3.3.62 \(\int \frac {F^{a+b (c+d x)^2}}{(c+d x)^3} \, dx\) [262]

Optimal. Leaf size=53 \[ -\frac {F^{a+b (c+d x)^2}}{2 d (c+d x)^2}+\frac {b F^a \text {Ei}\left (b (c+d x)^2 \log (F)\right ) \log (F)}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 2, 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 F^a \log (F) \text {Ei}\left (b (c+d x)^2 \log (F)\right )}{2 d}-\frac {F^{a+b (c+d x)^2}}{2 d (c+d x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*F^(a + b*(c + d*x)^2)/(d*(c + d*x)^2) + (b*F^a*ExpIntegralEi[b*(c + d*x)^2*Log[F]]*Log[F])/(2*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)^2}}{(c+d x)^3} \, dx &=-\frac {F^{a+b (c+d x)^2}}{2 d (c+d x)^2}+(b \log (F)) \int \frac {F^{a+b (c+d x)^2}}{c+d x} \, dx\\ &=-\frac {F^{a+b (c+d x)^2}}{2 d (c+d x)^2}+\frac {b F^a \text {Ei}\left (b (c+d x)^2 \log (F)\right ) \log (F)}{2 d}\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 47, normalized size = 0.89 \begin {gather*} \frac {F^a \left (-\frac {F^{b (c+d x)^2}}{(c+d x)^2}+b \text {Ei}\left (b (c+d x)^2 \log (F)\right ) \log (F)\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.07, size = 53, normalized size = 1.00

method	result	size
risch	\(-\frac {F^{b \left (d x +c \right )^{2}} F^{a}}{2 d \left (d x +c \right )^{2}}-\frac {b \ln \left (F \right ) F^{a} \expIntegral \left (1, -b \left (d x +c \right )^{2} \ln \left (F \right )\right )}{2 d}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^2)/(d*x+c)^3,x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (49) = 98\).
time = 0.39, size = 100, normalized size = 1.89 \begin {gather*} \frac {{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\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 ) - F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{2 \, {\left (d^{3} x^{2} + 2 \, c d^{2} x + c^{2} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {F^{a + b \left (c + d x\right )^{2}}}{\left (c + d x\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [B]
time = 4.68, size = 51, normalized size = 0.96 \begin {gather*} -\frac {F^a\,\left (F^{b\,{\left (c+d\,x\right )}^2}+b\,\ln \left (F\right )\,\mathrm {expint}\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2\right )\,{\left (c+d\,x\right )}^2\right )}{2\,d\,{\left (c+d\,x\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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