3.1.18 \(\int \frac {F^{c (a+b x)}}{d^5+5 d^4 e x+10 d^3 e^2 x^2+10 d^2 e^3 x^3+5 d e^4 x^4+e^5 x^5} \, dx\) [18]

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

[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 = 0.12, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 61, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {2218, 2208, 2209} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[F^(c*(a + b*x))/(d^5 + 5*d^4*e*x + 10*d^3*e^2*x^2 + 10*d^2*e^3*x^3 + 5*d*e^4*x^4 + e^5*x^5),x]

[Out]

-1/4*F^(c*(a + b*x))/(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 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 2218

Int[((a_.) + (b_.)*((F_)^((g_.)*(v_)))^(n_.))^(p_.)*(u_)^(m_.), x_Symbol] :> Int[NormalizePowerOfLinear[u, x]^
m*(a + b*(F^(g*ExpandToSum[v, x]))^n)^p, x] /; FreeQ[{F, a, b, g, n, p}, x] && LinearQ[v, x] && PowerOfLinearQ
[u, x] &&  !(LinearMatchQ[v, x] && PowerOfLinearMatchQ[u, x]) && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {F^{c (a+b x)}}{d^5+5 d^4 e x+10 d^3 e^2 x^2+10 d^2 e^3 x^3+5 d e^4 x^4+e^5 x^5} \, dx &=\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 = 0.02, size = 121, normalized size = 0.75 \begin {gather*} \frac {F^{a c} \left (b^4 c^4 F^{-\frac {b c d}{e}} \text {Ei}\left (\frac {b c (d+e x) \log (F)}{e}\right ) \log ^4(F)-\frac {e F^{b c x} \left (6 e^3+2 b c e^2 (d+e x) \log (F)+b^2 c^2 e (d+e x)^2 \log ^2(F)+b^3 c^3 (d+e x)^3 \log ^3(F)\right )}{(d+e x)^4}\right )}{24 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[F^(c*(a + b*x))/(d^5 + 5*d^4*e*x + 10*d^3*e^2*x^2 + 10*d^2*e^3*x^3 + 5*d*e^4*x^4 + e^5*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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Maple [A]
time = 0.08, size = 243, normalized size = 1.51

method	result	size
risch	\(-\frac {c^{4} b^{4} \ln \left (F \right )^{4} F^{b c x} F^{c a}}{4 e^{5} \left (b c x \ln \left (F \right )+\frac {\ln \left (F \right ) b c d}{e}\right )^{4}}-\frac {c^{4} b^{4} \ln \left (F \right )^{4} F^{b c x} F^{c a}}{12 e^{5} \left (b c x \ln \left (F \right )+\frac {\ln \left (F \right ) b c d}{e}\right )^{3}}-\frac {c^{4} b^{4} \ln \left (F \right )^{4} F^{b c x} F^{c a}}{24 e^{5} \left (b c x \ln \left (F \right )+\frac {\ln \left (F \right ) b c d}{e}\right )^{2}}-\frac {c^{4} b^{4} \ln \left (F \right )^{4} F^{b c x} F^{c a}}{24 e^{5} \left (b c x \ln \left (F \right )+\frac {\ln \left (F \right ) b c d}{e}\right )}-\frac {c^{4} b^{4} \ln \left (F \right )^{4} F^{\frac {c \left (a e -b d \right )}{e}} \expIntegral \left (1, -b c x \ln \left (F \right )-c a \ln \left (F \right )-\frac {-\ln \left (F \right ) a c e +\ln \left (F \right ) b c d}{e}\right )}{24 e^{5}}\)	\(243\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(b*x+a))/(e^5*x^5+5*d*e^4*x^4+10*d^2*e^3*x^3+10*d^3*e^2*x^2+5*d^4*e*x+d^5),x,method=_RETURNVERBOSE)

[Out]

-1/4*c^4*b^4*ln(F)^4/e^5*F^(b*c*x)*F^(c*a)/(b*c*x*ln(F)+1/e*ln(F)*b*c*d)^4-1/12*c^4*b^4*ln(F)^4/e^5*F^(b*c*x)*
F^(c*a)/(b*c*x*ln(F)+1/e*ln(F)*b*c*d)^3-1/24*c^4*b^4*ln(F)^4/e^5*F^(b*c*x)*F^(c*a)/(b*c*x*ln(F)+1/e*ln(F)*b*c*
d)^2-1/24*c^4*b^4*ln(F)^4/e^5*F^(b*c*x)*F^(c*a)/(b*c*x*ln(F)+1/e*ln(F)*b*c*d)-1/24*c^4*b^4*ln(F)^4/e^5*F^(c*(a
*e-b*d)/e)*Ei(1,-b*c*x*ln(F)-c*a*ln(F)-(-ln(F)*a*c*e+ln(F)*b*c*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^(c*(b*x+a))/(e^5*x^5+5*d*e^4*x^4+10*d^2*e^3*x^3+10*d^3*e^2*x^2+5*d^4*e*x+d^5),x, algorithm="maxima
")

[Out]

integrate(F^((b*x + a)*c)/(x^5*e^5 + 5*d*x^4*e^4 + 10*d^2*x^3*e^3 + 10*d^3*x^2*e^2 + 5*d^4*x*e + d^5), x)

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Fricas [A]
time = 0.37, size = 285, normalized size = 1.77 \begin {gather*} \frac {\frac {{\left (b^{4} c^{4} x^{4} e^{4} + 4 \, b^{4} c^{4} d x^{3} e^{3} + 6 \, b^{4} c^{4} d^{2} x^{2} e^{2} + 4 \, b^{4} c^{4} d^{3} x e + b^{4} c^{4} d^{4}\right )} {\rm Ei}\left ({\left (b c x e + b c d\right )} e^{\left (-1\right )} \log \left (F\right )\right ) \log \left (F\right )^{4}}{F^{{\left (b c d - a c e\right )} e^{\left (-1\right )}}} - {\left ({\left (b^{3} c^{3} x^{3} e^{4} + 3 \, b^{3} c^{3} d x^{2} e^{3} + 3 \, b^{3} c^{3} d^{2} x e^{2} + b^{3} c^{3} d^{3} e\right )} \log \left (F\right )^{3} + {\left (b^{2} c^{2} x^{2} e^{4} + 2 \, b^{2} c^{2} d x e^{3} + b^{2} c^{2} d^{2} e^{2}\right )} \log \left (F\right )^{2} + 2 \, {\left (b c x e^{4} + b c d e^{3}\right )} \log \left (F\right ) + 6 \, e^{4}\right )} F^{b c x + a c}}{24 \, {\left (x^{4} e^{9} + 4 \, d x^{3} e^{8} + 6 \, d^{2} x^{2} e^{7} + 4 \, d^{3} x e^{6} + d^{4} e^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(e^5*x^5+5*d*e^4*x^4+10*d^2*e^3*x^3+10*d^3*e^2*x^2+5*d^4*e*x+d^5),x, algorithm="fricas
")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(c*(b*x+a))/(e**5*x**5+5*d*e**4*x**4+10*d**2*e**3*x**3+10*d**3*e**2*x**2+5*d**4*e*x+d**5),x)

[Out]

Integral(F**(c*(a + b*x))/(d + e*x)**5, 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(F^(c*(b*x+a))/(e^5*x^5+5*d*e^4*x^4+10*d^2*e^3*x^3+10*d^3*e^2*x^2+5*d^4*e*x+d^5),x, algorithm="giac")

[Out]

integrate(F^((b*x + a)*c)/(x^5*e^5 + 5*d*x^4*e^4 + 10*d^2*x^3*e^3 + 10*d^3*x^2*e^2 + 5*d^4*x*e + d^5), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {F^{c\,\left (a+b\,x\right )}}{d^5+5\,d^4\,e\,x+10\,d^3\,e^2\,x^2+10\,d^2\,e^3\,x^3+5\,d\,e^4\,x^4+e^5\,x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(a + b*x))/(d^5 + e^5*x^5 + 5*d*e^4*x^4 + 10*d^3*e^2*x^2 + 10*d^2*e^3*x^3 + 5*d^4*e*x),x)

[Out]

int(F^(c*(a + b*x))/(d^5 + e^5*x^5 + 5*d*e^4*x^4 + 10*d^3*e^2*x^2 + 10*d^2*e^3*x^3 + 5*d^4*e*x), x)

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