3.1.54 \(\int \frac {c+d x}{(a+b (F^{g (e+f x)})^n)^2} \, dx\) [54]

Optimal. Leaf size=191 \[ \frac {(c+d x)^2}{2 a^2 d}-\frac {d x}{a^2 f g n \log (F)}+\frac {c+d x}{a f \left (a+b \left (F^{g (e+f x)}\right )^n\right ) g n \log (F)}+\frac {d \log \left (a+b \left (F^{g (e+f x)}\right )^n\right )}{a^2 f^2 g^2 n^2 \log ^2(F)}-\frac {(c+d x) \log \left (1+\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a^2 f g n \log (F)}-\frac {d \text {Li}_2\left (-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a^2 f^2 g^2 n^2 \log ^2(F)} \]

[Out]

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

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Rubi [A]
time = 0.24, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {2216, 2215, 2221, 2317, 2438, 2222, 2320, 272, 36, 29, 31} \begin {gather*} -\frac {d \text {PolyLog}\left (2,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a^2 f^2 g^2 n^2 \log ^2(F)}-\frac {(c+d x) \log \left (\frac {b \left (F^{g (e+f x)}\right )^n}{a}+1\right )}{a^2 f g n \log (F)}+\frac {d \log \left (a+b \left (F^{g (e+f x)}\right )^n\right )}{a^2 f^2 g^2 n^2 \log ^2(F)}+\frac {(c+d x)^2}{2 a^2 d}-\frac {d x}{a^2 f g n \log (F)}+\frac {c+d x}{a f g n \log (F) \left (a+b \left (F^{g (e+f x)}\right )^n\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 2216

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dis
t[1/a, Int[(c + d*x)^m*(a + b*(F^(g*(e + f*x)))^n)^(p + 1), x], x] - Dist[b/a, Int[(c + d*x)^m*(F^(g*(e + f*x)
))^n*(a + b*(F^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && ILtQ[p, 0] && 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 2222

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

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 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 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

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

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Mathematica [F]
time = 0.95, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {c+d x}{\left (a+b \left (F^{g (e+f x)}\right )^n\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(590\) vs. \(2(189)=378\).
time = 0.06, size = 591, normalized size = 3.09

method	result	size
risch	\(\frac {d x +c}{a f \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right ) g n \ln \left (F \right )}+\frac {d \ln \left (F^{g \left (f x +e \right )}\right )^{2}}{2 \ln \left (F \right )^{2} f^{2} g^{2} a^{2}}-\frac {d \ln \left (1+\frac {b \,F^{n g f x} F^{-n g f x} \left (F^{g \left (f x +e \right )}\right )^{n}}{a}\right ) \ln \left (F^{g \left (f x +e \right )}\right )}{\ln \left (F \right )^{2} f^{2} g^{2} n \,a^{2}}+\frac {d \ln \left (F^{n g f x} F^{-n g f x} \left (F^{g \left (f x +e \right )}\right )^{n}\right ) x}{\ln \left (F \right ) f g n \,a^{2}}-\frac {d \ln \left (F^{n g f x} F^{-n g f x} \left (F^{g \left (f x +e \right )}\right )^{n}\right ) \ln \left (F^{g \left (f x +e \right )}\right )}{\ln \left (F \right )^{2} f^{2} g^{2} n \,a^{2}}-\frac {d \ln \left (a +b \,F^{n g f x} F^{-n g f x} \left (F^{g \left (f x +e \right )}\right )^{n}\right ) x}{\ln \left (F \right ) f g n \,a^{2}}+\frac {d \ln \left (a +b \,F^{n g f x} F^{-n g f x} \left (F^{g \left (f x +e \right )}\right )^{n}\right ) \ln \left (F^{g \left (f x +e \right )}\right )}{\ln \left (F \right )^{2} f^{2} g^{2} n \,a^{2}}-\frac {d \polylog \left (2, -\frac {b \,F^{n g f x} F^{-n g f x} \left (F^{g \left (f x +e \right )}\right )^{n}}{a}\right )}{\ln \left (F \right )^{2} f^{2} g^{2} n^{2} a^{2}}-\frac {d \ln \left (F^{n g f x} F^{-n g f x} \left (F^{g \left (f x +e \right )}\right )^{n}\right )}{\ln \left (F \right )^{2} f^{2} g^{2} n^{2} a^{2}}+\frac {d \ln \left (a +b \,F^{n g f x} F^{-n g f x} \left (F^{g \left (f x +e \right )}\right )^{n}\right )}{\ln \left (F \right )^{2} f^{2} g^{2} n^{2} a^{2}}+\frac {c \ln \left (F^{n g f x} F^{-n g f x} \left (F^{g \left (f x +e \right )}\right )^{n}\right )}{\ln \left (F \right ) f g n \,a^{2}}-\frac {c \ln \left (a +b \,F^{n g f x} F^{-n g f x} \left (F^{g \left (f x +e \right )}\right )^{n}\right )}{\ln \left (F \right ) f g n \,a^{2}}\)	\(591\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 428 vs. \(2 (192) = 384\).
time = 0.36, size = 428, normalized size = 2.24 \begin {gather*} \frac {{\left (a d f^{2} g^{2} n^{2} x^{2} + 2 \, a c f^{2} g^{2} n^{2} x + 2 \, a c f g^{2} n^{2} e - a d g^{2} n^{2} e^{2}\right )} \log \left (F\right )^{2} + {\left ({\left (b d f^{2} g^{2} n^{2} x^{2} + 2 \, b c f^{2} g^{2} n^{2} x + 2 \, b c f g^{2} n^{2} e - b d g^{2} n^{2} e^{2}\right )} \log \left (F\right )^{2} - 2 \, {\left (b d f g n x + b d g n e\right )} \log \left (F\right )\right )} F^{f g n x + g n e} - 2 \, {\left (F^{f g n x + g n e} b d + a d\right )} {\rm Li}_2\left (-\frac {F^{f g n x + g n e} b + a}{a} + 1\right ) + 2 \, {\left ({\left (b d - {\left (b c f g n - b d g n e\right )} \log \left (F\right )\right )} F^{f g n x + g n e} + a d - {\left (a c f g n - a d g n e\right )} \log \left (F\right )\right )} \log \left (F^{f g n x + g n e} b + a\right ) + 2 \, {\left (a c f g n - a d g n e\right )} \log \left (F\right ) - 2 \, {\left ({\left (b d f g n x + b d g n e\right )} F^{f g n x + g n e} \log \left (F\right ) + {\left (a d f g n x + a d g n e\right )} \log \left (F\right )\right )} \log \left (\frac {F^{f g n x + g n e} b + a}{a}\right )}{2 \, {\left (F^{f g n x + g n e} a^{2} b f^{2} g^{2} n^{2} \log \left (F\right )^{2} + a^{3} f^{2} g^{2} n^{2} \log \left (F\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c + d*x)/(a**2*f*g*n*log(F) + a*b*f*g*n*(F**(g*(e + f*x)))**n*log(F)) + (Integral(-d/(a + b*exp(e*g*n*log(F))
*exp(f*g*n*x*log(F))), x) + Integral(c*f*g*n*log(F)/(a + b*exp(e*g*n*log(F))*exp(f*g*n*x*log(F))), x) + Integr
al(d*f*g*n*x*log(F)/(a + b*exp(e*g*n*log(F))*exp(f*g*n*x*log(F))), x))/(a*f*g*n*log(F))

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {c+d\,x}{{\left (a+b\,{\left (F^{g\,\left (e+f\,x\right )}\right )}^n\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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