3.2.53 \(\int (f x)^m (d+e x^n) (a+b x^n+c x^{2 n})^p \, dx\) [153]

Optimal. Leaf size=323 \[ \frac {d (f x)^{1+m} \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p F_1\left (\frac {1+m}{n};-p,-p;\frac {1+m+n}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{f (1+m)}+\frac {e x^{1+n} (f x)^m \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p F_1\left (\frac {1+m+n}{n};-p,-p;\frac {1+m+2 n}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{1+m+n} \]

[Out]

d*(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-(-4*a*c+b^2)^(1/2)),-2*c*x^n/
(b+(-4*a*c+b^2)^(1/2)))/f/(1+m)/((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)+
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-(-4*a*c+b^2)^(1/2)),-
2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))/(1+m+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)

________________________________________________________________________________________

Rubi [A]
time = 0.24, antiderivative size = 323, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1574, 1399, 524, 20} \begin {gather*} \frac {d (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 {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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(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) + (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)

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]

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 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 1574

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])

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.49, size = 273, normalized size = 0.85 \begin {gather*} \frac {x (f x)^m \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 (d (1+m+n) 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 )+e (1+m) x^n 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 )\right )}{(1+m) (1+m+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(f*x)^m*(a + x^n*(b + c*x^n))^p*(d*(1 + m + n)*AppellF1[(1 + m)/n, -p, -p, (1 + m + n)/n, (-2*c*x^n)/(b + S
qrt[b^2 - 4*a*c]), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])] + e*(1 + m)*x^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 + Sqrt[b^2 - 4*a*c])]))/((1 + m)*(1 + m + 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 = 0.02, size = 0, normalized size = 0.00 \[\int \left (f x \right )^{m} \left (d +e \,x^{n}\right ) \left (a +b \,x^{n}+c \,x^{2 n}\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((f*x)^m*(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((f*x)^m*(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*(f*x)^m, 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((f*x)^m*(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*(f*x)^m, 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((f*x)**m*(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((f*x)^m*(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*(f*x)^m, x)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________