3.37 \(\int \frac{F^{c (a+b x)} (e x \cos (d+e x)+(-2+b c x \log (F)) \sin (d+e x))}{x^3} \, dx\)

Optimal. Leaf size=20 \[ \frac{\sin (d+e x) F^{a c+b c x}}{x^2} \]

[Out]

(F^(a*c + b*c*x)*Sin[d + e*x])/x^2

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Rubi [A]  time = 1.94979, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {6741, 6742, 4468, 4467} \[ \frac{\sin (d+e x) F^{a c+b c x}}{x^2} \]

Antiderivative was successfully verified.

[In]

Int[(F^(c*(a + b*x))*(e*x*Cos[d + e*x] + (-2 + b*c*x*Log[F])*Sin[d + e*x]))/x^3,x]

[Out]

(F^(a*c + b*c*x)*Sin[d + e*x])/x^2

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 4468

Int[Cos[(d_.) + (e_.)*(x_)]*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_.)*(x_))^(m_), x_Symbol] :> Simp[((f*x)^(m +
1)*F^(c*(a + b*x))*Cos[d + e*x])/(f*(m + 1)), x] + (Dist[e/(f*(m + 1)), Int[(f*x)^(m + 1)*F^(c*(a + b*x))*Sin
[d + e*x], x], x] - Dist[(b*c*Log[F])/(f*(m + 1)), Int[(f*x)^(m + 1)*F^(c*(a + b*x))*Cos[d + e*x], x], x]) /;
FreeQ[{F, a, b, c, d, e, f, m}, x] && (LtQ[m, -1] || SumSimplerQ[m, 1])

Rule 4467

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_.)*(x_))^(m_)*Sin[(d_.) + (e_.)*(x_)], x_Symbol] :> Simp[((f*x)^(m +
1)*F^(c*(a + b*x))*Sin[d + e*x])/(f*(m + 1)), x] + (-Dist[e/(f*(m + 1)), Int[(f*x)^(m + 1)*F^(c*(a + b*x))*Co
s[d + e*x], x], x] - Dist[(b*c*Log[F])/(f*(m + 1)), Int[(f*x)^(m + 1)*F^(c*(a + b*x))*Sin[d + e*x], x], x]) /;
FreeQ[{F, a, b, c, d, e, f, m}, x] && (LtQ[m, -1] || SumSimplerQ[m, 1])

Rubi steps

\begin{align*} \int \frac{F^{c (a+b x)} (e x \cos (d+e x)+(-2+b c x \log (F)) \sin (d+e x))}{x^3} \, dx &=\int \frac{F^{a c+b c x} (e x \cos (d+e x)+(-2+b c x \log (F)) \sin (d+e x))}{x^3} \, dx\\ &=\int \left (\frac{e F^{a c+b c x} \cos (d+e x)}{x^2}+\frac{F^{a c+b c x} (-2+b c x \log (F)) \sin (d+e x)}{x^3}\right ) \, dx\\ &=e \int \frac{F^{a c+b c x} \cos (d+e x)}{x^2} \, dx+\int \frac{F^{a c+b c x} (-2+b c x \log (F)) \sin (d+e x)}{x^3} \, dx\\ &=-\frac{e F^{a c+b c x} \cos (d+e x)}{x}-e^2 \int \frac{F^{a c+b c x} \sin (d+e x)}{x} \, dx+(b c e \log (F)) \int \frac{F^{a c+b c x} \cos (d+e x)}{x} \, dx+\int \left (-\frac{2 F^{a c+b c x} \sin (d+e x)}{x^3}+\frac{b c F^{a c+b c x} \log (F) \sin (d+e x)}{x^2}\right ) \, dx\\ &=-\frac{e F^{a c+b c x} \cos (d+e x)}{x}-2 \int \frac{F^{a c+b c x} \sin (d+e x)}{x^3} \, dx-e^2 \int \frac{F^{a c+b c x} \sin (d+e x)}{x} \, dx+(b c \log (F)) \int \frac{F^{a c+b c x} \sin (d+e x)}{x^2} \, dx+(b c e \log (F)) \int \frac{F^{a c+b c x} \cos (d+e x)}{x} \, dx\\ &=-\frac{e F^{a c+b c x} \cos (d+e x)}{x}+\frac{F^{a c+b c x} \sin (d+e x)}{x^2}-\frac{b c F^{a c+b c x} \log (F) \sin (d+e x)}{x}-e \int \frac{F^{a c+b c x} \cos (d+e x)}{x^2} \, dx-e^2 \int \frac{F^{a c+b c x} \sin (d+e x)}{x} \, dx-(b c \log (F)) \int \frac{F^{a c+b c x} \sin (d+e x)}{x^2} \, dx+2 \left ((b c e \log (F)) \int \frac{F^{a c+b c x} \cos (d+e x)}{x} \, dx\right )+\left (b^2 c^2 \log ^2(F)\right ) \int \frac{F^{a c+b c x} \sin (d+e x)}{x} \, dx\\ &=\frac{F^{a c+b c x} \sin (d+e x)}{x^2}\\ \end{align*}

Mathematica [A]  time = 0.626507, size = 19, normalized size = 0.95 \[ \frac{\sin (d+e x) F^{c (a+b x)}}{x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(F^(c*(a + b*x))*(e*x*Cos[d + e*x] + (-2 + b*c*x*Log[F])*Sin[d + e*x]))/x^3,x]

[Out]

(F^(c*(a + b*x))*Sin[d + e*x])/x^2

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Maple [A]  time = 0.093, size = 40, normalized size = 2. \begin{align*} 2\,{\frac{{{\rm e}^{c \left ( bx+a \right ) \ln \left ( F \right ) }}\tan \left ( d/2+1/2\,ex \right ) }{ \left ( 1+ \left ( \tan \left ( d/2+1/2\,ex \right ) \right ) ^{2} \right ){x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(b*x+a))*(e*x*cos(e*x+d)+(-2+b*c*x*ln(F))*sin(e*x+d))/x^3,x)

[Out]

2*exp(c*(b*x+a)*ln(F))*tan(1/2*d+1/2*e*x)/(1+tan(1/2*d+1/2*e*x)^2)/x^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: IndexError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*(e*x*cos(e*x+d)+(-2+b*c*x*log(F))*sin(e*x+d))/x^3,x, algorithm="maxima")

[Out]

Exception raised: IndexError

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Fricas [A]  time = 0.471423, size = 46, normalized size = 2.3 \begin{align*} \frac{F^{b c x + a c} \sin \left (e x + d\right )}{x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

F^(b*c*x + a*c)*sin(e*x + d)/x^2

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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**(c*(b*x+a))*(e*x*cos(e*x+d)+(-2+b*c*x*ln(F))*sin(e*x+d))/x**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x \cos \left (e x + d\right ) +{\left (b c x \log \left (F\right ) - 2\right )} \sin \left (e x + d\right )\right )} F^{{\left (b x + a\right )} c}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*(e*x*cos(e*x+d)+(-2+b*c*x*log(F))*sin(e*x+d))/x^3,x, algorithm="giac")

[Out]

integrate((e*x*cos(e*x + d) + (b*c*x*log(F) - 2)*sin(e*x + d))*F^((b*x + a)*c)/x^3, x)