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

Optimal. Leaf size=96 \[ c x \left (a+b x^4\right )^p \left (\frac{b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{4},-p;\frac{5}{4};-\frac{b x^4}{a}\right )+\frac{1}{3} e x^3 \left (a+b x^4\right )^p \left (\frac{b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac{3}{4},-p;\frac{7}{4};-\frac{b x^4}{a}\right ) \]

[Out]

(c*x*(a + b*x^4)^p*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^4)/a)])/(1 + (b*x^4)/a)^p + (e*x^3*(a + b*x^4)^p*Hyp
ergeometric2F1[3/4, -p, 7/4, -((b*x^4)/a)])/(3*(1 + (b*x^4)/a)^p)

________________________________________________________________________________________

Rubi [A]  time = 0.0507824, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {1204, 246, 245, 365, 364} \[ c x \left (a+b x^4\right )^p \left (\frac{b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{4},-p;\frac{5}{4};-\frac{b x^4}{a}\right )+\frac{1}{3} e x^3 \left (a+b x^4\right )^p \left (\frac{b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac{3}{4},-p;\frac{7}{4};-\frac{b x^4}{a}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(c + e*x^2)*(a + b*x^4)^p,x]

[Out]

(c*x*(a + b*x^4)^p*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^4)/a)])/(1 + (b*x^4)/a)^p + (e*x^3*(a + b*x^4)^p*Hyp
ergeometric2F1[3/4, -p, 7/4, -((b*x^4)/a)])/(3*(1 + (b*x^4)/a)^p)

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 ) \left (a+b x^4\right )^p \, dx &=\int \left (c \left (a+b x^4\right )^p+e x^2 \left (a+b x^4\right )^p\right ) \, dx\\ &=c \int \left (a+b x^4\right )^p \, dx+e \int x^2 \left (a+b x^4\right )^p \, dx\\ &=\left (c \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+\left (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\\ &=c 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{1}{3} 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 = 0.0227502, size = 75, normalized size = 0.78 \[ \frac{1}{3} x \left (a+b x^4\right )^p \left (\frac{b x^4}{a}+1\right )^{-p} \left (3 c \, _2F_1\left (\frac{1}{4},-p;\frac{5}{4};-\frac{b x^4}{a}\right )+e x^2 \, _2F_1\left (\frac{3}{4},-p;\frac{7}{4};-\frac{b x^4}{a}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.028, size = 0, normalized size = 0. \begin{align*} \int \left ( e{x}^{2}+c \right ) \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)*(b*x^4+a)^p,x)

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Sympy [C]  time = 56.8967, size = 75, normalized size = 0.78 \begin{align*} \frac{a^{p} c 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} 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 )}}{4 \Gamma \left (\frac{7}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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