\(\int (b x+c x^2+d x^3)^n \, dx\) [31]

Optimal result

Integrand size = 16, antiderivative size = 132 \[ \int \left (b x+c x^2+d x^3\right )^n \, dx=\frac {x \left (1+\frac {2 d x}{c-\sqrt {c^2-4 b d}}\right )^{-n} \left (1+\frac {2 d x}{c+\sqrt {c^2-4 b d}}\right )^{-n} \left (b x+c x^2+d x^3\right )^n \operatorname {AppellF1}\left (1+n,-n,-n,2+n,-\frac {2 d x}{c-\sqrt {c^2-4 b d}},-\frac {2 d x}{c+\sqrt {c^2-4 b d}}\right )}{1+n} \]

[Out]

x*(d*x^3+c*x^2+b*x)^n*AppellF1(1+n,-n,-n,2+n,-2*d*x/(c-(-4*b*d+c^2)^(1/2)),-2*d*x/(c+(-4*b*d+c^2)^(1/2)))/(1+n
)/((1+2*d*x/(c-(-4*b*d+c^2)^(1/2)))^n)/((1+2*d*x/(c+(-4*b*d+c^2)^(1/2)))^n)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {1922, 773, 138} \[ \int \left (b x+c x^2+d x^3\right )^n \, dx=\frac {x \left (\frac {2 d x}{c-\sqrt {c^2-4 b d}}+1\right )^{-n} \left (\frac {2 d x}{\sqrt {c^2-4 b d}+c}+1\right )^{-n} \left (b x+c x^2+d x^3\right )^n \operatorname {AppellF1}\left (n+1,-n,-n,n+2,-\frac {2 d x}{c-\sqrt {c^2-4 b d}},-\frac {2 d x}{c+\sqrt {c^2-4 b d}}\right )}{n+1} \]

[In]

Int[(b*x + c*x^2 + d*x^3)^n,x]

[Out]

(x*(b*x + c*x^2 + d*x^3)^n*AppellF1[1 + n, -n, -n, 2 + n, (-2*d*x)/(c - Sqrt[c^2 - 4*b*d]), (-2*d*x)/(c + Sqrt
[c^2 - 4*b*d])])/((1 + n)*(1 + (2*d*x)/(c - Sqrt[c^2 - 4*b*d]))^n*(1 + (2*d*x)/(c + Sqrt[c^2 - 4*b*d]))^n)

Rule 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 773

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c,
2]}, Dist[(a + b*x + c*x^2)^p/(e*(1 - (d + e*x)/(d - e*((b - q)/(2*c))))^p*(1 - (d + e*x)/(d - e*((b + q)/(2*
c))))^p), Subst[Int[x^m*Simp[1 - x/(d - e*((b - q)/(2*c))), x]^p*Simp[1 - x/(d - e*((b + q)/(2*c))), x]^p, x],
x, d + e*x], x]] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &
& NeQ[2*c*d - b*e, 0] &&  !IntegerQ[p]

Rule 1922

Int[((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol] :> Dist[(a*x^q + b*x^n + c*x^(2*n
- q))^p/(x^(p*q)*(a + b*x^(n - q) + c*x^(2*(n - q)))^p), Int[x^(p*q)*(a + b*x^(n - q) + c*x^(2*(n - q)))^p, x
], x] /; FreeQ[{a, b, c, n, p, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q] &&  !IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \left (x^{-n} \left (b+c x+d x^2\right )^{-n} \left (b x+c x^2+d x^3\right )^n\right ) \int x^n \left (b+c x+d x^2\right )^n \, dx \\ & = \left (x^{-n} \left (1+\frac {2 d x}{c-\sqrt {c^2-4 b d}}\right )^{-n} \left (1+\frac {2 d x}{c+\sqrt {c^2-4 b d}}\right )^{-n} \left (b x+c x^2+d x^3\right )^n\right ) \text {Subst}\left (\int x^n \left (1+\frac {2 d x}{c-\sqrt {c^2-4 b d}}\right )^n \left (1+\frac {2 d x}{c+\sqrt {c^2-4 b d}}\right )^n \, dx,x,x\right ) \\ & = \frac {x \left (1+\frac {2 d x}{c-\sqrt {c^2-4 b d}}\right )^{-n} \left (1+\frac {2 d x}{c+\sqrt {c^2-4 b d}}\right )^{-n} \left (b x+c x^2+d x^3\right )^n F_1\left (1+n;-n,-n;2+n;-\frac {2 d x}{c-\sqrt {c^2-4 b d}},-\frac {2 d x}{c+\sqrt {c^2-4 b d}}\right )}{1+n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.98 \[ \int \left (b x+c x^2+d x^3\right )^n \, dx=\frac {x \left (1+\frac {2 d x}{c-\sqrt {c^2-4 b d}}\right )^{-n} \left (1+\frac {2 d x}{c+\sqrt {c^2-4 b d}}\right )^{-n} (x (b+x (c+d x)))^n \operatorname {AppellF1}\left (1+n,-n,-n,2+n,-\frac {2 d x}{c+\sqrt {c^2-4 b d}},\frac {2 d x}{-c+\sqrt {c^2-4 b d}}\right )}{1+n} \]

[In]

Integrate[(b*x + c*x^2 + d*x^3)^n,x]

[Out]

(x*(x*(b + x*(c + d*x)))^n*AppellF1[1 + n, -n, -n, 2 + n, (-2*d*x)/(c + Sqrt[c^2 - 4*b*d]), (2*d*x)/(-c + Sqrt
[c^2 - 4*b*d])])/((1 + n)*(1 + (2*d*x)/(c - Sqrt[c^2 - 4*b*d]))^n*(1 + (2*d*x)/(c + Sqrt[c^2 - 4*b*d]))^n)

Maple [F]

\[\int \left (x^{3} d +c \,x^{2}+b x \right )^{n}d x\]

[In]

int((d*x^3+c*x^2+b*x)^n,x)

[Out]

int((d*x^3+c*x^2+b*x)^n,x)

Fricas [F]

\[ \int \left (b x+c x^2+d x^3\right )^n \, dx=\int { {\left (d x^{3} + c x^{2} + b x\right )}^{n} \,d x } \]

[In]

integrate((d*x^3+c*x^2+b*x)^n,x, algorithm="fricas")

[Out]

integral((d*x^3 + c*x^2 + b*x)^n, x)

Sympy [F]

\[ \int \left (b x+c x^2+d x^3\right )^n \, dx=\int \left (b x + c x^{2} + d x^{3}\right )^{n}\, dx \]

[In]

integrate((d*x**3+c*x**2+b*x)**n,x)

[Out]

Integral((b*x + c*x**2 + d*x**3)**n, x)

Maxima [F]

\[ \int \left (b x+c x^2+d x^3\right )^n \, dx=\int { {\left (d x^{3} + c x^{2} + b x\right )}^{n} \,d x } \]

[In]

integrate((d*x^3+c*x^2+b*x)^n,x, algorithm="maxima")

[Out]

integrate((d*x^3 + c*x^2 + b*x)^n, x)

Giac [F]

\[ \int \left (b x+c x^2+d x^3\right )^n \, dx=\int { {\left (d x^{3} + c x^{2} + b x\right )}^{n} \,d x } \]

[In]

integrate((d*x^3+c*x^2+b*x)^n,x, algorithm="giac")

[Out]

integrate((d*x^3 + c*x^2 + b*x)^n, x)

Mupad [F(-1)]

Timed out. \[ \int \left (b x+c x^2+d x^3\right )^n \, dx=\int {\left (d\,x^3+c\,x^2+b\,x\right )}^n \,d x \]

[In]

int((b*x + c*x^2 + d*x^3)^n,x)

[Out]

int((b*x + c*x^2 + d*x^3)^n, x)