3.2.45 \(\int \frac {(f x)^m (d+e x^n)^q}{a+b x^n+c x^{2 n}} \, dx\) [145]

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

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

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Rubi [A]
time = 0.22, antiderivative size = 245, normalized size of antiderivative = 1.00, 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 = {1570, 525, 524} \begin {gather*} \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 )} \end {gather*}

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 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 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[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 1570

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]

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*}

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Mathematica [F]
time = 0.19, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(f x)^m \left (d+e x^n\right )^q}{a+b x^n+c x^{2 n}} \, dx \end {gather*}

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 = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (f x \right )^{m} \left (d +e \,x^{n}\right )^{q}}{a +b \,x^{n}+c \,x^{2 n}}\, dx\]

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 = 0.00, 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)^q/(a+b*x^n+c*x^(2*n)),x, algorithm="maxima")

[Out]

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

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

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: HeuristicGCDFailed} \end {gather*}

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]

Exception raised: HeuristicGCDFailed >> no luck

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

[Out]

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

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

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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