3.478 \(\int (\frac {(a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3})^p}{x^2}-\frac {2 b^3 (1-2 p) (1-p) p (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3})^p}{3 a^3 x}) \, dx\)

Optimal. Leaf size=146 \[ \frac {b (1-p) \left (a+b \sqrt [3]{x}\right ) \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{a^2 x^{2/3}}-\frac {\left (a+b \sqrt [3]{x}\right ) \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{a x}-\frac {b^2 (1-2 p) (1-p) \left (a+b \sqrt [3]{x}\right ) \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{a^3 \sqrt [3]{x}} \]

[Out]

-(a+b*x^(1/3))*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^p/a/x+b*(1-p)*(a+b*x^(1/3))*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^p/a
^2/x^(2/3)-b^2*(1-2*p)*(1-p)*(a+b*x^(1/3))*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^p/a^3/x^(1/3)

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Rubi [C]  time = 0.10, antiderivative size = 162, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 3, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {1356, 266, 65} \[ \frac {2 b^3 (1-2 p) (1-p) p \left (\frac {b \sqrt [3]{x}}{a}+1\right ) \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p \, _2F_1\left (1,2 p+1;2 (p+1);\frac {\sqrt [3]{x} b}{a}+1\right )}{a^3 (2 p+1)}+\frac {3 b^3 \left (\frac {b \sqrt [3]{x}}{a}+1\right ) \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p \, _2F_1\left (4,2 p+1;2 (p+1);\frac {\sqrt [3]{x} b}{a}+1\right )}{a^3 (2 p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

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

Rule 266

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

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]

Rubi steps

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

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Mathematica [C]  time = 0.08, size = 101, normalized size = 0.69 \[ \frac {b^3 \left (a+b \sqrt [3]{x}\right ) \left (\left (a+b \sqrt [3]{x}\right )^2\right )^p \left (2 p \left (2 p^2-3 p+1\right ) \, _2F_1\left (1,2 p+1;2 (p+1);\frac {\sqrt [3]{x} b}{a}+1\right )+3 \, _2F_1\left (4,2 p+1;2 (p+1);\frac {\sqrt [3]{x} b}{a}+1\right )\right )}{a^3 (2 a p+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.32, size = 82, normalized size = 0.56 \[ -\frac {{\left (a^{2} b p x^{\frac {1}{3}} + a^{3} + {\left (2 \, b^{3} p^{2} - 3 \, b^{3} p + b^{3}\right )} x + 2 \, {\left (a b^{2} p^{2} - a b^{2} p\right )} x^{\frac {2}{3}}\right )} {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p}}{a^{3} x} \]

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

[Out]

-(a^2*b*p*x^(1/3) + a^3 + (2*b^3*p^2 - 3*b^3*p + b^3)*x + 2*(a*b^2*p^2 - a*b^2*p)*x^(2/3))*(b^2*x^(2/3) + 2*a*
b*x^(1/3) + a^2)^p/(a^3*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {2 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} b^{3} {\left (2 \, p - 1\right )} {\left (p - 1\right )} p}{3 \, a^{3} x} + \frac {{\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p}}{x^{2}}\,{d x} \]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {2 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p} b^{3} {\left (2 \, p - 1\right )} {\left (p - 1\right )} p}{3 \, a^{3} x} + \frac {{\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p}}{x^{2}}\,{d x} \]

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

[Out]

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

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mupad [B]  time = 1.65, size = 69, normalized size = 0.47 \[ -\frac {{\left (a^2+b^2\,x^{2/3}+2\,a\,b\,x^{1/3}\right )}^p\,\left (\frac {b^3\,x\,\left (2\,p^2-3\,p+1\right )}{a^3}+\frac {b\,p\,x^{1/3}}{a}+\frac {2\,b^2\,p\,x^{2/3}\,\left (p-1\right )}{a^2}+1\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a^2 + b^2*x^(2/3) + 2*a*b*x^(1/3))^p*((b^3*x*(2*p^2 - 3*p + 1))/a^3 + (b*p*x^(1/3))/a + (2*b^2*p*x^(2/3)*(p
- 1))/a^2 + 1))/x

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \left (- \frac {3 a^{3} \left (a^{2} + 2 a b \sqrt [3]{x} + b^{2} x^{\frac {2}{3}}\right )^{p}}{x^{2}}\right )\, dx + \int \frac {2 b^{3} p \left (a^{2} + 2 a b \sqrt [3]{x} + b^{2} x^{\frac {2}{3}}\right )^{p}}{x}\, dx + \int \left (- \frac {6 b^{3} p^{2} \left (a^{2} + 2 a b \sqrt [3]{x} + b^{2} x^{\frac {2}{3}}\right )^{p}}{x}\right )\, dx + \int \frac {4 b^{3} p^{3} \left (a^{2} + 2 a b \sqrt [3]{x} + b^{2} x^{\frac {2}{3}}\right )^{p}}{x}\, dx}{3 a^{3}} \]

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/x**2-2/3*b**3*(1-2*p)*(1-p)*p*(a**2+2*a*b*x**(1/3)+b**2*x**(2
/3))**p/a**3/x,x)

[Out]

-(Integral(-3*a**3*(a**2 + 2*a*b*x**(1/3) + b**2*x**(2/3))**p/x**2, x) + Integral(2*b**3*p*(a**2 + 2*a*b*x**(1
/3) + b**2*x**(2/3))**p/x, x) + Integral(-6*b**3*p**2*(a**2 + 2*a*b*x**(1/3) + b**2*x**(2/3))**p/x, x) + Integ
ral(4*b**3*p**3*(a**2 + 2*a*b*x**(1/3) + b**2*x**(2/3))**p/x, x))/(3*a**3)

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