\(\int (c+d x)^m \cos (a+b x) \sin (a+b x) \, dx\) [1]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 20, antiderivative size = 137 \[ \int (c+d x)^m \cos (a+b x) \sin (a+b x) \, dx=-\frac {2^{-3-m} e^{2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 i b (c+d x)}{d}\right )}{b}-\frac {2^{-3-m} e^{-2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i b (c+d x)}{d}\right )}{b} \]

[Out]

-2^(-3-m)*exp(2*I*(a-b*c/d))*(d*x+c)^m*GAMMA(1+m,-2*I*b*(d*x+c)/d)/b/((-I*b*(d*x+c)/d)^m)-2^(-3-m)*(d*x+c)^m*G
AMMA(1+m,2*I*b*(d*x+c)/d)/b/exp(2*I*(a-b*c/d))/((I*b*(d*x+c)/d)^m)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4491, 12, 3389, 2212} \[ \int (c+d x)^m \cos (a+b x) \sin (a+b x) \, dx=-\frac {2^{-m-3} e^{2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {2 i b (c+d x)}{d}\right )}{b}-\frac {2^{-m-3} e^{-2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 i b (c+d x)}{d}\right )}{b} \]

[In]

Int[(c + d*x)^m*Cos[a + b*x]*Sin[a + b*x],x]

[Out]

-((2^(-3 - m)*E^((2*I)*(a - (b*c)/d))*(c + d*x)^m*Gamma[1 + m, ((-2*I)*b*(c + d*x))/d])/(b*(((-I)*b*(c + d*x))
/d)^m)) - (2^(-3 - m)*(c + d*x)^m*Gamma[1 + m, ((2*I)*b*(c + d*x))/d])/(b*E^((2*I)*(a - (b*c)/d))*((I*b*(c + d
*x))/d)^m)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2212

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-F^(g*(e - c*(f/d))))*((c
+ d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d))^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m
 + 1, ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 3389

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/E^(I*(e + f*x))
, x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]

Rule 4491

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{2} (c+d x)^m \sin (2 a+2 b x) \, dx \\ & = \frac {1}{2} \int (c+d x)^m \sin (2 a+2 b x) \, dx \\ & = \frac {1}{4} i \int e^{-i (2 a+2 b x)} (c+d x)^m \, dx-\frac {1}{4} i \int e^{i (2 a+2 b x)} (c+d x)^m \, dx \\ & = -\frac {2^{-3-m} e^{2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 i b (c+d x)}{d}\right )}{b}-\frac {2^{-3-m} e^{-2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i b (c+d x)}{d}\right )}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.01 \[ \int (c+d x)^m \cos (a+b x) \sin (a+b x) \, dx=-\frac {2^{-3-m} e^{-\frac {2 i (b c+a d)}{d}} (c+d x)^m \left (\frac {b^2 (c+d x)^2}{d^2}\right )^{-m} \left (e^{4 i a} \left (\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,-\frac {2 i b (c+d x)}{d}\right )+e^{\frac {4 i b c}{d}} \left (-\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,\frac {2 i b (c+d x)}{d}\right )\right )}{b} \]

[In]

Integrate[(c + d*x)^m*Cos[a + b*x]*Sin[a + b*x],x]

[Out]

-((2^(-3 - m)*(c + d*x)^m*(E^((4*I)*a)*((I*b*(c + d*x))/d)^m*Gamma[1 + m, ((-2*I)*b*(c + d*x))/d] + E^(((4*I)*
b*c)/d)*(((-I)*b*(c + d*x))/d)^m*Gamma[1 + m, ((2*I)*b*(c + d*x))/d]))/(b*E^(((2*I)*(b*c + a*d))/d)*((b^2*(c +
 d*x)^2)/d^2)^m))

Maple [F]

\[\int \left (d x +c \right )^{m} \cos \left (x b +a \right ) \sin \left (x b +a \right )d x\]

[In]

int((d*x+c)^m*cos(b*x+a)*sin(b*x+a),x)

[Out]

int((d*x+c)^m*cos(b*x+a)*sin(b*x+a),x)

Fricas [A] (verification not implemented)

none

Time = 0.10 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.70 \[ \int (c+d x)^m \cos (a+b x) \sin (a+b x) \, dx=-\frac {e^{\left (-\frac {d m \log \left (-\frac {2 i \, b}{d}\right ) + 2 i \, b c - 2 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (i \, b d x + i \, b c\right )}}{d}\right ) + e^{\left (-\frac {d m \log \left (\frac {2 i \, b}{d}\right ) - 2 i \, b c + 2 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right )}{8 \, b} \]

[In]

integrate((d*x+c)^m*cos(b*x+a)*sin(b*x+a),x, algorithm="fricas")

[Out]

-1/8*(e^(-(d*m*log(-2*I*b/d) + 2*I*b*c - 2*I*a*d)/d)*gamma(m + 1, -2*(I*b*d*x + I*b*c)/d) + e^(-(d*m*log(2*I*b
/d) - 2*I*b*c + 2*I*a*d)/d)*gamma(m + 1, -2*(-I*b*d*x - I*b*c)/d))/b

Sympy [F]

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

[In]

integrate((d*x+c)**m*cos(b*x+a)*sin(b*x+a),x)

[Out]

Integral((c + d*x)**m*sin(a + b*x)*cos(a + b*x), x)

Maxima [F]

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

[In]

integrate((d*x+c)^m*cos(b*x+a)*sin(b*x+a),x, algorithm="maxima")

[Out]

integrate((d*x + c)^m*cos(b*x + a)*sin(b*x + a), x)

Giac [F]

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

[In]

integrate((d*x+c)^m*cos(b*x+a)*sin(b*x+a),x, algorithm="giac")

[Out]

integrate((d*x + c)^m*cos(b*x + a)*sin(b*x + a), x)

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^m \cos (a+b x) \sin (a+b x) \, dx=\int \cos \left (a+b\,x\right )\,\sin \left (a+b\,x\right )\,{\left (c+d\,x\right )}^m \,d x \]

[In]

int(cos(a + b*x)*sin(a + b*x)*(c + d*x)^m,x)

[Out]

int(cos(a + b*x)*sin(a + b*x)*(c + d*x)^m, x)