\(\int \frac {(c+d x)^2}{a+b (F^{g (e+f x)})^n} \, dx\) [47]

Optimal result

Integrand size = 25, antiderivative size = 145 \[ \int \frac {(c+d x)^2}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\frac {(c+d x)^3}{3 a d}-\frac {(c+d x)^2 \log \left (1+\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f g n \log (F)}-\frac {2 d (c+d x) \operatorname {PolyLog}\left (2,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f^2 g^2 n^2 \log ^2(F)}+\frac {2 d^2 \operatorname {PolyLog}\left (3,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f^3 g^3 n^3 \log ^3(F)} \]

[Out]

1/3*(d*x+c)^3/a/d-(d*x+c)^2*ln(1+b*(F^(g*(f*x+e)))^n/a)/a/f/g/n/ln(F)-2*d*(d*x+c)*polylog(2,-b*(F^(g*(f*x+e)))
^n/a)/a/f^2/g^2/n^2/ln(F)^2+2*d^2*polylog(3,-b*(F^(g*(f*x+e)))^n/a)/a/f^3/g^3/n^3/ln(F)^3

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2215, 2221, 2611, 2320, 6724} \[ \int \frac {(c+d x)^2}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=-\frac {2 d (c+d x) \operatorname {PolyLog}\left (2,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f^2 g^2 n^2 \log ^2(F)}-\frac {(c+d x)^2 \log \left (\frac {b \left (F^{g (e+f x)}\right )^n}{a}+1\right )}{a f g n \log (F)}+\frac {2 d^2 \operatorname {PolyLog}\left (3,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f^3 g^3 n^3 \log ^3(F)}+\frac {(c+d x)^3}{3 a d} \]

[In]

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

[Out]

(c + d*x)^3/(3*a*d) - ((c + d*x)^2*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(a*f*g*n*Log[F]) - (2*d*(c + d*x)*PolyL
og[2, -((b*(F^(g*(e + f*x)))^n)/a)])/(a*f^2*g^2*n^2*Log[F]^2) + (2*d^2*PolyLog[3, -((b*(F^(g*(e + f*x)))^n)/a)
])/(a*f^3*g^3*n^3*Log[F]^3)

Rule 2215

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[(c
+ d*x)^(m + 1)/(a*d*(m + 1)), x] - Dist[b/a, Int[(c + d*x)^m*((F^(g*(e + f*x)))^n/(a + b*(F^(g*(e + f*x)))^n))
, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps \begin{align*} \text {integral}& = \frac {(c+d x)^3}{3 a d}-\frac {b \int \frac {\left (F^{g (e+f x)}\right )^n (c+d x)^2}{a+b \left (F^{g (e+f x)}\right )^n} \, dx}{a} \\ & = \frac {(c+d x)^3}{3 a d}-\frac {(c+d x)^2 \log \left (1+\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f g n \log (F)}+\frac {(2 d) \int (c+d x) \log \left (1+\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right ) \, dx}{a f g n \log (F)} \\ & = \frac {(c+d x)^3}{3 a d}-\frac {(c+d x)^2 \log \left (1+\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f g n \log (F)}-\frac {2 d (c+d x) \text {Li}_2\left (-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f^2 g^2 n^2 \log ^2(F)}+\frac {\left (2 d^2\right ) \int \text {Li}_2\left (-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right ) \, dx}{a f^2 g^2 n^2 \log ^2(F)} \\ & = \frac {(c+d x)^3}{3 a d}-\frac {(c+d x)^2 \log \left (1+\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f g n \log (F)}-\frac {2 d (c+d x) \text {Li}_2\left (-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f^2 g^2 n^2 \log ^2(F)}+\frac {\left (2 d^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {b x^n}{a}\right )}{x} \, dx,x,F^{g (e+f x)}\right )}{a f^3 g^3 n^2 \log ^3(F)} \\ & = \frac {(c+d x)^3}{3 a d}-\frac {(c+d x)^2 \log \left (1+\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f g n \log (F)}-\frac {2 d (c+d x) \text {Li}_2\left (-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f^2 g^2 n^2 \log ^2(F)}+\frac {2 d^2 \text {Li}_3\left (-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f^3 g^3 n^3 \log ^3(F)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.83 \[ \int \frac {(c+d x)^2}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\frac {-f^2 g^2 n^2 (c+d x)^2 \log ^2(F) \log \left (1+\frac {a \left (F^{g (e+f x)}\right )^{-n}}{b}\right )+2 d f g n (c+d x) \log (F) \operatorname {PolyLog}\left (2,-\frac {a \left (F^{g (e+f x)}\right )^{-n}}{b}\right )+2 d^2 \operatorname {PolyLog}\left (3,-\frac {a \left (F^{g (e+f x)}\right )^{-n}}{b}\right )}{a f^3 g^3 n^3 \log ^3(F)} \]

[In]

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

[Out]

(-(f^2*g^2*n^2*(c + d*x)^2*Log[F]^2*Log[1 + a/(b*(F^(g*(e + f*x)))^n)]) + 2*d*f*g*n*(c + d*x)*Log[F]*PolyLog[2
, -(a/(b*(F^(g*(e + f*x)))^n))] + 2*d^2*PolyLog[3, -(a/(b*(F^(g*(e + f*x)))^n))])/(a*f^3*g^3*n^3*Log[F]^3)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1123\) vs. \(2(143)=286\).

Time = 0.30 (sec) , antiderivative size = 1124, normalized size of antiderivative = 7.75

method	result	size
risch	\(\text {Expression too large to display}\)	\(1124\)

[In]

int((d*x+c)^2/(a+b*(F^(g*(f*x+e)))^n),x,method=_RETURNVERBOSE)

[Out]

-1/ln(F)/f/g/n*c^2/a*ln((F^(g*(f*x+e)))^n*F^(-n*g*f*x)*F^(n*g*f*x)*b+a)+1/ln(F)/f/g/n*c^2/a*ln(F^(n*g*f*x)*F^(
-n*g*f*x)*(F^(g*(f*x+e)))^n)-2/ln(F)^2/f^2/g^2/n*c*d/a*ln(F^(g*(f*x+e)))*ln(1+b*F^(n*g*f*x)*F^(-n*g*f*x)*(F^(g
*(f*x+e)))^n/a)-2/ln(F)^2/f^2/g^2/n*d^2/a*ln(F^(g*(f*x+e)))*ln(1+b*F^(n*g*f*x)*F^(-n*g*f*x)*(F^(g*(f*x+e)))^n/
a)*x+2/ln(F)^2/f^2/g^2/n*c*d/a*ln((F^(g*(f*x+e)))^n*F^(-n*g*f*x)*F^(n*g*f*x)*b+a)*ln(F^(g*(f*x+e)))+2/ln(F)/f/
g/n*c*d/a*ln(F^(n*g*f*x)*F^(-n*g*f*x)*(F^(g*(f*x+e)))^n)*x-2/ln(F)^2/f^2/g^2/n*c*d/a*ln(F^(n*g*f*x)*F^(-n*g*f*
x)*(F^(g*(f*x+e)))^n)*ln(F^(g*(f*x+e)))+1/ln(F)^2/f^2/g^2*d^2/a*ln(F^(g*(f*x+e)))^2*x-2/ln(F)^2/f^2/g^2/n^2*d^
2/a*polylog(2,-b*F^(n*g*f*x)*F^(-n*g*f*x)*(F^(g*(f*x+e)))^n/a)*x-1/ln(F)^3/f^3/g^3/n*d^2/a*ln((F^(g*(f*x+e)))^
n*F^(-n*g*f*x)*F^(n*g*f*x)*b+a)*ln(F^(g*(f*x+e)))^2+1/ln(F)/f/g/n*d^2/a*ln(F^(n*g*f*x)*F^(-n*g*f*x)*(F^(g*(f*x
+e)))^n)*x^2+1/ln(F)^3/f^3/g^3/n*d^2/a*ln(F^(n*g*f*x)*F^(-n*g*f*x)*(F^(g*(f*x+e)))^n)*ln(F^(g*(f*x+e)))^2-2/3/
ln(F)^3/f^3/g^3*d^2/a*ln(F^(g*(f*x+e)))^3+2/ln(F)^3/f^3/g^3/n^3*d^2/a*polylog(3,-b*F^(n*g*f*x)*F^(-n*g*f*x)*(F
^(g*(f*x+e)))^n/a)-1/ln(F)/f/g/n*d^2/a*ln((F^(g*(f*x+e)))^n*F^(-n*g*f*x)*F^(n*g*f*x)*b+a)*x^2+1/ln(F)^2/f^2/g^
2*c*d/a*ln(F^(g*(f*x+e)))^2-2/ln(F)^2/f^2/g^2/n^2*c*d/a*polylog(2,-b*F^(n*g*f*x)*F^(-n*g*f*x)*(F^(g*(f*x+e)))^
n/a)+1/ln(F)^3/f^3/g^3/n*d^2/a*ln(F^(g*(f*x+e)))^2*ln(1+b*F^(n*g*f*x)*F^(-n*g*f*x)*(F^(g*(f*x+e)))^n/a)+2/ln(F
)^2/f^2/g^2/n*d^2/a*ln((F^(g*(f*x+e)))^n*F^(-n*g*f*x)*F^(n*g*f*x)*b+a)*ln(F^(g*(f*x+e)))*x-2/ln(F)^2/f^2/g^2/n
*d^2/a*ln(F^(n*g*f*x)*F^(-n*g*f*x)*(F^(g*(f*x+e)))^n)*ln(F^(g*(f*x+e)))*x-2/ln(F)/f/g/n*c*d/a*ln((F^(g*(f*x+e)
))^n*F^(-n*g*f*x)*F^(n*g*f*x)*b+a)*x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.87 \[ \int \frac {(c+d x)^2}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=-\frac {3 \, {\left (d^{2} e^{2} - 2 \, c d e f + c^{2} f^{2}\right )} g^{2} n^{2} \log \left (F^{f g n x + e g n} b + a\right ) \log \left (F\right )^{2} - {\left (d^{2} f^{3} g^{3} n^{3} x^{3} + 3 \, c d f^{3} g^{3} n^{3} x^{2} + 3 \, c^{2} f^{3} g^{3} n^{3} x\right )} \log \left (F\right )^{3} + 3 \, {\left (d^{2} f^{2} g^{2} n^{2} x^{2} + 2 \, c d f^{2} g^{2} n^{2} x - {\left (d^{2} e^{2} - 2 \, c d e f\right )} g^{2} n^{2}\right )} \log \left (F\right )^{2} \log \left (\frac {F^{f g n x + e g n} b + a}{a}\right ) + 6 \, {\left (d^{2} f g n x + c d f g n\right )} {\rm Li}_2\left (-\frac {F^{f g n x + e g n} b + a}{a} + 1\right ) \log \left (F\right ) - 6 \, d^{2} {\rm polylog}\left (3, -\frac {F^{f g n x + e g n} b}{a}\right )}{3 \, a f^{3} g^{3} n^{3} \log \left (F\right )^{3}} \]

[In]

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

[Out]

-1/3*(3*(d^2*e^2 - 2*c*d*e*f + c^2*f^2)*g^2*n^2*log(F^(f*g*n*x + e*g*n)*b + a)*log(F)^2 - (d^2*f^3*g^3*n^3*x^3
+ 3*c*d*f^3*g^3*n^3*x^2 + 3*c^2*f^3*g^3*n^3*x)*log(F)^3 + 3*(d^2*f^2*g^2*n^2*x^2 + 2*c*d*f^2*g^2*n^2*x - (d^2
*e^2 - 2*c*d*e*f)*g^2*n^2)*log(F)^2*log((F^(f*g*n*x + e*g*n)*b + a)/a) + 6*(d^2*f*g*n*x + c*d*f*g*n)*dilog(-(F
^(f*g*n*x + e*g*n)*b + a)/a + 1)*log(F) - 6*d^2*polylog(3, -F^(f*g*n*x + e*g*n)*b/a))/(a*f^3*g^3*n^3*log(F)^3)

Sympy [F]

\[ \int \frac {(c+d x)^2}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\int \frac {\left (c + d x\right )^{2}}{a + b \left (F^{e g + f g x}\right )^{n}}\, dx \]

[In]

integrate((d*x+c)**2/(a+b*(F**(g*(f*x+e)))**n),x)

[Out]

Integral((c + d*x)**2/(a + b*(F**(e*g + f*g*x))**n), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 303 vs. \(2 (142) = 284\).

Time = 0.25 (sec) , antiderivative size = 303, normalized size of antiderivative = 2.09 \[ \int \frac {(c+d x)^2}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=c^{2} {\left (\frac {f g n x + e g n}{a f g n} - \frac {\log \left (F^{f g n x + e g n} b + a\right )}{a f g n \log \left (F\right )}\right )} - \frac {2 \, {\left (f g n x \log \left (\frac {F^{f g n x} F^{e g n} b}{a} + 1\right ) \log \left (F\right ) + {\rm Li}_2\left (-\frac {F^{f g n x} F^{e g n} b}{a}\right )\right )} c d}{a f^{2} g^{2} n^{2} \log \left (F\right )^{2}} - \frac {{\left (f^{2} g^{2} n^{2} x^{2} \log \left (\frac {F^{f g n x} F^{e g n} b}{a} + 1\right ) \log \left (F\right )^{2} + 2 \, f g n x {\rm Li}_2\left (-\frac {F^{f g n x} F^{e g n} b}{a}\right ) \log \left (F\right ) - 2 \, {\rm Li}_{3}(-\frac {F^{f g n x} F^{e g n} b}{a})\right )} d^{2}}{a f^{3} g^{3} n^{3} \log \left (F\right )^{3}} + \frac {d^{2} f^{3} g^{3} n^{3} x^{3} \log \left (F\right )^{3} + 3 \, c d f^{3} g^{3} n^{3} x^{2} \log \left (F\right )^{3}}{3 \, a f^{3} g^{3} n^{3} \log \left (F\right )^{3}} \]

[In]

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

[Out]

c^2*((f*g*n*x + e*g*n)/(a*f*g*n) - log(F^(f*g*n*x + e*g*n)*b + a)/(a*f*g*n*log(F))) - 2*(f*g*n*x*log(F^(f*g*n*
x)*F^(e*g*n)*b/a + 1)*log(F) + dilog(-F^(f*g*n*x)*F^(e*g*n)*b/a))*c*d/(a*f^2*g^2*n^2*log(F)^2) - (f^2*g^2*n^2*
x^2*log(F^(f*g*n*x)*F^(e*g*n)*b/a + 1)*log(F)^2 + 2*f*g*n*x*dilog(-F^(f*g*n*x)*F^(e*g*n)*b/a)*log(F) - 2*polyl
og(3, -F^(f*g*n*x)*F^(e*g*n)*b/a))*d^2/(a*f^3*g^3*n^3*log(F)^3) + 1/3*(d^2*f^3*g^3*n^3*x^3*log(F)^3 + 3*c*d*f^
3*g^3*n^3*x^2*log(F)^3)/(a*f^3*g^3*n^3*log(F)^3)

Giac [F]

\[ \int \frac {(c+d x)^2}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\int { \frac {{\left (d x + c\right )}^{2}}{{\left (F^{{\left (f x + e\right )} g}\right )}^{n} b + a} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^2}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\int \frac {{\left (c+d\,x\right )}^2}{a+b\,{\left (F^{g\,\left (e+f\,x\right )}\right )}^n} \,d x \]

[In]

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

[Out]

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