3.3.89 \(\int \frac {F^{a+b (c+d x)^3}}{(c+d x)^7} \, dx\) [289]

Optimal. Leaf size=87 \[ -\frac {F^{a+b (c+d x)^3}}{6 d (c+d x)^6}-\frac {b F^{a+b (c+d x)^3} \log (F)}{6 d (c+d x)^3}+\frac {b^2 F^a \text {Ei}\left (b (c+d x)^3 \log (F)\right ) \log ^2(F)}{6 d} \]

[Out]

-1/6*F^(a+b*(d*x+c)^3)/d/(d*x+c)^6-1/6*b*F^(a+b*(d*x+c)^3)*ln(F)/d/(d*x+c)^3+1/6*b^2*F^a*Ei(b*(d*x+c)^3*ln(F))
*ln(F)^2/d

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Rubi [A]
time = 0.13, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2245, 2241} \begin {gather*} \frac {b^2 F^a \log ^2(F) \text {Ei}\left (b (c+d x)^3 \log (F)\right )}{6 d}-\frac {F^{a+b (c+d x)^3}}{6 d (c+d x)^6}-\frac {b \log (F) F^{a+b (c+d x)^3}}{6 d (c+d x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[F^(a + b*(c + d*x)^3)/(c + d*x)^7,x]

[Out]

-1/6*F^(a + b*(c + d*x)^3)/(d*(c + d*x)^6) - (b*F^(a + b*(c + d*x)^3)*Log[F])/(6*d*(c + d*x)^3) + (b^2*F^a*Exp
IntegralEi[b*(c + d*x)^3*Log[F]]*Log[F]^2)/(6*d)

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 2245

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*(F^(a + b*(c + d*x)^n)/(d*(m + 1))), x] - Dist[b*n*(Log[F]/(m + 1)), Int[(c + d*x)^(m + n)*F^(a + b*(c +
d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/n)] && LtQ[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

Rubi steps

\begin {align*} \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^7} \, dx &=-\frac {F^{a+b (c+d x)^3}}{6 d (c+d x)^6}+\frac {1}{2} (b \log (F)) \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^4} \, dx\\ &=-\frac {F^{a+b (c+d x)^3}}{6 d (c+d x)^6}-\frac {b F^{a+b (c+d x)^3} \log (F)}{6 d (c+d x)^3}+\frac {1}{2} \left (b^2 \log ^2(F)\right ) \int \frac {F^{a+b (c+d x)^3}}{c+d x} \, dx\\ &=-\frac {F^{a+b (c+d x)^3}}{6 d (c+d x)^6}-\frac {b F^{a+b (c+d x)^3} \log (F)}{6 d (c+d x)^3}+\frac {b^2 F^a \text {Ei}\left (b (c+d x)^3 \log (F)\right ) \log ^2(F)}{6 d}\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 64, normalized size = 0.74 \begin {gather*} \frac {F^a \left (b^2 \text {Ei}\left (b (c+d x)^3 \log (F)\right ) \log ^2(F)-\frac {F^{b (c+d x)^3} \left (1+b (c+d x)^3 \log (F)\right )}{(c+d x)^6}\right )}{6 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b*(c + d*x)^3)/(c + d*x)^7,x]

[Out]

(F^a*(b^2*ExpIntegralEi[b*(c + d*x)^3*Log[F]]*Log[F]^2 - (F^(b*(c + d*x)^3)*(1 + b*(c + d*x)^3*Log[F]))/(c + d
*x)^6))/(6*d)

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {F^{a +b \left (d x +c \right )^{3}}}{\left (d x +c \right )^{7}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+b*(d*x+c)^3)/(d*x+c)^7,x)

[Out]

int(F^(a+b*(d*x+c)^3)/(d*x+c)^7,x)

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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)^3)/(d*x+c)^7,x, algorithm="maxima")

[Out]

integrate(F^((d*x + c)^3*b + a)/(d*x + c)^7, x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (81) = 162\).
time = 0.35, size = 269, normalized size = 3.09 \begin {gather*} \frac {{\left (b^{2} d^{6} x^{6} + 6 \, b^{2} c d^{5} x^{5} + 15 \, b^{2} c^{2} d^{4} x^{4} + 20 \, b^{2} c^{3} d^{3} x^{3} + 15 \, b^{2} c^{4} d^{2} x^{2} + 6 \, b^{2} c^{5} d x + b^{2} c^{6}\right )} F^{a} {\rm Ei}\left ({\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (F\right )\right ) \log \left (F\right )^{2} - {\left ({\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (F\right ) + 1\right )} F^{b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a}}{6 \, {\left (d^{7} x^{6} + 6 \, c d^{6} x^{5} + 15 \, c^{2} d^{5} x^{4} + 20 \, c^{3} d^{4} x^{3} + 15 \, c^{4} d^{3} x^{2} + 6 \, c^{5} d^{2} x + c^{6} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^3)/(d*x+c)^7,x, algorithm="fricas")

[Out]

1/6*((b^2*d^6*x^6 + 6*b^2*c*d^5*x^5 + 15*b^2*c^2*d^4*x^4 + 20*b^2*c^3*d^3*x^3 + 15*b^2*c^4*d^2*x^2 + 6*b^2*c^5
*d*x + b^2*c^6)*F^a*Ei((b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3)*log(F))*log(F)^2 - ((b*d^3*x^3 + 3*b*
c*d^2*x^2 + 3*b*c^2*d*x + b*c^3)*log(F) + 1)*F^(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a))/(d^7*x^6
+ 6*c*d^6*x^5 + 15*c^2*d^5*x^4 + 20*c^3*d^4*x^3 + 15*c^4*d^3*x^2 + 6*c^5*d^2*x + c^6*d)

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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)**3)/(d*x+c)**7,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)^3)/(d*x+c)^7,x, algorithm="giac")

[Out]

integrate(F^((d*x + c)^3*b + a)/(d*x + c)^7, x)

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Mupad [B]
time = 4.88, size = 76, normalized size = 0.87 \begin {gather*} -\frac {F^a\,b^2\,{\ln \left (F\right )}^2\,\left (\frac {\mathrm {expint}\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^3\right )}{2}+F^{b\,{\left (c+d\,x\right )}^3}\,\left (\frac {1}{2\,b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^3}+\frac {1}{2\,b^2\,{\ln \left (F\right )}^2\,{\left (c+d\,x\right )}^6}\right )\right )}{3\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a + b*(c + d*x)^3)/(c + d*x)^7,x)

[Out]

-(F^a*b^2*log(F)^2*(expint(-b*log(F)*(c + d*x)^3)/2 + F^(b*(c + d*x)^3)*(1/(2*b*log(F)*(c + d*x)^3) + 1/(2*b^2
*log(F)^2*(c + d*x)^6))))/(3*d)

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