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

Optimal. Leaf size=143 \[ \frac {c x \left (a+b x^4\right )^{p+1} \, _2F_1\left (1,p+\frac {5}{4};\frac {5}{4};-\frac {b x^4}{a}\right )}{a}+\frac {d x^2 \left (a+b x^4\right )^{p+1} \, _2F_1\left (1,p+\frac {3}{2};\frac {3}{2};-\frac {b x^4}{a}\right )}{2 a}+\frac {e x^3 \left (a+b x^4\right )^{p+1} \, _2F_1\left (1,p+\frac {7}{4};\frac {7}{4};-\frac {b x^4}{a}\right )}{3 a}+\frac {f \left (a+b x^4\right )^{p+1}}{4 b (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 170, normalized size of antiderivative = 1.19, number of steps used = 12, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {1885, 1204, 246, 245, 365, 364, 1248, 641} \[ c x \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac {1}{4},-p;\frac {5}{4};-\frac {b x^4}{a}\right )+\frac {1}{2} d x^2 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^4}{a}\right )+\frac {1}{3} e x^3 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac {3}{4},-p;\frac {7}{4};-\frac {b x^4}{a}\right )+\frac {f \left (a+b x^4\right )^{p+1}}{4 b (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f*(a + b*x^4)^(1 + p))/(4*b*(1 + p)) + (c*x*(a + b*x^4)^p*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^4)/a)])/(1 +
(b*x^4)/a)^p + (d*x^2*(a + b*x^4)^p*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^4)/a)])/(2*(1 + (b*x^4)/a)^p) + (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 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 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -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 1248

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

Rule 1885

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

Rubi steps

\begin {align*} \int \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^p \, dx &=\int \left (\left (c+e x^2\right ) \left (a+b x^4\right )^p+x \left (d+f x^2\right ) \left (a+b x^4\right )^p\right ) \, dx\\ &=\int \left (c+e x^2\right ) \left (a+b x^4\right )^p \, dx+\int x \left (d+f x^2\right ) \left (a+b x^4\right )^p \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int (d+f x) \left (a+b x^2\right )^p \, dx,x,x^2\right )+\int \left (c \left (a+b x^4\right )^p+e x^2 \left (a+b x^4\right )^p\right ) \, dx\\ &=\frac {f \left (a+b x^4\right )^{1+p}}{4 b (1+p)}+c \int \left (a+b x^4\right )^p \, dx+\frac {1}{2} d \operatorname {Subst}\left (\int \left (a+b x^2\right )^p \, dx,x,x^2\right )+e \int x^2 \left (a+b x^4\right )^p \, dx\\ &=\frac {f \left (a+b x^4\right )^{1+p}}{4 b (1+p)}+\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+\frac {1}{2} \left (d \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p}\right ) \operatorname {Subst}\left (\int \left (1+\frac {b x^2}{a}\right )^p \, dx,x,x^2\right )+\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\\ &=\frac {f \left (a+b x^4\right )^{1+p}}{4 b (1+p)}+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}{2} d x^2 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\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 = 0.12, size = 147, normalized size = 1.03 \[ \frac {1}{12} \left (a+b x^4\right )^p \left (12 c x \left (\frac {b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac {1}{4},-p;\frac {5}{4};-\frac {b x^4}{a}\right )+6 d x^2 \left (\frac {b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^4}{a}\right )+4 e x^3 \left (\frac {b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac {3}{4},-p;\frac {7}{4};-\frac {b x^4}{a}\right )+\frac {3 f \left (a+b x^4\right )}{b (p+1)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^4)^p*((3*f*(a + b*x^4))/(b*(1 + p)) + (12*c*x*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^4)/a)])/(1 + (b
*x^4)/a)^p + (6*d*x^2*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^4)/a)])/(1 + (b*x^4)/a)^p + (4*e*x^3*Hypergeometr
ic2F1[3/4, -p, 7/4, -((b*x^4)/a)])/(1 + (b*x^4)/a)^p))/12

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (f x^{3} + e x^{2} + d x + c\right )} {\left (b x^{4} + a\right )}^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)

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maple [F]  time = 0.47, size = 0, normalized size = 0.00 \[ \int \left (f \,x^{3}+e \,x^{2}+d x +c \right ) \left (b \,x^{4}+a \right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((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 \[ \int {\left (f x^{3} + e x^{2} + d x + c\right )} {\left (b x^{4} + a\right )}^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 49.57, size = 141, normalized size = 0.99 \[ \frac {a^{p} c x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {a^{p} d x^{2} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2} + \frac {a^{p} e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, - p \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + f \left (\begin {cases} \frac {a^{p} x^{4}}{4} & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x^{4}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + b x^{4} \right )} & \text {otherwise} \end {cases}}{4 b} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x**3+e*x**2+d*x+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*d*x**2*hyper((1/2
, -p), (3/2,), b*x**4*exp_polar(I*pi)/a)/2 + 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)) + f*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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