3.472 \(\int (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3})^p (d x)^m \, dx\)

Optimal. Leaf size=77 \[ \frac{x (d x)^m \left (\frac{b \sqrt [3]{x}}{a}+1\right )^{-2 p} \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p \, _2F_1\left (3 (m+1),-2 p;3 m+4;-\frac{b \sqrt [3]{x}}{a}\right )}{m+1} \]

[Out]

((a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^p*x*(d*x)^m*Hypergeometric2F1[3*(1 + m), -2*p, 4 + 3*m, -((b*x^(1/3))/a)]
)/((1 + m)*(1 + (b*x^(1/3))/a)^(2*p))

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Rubi [A]  time = 0.0367421, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {1356, 343, 341, 64} \[ \frac{x (d x)^m \left (\frac{b \sqrt [3]{x}}{a}+1\right )^{-2 p} \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p \, _2F_1\left (3 (m+1),-2 p;3 m+4;-\frac{b \sqrt [3]{x}}{a}\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^p*(d*x)^m,x]

[Out]

((a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^p*x*(d*x)^m*Hypergeometric2F1[3*(1 + m), -2*p, 4 + 3*m, -((b*x^(1/3))/a)]
)/((1 + m)*(1 + (b*x^(1/3))/a)^(2*p))

Rule 1356

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a
+ b*x^n + c*x^(2*n))^FracPart[p])/(1 + (2*c*x^n)/b)^(2*FracPart[p]), Int[(d*x)^m*(1 + (2*c*x^n)/b)^(2*p), x],
x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[2*p]

Rule 343

Int[((c_)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracPart[m])/x^FracP
art[m], Int[x^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && FractionQ[n]

Rule 341

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[Int[x^(k*(
m + 1) - 1)*(a + b*x^(k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, m, p}, x] && FractionQ[n]

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rubi steps

\begin{align*} \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p (d x)^m \, dx &=\left (\left (1+\frac{b \sqrt [3]{x}}{a}\right )^{-2 p} \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p\right ) \int \left (1+\frac{b \sqrt [3]{x}}{a}\right )^{2 p} (d x)^m \, dx\\ &=\left (\left (1+\frac{b \sqrt [3]{x}}{a}\right )^{-2 p} \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p x^{-m} (d x)^m\right ) \int \left (1+\frac{b \sqrt [3]{x}}{a}\right )^{2 p} x^m \, dx\\ &=\left (3 \left (1+\frac{b \sqrt [3]{x}}{a}\right )^{-2 p} \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p x^{-m} (d x)^m\right ) \operatorname{Subst}\left (\int x^{-1+3 (1+m)} \left (1+\frac{b x}{a}\right )^{2 p} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{\left (1+\frac{b \sqrt [3]{x}}{a}\right )^{-2 p} \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p x (d x)^m \, _2F_1\left (3 (1+m),-2 p;4+3 m;-\frac{b \sqrt [3]{x}}{a}\right )}{1+m}\\ \end{align*}

Mathematica [A]  time = 0.0301352, size = 68, normalized size = 0.88 \[ \frac{x (d x)^m \left (\left (a+b \sqrt [3]{x}\right )^2\right )^p \left (\frac{b \sqrt [3]{x}}{a}+1\right )^{-2 p} \, _2F_1\left (3 (m+1),-2 p;3 (m+1)+1;-\frac{b \sqrt [3]{x}}{a}\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^p*(d*x)^m,x]

[Out]

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

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Maple [F]  time = 0.033, size = 0, normalized size = 0. \begin{align*} \int \left ({a}^{2}+2\,ab\sqrt [3]{x}+{b}^{2}{x}^{{\frac{2}{3}}} \right ) ^{p} \left ( dx \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^p*(d*x)^m,x)

[Out]

int((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^p*(d*x)^m,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b^{2} x^{\frac{2}{3}} + 2 \, a b x^{\frac{1}{3}} + a^{2}\right )}^{p} \left (d x\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^p*(d*x)^m,x, algorithm="maxima")

[Out]

integrate((b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*(d*x)^m, x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^p*(d*x)^m,x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a**2+2*a*b*x**(1/3)+b**2*x**(2/3))**p*(d*x)**m,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b^{2} x^{\frac{2}{3}} + 2 \, a b x^{\frac{1}{3}} + a^{2}\right )}^{p} \left (d x\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^p*(d*x)^m,x, algorithm="giac")

[Out]

integrate((b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*(d*x)^m, x)