3.11 \(\int \frac {F^{c (a+b x)}}{(d+e x)^5} \, dx\)

Optimal. Leaf size=161 \[ \frac {b^4 c^4 \log ^4(F) F^{c \left (a-\frac {b d}{e}\right )} \text {Ei}\left (\frac {b c (d+e x) \log (F)}{e}\right )}{24 e^5}-\frac {b^3 c^3 \log ^3(F) F^{c (a+b x)}}{24 e^4 (d+e x)}-\frac {b^2 c^2 \log ^2(F) F^{c (a+b x)}}{24 e^3 (d+e x)^2}-\frac {b c \log (F) F^{c (a+b x)}}{12 e^2 (d+e x)^3}-\frac {F^{c (a+b x)}}{4 e (d+e x)^4} \]

[Out]

-1/4*F^(c*(b*x+a))/e/(e*x+d)^4-1/12*b*c*F^(c*(b*x+a))*ln(F)/e^2/(e*x+d)^3-1/24*b^2*c^2*F^(c*(b*x+a))*ln(F)^2/e
^3/(e*x+d)^2-1/24*b^3*c^3*F^(c*(b*x+a))*ln(F)^3/e^4/(e*x+d)+1/24*b^4*c^4*F^(c*(a-b*d/e))*Ei(b*c*(e*x+d)*ln(F)/
e)*ln(F)^4/e^5

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Rubi [A]  time = 0.13, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2177, 2178} \[ \frac {b^4 c^4 \log ^4(F) F^{c \left (a-\frac {b d}{e}\right )} \text {Ei}\left (\frac {b c (d+e x) \log (F)}{e}\right )}{24 e^5}-\frac {b^3 c^3 \log ^3(F) F^{c (a+b x)}}{24 e^4 (d+e x)}-\frac {b^2 c^2 \log ^2(F) F^{c (a+b x)}}{24 e^3 (d+e x)^2}-\frac {b c \log (F) F^{c (a+b x)}}{12 e^2 (d+e x)^3}-\frac {F^{c (a+b x)}}{4 e (d+e x)^4} \]

Antiderivative was successfully verified.

[In]

Int[F^(c*(a + b*x))/(d + e*x)^5,x]

[Out]

-F^(c*(a + b*x))/(4*e*(d + e*x)^4) - (b*c*F^(c*(a + b*x))*Log[F])/(12*e^2*(d + e*x)^3) - (b^2*c^2*F^(c*(a + b*
x))*Log[F]^2)/(24*e^3*(d + e*x)^2) - (b^3*c^3*F^(c*(a + b*x))*Log[F]^3)/(24*e^4*(d + e*x)) + (b^4*c^4*F^(c*(a
- (b*d)/e))*ExpIntegralEi[(b*c*(d + e*x)*Log[F])/e]*Log[F]^4)/(24*e^5)

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)^5} \, dx &=-\frac {F^{c (a+b x)}}{4 e (d+e x)^4}+\frac {(b c \log (F)) \int \frac {F^{c (a+b x)}}{(d+e x)^4} \, dx}{4 e}\\ &=-\frac {F^{c (a+b x)}}{4 e (d+e x)^4}-\frac {b c F^{c (a+b x)} \log (F)}{12 e^2 (d+e x)^3}+\frac {\left (b^2 c^2 \log ^2(F)\right ) \int \frac {F^{c (a+b x)}}{(d+e x)^3} \, dx}{12 e^2}\\ &=-\frac {F^{c (a+b x)}}{4 e (d+e x)^4}-\frac {b c F^{c (a+b x)} \log (F)}{12 e^2 (d+e x)^3}-\frac {b^2 c^2 F^{c (a+b x)} \log ^2(F)}{24 e^3 (d+e x)^2}+\frac {\left (b^3 c^3 \log ^3(F)\right ) \int \frac {F^{c (a+b x)}}{(d+e x)^2} \, dx}{24 e^3}\\ &=-\frac {F^{c (a+b x)}}{4 e (d+e x)^4}-\frac {b c F^{c (a+b x)} \log (F)}{12 e^2 (d+e x)^3}-\frac {b^2 c^2 F^{c (a+b x)} \log ^2(F)}{24 e^3 (d+e x)^2}-\frac {b^3 c^3 F^{c (a+b x)} \log ^3(F)}{24 e^4 (d+e x)}+\frac {\left (b^4 c^4 \log ^4(F)\right ) \int \frac {F^{c (a+b x)}}{d+e x} \, dx}{24 e^4}\\ &=-\frac {F^{c (a+b x)}}{4 e (d+e x)^4}-\frac {b c F^{c (a+b x)} \log (F)}{12 e^2 (d+e x)^3}-\frac {b^2 c^2 F^{c (a+b x)} \log ^2(F)}{24 e^3 (d+e x)^2}-\frac {b^3 c^3 F^{c (a+b x)} \log ^3(F)}{24 e^4 (d+e x)}+\frac {b^4 c^4 F^{c \left (a-\frac {b d}{e}\right )} \text {Ei}\left (\frac {b c (d+e x) \log (F)}{e}\right ) \log ^4(F)}{24 e^5}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 121, normalized size = 0.75 \[ \frac {F^{a c} \left (b^4 c^4 \log ^4(F) F^{-\frac {b c d}{e}} \text {Ei}\left (\frac {b c (d+e x) \log (F)}{e}\right )-\frac {e F^{b c x} \left (b^3 c^3 \log ^3(F) (d+e x)^3+b^2 c^2 e \log ^2(F) (d+e x)^2+2 b c e^2 \log (F) (d+e x)+6 e^3\right )}{(d+e x)^4}\right )}{24 e^5} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(c*(a + b*x))/(d + e*x)^5,x]

[Out]

(F^(a*c)*((b^4*c^4*ExpIntegralEi[(b*c*(d + e*x)*Log[F])/e]*Log[F]^4)/F^((b*c*d)/e) - (e*F^(b*c*x)*(6*e^3 + 2*b
*c*e^2*(d + e*x)*Log[F] + b^2*c^2*e*(d + e*x)^2*Log[F]^2 + b^3*c^3*(d + e*x)^3*Log[F]^3))/(d + e*x)^4))/(24*e^
5)

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fricas [A]  time = 0.45, size = 300, normalized size = 1.86 \[ \frac {\frac {{\left (b^{4} c^{4} e^{4} x^{4} + 4 \, b^{4} c^{4} d e^{3} x^{3} + 6 \, b^{4} c^{4} d^{2} e^{2} x^{2} + 4 \, b^{4} c^{4} d^{3} e x + b^{4} c^{4} d^{4}\right )} {\rm Ei}\left (\frac {{\left (b c e x + b c d\right )} \log \relax (F)}{e}\right ) \log \relax (F)^{4}}{F^{\frac {b c d - a c e}{e}}} - {\left (6 \, e^{4} + {\left (b^{3} c^{3} e^{4} x^{3} + 3 \, b^{3} c^{3} d e^{3} x^{2} + 3 \, b^{3} c^{3} d^{2} e^{2} x + b^{3} c^{3} d^{3} e\right )} \log \relax (F)^{3} + {\left (b^{2} c^{2} e^{4} x^{2} + 2 \, b^{2} c^{2} d e^{3} x + b^{2} c^{2} d^{2} e^{2}\right )} \log \relax (F)^{2} + 2 \, {\left (b c e^{4} x + b c d e^{3}\right )} \log \relax (F)\right )} F^{b c x + a c}}{24 \, {\left (e^{9} x^{4} + 4 \, d e^{8} x^{3} + 6 \, d^{2} e^{7} x^{2} + 4 \, d^{3} e^{6} x + d^{4} e^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(e*x+d)^5,x, algorithm="fricas")

[Out]

1/24*((b^4*c^4*e^4*x^4 + 4*b^4*c^4*d*e^3*x^3 + 6*b^4*c^4*d^2*e^2*x^2 + 4*b^4*c^4*d^3*e*x + b^4*c^4*d^4)*Ei((b*
c*e*x + b*c*d)*log(F)/e)*log(F)^4/F^((b*c*d - a*c*e)/e) - (6*e^4 + (b^3*c^3*e^4*x^3 + 3*b^3*c^3*d*e^3*x^2 + 3*
b^3*c^3*d^2*e^2*x + b^3*c^3*d^3*e)*log(F)^3 + (b^2*c^2*e^4*x^2 + 2*b^2*c^2*d*e^3*x + b^2*c^2*d^2*e^2)*log(F)^2
+ 2*(b*c*e^4*x + b*c*d*e^3)*log(F))*F^(b*c*x + a*c))/(e^9*x^4 + 4*d*e^8*x^3 + 6*d^2*e^7*x^2 + 4*d^3*e^6*x + d
^4*e^5)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{{\left (b x + a\right )} c}}{{\left (e x + d\right )}^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(e*x+d)^5,x, algorithm="giac")

[Out]

integrate(F^((b*x + a)*c)/(e*x + d)^5, x)

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maple [A]  time = 0.08, size = 243, normalized size = 1.51 \[ -\frac {b^{4} c^{4} F^{a c} F^{b c x} \ln \relax (F )^{4}}{4 \left (b c x \ln \relax (F )+\frac {b c d \ln \relax (F )}{e}\right )^{4} e^{5}}-\frac {b^{4} c^{4} F^{a c} F^{b c x} \ln \relax (F )^{4}}{12 \left (b c x \ln \relax (F )+\frac {b c d \ln \relax (F )}{e}\right )^{3} e^{5}}-\frac {b^{4} c^{4} F^{a c} F^{b c x} \ln \relax (F )^{4}}{24 \left (b c x \ln \relax (F )+\frac {b c d \ln \relax (F )}{e}\right )^{2} e^{5}}-\frac {b^{4} c^{4} F^{a c} F^{b c x} \ln \relax (F )^{4}}{24 \left (b c x \ln \relax (F )+\frac {b c d \ln \relax (F )}{e}\right ) e^{5}}-\frac {b^{4} c^{4} F^{\frac {\left (a e -b d \right ) c}{e}} \Ei \left (1, -b c x \ln \relax (F )-a c \ln \relax (F )-\frac {-a c e \ln \relax (F )+b c d \ln \relax (F )}{e}\right ) \ln \relax (F )^{4}}{24 e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^((b*x+a)*c)/(e*x+d)^5,x)

[Out]

-1/4*b^4*c^4*ln(F)^4/e^5*F^(b*c*x)*F^(a*c)/(b*c*x*ln(F)+b*c*d/e*ln(F))^4-1/12*b^4*c^4*ln(F)^4/e^5*F^(b*c*x)*F^
(a*c)/(b*c*x*ln(F)+b*c*d/e*ln(F))^3-1/24*b^4*c^4*ln(F)^4/e^5*F^(b*c*x)*F^(a*c)/(b*c*x*ln(F)+b*c*d/e*ln(F))^2-1
/24*b^4*c^4*ln(F)^4/e^5*F^(b*c*x)*F^(a*c)/(b*c*x*ln(F)+b*c*d/e*ln(F))-1/24*b^4*c^4*ln(F)^4/e^5*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 \[ \int \frac {F^{{\left (b x + a\right )} c}}{{\left (e x + d\right )}^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(e*x+d)^5,x, algorithm="maxima")

[Out]

integrate(F^((b*x + a)*c)/(e*x + d)^5, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {F^{c\,\left (a+b\,x\right )}}{{\left (d+e\,x\right )}^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(a + b*x))/(d + e*x)^5,x)

[Out]

int(F^(c*(a + b*x))/(d + e*x)^5, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{c \left (a + b x\right )}}{\left (d + e x\right )^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(c*(b*x+a))/(e*x+d)**5,x)

[Out]

Integral(F**(c*(a + b*x))/(d + e*x)**5, x)

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