3.4.5 \(\int F^{a+\frac {b}{c+d x}} (c+d x) \, dx\) [305]

Optimal. Leaf size=85 \[ \frac {F^{a+\frac {b}{c+d x}} (c+d x)^2}{2 d}+\frac {b F^{a+\frac {b}{c+d x}} (c+d x) \log (F)}{2 d}-\frac {b^2 F^a \text {Ei}\left (\frac {b \log (F)}{c+d x}\right ) \log ^2(F)}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2245, 2237, 2241} \begin {gather*} -\frac {b^2 F^a \log ^2(F) \text {Ei}\left (\frac {b \log (F)}{c+d x}\right )}{2 d}+\frac {(c+d x)^2 F^{a+\frac {b}{c+d x}}}{2 d}+\frac {b \log (F) (c+d x) F^{a+\frac {b}{c+d x}}}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[F^(a + b/(c + d*x))*(c + d*x),x]

[Out]

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

Rule 2237

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

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

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Mathematica [A]
time = 0.04, size = 58, normalized size = 0.68 \begin {gather*} \frac {F^a \left (-b^2 \text {Ei}\left (\frac {b \log (F)}{c+d x}\right ) \log ^2(F)+F^{\frac {b}{c+d x}} (c+d x) (c+d x+b \log (F))\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.07, size = 133, normalized size = 1.56

method	result	size
risch	\(\frac {d \,F^{a} F^{\frac {b}{d x +c}} x^{2}}{2}+F^{a} F^{\frac {b}{d x +c}} c x +\frac {F^{a} F^{\frac {b}{d x +c}} c^{2}}{2 d}+\frac {b \ln \left (F \right ) F^{a} F^{\frac {b}{d x +c}} x}{2}+\frac {b \ln \left (F \right ) F^{a} F^{\frac {b}{d x +c}} c}{2 d}+\frac {b^{2} \ln \left (F \right )^{2} F^{a} \expIntegral \left (1, -\frac {b \ln \left (F \right )}{d x +c}\right )}{2 d}\)	\(133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))*(d*x+c),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.38, size = 77, normalized size = 0.91 \begin {gather*} -\frac {F^{a} b^{2} {\rm Ei}\left (\frac {b \log \left (F\right )}{d x + c}\right ) \log \left (F\right )^{2} - {\left (d^{2} x^{2} + 2 \, c d x + c^{2} + {\left (b d x + b c\right )} \log \left (F\right )\right )} F^{\frac {a d x + a c + b}{d x + c}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [B]
time = 6.11, size = 82, normalized size = 0.96 \begin {gather*} \frac {F^a\,F^{\frac {b}{c+d\,x}}\,{\left (c+d\,x\right )}^2}{2\,d}+\frac {F^a\,b^2\,{\ln \left (F\right )}^2\,\mathrm {expint}\left (-\frac {b\,\ln \left (F\right )}{c+d\,x}\right )}{2\,d}+\frac {F^a\,F^{\frac {b}{c+d\,x}}\,b\,\ln \left (F\right )\,\left (c+d\,x\right )}{2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a + b/(c + d*x))*(c + d*x),x)

[Out]

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

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