3.2.77 \(\int (c+e x^2)^2 (a+b x^4)^p \, dx\) [177]

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

[Out]

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

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Rubi [A]
time = 0.09, 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 = {1221, 1218, 252, 251, 372, 371} \begin {gather*} 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)} \end {gather*}

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 251

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 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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 1218

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 1221

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]

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*}

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Mathematica [A]
time = 0.55, size = 106, normalized size = 0.71 \begin {gather*} \frac {1}{15} x \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\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 ) \end {gather*}

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

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 = 0.00, 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+a)^p,x, algorithm="maxima")

[Out]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 36.21, size = 119, normalized size = 0.79 \begin {gather*} \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 {gather*}

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 = 0.00, 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+a)^p,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (b\,x^4+a\right )}^p\,{\left (e\,x^2+c\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^p*(c + e*x^2)^2,x)

[Out]

int((a + b*x^4)^p*(c + e*x^2)^2, x)

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