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

Optimal result

Integrand size = 30, antiderivative size = 77 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p (d x)^m \, dx=\frac {\left (1+\frac {b \sqrt [3]{x}}{a}\right )^{-2 p} \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p x (d x)^m \operatorname {Hypergeometric2F1}\left (3 (1+m),-2 p,4+3 m,-\frac {b \sqrt [3]{x}}{a}\right )}{1+m} \]

[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))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1370, 350, 348, 66} \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p (d x)^m \, dx=\frac {x (d x)^m \left (\frac {b \sqrt [3]{x}}{a}+1\right )^{-2 p} \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p \operatorname {Hypergeometric2F1}\left (3 (m+1),-2 p,3 m+4,-\frac {b \sqrt [3]{x}}{a}\right )}{m+1} \]

[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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.88 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p (d x)^m \, dx=\frac {\left (\left (a+b \sqrt [3]{x}\right )^2\right )^p \left (1+\frac {b \sqrt [3]{x}}{a}\right )^{-2 p} x (d x)^m \operatorname {Hypergeometric2F1}\left (3 (1+m),-2 p,1+3 (1+m),-\frac {b \sqrt [3]{x}}{a}\right )}{1+m} \]

[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))

Maple [F]

\[\int \left (a^{2}+2 a b \,x^{\frac {1}{3}}+b^{2} x^{\frac {2}{3}}\right )^{p} \left (d x \right )^{m}d x\]

[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)

Fricas [F(-2)]

Exception generated. \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p (d x)^m \, dx=\text {Exception raised: TypeError} \]

[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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

\[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p (d x)^m \, dx=\int { {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} \left (d x\right )^{m} \,d x } \]

[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)

Giac [F]

\[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p (d x)^m \, dx=\int { {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} \left (d x\right )^{m} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p (d x)^m \, dx=\int {\left (d\,x\right )}^m\,{\left (a^2+b^2\,x^{2/3}+2\,a\,b\,x^{1/3}\right )}^p \,d x \]

[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)