\(\int (c+d x)^m \cos ^2(a+b x) \sin ^3(a+b x) \, dx\) [88]

Optimal result

Integrand size = 24, antiderivative size = 407 \[ \int (c+d x)^m \cos ^2(a+b x) \sin ^3(a+b x) \, dx=-\frac {e^{i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i b (c+d x)}{d}\right )}{16 b}-\frac {e^{-i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i b (c+d x)}{d}\right )}{16 b}-\frac {3^{-1-m} e^{3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {3 i b (c+d x)}{d}\right )}{32 b}-\frac {3^{-1-m} e^{-3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {3 i b (c+d x)}{d}\right )}{32 b}+\frac {5^{-1-m} e^{5 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {5 i b (c+d x)}{d}\right )}{32 b}+\frac {5^{-1-m} e^{-5 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {5 i b (c+d x)}{d}\right )}{32 b} \] Output:

-1/16*exp(I*(a-b*c/d))*(d*x+c)^m*GAMMA(1+m,-I*b*(d*x+c)/d)/b/((-I*b*(d*x+c 
)/d)^m)-1/16*(d*x+c)^m*GAMMA(1+m,I*b*(d*x+c)/d)/b/exp(I*(a-b*c/d))/((I*b*( 
d*x+c)/d)^m)-1/32*3^(-1-m)*exp(3*I*(a-b*c/d))*(d*x+c)^m*GAMMA(1+m,-3*I*b*( 
d*x+c)/d)/b/((-I*b*(d*x+c)/d)^m)-1/32*3^(-1-m)*(d*x+c)^m*GAMMA(1+m,3*I*b*( 
d*x+c)/d)/b/exp(3*I*(a-b*c/d))/((I*b*(d*x+c)/d)^m)+1/32*5^(-1-m)*exp(5*I*( 
a-b*c/d))*(d*x+c)^m*GAMMA(1+m,-5*I*b*(d*x+c)/d)/b/((-I*b*(d*x+c)/d)^m)+1/3 
2*5^(-1-m)*(d*x+c)^m*GAMMA(1+m,5*I*b*(d*x+c)/d)/b/exp(5*I*(a-b*c/d))/((I*b 
*(d*x+c)/d)^m)
 

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 376, normalized size of antiderivative = 0.92 \[ \int (c+d x)^m \cos ^2(a+b x) \sin ^3(a+b x) \, dx=\frac {e^{-\frac {5 i (b c+a d)}{d}} (c+d x)^m \left (30 e^{\frac {4 i (b c+a d)}{d}} \left (-e^{2 i a} \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i b (c+d x)}{d}\right )-e^{\frac {2 i b c}{d}} \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i b (c+d x)}{d}\right )\right )-5\ 3^{-m} e^{\frac {2 i (b c+a d)}{d}} \left (\frac {b^2 (c+d x)^2}{d^2}\right )^{-m} \left (e^{6 i a} \left (\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,-\frac {3 i b (c+d x)}{d}\right )+e^{\frac {6 i b c}{d}} \left (-\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,\frac {3 i b (c+d x)}{d}\right )\right )+3\ 5^{-m} \left (\frac {b^2 (c+d x)^2}{d^2}\right )^{-m} \left (e^{10 i a} \left (\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,-\frac {5 i b (c+d x)}{d}\right )+e^{\frac {10 i b c}{d}} \left (-\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,\frac {5 i b (c+d x)}{d}\right )\right )\right )}{480 b} \] Input:

Integrate[(c + d*x)^m*Cos[a + b*x]^2*Sin[a + b*x]^3,x]
 

Output:

((c + d*x)^m*(30*E^(((4*I)*(b*c + a*d))/d)*(-((E^((2*I)*a)*Gamma[1 + m, (( 
-I)*b*(c + d*x))/d])/(((-I)*b*(c + d*x))/d)^m) - (E^(((2*I)*b*c)/d)*Gamma[ 
1 + m, (I*b*(c + d*x))/d])/((I*b*(c + d*x))/d)^m) - (5*E^(((2*I)*(b*c + a* 
d))/d)*(E^((6*I)*a)*((I*b*(c + d*x))/d)^m*Gamma[1 + m, ((-3*I)*b*(c + d*x) 
)/d] + E^(((6*I)*b*c)/d)*(((-I)*b*(c + d*x))/d)^m*Gamma[1 + m, ((3*I)*b*(c 
 + d*x))/d]))/(3^m*((b^2*(c + d*x)^2)/d^2)^m) + (3*(E^((10*I)*a)*((I*b*(c 
+ d*x))/d)^m*Gamma[1 + m, ((-5*I)*b*(c + d*x))/d] + E^(((10*I)*b*c)/d)*((( 
-I)*b*(c + d*x))/d)^m*Gamma[1 + m, ((5*I)*b*(c + d*x))/d]))/(5^m*((b^2*(c 
+ d*x)^2)/d^2)^m)))/(480*b*E^(((5*I)*(b*c + a*d))/d))
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 407, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, 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.

\(\displaystyle \int \sin ^3(a+b x) \cos ^2(a+b x) (c+d x)^m \, dx\)

\(\Big \downarrow \) 4906

\(\displaystyle \int \left (\frac {1}{8} \sin (a+b x) (c+d x)^m+\frac {1}{16} \sin (3 a+3 b x) (c+d x)^m-\frac {1}{16} \sin (5 a+5 b x) (c+d x)^m\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {e^{i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {i b (c+d x)}{d}\right )}{16 b}-\frac {3^{-m-1} e^{3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {3 i b (c+d x)}{d}\right )}{32 b}+\frac {5^{-m-1} e^{5 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {5 i b (c+d x)}{d}\right )}{32 b}-\frac {e^{-i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {i b (c+d x)}{d}\right )}{16 b}-\frac {3^{-m-1} e^{-3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {3 i b (c+d x)}{d}\right )}{32 b}+\frac {5^{-m-1} e^{-5 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {5 i b (c+d x)}{d}\right )}{32 b}\)

Input:

Int[(c + d*x)^m*Cos[a + b*x]^2*Sin[a + b*x]^3,x]
 

Output:

-1/16*(E^(I*(a - (b*c)/d))*(c + d*x)^m*Gamma[1 + m, ((-I)*b*(c + d*x))/d]) 
/(b*(((-I)*b*(c + d*x))/d)^m) - ((c + d*x)^m*Gamma[1 + m, (I*b*(c + d*x))/ 
d])/(16*b*E^(I*(a - (b*c)/d))*((I*b*(c + d*x))/d)^m) - (3^(-1 - m)*E^((3*I 
)*(a - (b*c)/d))*(c + d*x)^m*Gamma[1 + m, ((-3*I)*b*(c + d*x))/d])/(32*b*( 
((-I)*b*(c + d*x))/d)^m) - (3^(-1 - m)*(c + d*x)^m*Gamma[1 + m, ((3*I)*b*( 
c + d*x))/d])/(32*b*E^((3*I)*(a - (b*c)/d))*((I*b*(c + d*x))/d)^m) + (5^(- 
1 - m)*E^((5*I)*(a - (b*c)/d))*(c + d*x)^m*Gamma[1 + m, ((-5*I)*b*(c + d*x 
))/d])/(32*b*(((-I)*b*(c + d*x))/d)^m) + (5^(-1 - m)*(c + d*x)^m*Gamma[1 + 
 m, ((5*I)*b*(c + d*x))/d])/(32*b*E^((5*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]

\[\int \left (d x +c \right )^{m} \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{3}d x\]

Input:

int((d*x+c)^m*cos(b*x+a)^2*sin(b*x+a)^3,x)
 

Output:

int((d*x+c)^m*cos(b*x+a)^2*sin(b*x+a)^3,x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 280, normalized size of antiderivative = 0.69 \[ \int (c+d x)^m \cos ^2(a+b x) \sin ^3(a+b x) \, dx=-\frac {30 \, e^{\left (-\frac {d m \log \left (\frac {i \, b}{d}\right ) - i \, b c + i \, a d}{d}\right )} \Gamma \left (m + 1, \frac {i \, b d x + i \, b c}{d}\right ) + 5 \, e^{\left (-\frac {d m \log \left (-\frac {3 i \, b}{d}\right ) + 3 i \, b c - 3 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {3 \, {\left (i \, b d x + i \, b c\right )}}{d}\right ) - 3 \, e^{\left (-\frac {d m \log \left (-\frac {5 i \, b}{d}\right ) + 5 i \, b c - 5 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {5 \, {\left (i \, b d x + i \, b c\right )}}{d}\right ) + 30 \, e^{\left (-\frac {d m \log \left (-\frac {i \, b}{d}\right ) + i \, b c - i \, a d}{d}\right )} \Gamma \left (m + 1, \frac {-i \, b d x - i \, b c}{d}\right ) + 5 \, e^{\left (-\frac {d m \log \left (\frac {3 i \, b}{d}\right ) - 3 i \, b c + 3 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {3 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right ) - 3 \, e^{\left (-\frac {d m \log \left (\frac {5 i \, b}{d}\right ) - 5 i \, b c + 5 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {5 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right )}{480 \, b} \] Input:

integrate((d*x+c)^m*cos(b*x+a)^2*sin(b*x+a)^3,x, algorithm="fricas")
 

Output:

-1/480*(30*e^(-(d*m*log(I*b/d) - I*b*c + I*a*d)/d)*gamma(m + 1, (I*b*d*x + 
 I*b*c)/d) + 5*e^(-(d*m*log(-3*I*b/d) + 3*I*b*c - 3*I*a*d)/d)*gamma(m + 1, 
 -3*(I*b*d*x + I*b*c)/d) - 3*e^(-(d*m*log(-5*I*b/d) + 5*I*b*c - 5*I*a*d)/d 
)*gamma(m + 1, -5*(I*b*d*x + I*b*c)/d) + 30*e^(-(d*m*log(-I*b/d) + I*b*c - 
 I*a*d)/d)*gamma(m + 1, (-I*b*d*x - I*b*c)/d) + 5*e^(-(d*m*log(3*I*b/d) - 
3*I*b*c + 3*I*a*d)/d)*gamma(m + 1, -3*(-I*b*d*x - I*b*c)/d) - 3*e^(-(d*m*l 
og(5*I*b/d) - 5*I*b*c + 5*I*a*d)/d)*gamma(m + 1, -5*(-I*b*d*x - I*b*c)/d)) 
/b
 

Sympy [F]

\[ \int (c+d x)^m \cos ^2(a+b x) \sin ^3(a+b x) \, dx=\int \left (c + d x\right )^{m} \sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}\, dx \] Input:

integrate((d*x+c)**m*cos(b*x+a)**2*sin(b*x+a)**3,x)
 

Output:

Integral((c + d*x)**m*sin(a + b*x)**3*cos(a + b*x)**2, x)
 

Maxima [F]

\[ \int (c+d x)^m \cos ^2(a+b x) \sin ^3(a+b x) \, dx=\int { {\left (d x + c\right )}^{m} \cos \left (b x + a\right )^{2} \sin \left (b x + a\right )^{3} \,d x } \] Input:

integrate((d*x+c)^m*cos(b*x+a)^2*sin(b*x+a)^3,x, algorithm="maxima")
 

Output:

integrate((d*x + c)^m*cos(b*x + a)^2*sin(b*x + a)^3, x)
 

Giac [F]

\[ \int (c+d x)^m \cos ^2(a+b x) \sin ^3(a+b x) \, dx=\int { {\left (d x + c\right )}^{m} \cos \left (b x + a\right )^{2} \sin \left (b x + a\right )^{3} \,d x } \] Input:

integrate((d*x+c)^m*cos(b*x+a)^2*sin(b*x+a)^3,x, algorithm="giac")
 

Output:

integrate((d*x + c)^m*cos(b*x + a)^2*sin(b*x + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^m \cos ^2(a+b x) \sin ^3(a+b x) \, dx=\int {\cos \left (a+b\,x\right )}^2\,{\sin \left (a+b\,x\right )}^3\,{\left (c+d\,x\right )}^m \,d x \] Input:

int(cos(a + b*x)^2*sin(a + b*x)^3*(c + d*x)^m,x)
 

Output:

int(cos(a + b*x)^2*sin(a + b*x)^3*(c + d*x)^m, x)
 

Reduce [F]

\[ \int (c+d x)^m \cos ^2(a+b x) \sin ^3(a+b x) \, dx=\int \left (d x +c \right )^{m} \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{3}d x \] Input:

int((d*x+c)^m*cos(b*x+a)^2*sin(b*x+a)^3,x)
 

Output:

int((c + d*x)**m*cos(a + b*x)**2*sin(a + b*x)**3,x)