\(\int (b x+d x^3)^n \, dx\) [30]

Optimal result

Integrand size = 11, antiderivative size = 53 \[ \int \left (b x+d x^3\right )^n \, dx=\frac {x \left (b+d x^2\right ) \left (b x+d x^3\right )^n \operatorname {Hypergeometric2F1}\left (1,\frac {3 (1+n)}{2},\frac {3+n}{2},-\frac {d x^2}{b}\right )}{b (1+n)} \]

[Out]

x*(d*x^2+b)*(d*x^3+b*x)^n*hypergeom([1, 3/2+3/2*n],[3/2+1/2*n],-d*x^2/b)/b/(1+n)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.11, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2036, 372, 371} \[ \int \left (b x+d x^3\right )^n \, dx=\frac {x \left (\frac {d x^2}{b}+1\right )^{-n} \left (b x+d x^3\right )^n \operatorname {Hypergeometric2F1}\left (-n,\frac {n+1}{2},\frac {n+3}{2},-\frac {d x^2}{b}\right )}{n+1} \]

[In]

Int[(b*x + d*x^3)^n,x]

[Out]

(x*(b*x + d*x^3)^n*Hypergeometric2F1[-n, (1 + n)/2, (3 + n)/2, -((d*x^2)/b)])/((1 + n)*(1 + (d*x^2)/b)^n)

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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 2036

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rubi steps \begin{align*} \text {integral}& = \left (x^{-n} \left (b+d x^2\right )^{-n} \left (b x+d x^3\right )^n\right ) \int x^n \left (b+d x^2\right )^n \, dx \\ & = \left (x^{-n} \left (1+\frac {d x^2}{b}\right )^{-n} \left (b x+d x^3\right )^n\right ) \int x^n \left (1+\frac {d x^2}{b}\right )^n \, dx \\ & = \frac {x \left (1+\frac {d x^2}{b}\right )^{-n} \left (b x+d x^3\right )^n \, _2F_1\left (-n,\frac {1+n}{2};\frac {3+n}{2};-\frac {d x^2}{b}\right )}{1+n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.15 \[ \int \left (b x+d x^3\right )^n \, dx=\frac {x \left (x \left (b+d x^2\right )\right )^n \left (1+\frac {d x^2}{b}\right )^{-n} \operatorname {Hypergeometric2F1}\left (-n,\frac {1+n}{2},1+\frac {1+n}{2},-\frac {d x^2}{b}\right )}{1+n} \]

[In]

Integrate[(b*x + d*x^3)^n,x]

[Out]

(x*(x*(b + d*x^2))^n*Hypergeometric2F1[-n, (1 + n)/2, 1 + (1 + n)/2, -((d*x^2)/b)])/((1 + n)*(1 + (d*x^2)/b)^n
)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((d*x^3 + b*x)^n, x)

Sympy [F]

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

[In]

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

[Out]

Integral((b*x + d*x**3)**n, x)

Maxima [F]

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

[In]

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

[Out]

integrate((d*x^3 + b*x)^n, x)

Giac [F]

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

[In]

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

[Out]

integrate((d*x^3 + b*x)^n, x)

Mupad [B] (verification not implemented)

Time = 10.15 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.06 \[ \int \left (b x+d x^3\right )^n \, dx=\frac {x\,{\left (d\,x^3+b\,x\right )}^n\,{{}}_2{\mathrm {F}}_1\left (\frac {n}{2}+\frac {1}{2},-n;\ \frac {n}{2}+\frac {3}{2};\ -\frac {d\,x^2}{b}\right )}{{\left (\frac {d\,x^2}{b}+1\right )}^n\,\left (n+1\right )} \]

[In]

int((b*x + d*x^3)^n,x)

[Out]

(x*(b*x + d*x^3)^n*hypergeom([n/2 + 1/2, -n], n/2 + 3/2, -(d*x^2)/b))/(((d*x^2)/b + 1)^n*(n + 1))