3.4.85 \(\int (a-b x^{n/2})^p (a+b x^{n/2})^p (\frac {a^2 d (1+p)}{b^2 (1+\frac {-1-2 n-n p}{n})}+d x^n)^{\frac {-1-2 n-n p}{n}} \, dx\) [385]

Optimal. Leaf size=96 \[ -\frac {b^2 (1+n+n p) x \left (a-b x^{n/2}\right )^{1+p} \left (a+b x^{n/2}\right )^{1+p} \left (-\frac {a^2 d n (1+p)}{b^2 (1+n+n p)}+d x^n\right )^{-\frac {1+n+n p}{n}}}{a^4 d n (1+p)} \]

[Out]

-b^2*(n*p+n+1)*x*(a-b*x^(1/2*n))^(1+p)*(a+b*x^(1/2*n))^(1+p)/a^4/d/n/(1+p)/((-a^2*d*n*(1+p)/b^2/(n*p+n+1)+d*x^
n)^((n*p+n+1)/n))

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Rubi [A]
time = 0.09, antiderivative size = 104, normalized size of antiderivative = 1.08, number of steps used = 2, number of rules used = 2, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {533, 389} \begin {gather*} -\frac {b^2 x (n p+n+1) \left (a^2-b^2 x^n\right ) \left (a-b x^{n/2}\right )^p \left (a+b x^{n/2}\right )^p \left (d x^n-\frac {a^2 d n (p+1)}{b^2 (n p+n+1)}\right )^{-\frac {n p+n+1}{n}}}{a^4 d n (p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2*(1 + n + n*p)*x*(a - b*x^(n/2))^p*(a + b*x^(n/2))^p*(a^2 - b^2*x^n))/(a^4*d*n*(1 + p)*(-((a^2*d*n*(1 +
p))/(b^2*(1 + n + n*p))) + d*x^n)^((1 + n + n*p)/n)))

Rule 389

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[x*(a + b*x^n)^(p + 1)*((c +
d*x^n)^(q + 1)/(a*c)), x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 2) + 1, 0
] && EqQ[a*d*(p + 1) + b*c*(q + 1), 0]

Rule 533

Int[(u_.)*((c_) + (d_.)*(x_)^(n_.))^(q_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_)*((a2_) + (b2_.)*(x_)^(non2_.))^(
p_), x_Symbol] :> Dist[(a1 + b1*x^(n/2))^FracPart[p]*((a2 + b2*x^(n/2))^FracPart[p]/(a1*a2 + b1*b2*x^n)^FracPa
rt[p]), Int[u*(a1*a2 + b1*b2*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a1, b1, a2, b2, c, d, n, p, q}, x] && EqQ[
non2, n/2] && EqQ[a2*b1 + a1*b2, 0] &&  !(EqQ[n, 2] && IGtQ[q, 0])

Rubi steps

\begin {align*} \int \left (a-b x^{n/2}\right )^p \left (a+b x^{n/2}\right )^p \left (\frac {a^2 d (1+p)}{b^2 \left (1+\frac {-1-2 n-n p}{n}\right )}+d x^n\right )^{\frac {-1-2 n-n p}{n}} \, dx &=\left (\left (a-b x^{n/2}\right )^p \left (a+b x^{n/2}\right )^p \left (a^2-b^2 x^n\right )^{-p}\right ) \int \left (a^2-b^2 x^n\right )^p \left (\frac {a^2 d (1+p)}{b^2 \left (1+\frac {-1-2 n-n p}{n}\right )}+d x^n\right )^{\frac {-1-2 n-n p}{n}} \, dx\\ &=-\frac {b^2 (1+n+n p) x \left (a-b x^{n/2}\right )^p \left (a+b x^{n/2}\right )^p \left (a^2-b^2 x^n\right ) \left (-\frac {a^2 d n (1+p)}{b^2 (1+n+n p)}+d x^n\right )^{-\frac {1+n+n p}{n}}}{a^4 d n (1+p)}\\ \end {align*}

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Mathematica [A]
time = 0.90, size = 103, normalized size = 1.07 \begin {gather*} -\frac {b^2 (1+n+n p) x \left (a-b x^{n/2}\right )^p \left (a+b x^{n/2}\right )^p \left (d \left (-\frac {a^2 n (1+p)}{b^2 (1+n+n p)}+x^n\right )\right )^{-\frac {1+n+n p}{n}} \left (a^2-b^2 x^n\right )}{a^4 d n (1+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2*(1 + n + n*p)*x*(a - b*x^(n/2))^p*(a + b*x^(n/2))^p*(a^2 - b^2*x^n))/(a^4*d*n*(1 + p)*(d*(-((a^2*n*(1 +
p))/(b^2*(1 + n + n*p))) + x^n))^((1 + n + n*p)/n)))

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Maple [F]
time = 0.39, size = 0, normalized size = 0.00 \[\int \left (a -b \,x^{\frac {n}{2}}\right )^{p} \left (a +b \,x^{\frac {n}{2}}\right )^{p} \left (\frac {a^{2} d \left (1+p \right )}{b^{2} \left (1+\frac {-n p -2 n -1}{n}\right )}+d \,x^{n}\right )^{\frac {-n p -2 n -1}{n}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a-b*x^(1/2*n))^p*(a+b*x^(1/2*n))^p*(a^2*d*(1+p)/b^2/(1+(-n*p-2*n-1)/n)+d*x^n)^((-n*p-2*n-1)/n),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-b*x^(1/2*n))^p*(a+b*x^(1/2*n))^p*(a^2*d*(1+p)/b^2/(1+(-n*p-2*n-1)/n)+d*x^n)^((-n*p-2*n-1)/n),x, a
lgorithm="maxima")

[Out]

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

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Fricas [A]
time = 3.27, size = 180, normalized size = 1.88 \begin {gather*} \frac {{\left ({\left (b^{4} n p + b^{4} n + b^{4}\right )} x x^{2 \, n} - {\left (2 \, a^{2} b^{2} n p + 2 \, a^{2} b^{2} n + a^{2} b^{2}\right )} x x^{n} + {\left (a^{4} n p + a^{4} n\right )} x\right )} {\left (b x^{\frac {1}{2} \, n} + a\right )}^{p} {\left (-b x^{\frac {1}{2} \, n} + a\right )}^{p}}{{\left (a^{4} n p + a^{4} n\right )} \left (-\frac {a^{2} d n p + a^{2} d n - {\left (b^{2} d n p + b^{2} d n + b^{2} d\right )} x^{n}}{b^{2} n p + b^{2} n + b^{2}}\right )^{\frac {n p + 2 \, n + 1}{n}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-b*x^(1/2*n))^p*(a+b*x^(1/2*n))^p*(a^2*d*(1+p)/b^2/(1+(-n*p-2*n-1)/n)+d*x^n)^((-n*p-2*n-1)/n),x, a
lgorithm="fricas")

[Out]

((b^4*n*p + b^4*n + b^4)*x*x^(2*n) - (2*a^2*b^2*n*p + 2*a^2*b^2*n + a^2*b^2)*x*x^n + (a^4*n*p + a^4*n)*x)*(b*x
^(1/2*n) + a)^p*(-b*x^(1/2*n) + a)^p/((a^4*n*p + a^4*n)*(-(a^2*d*n*p + a^2*d*n - (b^2*d*n*p + b^2*d*n + b^2*d)
*x^n)/(b^2*n*p + b^2*n + b^2))^((n*p + 2*n + 1)/n))

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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-b*x**(1/2*n))**p*(a+b*x**(1/2*n))**p*(a**2*d*(1+p)/b**2/(1+(-n*p-2*n-1)/n)+d*x**n)**((-n*p-2*n-1)
/n),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-b*x^(1/2*n))^p*(a+b*x^(1/2*n))^p*(a^2*d*(1+p)/b^2/(1+(-n*p-2*n-1)/n)+d*x^n)^((-n*p-2*n-1)/n),x, a
lgorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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