\(\int (a+b x^3)^m \, dx\) [141]

Optimal result

Integrand size = 9, antiderivative size = 44 \[ \int \left (a+b x^3\right )^m \, dx=x \left (a+b x^3\right )^m \left (1+\frac {b x^3}{a}\right )^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},-m,\frac {4}{3},-\frac {b x^3}{a}\right ) \]

[Out]

x*(b*x^3+a)^m*hypergeom([1/3, -m],[4/3],-b*x^3/a)/((1+b*x^3/a)^m)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {252, 251} \[ \int \left (a+b x^3\right )^m \, dx=x \left (a+b x^3\right )^m \left (\frac {b x^3}{a}+1\right )^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},-m,\frac {4}{3},-\frac {b x^3}{a}\right ) \]

[In]

Int[(a + b*x^3)^m,x]

[Out]

(x*(a + b*x^3)^m*Hypergeometric2F1[1/3, -m, 4/3, -((b*x^3)/a)])/(1 + (b*x^3)/a)^m

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 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[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])

Rubi steps \begin{align*} \text {integral}& = \left (\left (a+b x^3\right )^m \left (1+\frac {b x^3}{a}\right )^{-m}\right ) \int \left (1+\frac {b x^3}{a}\right )^m \, dx \\ & = x \left (a+b x^3\right )^m \left (1+\frac {b x^3}{a}\right )^{-m} \, _2F_1\left (\frac {1}{3},-m;\frac {4}{3};-\frac {b x^3}{a}\right ) \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 0.18 (sec) , antiderivative size = 203, normalized size of antiderivative = 4.61 \[ \int \left (a+b x^3\right )^m \, dx=\frac {2^{-m} \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (\frac {\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x}{\left (1+\sqrt [3]{-1}\right ) \sqrt [3]{a}}\right )^{-m} \left (\frac {i \left (1+\frac {\sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{3 i+\sqrt {3}}\right )^{-m} \left (a+b x^3\right )^m \operatorname {AppellF1}\left (1+m,-m,-m,2+m,-\frac {(-1)^{2/3} \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (1+\sqrt [3]{-1}\right ) \sqrt [3]{a}},\frac {i+\sqrt {3}-\frac {2 i \sqrt [3]{b} x}{\sqrt [3]{a}}}{3 i+\sqrt {3}}\right )}{\sqrt [3]{b} (1+m)} \]

[In]

Integrate[(a + b*x^3)^m,x]

[Out]

(((-1)^(2/3)*a^(1/3) + b^(1/3)*x)*(a + b*x^3)^m*AppellF1[1 + m, -m, -m, 2 + m, -(((-1)^(2/3)*((-1)^(2/3)*a^(1/
3) + b^(1/3)*x))/((1 + (-1)^(1/3))*a^(1/3))), (I + Sqrt[3] - ((2*I)*b^(1/3)*x)/a^(1/3))/(3*I + Sqrt[3])])/(2^m
*b^(1/3)*(1 + m)*((a^(1/3) + (-1)^(2/3)*b^(1/3)*x)/((1 + (-1)^(1/3))*a^(1/3)))^m*((I*(1 + (b^(1/3)*x)/a^(1/3))
)/(3*I + Sqrt[3]))^m)

Maple [F]

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

[In]

int((b*x^3+a)^m,x)

[Out]

int((b*x^3+a)^m,x)

Fricas [F]

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

[In]

integrate((b*x^3+a)^m,x, algorithm="fricas")

[Out]

integral((b*x^3 + a)^m, x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 5.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.77 \[ \int \left (a+b x^3\right )^m \, dx=\frac {a^{m} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, - m \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} \]

[In]

integrate((b*x**3+a)**m,x)

[Out]

a**m*x*gamma(1/3)*hyper((1/3, -m), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(4/3))

Maxima [F]

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

[In]

integrate((b*x^3+a)^m,x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^m, x)

Giac [F]

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

[In]

integrate((b*x^3+a)^m,x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^m, x)

Mupad [B] (verification not implemented)

Time = 5.43 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93 \[ \int \left (a+b x^3\right )^m \, dx=\frac {x\,{\left (b\,x^3+a\right )}^m\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},-m;\ \frac {4}{3};\ -\frac {b\,x^3}{a}\right )}{{\left (\frac {b\,x^3}{a}+1\right )}^m} \]

[In]

int((a + b*x^3)^m,x)

[Out]

(x*(a + b*x^3)^m*hypergeom([1/3, -m], 4/3, -(b*x^3)/a))/((b*x^3)/a + 1)^m