3.5.72 \(\int (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3})^p (d x)^m \, dx\) [472]

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

[Out]

(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^p*x*(d*x)^m*hypergeom([-2*p, 3+3*m],[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 = 0.03, antiderivative size = 77, normalized size of antiderivative = 1.00, 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 = {1370, 350, 348, 66} \begin {gather*} \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} \end {gather*}

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 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], 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])))

Rule 348

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 350

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

Rule 1370

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]

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 ) \text {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*}

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Mathematica [A]
time = 0.07, size = 68, normalized size = 0.88 \begin {gather*} \frac {\left (\left (a+b \sqrt [3]{x}\right )^2\right )^p \left (1+\frac {b \sqrt [3]{x}}{a}\right )^{-2 p} x (d x)^m \, _2F_1\left (3 (1+m),-2 p;1+3 (1+m);-\frac {b \sqrt [3]{x}}{a}\right )}{1+m} \end {gather*}

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 = 0.01, size = 0, normalized size = 0.00 \[\int \left (a^{2}+2 a b \,x^{\frac {1}{3}}+b^{2} x^{\frac {2}{3}}\right )^{p} \left (d x \right )^{m}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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: TypeError >>  Error detected within library code:   alglogextint: unimplemented

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (d\,x\right )}^m\,{\left (a^2+b^2\,x^{2/3}+2\,a\,b\,x^{1/3}\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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