### 3.152 $$\int (f x)^m (d+e x^n)^2 (a+b x^n+c x^{2 n})^p \, dx$$

Optimal. Leaf size=498 $\frac{d^2 (f x)^{m+1} \left (\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}+1\right )^{-p} \left (\frac{2 c x^n}{\sqrt{b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p F_1\left (\frac{m+1}{n};-p,-p;\frac{m+n+1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{f (m+1)}+\frac{2 d e x^{n+1} (f x)^m \left (\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}+1\right )^{-p} \left (\frac{2 c x^n}{\sqrt{b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p F_1\left (\frac{m+n+1}{n};-p,-p;\frac{m+2 n+1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{m+n+1}+\frac{e^2 x^{2 n+1} (f x)^m \left (\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}+1\right )^{-p} \left (\frac{2 c x^n}{\sqrt{b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p F_1\left (\frac{m+2 n+1}{n};-p,-p;\frac{m+3 n+1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{m+2 n+1}$

[Out]

(d^2*(f*x)^(1 + m)*(a + b*x^n + c*x^(2*n))^p*AppellF1[(1 + m)/n, -p, -p, (1 + m + n)/n, (-2*c*x^n)/(b - Sqrt[b
^2 - 4*a*c]), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(f*(1 + 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) + (2*d*e*x^(1 + n)*(f*x)^m*(a + b*x^n + c*x^(2*n))^p*AppellF1[(1 + m + n)
/n, -p, -p, (1 + m + 2*n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/((1 + m
+ 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) + (e^2*x^(1 + 2*n)*(
f*x)^m*(a + b*x^n + c*x^(2*n))^p*AppellF1[(1 + m + 2*n)/n, -p, -p, (1 + m + 3*n)/n, (-2*c*x^n)/(b - Sqrt[b^2 -
4*a*c]), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/((1 + m + 2*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)

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Rubi [A]  time = 0.609488, antiderivative size = 498, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 4, integrand size = 31, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.129, Rules used = {1560, 1385, 510, 20} $\frac{d^2 (f x)^{m+1} \left (\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}+1\right )^{-p} \left (\frac{2 c x^n}{\sqrt{b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p F_1\left (\frac{m+1}{n};-p,-p;\frac{m+n+1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{f (m+1)}+\frac{2 d e x^{n+1} (f x)^m \left (\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}+1\right )^{-p} \left (\frac{2 c x^n}{\sqrt{b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p F_1\left (\frac{m+n+1}{n};-p,-p;\frac{m+2 n+1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{m+n+1}+\frac{e^2 x^{2 n+1} (f x)^m \left (\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}+1\right )^{-p} \left (\frac{2 c x^n}{\sqrt{b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p F_1\left (\frac{m+2 n+1}{n};-p,-p;\frac{m+3 n+1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{m+2 n+1}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(d^2*(f*x)^(1 + m)*(a + b*x^n + c*x^(2*n))^p*AppellF1[(1 + m)/n, -p, -p, (1 + m + n)/n, (-2*c*x^n)/(b - Sqrt[b
^2 - 4*a*c]), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(f*(1 + 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) + (2*d*e*x^(1 + n)*(f*x)^m*(a + b*x^n + c*x^(2*n))^p*AppellF1[(1 + m + n)
/n, -p, -p, (1 + m + 2*n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/((1 + m
+ 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) + (e^2*x^(1 + 2*n)*(
f*x)^m*(a + b*x^n + c*x^(2*n))^p*AppellF1[(1 + m + 2*n)/n, -p, -p, (1 + m + 3*n)/n, (-2*c*x^n)/(b - Sqrt[b^2 -
4*a*c]), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/((1 + m + 2*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)

Rule 1560

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Sy
mbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^n)^q*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, e,
f, m, n, p, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && (IGtQ[p, 0] || IGtQ[q, 0])

Rule 1385

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 510

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)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), 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 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(b^IntPart[n]*(b*v)^FracPart[n])/(a^IntPart[n
]*(a*v)^FracPart[n]), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

Rubi steps

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

Mathematica [A]  time = 1.16659, size = 391, normalized size = 0.79 $\frac{x (f x)^m \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x^n}{b-\sqrt{b^2-4 a c}}\right )^{-p} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x^n}{\sqrt{b^2-4 a c}+b}\right )^{-p} \left (a+x^n \left (b+c x^n\right )\right )^p \left (d^2 \left (m^2+m (3 n+2)+2 n^2+3 n+1\right ) F_1\left (\frac{m+1}{n};-p,-p;\frac{m+n+1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}},\frac{2 c x^n}{\sqrt{b^2-4 a c}-b}\right )+e (m+1) x^n \left (2 d (m+2 n+1) F_1\left (\frac{m+n+1}{n};-p,-p;\frac{m+2 n+1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}},\frac{2 c x^n}{\sqrt{b^2-4 a c}-b}\right )+e (m+n+1) x^n F_1\left (\frac{m+2 n+1}{n};-p,-p;\frac{m+3 n+1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}},\frac{2 c x^n}{\sqrt{b^2-4 a c}-b}\right )\right )\right )}{(m+1) (m+n+1) (m+2 n+1)}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(f*x)^m*(a + x^n*(b + c*x^n))^p*(d^2*(1 + m^2 + 3*n + 2*n^2 + m*(2 + 3*n))*AppellF1[(1 + m)/n, -p, -p, (1 +
m + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])] + e*(1 + m)*x^n*(2*d*(1 + m
+ 2*n)*AppellF1[(1 + m + n)/n, -p, -p, (1 + m + 2*n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^n)/(-b + S
qrt[b^2 - 4*a*c])] + e*(1 + m + n)*x^n*AppellF1[(1 + m + 2*n)/n, -p, -p, (1 + m + 3*n)/n, (-2*c*x^n)/(b + Sqrt
[b^2 - 4*a*c]), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])])))/((1 + m)*(1 + m + n)*(1 + m + 2*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)

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Maple [F]  time = 0.054, size = 0, normalized size = 0. \begin{align*} \int \left ( fx \right ) ^{m} \left ( d+e{x}^{n} \right ) ^{2} \left ( a+b{x}^{n}+c{x}^{2\,n} \right ) ^{p}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x^{n} + d\right )}^{2}{\left (c x^{2 \, n} + b x^{n} + a\right )}^{p} \left (f x\right )^{m}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (e^{2} x^{2 \, n} + 2 \, d e x^{n} + d^{2}\right )}{\left (c x^{2 \, n} + b x^{n} + a\right )}^{p} \left (f x\right )^{m}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError