### 3.148 $$\int \frac{(d+e x^n)^q}{a+b x^n+c x^{2 n}} \, dx$$

Optimal. Leaf size=194 $-\frac{2 c x \left (d+e x^n\right )^q \left (\frac{e x^n}{d}+1\right )^{-q} F_1\left (\frac{1}{n};1,-q;1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}},-\frac{e x^n}{d}\right )}{-b \sqrt{b^2-4 a c}-4 a c+b^2}-\frac{2 c x \left (d+e x^n\right )^q \left (\frac{e x^n}{d}+1\right )^{-q} F_1\left (\frac{1}{n};1,-q;1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}},-\frac{e x^n}{d}\right )}{b \sqrt{b^2-4 a c}-4 a c+b^2}$

[Out]

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

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Rubi [A]  time = 0.299677, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.115, Rules used = {1428, 430, 429} $-\frac{2 c x \left (d+e x^n\right )^q \left (\frac{e x^n}{d}+1\right )^{-q} F_1\left (\frac{1}{n};1,-q;1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}},-\frac{e x^n}{d}\right )}{-b \sqrt{b^2-4 a c}-4 a c+b^2}-\frac{2 c x \left (d+e x^n\right )^q \left (\frac{e x^n}{d}+1\right )^{-q} F_1\left (\frac{1}{n};1,-q;1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}},-\frac{e x^n}{d}\right )}{b \sqrt{b^2-4 a c}-4 a c+b^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 1428

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

Rule 430

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

Rule 429

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])

Rubi steps

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

Mathematica [F]  time = 0.0837886, size = 0, normalized size = 0. $\int \frac{\left (d+e x^n\right )^q}{a+b x^n+c x^{2 n}} \, dx$

Veriﬁcation is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.052, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \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((d+e*x^n)^q/(a+b*x^n+c*x^(2*n)),x)

[Out]

int((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}}{c x^{2 \, n} + b x^{n} + a}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x^n + d)^q/(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}}{c x^{2 \, n} + b x^{n} + a}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x^n + d)^q/(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((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}}{c x^{2 \, n} + b x^{n} + a}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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