### 3.145 $$\int \frac{(f x)^m (d+e x^n)^q}{a+b x^n+c x^{2 n}} \, dx$$

Optimal. Leaf size=245 $\frac{2 c (f x)^{m+1} \left (d+e x^n\right )^q \left (\frac{e x^n}{d}+1\right )^{-q} F_1\left (\frac{m+1}{n};1,-q;\frac{m+n+1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}},-\frac{e x^n}{d}\right )}{f (m+1) \sqrt{b^2-4 a c} \left (b-\sqrt{b^2-4 a c}\right )}-\frac{2 c (f x)^{m+1} \left (d+e x^n\right )^q \left (\frac{e x^n}{d}+1\right )^{-q} F_1\left (\frac{m+1}{n};1,-q;\frac{m+n+1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}},-\frac{e x^n}{d}\right )}{f (m+1) \sqrt{b^2-4 a c} \left (\sqrt{b^2-4 a c}+b\right )}$

[Out]

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

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Rubi [A]  time = 0.539941, antiderivative size = 245, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 31, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.097, Rules used = {1556, 511, 510} $\frac{2 c (f x)^{m+1} \left (d+e x^n\right )^q \left (\frac{e x^n}{d}+1\right )^{-q} F_1\left (\frac{m+1}{n};1,-q;\frac{m+n+1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}},-\frac{e x^n}{d}\right )}{f (m+1) \sqrt{b^2-4 a c} \left (b-\sqrt{b^2-4 a c}\right )}-\frac{2 c (f x)^{m+1} \left (d+e x^n\right )^q \left (\frac{e x^n}{d}+1\right )^{-q} F_1\left (\frac{m+1}{n};1,-q;\frac{m+n+1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}},-\frac{e x^n}{d}\right )}{f (m+1) \sqrt{b^2-4 a c} \left (\sqrt{b^2-4 a c}+b\right )}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 1556

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))^(q_))/((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_)), x_Symbol]
:> With[{r = Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/r, Int[((f*x)^m*(d + e*x^n)^q)/(b - r + 2*c*x^n), x], x] - Dist[
(2*c)/r, Int[((f*x)^m*(d + e*x^n)^q)/(b + r + 2*c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, f, m, n, q}, x] && Eq
Q[n2, 2*n] && NeQ[b^2 - 4*a*c, 0]

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{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])

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])

Rubi steps

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

Mathematica [F]  time = 0.174706, size = 0, normalized size = 0. $\int \frac{(f x)^m \left (d+e x^n\right )^q}{a+b x^n+c x^{2 n}} \, dx$

Veriﬁcation is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.076, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx \right ) ^{m} \left ( d+e{x}^{n} \right ) ^{q}}{a+b{x}^{n}+c{x}^{2\,n}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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