3.31 \(\int (b x+c x^2+d x^3)^n \, dx\)

Optimal. Leaf size=132 \[ \frac{x \left (\frac{2 d x}{c-\sqrt{c^2-4 b d}}+1\right )^{-n} \left (\frac{2 d x}{\sqrt{c^2-4 b d}+c}+1\right )^{-n} \left (b x+c x^2+d x^3\right )^n F_1\left (n+1;-n,-n;n+2;-\frac{2 d x}{c-\sqrt{c^2-4 b d}},-\frac{2 d x}{c+\sqrt{c^2-4 b d}}\right )}{n+1} \]

[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)

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Rubi [A]  time = 0.158354, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {1908, 759, 133} \[ \frac{x \left (\frac{2 d x}{c-\sqrt{c^2-4 b d}}+1\right )^{-n} \left (\frac{2 d x}{\sqrt{c^2-4 b d}+c}+1\right )^{-n} \left (b x+c x^2+d x^3\right )^n F_1\left (n+1;-n,-n;n+2;-\frac{2 d x}{c-\sqrt{c^2-4 b d}},-\frac{2 d x}{c+\sqrt{c^2-4 b d}}\right )}{n+1} \]

Antiderivative was successfully verified.

[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 1908

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]

Rule 759

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 133

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

Rubi steps

\begin{align*} \int \left (b x+c x^2+d x^3\right )^n \, dx &=\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 ) \operatorname{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]  time = 0.283606, size = 157, normalized size = 1.19 \[ \frac{x \left (\frac{-\sqrt{c^2-4 b d}+c+2 d x}{c-\sqrt{c^2-4 b d}}\right )^{-n} \left (\frac{\sqrt{c^2-4 b d}+c+2 d x}{\sqrt{c^2-4 b d}+c}\right )^{-n} (x (b+x (c+d x)))^n F_1\left (n+1;-n,-n;n+2;-\frac{2 d x}{c+\sqrt{c^2-4 b d}},\frac{2 d x}{\sqrt{c^2-4 b d}-c}\right )}{n+1} \]

Warning: Unable to verify antiderivative.

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

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Maple [F]  time = 0.02, size = 0, normalized size = 0. \begin{align*} \int \left ( d{x}^{3}+c{x}^{2}+bx \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x^{3} + c x^{2} + b x\right )}^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (d x^{3} + c x^{2} + b x\right )}^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b x + c x^{2} + d x^{3}\right )^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x^{3} + c x^{2} + b x\right )}^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)