\(\int (d x)^m (a+b x^n+c x^{2 n})^p \, dx\) [606]

Optimal result

Integrand size = 22, antiderivative size = 158 \[ \int (d x)^m \left (a+b x^n+c x^{2 n}\right )^p \, dx=\frac {(d x)^{1+m} \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p \operatorname {AppellF1}\left (\frac {1+m}{n},-p,-p,\frac {1+m+n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{d (1+m)} \]

[Out]

(d*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)))/d/(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)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1399, 524} \[ \int (d x)^m \left (a+b x^n+c x^{2 n}\right )^p \, dx=\frac {(d x)^{m+1} \left (\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p \operatorname {AppellF1}\left (\frac {m+1}{n},-p,-p,\frac {m+n+1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{d (m+1)} \]

[In]

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

[Out]

((d*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])])/(d*(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)

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]

Rubi steps \begin{align*} \text {integral}& = \left (\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 (d 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 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 )}{d (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.15 \[ \int (d x)^m \left (a+b x^n+c x^{2 n}\right )^p \, dx=\frac {x (d x)^m \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+x^n \left (b+c x^n\right )\right )^p \operatorname {AppellF1}\left (\frac {1+m}{n},-p,-p,\frac {1+m+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )}{1+m} \]

[In]

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

[Out]

(x*(d*x)^m*(a + x^n*(b + c*x^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])])/((1 + m)*((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]

\[\int \left (d x \right )^{m} \left (a +b \,x^{n}+c \,x^{2 n}\right )^{p}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int (d x)^m \left (a+b x^n+c x^{2 n}\right )^p \, dx=\int { {\left (c x^{2 \, n} + b x^{n} + a\right )}^{p} \left (d x\right )^{m} \,d x } \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int (d x)^m \left (a+b x^n+c x^{2 n}\right )^p \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int (d x)^m \left (a+b x^n+c x^{2 n}\right )^p \, dx=\int { {\left (c x^{2 \, n} + b x^{n} + a\right )}^{p} \left (d x\right )^{m} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int (d x)^m \left (a+b x^n+c x^{2 n}\right )^p \, dx=\int { {\left (c x^{2 \, n} + b x^{n} + a\right )}^{p} \left (d x\right )^{m} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (d x)^m \left (a+b x^n+c x^{2 n}\right )^p \, dx=\int {\left (d\,x\right )}^m\,{\left (a+b\,x^n+c\,x^{2\,n}\right )}^p \,d x \]

[In]

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

[Out]

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