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*(b*x^4+a)^p*hypergeom([1/4, -p],[5/4],-b*x^4/a)/((1+b*x^4/a)^p)+1/3*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 = 0.05, antiderivative size = 96, normalized size of antiderivative = 1.00, 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 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 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.02, 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)

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e x^{2} + c\right )} {\left (b x^{4} + a\right )}^{p}, x\right ) \]

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)

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

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)

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \left (e \,x^{2}+c \right ) \left (b \,x^{4}+a \right )^{p}\, dx \]

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)

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

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^p*(c + e*x^2),x)

[Out]

int((a + b*x^4)^p*(c + e*x^2), x)

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sympy [C]  time = 42.84, size = 75, normalized size = 0.78 \[ \frac {a^{p} c x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {a^{p} e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, - p \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} \]

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))

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