\(\int \frac {F^{c (a+b x)} (e x \cos (d+e x)+(-1+b c x \log (F)) \sin (d+e x))}{x^2} \, dx\) [36]

Optimal result

Integrand size = 38, antiderivative size = 20 \[ \int \frac {F^{c (a+b x)} (e x \cos (d+e x)+(-1+b c x \log (F)) \sin (d+e x))}{x^2} \, dx=\frac {F^{a c+b c x} \sin (d+e x)}{x} \]

[Out]

F^(b*c*x+a*c)*sin(e*x+d)/x

Rubi [A] (verified)

Time = 2.17 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {6873, 6874, 4555} \[ \int \frac {F^{c (a+b x)} (e x \cos (d+e x)+(-1+b c x \log (F)) \sin (d+e x))}{x^2} \, dx=\frac {\sin (d+e x) F^{a c+b c x}}{x} \]

[In]

Int[(F^(c*(a + b*x))*(e*x*Cos[d + e*x] + (-1 + b*c*x*Log[F])*Sin[d + e*x]))/x^2,x]

[Out]

(F^(a*c + b*c*x)*Sin[d + e*x])/x

Rule 4555

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_.)*(x_))^(m_)*Sin[(d_.) + (e_.)*(x_)], x_Symbol] :> Simp[((f*x)^(m +
1)/(f*(m + 1)))*F^(c*(a + b*x))*Sin[d + e*x], 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])

Rule 6873

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {F^{a c+b c x} (e x \cos (d+e x)+(-1+b c x \log (F)) \sin (d+e x))}{x^2} \, dx \\ & = \int \left (\frac {e F^{a c+b c x} \cos (d+e x)}{x}+\frac {F^{a c+b c x} (-1+b c x \log (F)) \sin (d+e x)}{x^2}\right ) \, dx \\ & = e \int \frac {F^{a c+b c x} \cos (d+e x)}{x} \, dx+\int \frac {F^{a c+b c x} (-1+b c x \log (F)) \sin (d+e x)}{x^2} \, dx \\ & = e \int \frac {F^{a c+b c x} \cos (d+e x)}{x} \, dx+\int \left (-\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}\right ) \, dx \\ & = e \int \frac {F^{a c+b c x} \cos (d+e x)}{x} \, dx+(b c \log (F)) \int \frac {F^{a c+b c x} \sin (d+e x)}{x} \, dx-\int \frac {F^{a c+b c x} \sin (d+e x)}{x^2} \, dx \\ & = \frac {F^{a c+b c x} \sin (d+e x)}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {F^{c (a+b x)} (e x \cos (d+e x)+(-1+b c x \log (F)) \sin (d+e x))}{x^2} \, dx=\frac {F^{a c+b c x} \sin (d+e x)}{x} \]

[In]

Integrate[(F^(c*(a + b*x))*(e*x*Cos[d + e*x] + (-1 + b*c*x*Log[F])*Sin[d + e*x]))/x^2,x]

[Out]

(F^(a*c + b*c*x)*Sin[d + e*x])/x

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {F^{c \left (x b +a \right )} \sin \left (e x +d \right )}{x}\)	\(20\)
parallelrisch	\(\frac {F^{c \left (x b +a \right )} \sin \left (e x +d \right )}{x}\)	\(20\)
norman	\(\frac {2 \,{\mathrm e}^{c \left (x b +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{\left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}\right ) x}\)	\(40\)

[In]

int(F^(c*(b*x+a))*(e*x*cos(e*x+d)+(-1+b*c*x*ln(F))*sin(e*x+d))/x^2,x,method=_RETURNVERBOSE)

[Out]

1/x*F^(c*(b*x+a))*sin(e*x+d)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {F^{c (a+b x)} (e x \cos (d+e x)+(-1+b c x \log (F)) \sin (d+e x))}{x^2} \, dx=\frac {F^{b c x + a c} \sin \left (e x + d\right )}{x} \]

[In]

integrate(F^(c*(b*x+a))*(e*x*cos(e*x+d)+(-1+b*c*x*log(F))*sin(e*x+d))/x^2,x, algorithm="fricas")

[Out]

F^(b*c*x + a*c)*sin(e*x + d)/x

Sympy [F]

\[ \int \frac {F^{c (a+b x)} (e x \cos (d+e x)+(-1+b c x \log (F)) \sin (d+e x))}{x^2} \, dx=\int \frac {F^{c \left (a + b x\right )} \left (b c x \log {\left (F \right )} \sin {\left (d + e x \right )} + e x \cos {\left (d + e x \right )} - \sin {\left (d + e x \right )}\right )}{x^{2}}\, dx \]

[In]

integrate(F**(c*(b*x+a))*(e*x*cos(e*x+d)+(-1+b*c*x*ln(F))*sin(e*x+d))/x**2,x)

[Out]

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

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.76 (sec) , antiderivative size = 564, normalized size of antiderivative = 28.20 \[ \int \frac {F^{c (a+b x)} (e x \cos (d+e x)+(-1+b c x \log (F)) \sin (d+e x))}{x^2} \, dx=\text {Too large to display} \]

[In]

integrate(F^(c*(b*x+a))*(e*x*cos(e*x+d)+(-1+b*c*x*log(F))*sin(e*x+d))/x^2,x, algorithm="maxima")

[Out]

-1/4*F^(a*c)*b*c*(I*conjugate(gamma(-1, -(b*c*log(F) + I*e)*x)) - I*conjugate(gamma(-1, -(b*c*log(F) - I*e)*x)
) - I*gamma(-1, -(b*c*log(F) + I*e)*x) + I*gamma(-1, -(b*c*log(F) - I*e)*x))*cos(d)*log(F) - 1/4*F^(a*c)*b*c*(
conjugate(gamma(-1, -(b*c*log(F) + I*e)*x)) + conjugate(gamma(-1, -(b*c*log(F) - I*e)*x)) + gamma(-1, -(b*c*lo
g(F) + I*e)*x) + gamma(-1, -(b*c*log(F) - I*e)*x))*log(F)*sin(d) - 1/4*(F^(a*c)*(I*Ei((b*c*log(F) + I*e)*x) -
I*Ei((b*c*log(F) - I*e)*x) - I*conjugate(Ei((b*c*log(F) + I*e)*x)) + I*conjugate(Ei((b*c*log(F) - I*e)*x)))*co
s(d) - F^(a*c)*(Ei((b*c*log(F) + I*e)*x) + Ei((b*c*log(F) - I*e)*x) + conjugate(Ei((b*c*log(F) + I*e)*x)) + co
njugate(Ei((b*c*log(F) - I*e)*x)))*sin(d))*b*c*log(F) + 1/4*(F^(a*c)*(Ei((b*c*log(F) + I*e)*x) + Ei((b*c*log(F
) - I*e)*x) + conjugate(Ei((b*c*log(F) + I*e)*x)) + conjugate(Ei((b*c*log(F) - I*e)*x)))*cos(d) - F^(a*c)*(-I*
Ei((b*c*log(F) + I*e)*x) + I*Ei((b*c*log(F) - I*e)*x) + I*conjugate(Ei((b*c*log(F) + I*e)*x)) - I*conjugate(Ei
((b*c*log(F) - I*e)*x)))*sin(d))*e - 1/4*(F^(a*c)*(conjugate(gamma(-1, -(b*c*log(F) + I*e)*x)) + conjugate(gam
ma(-1, -(b*c*log(F) - I*e)*x)) + gamma(-1, -(b*c*log(F) + I*e)*x) + gamma(-1, -(b*c*log(F) - I*e)*x))*cos(d) +
F^(a*c)*(-I*conjugate(gamma(-1, -(b*c*log(F) + I*e)*x)) + I*conjugate(gamma(-1, -(b*c*log(F) - I*e)*x)) + I*g
amma(-1, -(b*c*log(F) + I*e)*x) - I*gamma(-1, -(b*c*log(F) - I*e)*x))*sin(d))*e

Giac [F]

\[ \int \frac {F^{c (a+b x)} (e x \cos (d+e x)+(-1+b c x \log (F)) \sin (d+e x))}{x^2} \, dx=\int { \frac {{\left (e x \cos \left (e x + d\right ) + {\left (b c x \log \left (F\right ) - 1\right )} \sin \left (e x + d\right )\right )} F^{{\left (b x + a\right )} c}}{x^{2}} \,d x } \]

[In]

integrate(F^(c*(b*x+a))*(e*x*cos(e*x+d)+(-1+b*c*x*log(F))*sin(e*x+d))/x^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 27.72 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {F^{c (a+b x)} (e x \cos (d+e x)+(-1+b c x \log (F)) \sin (d+e x))}{x^2} \, dx=\frac {F^{c\,\left (a+b\,x\right )}\,\sin \left (d+e\,x\right )}{x} \]

[In]

int((F^(c*(a + b*x))*(sin(d + e*x)*(b*c*x*log(F) - 1) + e*x*cos(d + e*x)))/x^2,x)

[Out]

(F^(c*(a + b*x))*sin(d + e*x))/x