3.553 \(\int x^3 (c+d x+e x^2+f x^3) (a+b x^4)^p \, dx\)

Optimal. Leaf size=175 \[ \frac {c \left (a+b x^4\right )^{p+1}}{4 b (p+1)}+\frac {1}{5} d x^5 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac {5}{4},-p;\frac {9}{4};-\frac {b x^4}{a}\right )+\frac {1}{6} e x^6 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^4}{a}\right )+\frac {1}{7} f x^7 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac {7}{4},-p;\frac {11}{4};-\frac {b x^4}{a}\right ) \]

[Out]

1/4*c*(b*x^4+a)^(1+p)/b/(1+p)+1/5*d*x^5*(b*x^4+a)^p*hypergeom([5/4, -p],[9/4],-b*x^4/a)/((1+b*x^4/a)^p)+1/6*e*
x^6*(b*x^4+a)^p*hypergeom([3/2, -p],[5/2],-b*x^4/a)/((1+b*x^4/a)^p)+1/7*f*x^7*(b*x^4+a)^p*hypergeom([7/4, -p],
[11/4],-b*x^4/a)/((1+b*x^4/a)^p)

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Rubi [A]  time = 0.18, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1833, 1252, 764, 261, 365, 364, 1336} \[ \frac {c \left (a+b x^4\right )^{p+1}}{4 b (p+1)}+\frac {1}{5} d x^5 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac {5}{4},-p;\frac {9}{4};-\frac {b x^4}{a}\right )+\frac {1}{6} e x^6 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^4}{a}\right )+\frac {1}{7} f x^7 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac {7}{4},-p;\frac {11}{4};-\frac {b x^4}{a}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^3*(c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^p,x]

[Out]

(c*(a + b*x^4)^(1 + p))/(4*b*(1 + p)) + (d*x^5*(a + b*x^4)^p*Hypergeometric2F1[5/4, -p, 9/4, -((b*x^4)/a)])/(5
*(1 + (b*x^4)/a)^p) + (e*x^6*(a + b*x^4)^p*Hypergeometric2F1[3/2, -p, 5/2, -((b*x^4)/a)])/(6*(1 + (b*x^4)/a)^p
) + (f*x^7*(a + b*x^4)^p*Hypergeometric2F1[7/4, -p, 11/4, -((b*x^4)/a)])/(7*(1 + (b*x^4)/a)^p)

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

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 764

Int[(x_)^(m_.)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[f, Int[x^m*(a + c*x^2)^p, x]
, x] + Dist[g, Int[x^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && IntegerQ[m] &&  !IntegerQ[2
*p]

Rule 1252

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m
- 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]

Rule 1336

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(f*x)^m*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a, c, d, e, f, m, p, q}, x] && (IGtQ[p, 0] || IGtQ[q,
0] || IntegersQ[m, q])

Rule 1833

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[
Sum[((c*x)^(m + j)*Sum[Coeff[Pq, x, j + (k*n)/2]*x^((k*n)/2), {k, 0, (2*(q - j))/n + 1}]*(a + b*x^n)^p)/c^j, {
j, 0, n/2 - 1}], x]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] &&  !PolyQ[Pq, x^(n/2)]

Rubi steps

\begin {align*} \int x^3 \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^p \, dx &=\int \left (x^3 \left (c+e x^2\right ) \left (a+b x^4\right )^p+x^4 \left (d+f x^2\right ) \left (a+b x^4\right )^p\right ) \, dx\\ &=\int x^3 \left (c+e x^2\right ) \left (a+b x^4\right )^p \, dx+\int x^4 \left (d+f x^2\right ) \left (a+b x^4\right )^p \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int x (c+e x) \left (a+b x^2\right )^p \, dx,x,x^2\right )+\int \left (d x^4 \left (a+b x^4\right )^p+f x^6 \left (a+b x^4\right )^p\right ) \, dx\\ &=\frac {1}{2} c \operatorname {Subst}\left (\int x \left (a+b x^2\right )^p \, dx,x,x^2\right )+d \int x^4 \left (a+b x^4\right )^p \, dx+\frac {1}{2} e \operatorname {Subst}\left (\int x^2 \left (a+b x^2\right )^p \, dx,x,x^2\right )+f \int x^6 \left (a+b x^4\right )^p \, dx\\ &=\frac {c \left (a+b x^4\right )^{1+p}}{4 b (1+p)}+\left (d \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p}\right ) \int x^4 \left (1+\frac {b x^4}{a}\right )^p \, dx+\frac {1}{2} \left (e \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p}\right ) \operatorname {Subst}\left (\int x^2 \left (1+\frac {b x^2}{a}\right )^p \, dx,x,x^2\right )+\left (f \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p}\right ) \int x^6 \left (1+\frac {b x^4}{a}\right )^p \, dx\\ &=\frac {c \left (a+b x^4\right )^{1+p}}{4 b (1+p)}+\frac {1}{5} d x^5 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \, _2F_1\left (\frac {5}{4},-p;\frac {9}{4};-\frac {b x^4}{a}\right )+\frac {1}{6} e x^6 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^4}{a}\right )+\frac {1}{7} f x^7 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \, _2F_1\left (\frac {7}{4},-p;\frac {11}{4};-\frac {b x^4}{a}\right )\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 145, normalized size = 0.83 \[ \frac {\left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \left (105 c \left (a+b x^4\right ) \left (\frac {b x^4}{a}+1\right )^p+84 b d (p+1) x^5 \, _2F_1\left (\frac {5}{4},-p;\frac {9}{4};-\frac {b x^4}{a}\right )+70 b e (p+1) x^6 \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^4}{a}\right )+60 b f (p+1) x^7 \, _2F_1\left (\frac {7}{4},-p;\frac {11}{4};-\frac {b x^4}{a}\right )\right )}{420 b (p+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^p,x]

[Out]

((a + b*x^4)^p*(105*c*(a + b*x^4)*(1 + (b*x^4)/a)^p + 84*b*d*(1 + p)*x^5*Hypergeometric2F1[5/4, -p, 9/4, -((b*
x^4)/a)] + 70*b*e*(1 + p)*x^6*Hypergeometric2F1[3/2, -p, 5/2, -((b*x^4)/a)] + 60*b*f*(1 + p)*x^7*Hypergeometri
c2F1[7/4, -p, 11/4, -((b*x^4)/a)]))/(420*b*(1 + p)*(1 + (b*x^4)/a)^p)

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (f x^{6} + e x^{5} + d x^{4} + c x^{3}\right )} {\left (b x^{4} + a\right )}^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(f*x^3+e*x^2+d*x+c)*(b*x^4+a)^p,x, algorithm="fricas")

[Out]

integral((f*x^6 + e*x^5 + d*x^4 + c*x^3)*(b*x^4 + a)^p, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (f x^{3} + e x^{2} + d x + c\right )} {\left (b x^{4} + a\right )}^{p} x^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(f*x^3+e*x^2+d*x+c)*(b*x^4+a)^p,x, algorithm="giac")

[Out]

integrate((f*x^3 + e*x^2 + d*x + c)*(b*x^4 + a)^p*x^3, x)

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maple [F]  time = 0.48, size = 0, normalized size = 0.00 \[ \int \left (f \,x^{3}+e \,x^{2}+d x +c \right ) x^{3} \left (b \,x^{4}+a \right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(f*x^3+e*x^2+d*x+c)*(b*x^4+a)^p,x)

[Out]

int(x^3*(f*x^3+e*x^2+d*x+c)*(b*x^4+a)^p,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (b x^{4} + a\right )}^{p + 1} c}{4 \, b {\left (p + 1\right )}} + \int {\left (f x^{6} + e x^{5} + d x^{4}\right )} {\left (b x^{4} + a\right )}^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(f*x^3+e*x^2+d*x+c)*(b*x^4+a)^p,x, algorithm="maxima")

[Out]

1/4*(b*x^4 + a)^(p + 1)*c/(b*(p + 1)) + integrate((f*x^6 + e*x^5 + d*x^4)*(b*x^4 + a)^p, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,{\left (b\,x^4+a\right )}^p\,\left (f\,x^3+e\,x^2+d\,x+c\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*x^4)^p*(c + d*x + e*x^2 + f*x^3),x)

[Out]

int(x^3*(a + b*x^4)^p*(c + d*x + e*x^2 + f*x^3), x)

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sympy [A]  time = 111.18, size = 143, normalized size = 0.82 \[ \frac {a^{p} d x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, - p \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} + \frac {a^{p} e x^{6} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{6} + \frac {a^{p} f x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, - p \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} + c \left (\begin {cases} \frac {a^{p} x^{4}}{4} & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x^{4}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + b x^{4} \right )} & \text {otherwise} \end {cases}}{4 b} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(f*x**3+e*x**2+d*x+c)*(b*x**4+a)**p,x)

[Out]

a**p*d*x**5*gamma(5/4)*hyper((5/4, -p), (9/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(9/4)) + a**p*e*x**6*hyper((
3/2, -p), (5/2,), b*x**4*exp_polar(I*pi)/a)/6 + a**p*f*x**7*gamma(7/4)*hyper((7/4, -p), (11/4,), b*x**4*exp_po
lar(I*pi)/a)/(4*gamma(11/4)) + c*Piecewise((a**p*x**4/4, Eq(b, 0)), (Piecewise(((a + b*x**4)**(p + 1)/(p + 1),
Ne(p, -1)), (log(a + b*x**4), True))/(4*b), True))

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