3.264 \(\int \frac{F^{a+b (c+d x)^2}}{(c+d x)^7} \, dx\)

Optimal. Leaf size=121 \[ \frac{b^3 F^a \log ^3(F) \text{Ei}\left (b (c+d x)^2 \log (F)\right )}{12 d}-\frac{b^2 \log ^2(F) F^{a+b (c+d x)^2}}{12 d (c+d x)^2}-\frac{F^{a+b (c+d x)^2}}{6 d (c+d x)^6}-\frac{b \log (F) F^{a+b (c+d x)^2}}{12 d (c+d x)^4} \]

[Out]

-F^(a + b*(c + d*x)^2)/(6*d*(c + d*x)^6) - (b*F^(a + b*(c + d*x)^2)*Log[F])/(12*d*(c + d*x)^4) - (b^2*F^(a + b
*(c + d*x)^2)*Log[F]^2)/(12*d*(c + d*x)^2) + (b^3*F^a*ExpIntegralEi[b*(c + d*x)^2*Log[F]]*Log[F]^3)/(12*d)

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Rubi [A]  time = 0.25758, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2214, 2210} \[ \frac{b^3 F^a \log ^3(F) \text{Ei}\left (b (c+d x)^2 \log (F)\right )}{12 d}-\frac{b^2 \log ^2(F) F^{a+b (c+d x)^2}}{12 d (c+d x)^2}-\frac{F^{a+b (c+d x)^2}}{6 d (c+d x)^6}-\frac{b \log (F) F^{a+b (c+d x)^2}}{12 d (c+d x)^4} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b*(c + d*x)^2)/(c + d*x)^7,x]

[Out]

-F^(a + b*(c + d*x)^2)/(6*d*(c + d*x)^6) - (b*F^(a + b*(c + d*x)^2)*Log[F])/(12*d*(c + d*x)^4) - (b^2*F^(a + b
*(c + d*x)^2)*Log[F]^2)/(12*d*(c + d*x)^2) + (b^3*F^a*ExpIntegralEi[b*(c + d*x)^2*Log[F]]*Log[F]^3)/(12*d)

Rule 2214

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]))

Rule 2210

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]

Rubi steps

\begin{align*} \int \frac{F^{a+b (c+d x)^2}}{(c+d x)^7} \, dx &=-\frac{F^{a+b (c+d x)^2}}{6 d (c+d x)^6}+\frac{1}{3} (b \log (F)) \int \frac{F^{a+b (c+d x)^2}}{(c+d x)^5} \, dx\\ &=-\frac{F^{a+b (c+d x)^2}}{6 d (c+d x)^6}-\frac{b F^{a+b (c+d x)^2} \log (F)}{12 d (c+d x)^4}+\frac{1}{6} \left (b^2 \log ^2(F)\right ) \int \frac{F^{a+b (c+d x)^2}}{(c+d x)^3} \, dx\\ &=-\frac{F^{a+b (c+d x)^2}}{6 d (c+d x)^6}-\frac{b F^{a+b (c+d x)^2} \log (F)}{12 d (c+d x)^4}-\frac{b^2 F^{a+b (c+d x)^2} \log ^2(F)}{12 d (c+d x)^2}+\frac{1}{6} \left (b^3 \log ^3(F)\right ) \int \frac{F^{a+b (c+d x)^2}}{c+d x} \, dx\\ &=-\frac{F^{a+b (c+d x)^2}}{6 d (c+d x)^6}-\frac{b F^{a+b (c+d x)^2} \log (F)}{12 d (c+d x)^4}-\frac{b^2 F^{a+b (c+d x)^2} \log ^2(F)}{12 d (c+d x)^2}+\frac{b^3 F^a \text{Ei}\left (b (c+d x)^2 \log (F)\right ) \log ^3(F)}{12 d}\\ \end{align*}

Mathematica [A]  time = 0.0942973, size = 79, normalized size = 0.65 \[ \frac{F^a \left (b^3 \log ^3(F) \text{Ei}\left (b (c+d x)^2 \log (F)\right )-\frac{F^{b (c+d x)^2} \left (b^2 \log ^2(F) (c+d x)^4+b \log (F) (c+d x)^2+2\right )}{(c+d x)^6}\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b*(c + d*x)^2)/(c + d*x)^7,x]

[Out]

(F^a*(b^3*ExpIntegralEi[b*(c + d*x)^2*Log[F]]*Log[F]^3 - (F^(b*(c + d*x)^2)*(2 + b*(c + d*x)^2*Log[F] + b^2*(c
+ d*x)^4*Log[F]^2))/(c + d*x)^6))/(12*d)

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Maple [A]  time = 0.064, size = 119, normalized size = 1. \begin{align*} -{\frac{{F}^{b \left ( dx+c \right ) ^{2}}{F}^{a}}{6\,d \left ( dx+c \right ) ^{6}}}-{\frac{b\ln \left ( F \right ){F}^{b \left ( dx+c \right ) ^{2}}{F}^{a}}{12\,d \left ( dx+c \right ) ^{4}}}-{\frac{{b}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}{F}^{b \left ( dx+c \right ) ^{2}}{F}^{a}}{12\,d \left ( dx+c \right ) ^{2}}}-{\frac{{b}^{3} \left ( \ln \left ( F \right ) \right ) ^{3}{F}^{a}{\it Ei} \left ( 1,-b \left ( dx+c \right ) ^{2}\ln \left ( F \right ) \right ) }{12\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+b*(d*x+c)^2)/(d*x+c)^7,x)

[Out]

-1/6/d/(d*x+c)^6*F^(b*(d*x+c)^2)*F^a-1/12/d*b*ln(F)/(d*x+c)^4*F^(b*(d*x+c)^2)*F^a-1/12/d*b^2*ln(F)^2/(d*x+c)^2
*F^(b*(d*x+c)^2)*F^a-1/12/d*b^3*ln(F)^3*F^a*Ei(1,-b*(d*x+c)^2*ln(F))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{7}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^7,x, algorithm="maxima")

[Out]

integrate(F^((d*x + c)^2*b + a)/(d*x + c)^7, x)

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Fricas [B]  time = 1.52452, size = 616, normalized size = 5.09 \begin{align*} \frac{{\left (b^{3} d^{6} x^{6} + 6 \, b^{3} c d^{5} x^{5} + 15 \, b^{3} c^{2} d^{4} x^{4} + 20 \, b^{3} c^{3} d^{3} x^{3} + 15 \, b^{3} c^{4} d^{2} x^{2} + 6 \, b^{3} c^{5} d x + b^{3} c^{6}\right )} F^{a}{\rm Ei}\left ({\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right )\right ) \log \left (F\right )^{3} -{\left ({\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} x^{2} + 4 \, b^{2} c^{3} d x + b^{2} c^{4}\right )} \log \left (F\right )^{2} +{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right ) + 2\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{12 \,{\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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^7,x, algorithm="fricas")

[Out]

1/12*((b^3*d^6*x^6 + 6*b^3*c*d^5*x^5 + 15*b^3*c^2*d^4*x^4 + 20*b^3*c^3*d^3*x^3 + 15*b^3*c^4*d^2*x^2 + 6*b^3*c^
5*d*x + b^3*c^6)*F^a*Ei((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*log(F))*log(F)^3 - ((b^2*d^4*x^4 + 4*b^2*c*d^3*x^3 + 6
*b^2*c^2*d^2*x^2 + 4*b^2*c^3*d*x + b^2*c^4)*log(F)^2 + (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*log(F) + 2)*F^(b*d^2*x^
2 + 2*b*c*d*x + b*c^2 + 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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(a+b*(d*x+c)**2)/(d*x+c)**7,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{7}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^7,x, algorithm="giac")

[Out]

integrate(F^((d*x + c)^2*b + a)/(d*x + c)^7, x)