\(\int \frac {F^{c (a+b x)} (f x)^m (e x \cos (d+e x)+(m+b c x \log (F)) \sin (d+e x))}{x} \, dx\) [33]

Optimal result

Integrand size = 43, antiderivative size = 22 \[ \int \frac {F^{c (a+b x)} (f x)^m (e x \cos (d+e x)+(m+b c x \log (F)) \sin (d+e x))}{x} \, dx=F^{a c+b c x} (f x)^m \sin (d+e x) \]

[Out]

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

Rubi [A] (verified)

Time = 2.99 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {16, 6873, 6874, 4555} \[ \int \frac {F^{c (a+b x)} (f x)^m (e x \cos (d+e x)+(m+b c x \log (F)) \sin (d+e x))}{x} \, dx=(f x)^m \sin (d+e x) F^{a c+b c x} \]

[In]

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

[Out]

F^(a*c + b*c*x)*(f*x)^m*Sin[d + e*x]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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}& = f \int F^{c (a+b x)} (f x)^{-1+m} (e x \cos (d+e x)+(m+b c x \log (F)) \sin (d+e x)) \, dx \\ & = f \int F^{a c+b c x} (f x)^{-1+m} (e x \cos (d+e x)+(m+b c x \log (F)) \sin (d+e x)) \, dx \\ & = f \int \left (\frac {e F^{a c+b c x} (f x)^m \cos (d+e x)}{f}+F^{a c+b c x} (f x)^{-1+m} (m+b c x \log (F)) \sin (d+e x)\right ) \, dx \\ & = e \int F^{a c+b c x} (f x)^m \cos (d+e x) \, dx+f \int F^{a c+b c x} (f x)^{-1+m} (m+b c x \log (F)) \sin (d+e x) \, dx \\ & = e \int F^{a c+b c x} (f x)^m \cos (d+e x) \, dx+f \int \left (F^{a c+b c x} m (f x)^{-1+m} \sin (d+e x)+\frac {b c F^{a c+b c x} (f x)^m \log (F) \sin (d+e x)}{f}\right ) \, dx \\ & = e \int F^{a c+b c x} (f x)^m \cos (d+e x) \, dx+(f m) \int F^{a c+b c x} (f x)^{-1+m} \sin (d+e x) \, dx+(b c \log (F)) \int F^{a c+b c x} (f x)^m \sin (d+e x) \, dx \\ & = F^{a c+b c x} (f x)^m \sin (d+e x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {F^{c (a+b x)} (f x)^m (e x \cos (d+e x)+(m+b c x \log (F)) \sin (d+e x))}{x} \, dx=F^{a c+b c x} (f x)^m \sin (d+e x) \]

[In]

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

[Out]

F^(a*c + b*c*x)*(f*x)^m*Sin[d + e*x]

Maple [A] (verified)

Time = 5.87 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(F^{c \left (x b +a \right )} \left (f x \right )^{m} \sin \left (e x +d \right )\)	\(22\)
risch	\(-\frac {i x^{m} f^{m} F^{c \left (x b +a \right )} \left ({\mathrm e}^{i e x} {\mathrm e}^{i d} {\mathrm e}^{-\frac {i \pi \operatorname {csgn}\left (i f x \right )^{3} m}{2}} {\mathrm e}^{\frac {i \pi \operatorname {csgn}\left (i f x \right )^{2} \operatorname {csgn}\left (i f \right ) m}{2}} {\mathrm e}^{\frac {i \pi \operatorname {csgn}\left (i f x \right )^{2} \operatorname {csgn}\left (i x \right ) m}{2}} {\mathrm e}^{-\frac {i \pi \,\operatorname {csgn}\left (i f x \right ) \operatorname {csgn}\left (i f \right ) \operatorname {csgn}\left (i x \right ) m}{2}}-{\mathrm e}^{-i e x} {\mathrm e}^{-i d} {\mathrm e}^{-\frac {i \pi \operatorname {csgn}\left (i f x \right )^{3} m}{2}} {\mathrm e}^{\frac {i \pi \operatorname {csgn}\left (i f x \right )^{2} \operatorname {csgn}\left (i f \right ) m}{2}} {\mathrm e}^{\frac {i \pi \operatorname {csgn}\left (i f x \right )^{2} \operatorname {csgn}\left (i x \right ) m}{2}} {\mathrm e}^{-\frac {i \pi \,\operatorname {csgn}\left (i f x \right ) \operatorname {csgn}\left (i f \right ) \operatorname {csgn}\left (i x \right ) m}{2}}\right )}{2}\)	\(193\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {F^{c (a+b x)} (f x)^m (e x \cos (d+e x)+(m+b c x \log (F)) \sin (d+e x))}{x} \, dx=\left (f x\right )^{m} F^{b c x + a c} \sin \left (e x + d\right ) \]

[In]

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

[Out]

(f*x)^m*F^(b*c*x + a*c)*sin(e*x + d)

Sympy [F]

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

[In]

integrate(F**(c*(b*x+a))*(f*x)**m*(e*x*cos(e*x+d)+(m+b*c*x*ln(F))*sin(e*x+d))/x,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {F^{c (a+b x)} (f x)^m (e x \cos (d+e x)+(m+b c x \log (F)) \sin (d+e x))}{x} \, dx=F^{a c} f^{m} e^{\left (b c x \log \left (F\right ) + m \log \left (x\right )\right )} \sin \left (e x + d\right ) \]

[In]

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

[Out]

F^(a*c)*f^m*e^(b*c*x*log(F) + m*log(x))*sin(e*x + d)

Giac [F]

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

[In]

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

[Out]

integrate((e*x*cos(e*x + d) + (b*c*x*log(F) + m)*sin(e*x + d))*(f*x)^m*F^((b*x + a)*c)/x, x)

Mupad [B] (verification not implemented)

Time = 27.98 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {F^{c (a+b x)} (f x)^m (e x \cos (d+e x)+(m+b c x \log (F)) \sin (d+e x))}{x} \, dx=F^{c\,\left (a+b\,x\right )}\,\sin \left (d+e\,x\right )\,{\left (f\,x\right )}^m \]

[In]

int((F^(c*(a + b*x))*(f*x)^m*(sin(d + e*x)*(m + b*c*x*log(F)) + e*x*cos(d + e*x)))/x,x)

[Out]

F^(c*(a + b*x))*sin(d + e*x)*(f*x)^m