, 5/4, -((b*x^4)/a)])/(b*(5 + 4*p)*(1 + (b*x^4)/a)^p) + (2*c*e*x^3*(a + b*x^4)^p*Hypergeometric2F1[3/4, -p,
7/4, -((b*x^4)/a)])/(3*(1 + (b*x^4)/a)^p)

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Rubi [A]  time = 0.131657, antiderivative size = 142, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {1207, 1204, 246, 245, 365, 364} \[ x \left (a+b x^4\right )^p \left (\frac{b x^4}{a}+1\right )^{-p} \left (c^2-\frac{a e^2}{4 b p+5 b}\right ) \, _2F_1\left (\frac{1}{4},-p;\frac{5}{4};-\frac{b x^4}{a}\right )+\frac{2}{3} c e x^3 \left (a+b x^4\right )^p \left (\frac{b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac{3}{4},-p;\frac{7}{4};-\frac{b x^4}{a}\right )+\frac{e^2 x \left (a+b x^4\right )^{p+1}}{b (4 p+5)} \]

Antiderivative was successfully verified.

[In]

Int[(c + e*x^2)^2*(a + b*x^4)^p,x]

[Out]

(e^2*x*(a + b*x^4)^(1 + p))/(b*(5 + 4*p)) + ((c^2 - (a*e^2)/(5*b + 4*b*p))*x*(a + b*x^4)^p*Hypergeometric2F1[1
/4, -p, 5/4, -((b*x^4)/a)])/(1 + (b*x^4)/a)^p + (2*c*e*x^3*(a + b*x^4)^p*Hypergeometric2F1[3/4, -p, 7/4, -((b*
x^4)/a)])/(3*(1 + (b*x^4)/a)^p)

Rule 1207

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

Rule 1204

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

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \left (c+e x^2\right )^2 \left (a+b x^4\right )^p \, dx &=\frac{e^2 x \left (a+b x^4\right )^{1+p}}{b (5+4 p)}+\frac{\int \left (-a e^2+b c^2 (5+4 p)+2 b c e (5+4 p) x^2\right ) \left (a+b x^4\right )^p \, dx}{b (5+4 p)}\\ &=\frac{e^2 x \left (a+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+b x^4\right )^p+2 b c e (5+4 p) x^2 \left (a+b x^4\right )^p\right ) \, dx}{b (5+4 p)}\\ &=\frac{e^2 x \left (a+b x^4\right )^{1+p}}{b (5+4 p)}+(2 c e) \int x^2 \left (a+b x^4\right )^p \, dx-\left (-c^2+\frac{a e^2}{5 b+4 b p}\right ) \int \left (a+b x^4\right )^p \, dx\\ &=\frac{e^2 x \left (a+b x^4\right )^{1+p}}{b (5+4 p)}+\left (2 c e \left (a+b x^4\right )^p \left (1+\frac{b x^4}{a}\right )^{-p}\right ) \int x^2 \left (1+\frac{b x^4}{a}\right )^p \, dx-\left (\left (-c^2+\frac{a e^2}{5 b+4 b p}\right ) \left (a+b x^4\right )^p \left (1+\frac{b x^4}{a}\right )^{-p}\right ) \int \left (1+\frac{b x^4}{a}\right )^p \, dx\\ &=\frac{e^2 x \left (a+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 (a+b x^4\right )^p \left (1+\frac{b x^4}{a}\right )^{-p} \, _2F_1\left (\frac{1}{4},-p;\frac{5}{4};-\frac{b x^4}{a}\right )+\frac{2}{3} c e x^3 \left (a+b x^4\right )^p \left (1+\frac{b x^4}{a}\right )^{-p} \, _2F_1\left (\frac{3}{4},-p;\frac{7}{4};-\frac{b x^4}{a}\right )\\ \end{align*}

Mathematica [A]  time = 0.0412566, size = 106, normalized size = 0.71 \[ \frac{1}{15} x \left (a+b x^4\right )^p \left (\frac{b x^4}{a}+1\right )^{-p} \left (15 c^2 \, _2F_1\left (\frac{1}{4},-p;\frac{5}{4};-\frac{b x^4}{a}\right )+e x^2 \left (10 c \, _2F_1\left (\frac{3}{4},-p;\frac{7}{4};-\frac{b x^4}{a}\right )+3 e x^2 \, _2F_1\left (\frac{5}{4},-p;\frac{9}{4};-\frac{b x^4}{a}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*x^4)^p*(15*c^2*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^4)/a)] + e*x^2*(10*c*Hypergeometric2F1[3/4, -p
, 7/4, -((b*x^4)/a)] + 3*e*x^2*Hypergeometric2F1[5/4, -p, 9/4, -((b*x^4)/a)])))/(15*(1 + (b*x^4)/a)^p)

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Maple [F]  time = 0.04, size = 0, normalized size = 0. \begin{align*} \int \left ( e{x}^{2}+c \right ) ^{2} \left ( b{x}^{4}+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+a)^p,x)

[Out]

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

[Out]

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

[Out]

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

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Sympy [C]  time = 118.77, size = 119, normalized size = 0.79 \begin{align*} \frac{a^{p} c^{2} x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, - p \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{5}{4}\right )} + \frac{a^{p} c e x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{4}, - p \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac{7}{4}\right )} + \frac{a^{p} e^{2} x^{5} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{5}{4}, - p \\ \frac{9}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{9}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**p*c**2*x*gamma(1/4)*hyper((1/4, -p), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(5/4)) + a**p*c*e*x**3*gamma
(3/4)*hyper((3/4, -p), (7/4,), b*x**4*exp_polar(I*pi)/a)/(2*gamma(7/4)) + a**p*e**2*x**5*gamma(5/4)*hyper((5/4
, -p), (9/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(9/4))

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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} + 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+a)^p,x, algorithm="giac")

[Out]

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