3.60 \(\int (d+e x^n)^3 (a+c x^{2 n})^p \, dx\)

Optimal. Leaf size=299 \[ d^3 x \left (a+c x^{2 n}\right )^p \left (\frac {c x^{2 n}}{a}+1\right )^{-p} \, _2F_1\left (\frac {1}{2 n},-p;\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )+\frac {3 d^2 e x^{n+1} \left (a+c x^{2 n}\right )^p \left (\frac {c x^{2 n}}{a}+1\right )^{-p} \, _2F_1\left (\frac {n+1}{2 n},-p;\frac {1}{2} \left (3+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{n+1}+\frac {3 d e^2 x^{2 n+1} \left (a+c x^{2 n}\right )^p \left (\frac {c x^{2 n}}{a}+1\right )^{-p} \, _2F_1\left (\frac {1}{2} \left (2+\frac {1}{n}\right ),-p;\frac {1}{2} \left (4+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{2 n+1}+\frac {e^3 x^{3 n+1} \left (a+c x^{2 n}\right )^p \left (\frac {c x^{2 n}}{a}+1\right )^{-p} \, _2F_1\left (\frac {1}{2} \left (3+\frac {1}{n}\right ),-p;\frac {1}{2} \left (5+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{3 n+1} \]

[Out]

3*d*e^2*x^(1+2*n)*(a+c*x^(2*n))^p*hypergeom([-p, 1+1/2/n],[2+1/2/n],-c*x^(2*n)/a)/(1+2*n)/((1+c*x^(2*n)/a)^p)+
e^3*x^(1+3*n)*(a+c*x^(2*n))^p*hypergeom([-p, 3/2+1/2/n],[5/2+1/2/n],-c*x^(2*n)/a)/(1+3*n)/((1+c*x^(2*n)/a)^p)+
d^3*x*(a+c*x^(2*n))^p*hypergeom([-p, 1/2/n],[1+1/2/n],-c*x^(2*n)/a)/((1+c*x^(2*n)/a)^p)+3*d^2*e*x^(1+n)*(a+c*x
^(2*n))^p*hypergeom([-p, 1/2*(1+n)/n],[3/2+1/2/n],-c*x^(2*n)/a)/(1+n)/((1+c*x^(2*n)/a)^p)

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Rubi [A]  time = 0.16, antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1437, 246, 245, 365, 364} \[ \frac {3 d^2 e x^{n+1} \left (a+c x^{2 n}\right )^p \left (\frac {c x^{2 n}}{a}+1\right )^{-p} \, _2F_1\left (\frac {n+1}{2 n},-p;\frac {1}{2} \left (3+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{n+1}+d^3 x \left (a+c x^{2 n}\right )^p \left (\frac {c x^{2 n}}{a}+1\right )^{-p} \, _2F_1\left (\frac {1}{2 n},-p;\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )+\frac {3 d e^2 x^{2 n+1} \left (a+c x^{2 n}\right )^p \left (\frac {c x^{2 n}}{a}+1\right )^{-p} \, _2F_1\left (\frac {1}{2} \left (2+\frac {1}{n}\right ),-p;\frac {1}{2} \left (4+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{2 n+1}+\frac {e^3 x^{3 n+1} \left (a+c x^{2 n}\right )^p \left (\frac {c x^{2 n}}{a}+1\right )^{-p} \, _2F_1\left (\frac {1}{2} \left (3+\frac {1}{n}\right ),-p;\frac {1}{2} \left (5+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{3 n+1} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^n)^3*(a + c*x^(2*n))^p,x]

[Out]

(3*d*e^2*x^(1 + 2*n)*(a + c*x^(2*n))^p*Hypergeometric2F1[(2 + n^(-1))/2, -p, (4 + n^(-1))/2, -((c*x^(2*n))/a)]
)/((1 + 2*n)*(1 + (c*x^(2*n))/a)^p) + (e^3*x^(1 + 3*n)*(a + c*x^(2*n))^p*Hypergeometric2F1[(3 + n^(-1))/2, -p,
(5 + n^(-1))/2, -((c*x^(2*n))/a)])/((1 + 3*n)*(1 + (c*x^(2*n))/a)^p) + (d^3*x*(a + c*x^(2*n))^p*Hypergeometri
c2F1[1/(2*n), -p, (2 + n^(-1))/2, -((c*x^(2*n))/a)])/(1 + (c*x^(2*n))/a)^p + (3*d^2*e*x^(1 + n)*(a + c*x^(2*n)
)^p*Hypergeometric2F1[(1 + n)/(2*n), -p, (3 + n^(-1))/2, -((c*x^(2*n))/a)])/((1 + n)*(1 + (c*x^(2*n))/a)^p)

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 1437

Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^n)
^q*(a + c*x^(2*n))^p, x], x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && NeQ[c*d^2 + a*e^2, 0] && ((
IntegersQ[p, q] &&  !IntegerQ[n]) || IGtQ[p, 0] || (IGtQ[q, 0] &&  !IntegerQ[n]))

Rubi steps

\begin {align*} \int \left (d+e x^n\right )^3 \left (a+c x^{2 n}\right )^p \, dx &=\int \left (d^3 \left (a+c x^{2 n}\right )^p+3 d^2 e x^n \left (a+c x^{2 n}\right )^p+3 d e^2 x^{2 n} \left (a+c x^{2 n}\right )^p+e^3 x^{3 n} \left (a+c x^{2 n}\right )^p\right ) \, dx\\ &=d^3 \int \left (a+c x^{2 n}\right )^p \, dx+\left (3 d^2 e\right ) \int x^n \left (a+c x^{2 n}\right )^p \, dx+\left (3 d e^2\right ) \int x^{2 n} \left (a+c x^{2 n}\right )^p \, dx+e^3 \int x^{3 n} \left (a+c x^{2 n}\right )^p \, dx\\ &=\left (d^3 \left (a+c x^{2 n}\right )^p \left (1+\frac {c x^{2 n}}{a}\right )^{-p}\right ) \int \left (1+\frac {c x^{2 n}}{a}\right )^p \, dx+\left (3 d^2 e \left (a+c x^{2 n}\right )^p \left (1+\frac {c x^{2 n}}{a}\right )^{-p}\right ) \int x^n \left (1+\frac {c x^{2 n}}{a}\right )^p \, dx+\left (3 d e^2 \left (a+c x^{2 n}\right )^p \left (1+\frac {c x^{2 n}}{a}\right )^{-p}\right ) \int x^{2 n} \left (1+\frac {c x^{2 n}}{a}\right )^p \, dx+\left (e^3 \left (a+c x^{2 n}\right )^p \left (1+\frac {c x^{2 n}}{a}\right )^{-p}\right ) \int x^{3 n} \left (1+\frac {c x^{2 n}}{a}\right )^p \, dx\\ &=\frac {3 d e^2 x^{1+2 n} \left (a+c x^{2 n}\right )^p \left (1+\frac {c x^{2 n}}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2} \left (2+\frac {1}{n}\right ),-p;\frac {1}{2} \left (4+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{1+2 n}+\frac {e^3 x^{1+3 n} \left (a+c x^{2 n}\right )^p \left (1+\frac {c x^{2 n}}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2} \left (3+\frac {1}{n}\right ),-p;\frac {1}{2} \left (5+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{1+3 n}+d^3 x \left (a+c x^{2 n}\right )^p \left (1+\frac {c x^{2 n}}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2 n},-p;\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )+\frac {3 d^2 e x^{1+n} \left (a+c x^{2 n}\right )^p \left (1+\frac {c x^{2 n}}{a}\right )^{-p} \, _2F_1\left (\frac {1+n}{2 n},-p;\frac {1}{2} \left (3+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{1+n}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 213, normalized size = 0.71 \[ x \left (a+c x^{2 n}\right )^p \left (\frac {c x^{2 n}}{a}+1\right )^{-p} \left (d^2 \left (d \, _2F_1\left (\frac {1}{2 n},-p;\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )+\frac {3 e x^n \, _2F_1\left (\frac {n+1}{2 n},-p;\frac {1}{2} \left (3+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{n+1}\right )+\frac {3 d e^2 x^{2 n} \, _2F_1\left (\frac {1}{2} \left (2+\frac {1}{n}\right ),-p;\frac {1}{2} \left (4+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{2 n+1}+\frac {e^3 x^{3 n} \, _2F_1\left (\frac {1}{2} \left (3+\frac {1}{n}\right ),-p;\frac {1}{2} \left (5+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{3 n+1}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x^n)^3*(a + c*x^(2*n))^p,x]

[Out]

(x*(a + c*x^(2*n))^p*((3*d*e^2*x^(2*n)*Hypergeometric2F1[(2 + n^(-1))/2, -p, (4 + n^(-1))/2, -((c*x^(2*n))/a)]
)/(1 + 2*n) + (e^3*x^(3*n)*Hypergeometric2F1[(3 + n^(-1))/2, -p, (5 + n^(-1))/2, -((c*x^(2*n))/a)])/(1 + 3*n)
+ d^2*(d*Hypergeometric2F1[1/(2*n), -p, (2 + n^(-1))/2, -((c*x^(2*n))/a)] + (3*e*x^n*Hypergeometric2F1[(1 + n)
/(2*n), -p, (3 + n^(-1))/2, -((c*x^(2*n))/a)])/(1 + n))))/(1 + (c*x^(2*n))/a)^p

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fricas [F]  time = 0.98, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e^{3} x^{3 \, n} + 3 \, d e^{2} x^{2 \, n} + 3 \, d^{2} e x^{n} + d^{3}\right )} {\left (c x^{2 \, n} + a\right )}^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+e*x^n)^3*(a+c*x^(2*n))^p,x, algorithm="fricas")

[Out]

integral((e^3*x^(3*n) + 3*d*e^2*x^(2*n) + 3*d^2*e*x^n + d^3)*(c*x^(2*n) + a)^p, x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+e*x^n)^3*(a+c*x^(2*n))^p,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to divide, perhaps due to rounding error%%%{-16,[1,0,6,3,2,4,4,1]%%%}+%%%{-64,[1,0,6,3,2,3,4,1]%%%}+%%%{-96
,[1,0,6,3,2,2,4,1]%%%}+%%%{-64,[1,0,6,3,2,1,4,1]%%%}+%%%{-16,[1,0,6,3,2,0,4,1]%%%} / %%%{16,[0,0,6,4,2,4,4,0]%
%%}+%%%{64,[0,0,6,4,2,3,4,0]%%%}+%%%{96,[0,0,6,4,2,2,4,0]%%%}+%%%{64,[0,0,6,4,2,1,4,0]%%%}+%%%{16,[0,0,6,4,2,0
,4,0]%%%} Error: Bad Argument Value

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maple [F]  time = 0.13, size = 0, normalized size = 0.00 \[ \int \left (e \,x^{n}+d \right )^{3} \left (c \,x^{2 n}+a \right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^n+d)^3*(c*x^(2*n)+a)^p,x)

[Out]

int((e*x^n+d)^3*(c*x^(2*n)+a)^p,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x^{n} + d\right )}^{3} {\left (c x^{2 \, n} + a\right )}^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+e*x^n)^3*(a+c*x^(2*n))^p,x, algorithm="maxima")

[Out]

integrate((e*x^n + d)^3*(c*x^(2*n) + a)^p, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+c\,x^{2\,n}\right )}^p\,{\left (d+e\,x^n\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^(2*n))^p*(d + e*x^n)^3,x)

[Out]

int((a + c*x^(2*n))^p*(d + e*x^n)^3, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+e*x**n)**3*(a+c*x**(2*n))**p,x)

[Out]

Timed out

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