3.2722 \(\int x^{-1+3 n} (a+b x^n)^p \, dx\)

Optimal. Leaf size=75 \[ \frac{a^2 \left (a+b x^n\right )^{p+1}}{b^3 n (p+1)}-\frac{2 a \left (a+b x^n\right )^{p+2}}{b^3 n (p+2)}+\frac{\left (a+b x^n\right )^{p+3}}{b^3 n (p+3)} \]

[Out]

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

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Rubi [A]  time = 0.0448444, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {266, 43} \[ \frac{a^2 \left (a+b x^n\right )^{p+1}}{b^3 n (p+1)}-\frac{2 a \left (a+b x^n\right )^{p+2}}{b^3 n (p+2)}+\frac{\left (a+b x^n\right )^{p+3}}{b^3 n (p+3)} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 + 3*n)*(a + b*x^n)^p,x]

[Out]

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

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int x^{-1+3 n} \left (a+b x^n\right )^p \, dx &=\frac{\operatorname{Subst}\left (\int x^2 (a+b x)^p \, dx,x,x^n\right )}{n}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{a^2 (a+b x)^p}{b^2}-\frac{2 a (a+b x)^{1+p}}{b^2}+\frac{(a+b x)^{2+p}}{b^2}\right ) \, dx,x,x^n\right )}{n}\\ &=\frac{a^2 \left (a+b x^n\right )^{1+p}}{b^3 n (1+p)}-\frac{2 a \left (a+b x^n\right )^{2+p}}{b^3 n (2+p)}+\frac{\left (a+b x^n\right )^{3+p}}{b^3 n (3+p)}\\ \end{align*}

Mathematica [A]  time = 0.0370347, size = 66, normalized size = 0.88 \[ \frac{\left (a+b x^n\right )^{p+1} \left (2 a^2-2 a b (p+1) x^n+b^2 \left (p^2+3 p+2\right ) x^{2 n}\right )}{b^3 n (p+1) (p+2) (p+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.06, size = 105, normalized size = 1.4 \begin{align*}{\frac{ \left ({b}^{3}{p}^{2} \left ({x}^{n} \right ) ^{3}+a{b}^{2}{p}^{2} \left ({x}^{n} \right ) ^{2}+3\,{b}^{3}p \left ({x}^{n} \right ) ^{3}+ap \left ({x}^{n} \right ) ^{2}{b}^{2}+2\, \left ({x}^{n} \right ) ^{3}{b}^{3}-2\,{a}^{2}p{x}^{n}b+2\,{a}^{3} \right ) \left ( a+b{x}^{n} \right ) ^{p}}{ \left ( 2+p \right ) \left ( 3+p \right ) \left ( 1+p \right ) n{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.02099, size = 107, normalized size = 1.43 \begin{align*} \frac{{\left ({\left (p^{2} + 3 \, p + 2\right )} b^{3} x^{3 \, n} +{\left (p^{2} + p\right )} a b^{2} x^{2 \, n} - 2 \, a^{2} b p x^{n} + 2 \, a^{3}\right )}{\left (b x^{n} + a\right )}^{p}}{{\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{3} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((p^2 + 3*p + 2)*b^3*x^(3*n) + (p^2 + p)*a*b^2*x^(2*n) - 2*a^2*b*p*x^n + 2*a^3)*(b*x^n + a)^p/((p^3 + 6*p^2 +
11*p + 6)*b^3*n)

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Fricas [A]  time = 1.43983, size = 215, normalized size = 2.87 \begin{align*} -\frac{{\left (2 \, a^{2} b p x^{n} - 2 \, a^{3} -{\left (b^{3} p^{2} + 3 \, b^{3} p + 2 \, b^{3}\right )} x^{3 \, n} -{\left (a b^{2} p^{2} + a b^{2} p\right )} x^{2 \, n}\right )}{\left (b x^{n} + a\right )}^{p}}{b^{3} n p^{3} + 6 \, b^{3} n p^{2} + 11 \, b^{3} n p + 6 \, b^{3} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a^2*b*p*x^n - 2*a^3 - (b^3*p^2 + 3*b^3*p + 2*b^3)*x^(3*n) - (a*b^2*p^2 + a*b^2*p)*x^(2*n))*(b*x^n + a)^p/(
b^3*n*p^3 + 6*b^3*n*p^2 + 11*b^3*n*p + 6*b^3*n)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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