3.4.98 \(\int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^2} \, dx\) [398]

Optimal. Leaf size=116 \[ \frac {d F^{a+\frac {b}{c+d x}}}{f (d e-c f)}-\frac {F^{a+\frac {b}{c+d x}}}{f (e+f x)}-\frac {b d F^{a-\frac {b f}{d e-c f}} \text {Ei}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right ) \log (F)}{(d e-c f)^2} \]

[Out]

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

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Rubi [A]
time = 0.67, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2255, 6874, 2240, 2241, 2254, 2260, 2209} \begin {gather*} -\frac {b d \log (F) F^{a-\frac {b f}{d e-c f}} \text {Ei}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right )}{(d e-c f)^2}+\frac {d F^{a+\frac {b}{c+d x}}}{f (d e-c f)}-\frac {F^{a+\frac {b}{c+d x}}}{f (e+f x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*F^(a + b/(c + d*x)))/(f*(d*e - c*f)) - F^(a + b/(c + d*x))/(f*(e + f*x)) - (b*d*F^(a - (b*f)/(d*e - c*f))*E
xpIntegralEi[(b*d*(e + f*x)*Log[F])/((d*e - c*f)*(c + d*x))]*Log[F])/(d*e - c*f)^2

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 2240

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

Rule 2241

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 2254

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

Rule 2255

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

Rule 2260

Int[(F_)^((a_.) + (b_.)/((c_.) + (d_.)*(x_)))/(((e_.) + (f_.)*(x_))*((g_.) + (h_.)*(x_))), x_Symbol] :> Dist[-
d/(f*(d*g - c*h)), Subst[Int[F^(a - b*(h/(d*g - c*h)) + d*b*(x/(d*g - c*h)))/x, x], x, (g + h*x)/(c + d*x)], x
] /; FreeQ[{F, a, b, c, d, e, f}, x] && EqQ[d*e - c*f, 0]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]
time = 0.19, size = 116, normalized size = 1.00 \begin {gather*} \frac {d F^{a+\frac {b}{c+d x}}}{f (d e-c f)}-\frac {F^{a+\frac {b}{c+d x}}}{f (e+f x)}-\frac {b d F^{a+\frac {b f}{-d e+c f}} \text {Ei}\left (-\frac {b f \log (F)}{-d e+c f}+\frac {b \log (F)}{c+d x}\right ) \log (F)}{(d e-c f)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.11, size = 191, normalized size = 1.65

method	result	size
risch	\(\frac {d \ln \left (F \right ) b \,F^{a} F^{\frac {b}{d x +c}}}{\left (c f -e d \right )^{2} \left (\frac {b \ln \left (F \right )}{d x +c}+\ln \left (F \right ) a -\frac {\ln \left (F \right ) a c f}{c f -e d}+\frac {\ln \left (F \right ) a d e}{c f -e d}-\frac {\ln \left (F \right ) b f}{c f -e d}\right )}+\frac {d \ln \left (F \right ) b \,F^{\frac {a c f -a d e +b f}{c f -e d}} \expIntegral \left (1, -\frac {b \ln \left (F \right )}{d x +c}-\ln \left (F \right ) a -\frac {-\ln \left (F \right ) a c f +\ln \left (F \right ) a d e -\ln \left (F \right ) b f}{c f -e d}\right )}{\left (c f -e d \right )^{2}}\)	\(191\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [A]
time = 0.36, size = 186, normalized size = 1.60 \begin {gather*} -\frac {{\left (c d f x + c^{2} f - {\left (d^{2} x + c d\right )} e\right )} F^{\frac {a d x + a c + b}{d x + c}} + \frac {{\left (b d f x + b d e\right )} {\rm Ei}\left (-\frac {{\left (b d f x + b d e\right )} \log \left (F\right )}{c d f x + c^{2} f - {\left (d^{2} x + c d\right )} e}\right ) \log \left (F\right )}{F^{\frac {a d e - {\left (a c + b\right )} f}{c f - d e}}}}{c^{2} f^{3} x + d^{2} e^{3} + {\left (d^{2} f x - 2 \, c d f\right )} e^{2} - {\left (2 \, c d f^{2} x - c^{2} f^{2}\right )} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((c*d*f*x + c^2*f - (d^2*x + c*d)*e)*F^((a*d*x + a*c + b)/(d*x + c)) + (b*d*f*x + b*d*e)*Ei(-(b*d*f*x + b*d*e
)*log(F)/(c*d*f*x + c^2*f - (d^2*x + c*d)*e))*log(F)/F^((a*d*e - (a*c + b)*f)/(c*f - d*e)))/(c^2*f^3*x + d^2*e
^3 + (d^2*f*x - 2*c*d*f)*e^2 - (2*c*d*f^2*x - c^2*f^2)*e)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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