3.9 \(\int \frac {F^{c (a+b x)}}{(d+e x)^3} \, dx\)

Optimal. Leaf size=95 \[ \frac {b^2 c^2 \log ^2(F) F^{c \left (a-\frac {b d}{e}\right )} \text {Ei}\left (\frac {b c (d+e x) \log (F)}{e}\right )}{2 e^3}-\frac {b c \log (F) F^{c (a+b x)}}{2 e^2 (d+e x)}-\frac {F^{c (a+b x)}}{2 e (d+e x)^2} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2177, 2178} \[ \frac {b^2 c^2 \log ^2(F) F^{c \left (a-\frac {b d}{e}\right )} \text {Ei}\left (\frac {b c (d+e x) \log (F)}{e}\right )}{2 e^3}-\frac {b c \log (F) F^{c (a+b x)}}{2 e^2 (d+e x)}-\frac {F^{c (a+b x)}}{2 e (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

Int[F^(c*(a + b*x))/(d + e*x)^3,x]

[Out]

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

Rule 2177

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] &&  !$UseGamma ===
True

Rule 2178

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

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 88, normalized size = 0.93 \[ \frac {F^{c \left (a-\frac {b d}{e}\right )} \left (b^2 c^2 \log ^2(F) (d+e x)^2 \text {Ei}\left (\frac {b c (d+e x) \log (F)}{e}\right )-e F^{\frac {b c (d+e x)}{e}} (b c \log (F) (d+e x)+e)\right )}{2 e^3 (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 134, normalized size = 1.41 \[ \frac {\frac {{\left (b^{2} c^{2} e^{2} x^{2} + 2 \, b^{2} c^{2} d e x + b^{2} c^{2} d^{2}\right )} {\rm Ei}\left (\frac {{\left (b c e x + b c d\right )} \log \relax (F)}{e}\right ) \log \relax (F)^{2}}{F^{\frac {b c d - a c e}{e}}} - {\left (e^{2} + {\left (b c e^{2} x + b c d e\right )} \log \relax (F)\right )} F^{b c x + a c}}{2 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 155, normalized size = 1.63 \[ -\frac {b^{2} c^{2} F^{a c} F^{b c x} \ln \relax (F )^{2}}{2 \left (b c x \ln \relax (F )+\frac {b c d \ln \relax (F )}{e}\right )^{2} e^{3}}-\frac {b^{2} c^{2} F^{a c} F^{b c x} \ln \relax (F )^{2}}{2 \left (b c x \ln \relax (F )+\frac {b c d \ln \relax (F )}{e}\right ) e^{3}}-\frac {b^{2} c^{2} F^{\frac {\left (a e -b d \right ) c}{e}} \Ei \left (1, -b c x \ln \relax (F )-a c \ln \relax (F )-\frac {-a c e \ln \relax (F )+b c d \ln \relax (F )}{e}\right ) \ln \relax (F )^{2}}{2 e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(e*x+d)^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {F^{c\,\left (a+b\,x\right )}}{{\left (d+e\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(a + b*x))/(d + e*x)^3,x)

[Out]

int(F^(c*(a + b*x))/(d + e*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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