3.376 \(\int F^{a+b (c+d x)^n} (c+d x)^{-1-2 n} \, dx\)

Optimal. Leaf size=100 \[ \frac {b^2 F^a \log ^2(F) \text {Ei}\left (b (c+d x)^n \log (F)\right )}{2 d n}-\frac {(c+d x)^{-2 n} F^{a+b (c+d x)^n}}{2 d n}-\frac {b \log (F) (c+d x)^{-n} F^{a+b (c+d x)^n}}{2 d n} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2215, 2210} \[ \frac {b^2 F^a \log ^2(F) \text {Ei}\left (b (c+d x)^n \log (F)\right )}{2 d n}-\frac {(c+d x)^{-2 n} F^{a+b (c+d x)^n}}{2 d n}-\frac {b \log (F) (c+d x)^{-n} F^{a+b (c+d x)^n}}{2 d n} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b*(c + d*x)^n)*(c + d*x)^(-1 - 2*n),x]

[Out]

-F^(a + b*(c + d*x)^n)/(2*d*n*(c + d*x)^(2*n)) - (b*F^(a + b*(c + d*x)^n)*Log[F])/(2*d*n*(c + d*x)^n) + (b^2*F
^a*ExpIntegralEi[b*(c + d*x)^n*Log[F]]*Log[F]^2)/(2*d*n)

Rule 2210

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

Rule 2215

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*F^(a + b*(c + d*x)^n))/(d*(m + 1)), x] - Dist[(b*n*Log[F])/(m + 1), Int[(c + d*x)^Simplify[m + n]*F^(a +
b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d, m, n}, x] && IntegerQ[2*Simplify[(m + 1)/n]] && LtQ[-4, Simpl
ify[(m + 1)/n], 5] &&  !RationalQ[m] && SumSimplerQ[m, n]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 32, normalized size = 0.32 \[ -\frac {b^2 F^a \log ^2(F) \Gamma \left (-2,-b (c+d x)^n \log (F)\right )}{d n} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b*(c + d*x)^n)*(c + d*x)^(-1 - 2*n),x]

[Out]

-((b^2*F^a*Gamma[-2, -(b*(c + d*x)^n*Log[F])]*Log[F]^2)/(d*n))

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fricas [A]  time = 0.43, size = 84, normalized size = 0.84 \[ \frac {{\left (d x + c\right )}^{2 \, n} F^{a} b^{2} {\rm Ei}\left ({\left (d x + c\right )}^{n} b \log \relax (F)\right ) \log \relax (F)^{2} - {\left ({\left (d x + c\right )}^{n} b \log \relax (F) + 1\right )} e^{\left ({\left (d x + c\right )}^{n} b \log \relax (F) + a \log \relax (F)\right )}}{2 \, {\left (d x + c\right )}^{2 \, n} d n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1-2*n),x, algorithm="fricas")

[Out]

1/2*((d*x + c)^(2*n)*F^a*b^2*Ei((d*x + c)^n*b*log(F))*log(F)^2 - ((d*x + c)^n*b*log(F) + 1)*e^((d*x + c)^n*b*l
og(F) + a*log(F)))/((d*x + c)^(2*n)*d*n)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{-2 \, n - 1} F^{{\left (d x + c\right )}^{n} b + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1-2*n),x, algorithm="giac")

[Out]

integrate((d*x + c)^(-2*n - 1)*F^((d*x + c)^n*b + a), x)

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maple [A]  time = 0.17, size = 99, normalized size = 0.99 \[ -\frac {b^{2} F^{a} \Ei \left (1, -b \left (d x +c \right )^{n} \ln \relax (F )\right ) \ln \relax (F )^{2}}{2 d n}-\frac {b \,F^{a} F^{b \left (d x +c \right )^{n}} \left (d x +c \right )^{-n} \ln \relax (F )}{2 d n}-\frac {F^{a} F^{b \left (d x +c \right )^{n}} \left (d x +c \right )^{-2 n}}{2 d n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(b*(d*x+c)^n+a)*(d*x+c)^(-1-2*n),x)

[Out]

-1/2/n/d*F^(b*(d*x+c)^n)*F^a/((d*x+c)^n)^2-1/2/n/d*ln(F)*b*F^(b*(d*x+c)^n)*F^a/((d*x+c)^n)-1/2/n/d*ln(F)^2*b^2
*F^a*Ei(1,-b*(d*x+c)^n*ln(F))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{-2 \, n - 1} F^{{\left (d x + c\right )}^{n} b + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1-2*n),x, algorithm="maxima")

[Out]

integrate((d*x + c)^(-2*n - 1)*F^((d*x + c)^n*b + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {F^{a+b\,{\left (c+d\,x\right )}^n}}{{\left (c+d\,x\right )}^{2\,n+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a + b*(c + d*x)^n)/(c + d*x)^(2*n + 1),x)

[Out]

int(F^(a + b*(c + d*x)^n)/(c + d*x)^(2*n + 1), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(a+b*(d*x+c)**n)*(d*x+c)**(-1-2*n),x)

[Out]

Timed out

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