3.1.16 \(\int \frac {F^{c (a+b x)}}{d^3+3 d^2 e x+3 d e^2 x^2+e^3 x^3} \, dx\) [16]

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

[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 = 4, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2218, 2208, 2209} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*F^(c*(a + b*x))/(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 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^3+3 d^2 e x+3 d e^2 x^2+e^3 x^3} \, dx &=\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 = 0.04, size = 88, normalized size = 0.93 \begin {gather*} \frac {F^{c \left (a-\frac {b d}{e}\right )} \left (b^2 c^2 (d+e x)^2 \text {Ei}\left (\frac {b c (d+e x) \log (F)}{e}\right ) \log ^2(F)-e F^{\frac {b c (d+e x)}{e}} (e+b c (d+e x) \log (F))\right )}{2 e^3 (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[F^(c*(a + b*x))/(d^3 + 3*d^2*e*x + 3*d*e^2*x^2 + e^3*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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Maple [A]
time = 0.05, size = 155, normalized size = 1.63

method	result	size
risch	\(-\frac {c^{2} b^{2} \ln \left (F \right )^{2} F^{b c x} F^{c a}}{2 e^{3} \left (b c x \ln \left (F \right )+\frac {\ln \left (F \right ) b c d}{e}\right )^{2}}-\frac {c^{2} b^{2} \ln \left (F \right )^{2} F^{b c x} F^{c a}}{2 e^{3} \left (b c x \ln \left (F \right )+\frac {\ln \left (F \right ) b c d}{e}\right )}-\frac {c^{2} b^{2} \ln \left (F \right )^{2} F^{\frac {c \left (a e -b d \right )}{e}} \expIntegral \left (1, -b c x \ln \left (F \right )-c a \ln \left (F \right )-\frac {-\ln \left (F \right ) a c e +\ln \left (F \right ) b c d}{e}\right )}{2 e^{3}}\)	\(155\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c^2*b^2*ln(F)^2/e^3*F^(b*c*x)*F^(c*a)/(b*c*x*ln(F)+1/e*ln(F)*b*c*d)^2-1/2*c^2*b^2*ln(F)^2/e^3*F^(b*c*x)*F
^(c*a)/(b*c*x*ln(F)+1/e*ln(F)*b*c*d)-1/2*c^2*b^2*ln(F)^2/e^3*F^(c*(a*e-b*d)/e)*Ei(1,-b*c*x*ln(F)-c*a*ln(F)-(-l
n(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^3*x^3+3*d*e^2*x^2+3*d^2*e*x+d^3),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 130, normalized size = 1.37 \begin {gather*} \frac {\frac {{\left (b^{2} c^{2} x^{2} e^{2} + 2 \, b^{2} c^{2} d x e + b^{2} c^{2} d^{2}\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 )^{2}}{F^{{\left (b c d - a c e\right )} e^{\left (-1\right )}}} - {\left ({\left (b c x e^{2} + b c d e\right )} \log \left (F\right ) + e^{2}\right )} F^{b c x + a c}}{2 \, {\left (x^{2} e^{5} + 2 \, d x e^{4} + d^{2} e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(F^((b*x + a)*c)/(x^3*e^3 + 3*d*x^2*e^2 + 3*d^2*x*e + d^3), 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^3+3\,d^2\,e\,x+3\,d\,e^2\,x^2+e^3\,x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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