3.263 \(\int \frac {F^{a+b (c+d x)^2}}{(c+d x)^5} \, dx\)

Optimal. Leaf size=87 \[ \frac {b^2 F^a \log ^2(F) \text {Ei}\left (b (c+d x)^2 \log (F)\right )}{4 d}-\frac {F^{a+b (c+d x)^2}}{4 d (c+d x)^4}-\frac {b \log (F) F^{a+b (c+d x)^2}}{4 d (c+d x)^2} \]

[Out]

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

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Rubi [A]  time = 0.20, 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 = {2214, 2210} \[ \frac {b^2 F^a \log ^2(F) \text {Ei}\left (b (c+d x)^2 \log (F)\right )}{4 d}-\frac {F^{a+b (c+d x)^2}}{4 d (c+d x)^4}-\frac {b \log (F) F^{a+b (c+d x)^2}}{4 d (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-F^(a + b*(c + d*x)^2)/(4*d*(c + d*x)^4) - (b*F^(a + b*(c + d*x)^2)*Log[F])/(4*d*(c + d*x)^2) + (b^2*F^a*ExpIn
tegralEi[b*(c + d*x)^2*Log[F]]*Log[F]^2)/(4*d)

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]

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]))

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 64, normalized size = 0.74 \[ \frac {F^a \left (b^2 \log ^2(F) \text {Ei}\left (b (c+d x)^2 \log (F)\right )-\frac {F^{b (c+d x)^2} \left (b \log (F) (c+d x)^2+1\right )}{(c+d x)^4}\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b*(c + d*x)^2)/(c + d*x)^5,x]

[Out]

(F^a*(b^2*ExpIntegralEi[b*(c + d*x)^2*Log[F]]*Log[F]^2 - (F^(b*(c + d*x)^2)*(1 + b*(c + d*x)^2*Log[F]))/(c + d
*x)^4))/(4*d)

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fricas [B]  time = 0.42, size = 183, normalized size = 2.10 \[ \frac {{\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} x^{2} + 4 \, b^{2} c^{3} d x + b^{2} c^{4}\right )} F^{a} {\rm Ei}\left ({\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \relax (F)\right ) \log \relax (F)^{2} - {\left ({\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \relax (F) + 1\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{4 \, {\left (d^{5} x^{4} + 4 \, c d^{4} x^{3} + 6 \, c^{2} d^{3} x^{2} + 4 \, c^{3} d^{2} x + c^{4} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^5,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 86, normalized size = 0.99 \[ -\frac {b^{2} F^{a} \Ei \left (1, -\left (d x +c \right )^{2} b \ln \relax (F )\right ) \ln \relax (F )^{2}}{4 d}-\frac {b \,F^{a} F^{\left (d x +c \right )^{2} b} \ln \relax (F )}{4 \left (d x +c \right )^{2} d}-\frac {F^{a} F^{\left (d x +c \right )^{2} b}}{4 \left (d x +c \right )^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^5,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.76, size = 76, normalized size = 0.87 \[ -\frac {F^a\,b^2\,{\ln \relax (F)}^2\,\left (\frac {\mathrm {expint}\left (-b\,\ln \relax (F)\,{\left (c+d\,x\right )}^2\right )}{2}+F^{b\,{\left (c+d\,x\right )}^2}\,\left (\frac {1}{2\,b\,\ln \relax (F)\,{\left (c+d\,x\right )}^2}+\frac {1}{2\,b^2\,{\ln \relax (F)}^2\,{\left (c+d\,x\right )}^4}\right )\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(F^a*b^2*log(F)^2*(expint(-b*log(F)*(c + d*x)^2)/2 + F^(b*(c + d*x)^2)*(1/(2*b*log(F)*(c + d*x)^2) + 1/(2*b^2
*log(F)^2*(c + d*x)^4))))/(2*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(a+b*(d*x+c)**2)/(d*x+c)**5,x)

[Out]

Timed out

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