\(\int (c+d x)^m \cos (a+b x) \sin ^3(a+b x) \, dx\) [22]

Optimal result

Integrand size = 22, antiderivative size = 271 \[ \int (c+d x)^m \cos (a+b x) \sin ^3(a+b x) \, dx=-\frac {2^{-4-m} e^{2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 i b (c+d x)}{d}\right )}{b}-\frac {2^{-4-m} e^{-2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i b (c+d x)}{d}\right )}{b}+\frac {2^{-2 (3+m)} e^{4 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {4 i b (c+d x)}{d}\right )}{b}+\frac {2^{-2 (3+m)} e^{-4 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {4 i b (c+d x)}{d}\right )}{b} \] Output:

-2^(-4-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^(-4-m)*(d*x+c)^m*GAMMA(1+m,2*I*b*(d*x+c)/d)/b/exp(2*I*(a 
-b*c/d))/((I*b*(d*x+c)/d)^m)+exp(4*I*(a-b*c/d))*(d*x+c)^m*GAMMA(1+m,-4*I*b 
*(d*x+c)/d)/(2^(6+2*m))/b/((-I*b*(d*x+c)/d)^m)+(d*x+c)^m*GAMMA(1+m,4*I*b*( 
d*x+c)/d)/(2^(6+2*m))/b/exp(4*I*(a-b*c/d))/((I*b*(d*x+c)/d)^m)
 

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.91 \[ \int (c+d x)^m \cos (a+b x) \sin ^3(a+b x) \, dx=\frac {4^{-3-m} e^{-\frac {4 i (b c+a d)}{d}} (c+d x)^m \left (\frac {b^2 (c+d x)^2}{d^2}\right )^{-m} \left (-2^{2+m} e^{2 i \left (3 a+\frac {b c}{d}\right )} \left (\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,-\frac {2 i b (c+d x)}{d}\right )-2^{2+m} e^{2 i \left (a+\frac {3 b c}{d}\right )} \left (-\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,\frac {2 i b (c+d x)}{d}\right )+e^{8 i a} \left (\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,-\frac {4 i b (c+d x)}{d}\right )+e^{\frac {8 i b c}{d}} \left (-\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,\frac {4 i b (c+d x)}{d}\right )\right )}{b} \] Input:

Integrate[(c + d*x)^m*Cos[a + b*x]*Sin[a + b*x]^3,x]
 

Output:

(4^(-3 - m)*(c + d*x)^m*(-(2^(2 + m)*E^((2*I)*(3*a + (b*c)/d))*((I*b*(c + 
d*x))/d)^m*Gamma[1 + m, ((-2*I)*b*(c + d*x))/d]) - 2^(2 + m)*E^((2*I)*(a + 
 (3*b*c)/d))*(((-I)*b*(c + d*x))/d)^m*Gamma[1 + m, ((2*I)*b*(c + d*x))/d] 
+ E^((8*I)*a)*((I*b*(c + d*x))/d)^m*Gamma[1 + m, ((-4*I)*b*(c + d*x))/d] + 
 E^(((8*I)*b*c)/d)*(((-I)*b*(c + d*x))/d)^m*Gamma[1 + m, ((4*I)*b*(c + d*x 
))/d]))/(b*E^(((4*I)*(b*c + a*d))/d)*((b^2*(c + d*x)^2)/d^2)^m)
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4906, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sin ^3(a+b x) \cos (a+b x) (c+d x)^m \, dx\)

\(\Big \downarrow \) 4906

\(\displaystyle \int \left (\frac {1}{4} \sin (2 a+2 b x) (c+d x)^m-\frac {1}{8} \sin (4 a+4 b x) (c+d x)^m\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2^{-m-4} e^{2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {2 i b (c+d x)}{d}\right )}{b}+\frac {2^{-2 (m+3)} e^{4 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {4 i b (c+d x)}{d}\right )}{b}-\frac {2^{-m-4} e^{-2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 i b (c+d x)}{d}\right )}{b}+\frac {2^{-2 (m+3)} e^{-4 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {4 i b (c+d x)}{d}\right )}{b}\)

Input:

Int[(c + d*x)^m*Cos[a + b*x]*Sin[a + b*x]^3,x]
 

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4906

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b 
_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(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] && IG 
tQ[p, 0]
 

Maple [F]

\[\int \left (d x +c \right )^{m} \cos \left (b x +a \right ) \sin \left (b x +a \right )^{3}d x\]

Input:

int((d*x+c)^m*cos(b*x+a)*sin(b*x+a)^3,x)
 

Output:

int((d*x+c)^m*cos(b*x+a)*sin(b*x+a)^3,x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.70 \[ \int (c+d x)^m \cos (a+b x) \sin ^3(a+b x) \, dx=-\frac {4 \, e^{\left (-\frac {d m \log \left (-\frac {2 i \, b}{d}\right ) + 2 i \, b c - 2 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (i \, b d x + i \, b c\right )}}{d}\right ) - e^{\left (-\frac {d m \log \left (-\frac {4 i \, b}{d}\right ) + 4 i \, b c - 4 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {4 \, {\left (i \, b d x + i \, b c\right )}}{d}\right ) + 4 \, e^{\left (-\frac {d m \log \left (\frac {2 i \, b}{d}\right ) - 2 i \, b c + 2 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right ) - e^{\left (-\frac {d m \log \left (\frac {4 i \, b}{d}\right ) - 4 i \, b c + 4 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {4 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right )}{64 \, b} \] Input:

integrate((d*x+c)^m*cos(b*x+a)*sin(b*x+a)^3,x, algorithm="fricas")
 

Output:

-1/64*(4*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(-4*I*b/d) + 4*I*b*c - 4*I*a*d)/d)*gamma( 
m + 1, -4*(I*b*d*x + I*b*c)/d) + 4*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(4*I*b/d) - 4*I 
*b*c + 4*I*a*d)/d)*gamma(m + 1, -4*(-I*b*d*x - I*b*c)/d))/b
 

Sympy [F(-2)]

Exception generated. \[ \int (c+d x)^m \cos (a+b x) \sin ^3(a+b x) \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:

integrate((d*x+c)**m*cos(b*x+a)*sin(b*x+a)**3,x)
 

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

\[ \int (c+d x)^m \cos (a+b x) \sin ^3(a+b x) \, dx=\int { {\left (d x + c\right )}^{m} \cos \left (b x + a\right ) \sin \left (b x + a\right )^{3} \,d x } \] Input:

integrate((d*x+c)^m*cos(b*x+a)*sin(b*x+a)^3,x, algorithm="maxima")
 

Output:

integrate((d*x + c)^m*cos(b*x + a)*sin(b*x + a)^3, x)
 

Giac [F]

\[ \int (c+d x)^m \cos (a+b x) \sin ^3(a+b x) \, dx=\int { {\left (d x + c\right )}^{m} \cos \left (b x + a\right ) \sin \left (b x + a\right )^{3} \,d x } \] Input:

integrate((d*x+c)^m*cos(b*x+a)*sin(b*x+a)^3,x, algorithm="giac")
 

Output:

integrate((d*x + c)^m*cos(b*x + a)*sin(b*x + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^m \cos (a+b x) \sin ^3(a+b x) \, dx=\int \cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^3\,{\left (c+d\,x\right )}^m \,d x \] Input:

int(cos(a + b*x)*sin(a + b*x)^3*(c + d*x)^m,x)
 

Output:

int(cos(a + b*x)*sin(a + b*x)^3*(c + d*x)^m, x)
 

Reduce [F]

\[ \int (c+d x)^m \cos (a+b x) \sin ^3(a+b x) \, dx=\int \left (d x +c \right )^{m} \cos \left (b x +a \right ) \sin \left (b x +a \right )^{3}d x \] Input:

int((d*x+c)^m*cos(b*x+a)*sin(b*x+a)^3,x)
 

Output:

int((c + d*x)**m*cos(a + b*x)*sin(a + b*x)**3,x)