3.141 \(\int (a+b x^3)^m \, dx\)

Optimal. Leaf size=44 \[ x \left (a+b x^3\right )^m \left (\frac {b x^3}{a}+1\right )^{-m} \, _2F_1\left (\frac {1}{3},-m;\frac {4}{3};-\frac {b x^3}{a}\right ) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {246, 245} \[ x \left (a+b x^3\right )^m \left (\frac {b x^3}{a}+1\right )^{-m} \, _2F_1\left (\frac {1}{3},-m;\frac {4}{3};-\frac {b x^3}{a}\right ) \]

Antiderivative was successfully verified.

[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 245

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 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[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*} \int \left (a+b x^3\right )^m \, dx &=\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*}

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Mathematica [C]  time = 0.20, size = 196, normalized size = 4.45 \[ \frac {2^{-m} \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (\frac {\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x}{\left (1+\sqrt [3]{-1}\right ) \sqrt [3]{a}}\right )^{-m} \left (\frac {i \left (\frac {\sqrt [3]{b} x}{\sqrt [3]{a}}+1\right )}{\sqrt {3}+3 i}\right )^{-m} \left (a+b x^3\right )^m F_1\left (m+1;-m,-m;m+2;-\frac {i \left (\sqrt [3]{b} x+(-1)^{2/3} \sqrt [3]{a}\right )}{\sqrt {3} \sqrt [3]{a}},\frac {-\frac {2 i \sqrt [3]{b} x}{\sqrt [3]{a}}+\sqrt {3}+i}{3 i+\sqrt {3}}\right )}{\sqrt [3]{b} (m+1)} \]

Warning: Unable to verify antiderivative.

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

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fricas [F]  time = 1.26, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b x^{3} + a\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^3 + a)^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{3} + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a)^m, x)

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maple [F]  time = 0.36, size = 0, normalized size = 0.00 \[ \int \left (b \,x^{3}+a \right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{3} + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a)^m, x)

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mupad [B]  time = 1.35, size = 41, normalized size = 0.93 \[ \frac {x\,{\left (b\,x^3+a\right )}^m\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},-m;\ \frac {4}{3};\ -\frac {b\,x^3}{a}\right )}{{\left (\frac {b\,x^3}{a}+1\right )}^m} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [C]  time = 15.56, size = 34, normalized size = 0.77 \[ \frac {a^{m} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, - m \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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