3.1.93 \(\int (d+e x^n) (a+b x^n+c x^{2 n})^p \, dx\) [93]

Optimal. Leaf size=288 \[ \frac {e x^{1+n} \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p F_1\left (1+\frac {1}{n};-p,-p;2+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{1+n}+d x \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p F_1\left (\frac {1}{n};-p,-p;1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) \]

[Out]

e*x^(1+n)*(a+b*x^n+c*x^(2*n))^p*AppellF1(1+1/n,-p,-p,2+1/n,-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)),-2*c*x^n/(b+(-4*a*c
+b^2)^(1/2)))/(1+n)/((1+2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))^p)/((1+2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))^p)+d*x*(a+b*x^n
+c*x^(2*n))^p*AppellF1(1/n,-p,-p,1+1/n,-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)),-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))/((1+2*
c*x^n/(b-(-4*a*c+b^2)^(1/2)))^p)/((1+2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))^p)

________________________________________________________________________________________

Rubi [A]
time = 0.19, antiderivative size = 288, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1446, 1362, 440, 1399, 524} \begin {gather*} d x \left (\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p F_1\left (\frac {1}{n};-p,-p;1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )+\frac {e x^{n+1} \left (\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p F_1\left (1+\frac {1}{n};-p,-p;2+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{n+1} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*x^(1 + n)*(a + b*x^n + c*x^(2*n))^p*AppellF1[1 + n^(-1), -p, -p, 2 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a
*c]), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/((1 + n)*(1 + (2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]))^p*(1 + (2*c*x^n)/(
b + Sqrt[b^2 - 4*a*c]))^p) + (d*x*(a + b*x^n + c*x^(2*n))^p*AppellF1[n^(-1), -p, -p, 1 + n^(-1), (-2*c*x^n)/(b
- Sqrt[b^2 - 4*a*c]), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/((1 + (2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]))^p*(1 + (2
*c*x^n)/(b + Sqrt[b^2 - 4*a*c]))^p)

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 1362

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n + c*x^(2*n))
^FracPart[p]/((1 + 2*c*(x^n/(b + Rt[b^2 - 4*a*c, 2])))^FracPart[p]*(1 + 2*c*(x^n/(b - Rt[b^2 - 4*a*c, 2])))^Fr
acPart[p])), Int[(1 + 2*c*(x^n/(b + Sqrt[b^2 - 4*a*c])))^p*(1 + 2*c*(x^n/(b - Sqrt[b^2 - 4*a*c])))^p, x], x] /
; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]

Rule 1399

Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a +
b*x^n + c*x^(2*n))^FracPart[p]/((1 + 2*c*(x^n/(b + Rt[b^2 - 4*a*c, 2])))^FracPart[p]*(1 + 2*c*(x^n/(b - Rt[b^
2 - 4*a*c, 2])))^FracPart[p])), Int[(d*x)^m*(1 + 2*c*(x^n/(b + Sqrt[b^2 - 4*a*c])))^p*(1 + 2*c*(x^n/(b - Sqrt[
b^2 - 4*a*c])))^p, x], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n]

Rule 1446

Int[((d_) + (e_.)*(x_)^(n_))*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Int[ExpandIntegran
d[(d + e*x^n)*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4
*a*c, 0]

Rubi steps

\begin {align*} \int \left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )^p \, dx &=\int \left (d \left (a+b x^n+c x^{2 n}\right )^p+e x^n \left (a+b x^n+c x^{2 n}\right )^p\right ) \, dx\\ &=d \int \left (a+b x^n+c x^{2 n}\right )^p \, dx+e \int x^n \left (a+b x^n+c x^{2 n}\right )^p \, dx\\ &=\left (d \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p\right ) \int \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^p \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^p \, dx+\left (e \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p\right ) \int x^n \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^p \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^p \, dx\\ &=\frac {e x^{1+n} \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p F_1\left (1+\frac {1}{n};-p,-p;2+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{1+n}+d x \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p F_1\left (\frac {1}{n};-p,-p;1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.36, size = 243, normalized size = 0.84 \begin {gather*} \frac {x \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+x^n \left (b+c x^n\right )\right )^p \left (e x^n F_1\left (1+\frac {1}{n};-p,-p;2+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )+d (1+n) F_1\left (\frac {1}{n};-p,-p;1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )\right )}{1+n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + x^n*(b + c*x^n))^p*(e*x^n*AppellF1[1 + n^(-1), -p, -p, 2 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]),
(2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])] + d*(1 + n)*AppellF1[n^(-1), -p, -p, 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2
- 4*a*c]), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])]))/((1 + n)*((b - Sqrt[b^2 - 4*a*c] + 2*c*x^n)/(b - Sqrt[b^2 -
4*a*c]))^p*((b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]))^p)

________________________________________________________________________________________

Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \left (d +e \,x^{n}\right ) \left (a +b \,x^{n}+c \,x^{2 n}\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [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((d+e*x^n)*(a+b*x^n+c*x^(2*n))^p,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________