3.1.31 \(\int \frac {a+b (F^{g (e+f x)})^n}{(c+d x)^3} \, dx\) [31]

Optimal. Leaf size=147 \[ -\frac {a}{2 d (c+d x)^2}-\frac {b \left (F^{e g+f g x}\right )^n}{2 d (c+d x)^2}-\frac {b f \left (F^{e g+f g x}\right )^n g n \log (F)}{2 d^2 (c+d x)}+\frac {b f^2 F^{\left (e-\frac {c f}{d}\right ) g n-g n (e+f x)} \left (F^{e g+f g x}\right )^n g^2 n^2 \text {Ei}\left (\frac {f g n (c+d x) \log (F)}{d}\right ) \log ^2(F)}{2 d^3} \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2214, 2208, 2213, 2209} \begin {gather*} -\frac {a}{2 d (c+d x)^2}+\frac {b f^2 g^2 n^2 \log ^2(F) \left (F^{e g+f g x}\right )^n F^{g n \left (e-\frac {c f}{d}\right )-g n (e+f x)} \text {Ei}\left (\frac {f g n (c+d x) \log (F)}{d}\right )}{2 d^3}-\frac {b f g n \log (F) \left (F^{e g+f g x}\right )^n}{2 d^2 (c+d x)}-\frac {b \left (F^{e g+f g x}\right )^n}{2 d (c+d x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*(F^(g*(e + f*x)))^n)/(c + d*x)^3,x]

[Out]

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

Rule 2208

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*((b*F^(g*(e + f*x)))^n/(d*(m + 1))), x] - Dist[f*g*n*(Log[F]/(d*(m + 1))), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !TrueQ[$UseGamm
a]

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2213

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

Rule 2214

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

Rubi steps

\begin {align*} \int \frac {a+b \left (F^{g (e+f x)}\right )^n}{(c+d x)^3} \, dx &=\int \left (\frac {a}{(c+d x)^3}+\frac {b \left (F^{e g+f g x}\right )^n}{(c+d x)^3}\right ) \, dx\\ &=-\frac {a}{2 d (c+d x)^2}+b \int \frac {\left (F^{e g+f g x}\right )^n}{(c+d x)^3} \, dx\\ &=-\frac {a}{2 d (c+d x)^2}-\frac {b \left (F^{e g+f g x}\right )^n}{2 d (c+d x)^2}+\frac {(b f g n \log (F)) \int \frac {\left (F^{e g+f g x}\right )^n}{(c+d x)^2} \, dx}{2 d}\\ &=-\frac {a}{2 d (c+d x)^2}-\frac {b \left (F^{e g+f g x}\right )^n}{2 d (c+d x)^2}-\frac {b f \left (F^{e g+f g x}\right )^n g n \log (F)}{2 d^2 (c+d x)}+\frac {\left (b f^2 g^2 n^2 \log ^2(F)\right ) \int \frac {\left (F^{e g+f g x}\right )^n}{c+d x} \, dx}{2 d^2}\\ &=-\frac {a}{2 d (c+d x)^2}-\frac {b \left (F^{e g+f g x}\right )^n}{2 d (c+d x)^2}-\frac {b f \left (F^{e g+f g x}\right )^n g n \log (F)}{2 d^2 (c+d x)}+\frac {\left (b f^2 F^{-n (e g+f g x)} \left (F^{e g+f g x}\right )^n g^2 n^2 \log ^2(F)\right ) \int \frac {F^{n (e g+f g x)}}{c+d x} \, dx}{2 d^2}\\ &=-\frac {a}{2 d (c+d x)^2}-\frac {b \left (F^{e g+f g x}\right )^n}{2 d (c+d x)^2}-\frac {b f \left (F^{e g+f g x}\right )^n g n \log (F)}{2 d^2 (c+d x)}+\frac {b f^2 F^{\left (e-\frac {c f}{d}\right ) g n-g n (e+f x)} \left (F^{e g+f g x}\right )^n g^2 n^2 \text {Ei}\left (\frac {f g n (c+d x) \log (F)}{d}\right ) \log ^2(F)}{2 d^3}\\ \end {align*}

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Mathematica [A]
time = 0.30, size = 111, normalized size = 0.76 \begin {gather*} -\frac {a d^2-b f^2 F^{-\frac {f g n (c+d x)}{d}} \left (F^{g (e+f x)}\right )^n g^2 n^2 (c+d x)^2 \text {Ei}\left (\frac {f g n (c+d x) \log (F)}{d}\right ) \log ^2(F)+b d \left (F^{g (e+f x)}\right )^n (d+f g n (c+d x) \log (F))}{2 d^3 (c+d x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*(F^(g*(e + f*x)))^n)/(c + d*x)^3,x]

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {a +b \left (F^{g \left (f x +e \right )}\right )^{n}}{\left (d x +c \right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*(F^(g*(f*x+e)))^n)/(d*x+c)^3,x)

[Out]

int((a+b*(F^(g*(f*x+e)))^n)/(d*x+c)^3,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((a+b*(F^(g*(f*x+e)))^n)/(d*x+c)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.44, size = 165, normalized size = 1.12 \begin {gather*} -\frac {a d^{2} - \frac {{\left (b d^{2} f^{2} g^{2} n^{2} x^{2} + 2 \, b c d f^{2} g^{2} n^{2} x + b c^{2} f^{2} g^{2} n^{2}\right )} {\rm Ei}\left (\frac {{\left (d f g n x + c f g n\right )} \log \left (F\right )}{d}\right ) \log \left (F\right )^{2}}{F^{\frac {c f g n - d g n e}{d}}} + {\left (b d^{2} + {\left (b d^{2} f g n x + b c d f g n\right )} \log \left (F\right )\right )} F^{f g n x + g n e}}{2 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(F^(g*(f*x+e)))^n)/(d*x+c)^3,x, algorithm="fricas")

[Out]

-1/2*(a*d^2 - (b*d^2*f^2*g^2*n^2*x^2 + 2*b*c*d*f^2*g^2*n^2*x + b*c^2*f^2*g^2*n^2)*Ei((d*f*g*n*x + c*f*g*n)*log
(F)/d)*log(F)^2/F^((c*f*g*n - d*g*n*e)/d) + (b*d^2 + (b*d^2*f*g*n*x + b*c*d*f*g*n)*log(F))*F^(f*g*n*x + g*n*e)
)/(d^5*x^2 + 2*c*d^4*x + c^2*d^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \left (F^{e g} F^{f g x}\right )^{n}}{\left (c + d x\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(F**(g*(f*x+e)))**n)/(d*x+c)**3,x)

[Out]

Integral((a + b*(F**(e*g)*F**(f*g*x))**n)/(c + d*x)**3, x)

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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((a+b*(F^(g*(f*x+e)))^n)/(d*x+c)^3,x, algorithm="giac")

[Out]

integrate(((F^((f*x + e)*g))^n*b + a)/(d*x + c)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,{\left (F^{g\,\left (e+f\,x\right )}\right )}^n}{{\left (c+d\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*(F^(g*(e + f*x)))^n)/(c + d*x)^3,x)

[Out]

int((a + b*(F^(g*(e + f*x)))^n)/(c + d*x)^3, x)

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