\(\int (3+4 x+5 x^2)^p \, dx\) [133]

Optimal result

Integrand size = 12, antiderivative size = 37 \[ \int \left (3+4 x+5 x^2\right )^p \, dx=5^{-1-p} 11^p (2+5 x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {1}{11} (2+5 x)^2\right ) \]

[Out]

5^(-1-p)*11^p*(2+5*x)*hypergeom([1/2, -p],[3/2],-1/11*(2+5*x)^2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {633, 251} \[ \int \left (3+4 x+5 x^2\right )^p \, dx=5^{-p-1} 11^p (5 x+2) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {1}{11} (5 x+2)^2\right ) \]

[In]

Int[(3 + 4*x + 5*x^2)^p,x]

[Out]

5^(-1 - p)*11^p*(2 + 5*x)*Hypergeometric2F1[1/2, -p, 3/2, -1/11*(2 + 5*x)^2]

Rule 251

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 633

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*(-4*(c/(b^2 - 4*a*c)))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \left (5^{-1-p} 11^p\right ) \text {Subst}\left (\int \left (1+\frac {x^2}{44}\right )^p \, dx,x,4+10 x\right ) \\ & = 5^{-1-p} 11^p (2+5 x) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {1}{11} (2+5 x)^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int \left (3+4 x+5 x^2\right )^p \, dx=5^{-1-p} 11^p (2+5 x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {1}{11} (2+5 x)^2\right ) \]

[In]

Integrate[(3 + 4*x + 5*x^2)^p,x]

[Out]

5^(-1 - p)*11^p*(2 + 5*x)*Hypergeometric2F1[1/2, -p, 3/2, -1/11*(2 + 5*x)^2]

Maple [F]

\[\int \left (5 x^{2}+4 x +3\right )^{p}d x\]

[In]

int((5*x^2+4*x+3)^p,x)

[Out]

int((5*x^2+4*x+3)^p,x)

Fricas [F]

\[ \int \left (3+4 x+5 x^2\right )^p \, dx=\int { {\left (5 \, x^{2} + 4 \, x + 3\right )}^{p} \,d x } \]

[In]

integrate((5*x^2+4*x+3)^p,x, algorithm="fricas")

[Out]

integral((5*x^2 + 4*x + 3)^p, x)

Sympy [F]

\[ \int \left (3+4 x+5 x^2\right )^p \, dx=\int \left (5 x^{2} + 4 x + 3\right )^{p}\, dx \]

[In]

integrate((5*x**2+4*x+3)**p,x)

[Out]

Integral((5*x**2 + 4*x + 3)**p, x)

Maxima [F]

\[ \int \left (3+4 x+5 x^2\right )^p \, dx=\int { {\left (5 \, x^{2} + 4 \, x + 3\right )}^{p} \,d x } \]

[In]

integrate((5*x^2+4*x+3)^p,x, algorithm="maxima")

[Out]

integrate((5*x^2 + 4*x + 3)^p, x)

Giac [F]

\[ \int \left (3+4 x+5 x^2\right )^p \, dx=\int { {\left (5 \, x^{2} + 4 \, x + 3\right )}^{p} \,d x } \]

[In]

integrate((5*x^2+4*x+3)^p,x, algorithm="giac")

[Out]

integrate((5*x^2 + 4*x + 3)^p, x)

Mupad [F(-1)]

Timed out. \[ \int \left (3+4 x+5 x^2\right )^p \, dx=\int {\left (5\,x^2+4\,x+3\right )}^p \,d x \]

[In]

int((4*x + 5*x^2 + 3)^p,x)

[Out]

int((4*x + 5*x^2 + 3)^p, x)