\(\int (c+d x)^m \cos ^3(a+b x) \, dx\) [99]

Optimal result

Integrand size = 16, antiderivative size = 275 \[ \int (c+d x)^m \cos ^3(a+b x) \, dx=-\frac {3 i e^{i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i b (c+d x)}{d}\right )}{8 b}+\frac {3 i e^{-i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i b (c+d x)}{d}\right )}{8 b}-\frac {i 3^{-1-m} e^{3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {3 i b (c+d x)}{d}\right )}{8 b}+\frac {i 3^{-1-m} e^{-3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {3 i b (c+d x)}{d}\right )}{8 b} \] Output:

-3/8*I*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)+3/8*I*(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)-1/8*I*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)+1/8*I*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)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 253, normalized size of antiderivative = 0.92 \[ \int (c+d x)^m \cos ^3(a+b x) \, dx=\frac {i 3^{-1-m} e^{-\frac {3 i (b c+a d)}{d}} (c+d x)^m \left (\frac {b^2 (c+d x)^2}{d^2}\right )^{-m} \left (-3^{2+m} e^{2 i \left (2 a+\frac {b c}{d}\right )} \left (\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,-\frac {i b (c+d x)}{d}\right )+3^{2+m} e^{2 i a+\frac {4 i b c}{d}} \left (-\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,\frac {i b (c+d x)}{d}\right )-e^{6 i a} \left (\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,-\frac {3 i b (c+d x)}{d}\right )+e^{\frac {6 i b c}{d}} \left (-\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,\frac {3 i b (c+d x)}{d}\right )\right )}{8 b} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3042, 3793, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin \left (a+b x+\frac {\pi }{2}\right )^3 (c+d x)^mdx\)

\(\Big \downarrow \) 3793

\(\displaystyle \int \left (\frac {3}{4} \cos (a+b x) (c+d x)^m+\frac {1}{4} \cos (3 a+3 b x) (c+d x)^m\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3 i e^{i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {i b (c+d x)}{d}\right )}{8 b}-\frac {i 3^{-m-1} e^{3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {3 i b (c+d x)}{d}\right )}{8 b}+\frac {3 i e^{-i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {i b (c+d x)}{d}\right )}{8 b}+\frac {i 3^{-m-1} e^{-3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {3 i b (c+d x)}{d}\right )}{8 b}\)

Input:

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

Output:

(((-3*I)/8)*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) + (((3*I)/8)*(c + d*x)^m*Gamma[1 + m, ( 
I*b*(c + d*x))/d])/(b*E^(I*(a - (b*c)/d))*((I*b*(c + d*x))/d)^m) - ((I/8)* 
3^(-1 - m)*E^((3*I)*(a - (b*c)/d))*(c + d*x)^m*Gamma[1 + m, ((-3*I)*b*(c + 
 d*x))/d])/(b*(((-I)*b*(c + d*x))/d)^m) + ((I/8)*3^(-1 - m)*(c + d*x)^m*Ga 
mma[1 + m, ((3*I)*b*(c + d*x))/d])/(b*E^((3*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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3793

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In 
t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f 
, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.68 \[ \int (c+d x)^m \cos ^3(a+b x) \, dx=\frac {9 i \, e^{\left (-\frac {d m \log \left (\frac {i \, b}{d}\right ) - i \, b c + i \, a d}{d}\right )} \Gamma \left (m + 1, \frac {i \, b d x + i \, b c}{d}\right ) - i \, e^{\left (-\frac {d m \log \left (-\frac {3 i \, b}{d}\right ) + 3 i \, b c - 3 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {3 \, {\left (i \, b d x + i \, b c\right )}}{d}\right ) - 9 i \, e^{\left (-\frac {d m \log \left (-\frac {i \, b}{d}\right ) + i \, b c - i \, a d}{d}\right )} \Gamma \left (m + 1, \frac {-i \, b d x - i \, b c}{d}\right ) + i \, e^{\left (-\frac {d m \log \left (\frac {3 i \, b}{d}\right ) - 3 i \, b c + 3 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {3 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right )}{24 \, b} \] Input:

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

Output:

1/24*(9*I*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) - I*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) - 9*I*e^(-(d*m*log(-I*b/d) + I*b*c - I*a*d)/d)*gam 
ma(m + 1, (-I*b*d*x - I*b*c)/d) + I*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))/b
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (c+d x)^m \cos ^3(a+b x) \, dx=\text {too large to display} \] Input:

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

Output:

( - 9*(c + d*x)**m*cos(a + b*x)*sin(a + b*x)*tan((a + b*x)/2)**6*d*m - 9*( 
c + d*x)**m*cos(a + b*x)*sin(a + b*x)*tan((a + b*x)/2)**6*d - 27*(c + d*x) 
**m*cos(a + b*x)*sin(a + b*x)*tan((a + b*x)/2)**4*d*m - 27*(c + d*x)**m*co 
s(a + b*x)*sin(a + b*x)*tan((a + b*x)/2)**4*d - 27*(c + d*x)**m*cos(a + b* 
x)*sin(a + b*x)*tan((a + b*x)/2)**2*d*m - 27*(c + d*x)**m*cos(a + b*x)*sin 
(a + b*x)*tan((a + b*x)/2)**2*d - 9*(c + d*x)**m*cos(a + b*x)*sin(a + b*x) 
*d*m - 9*(c + d*x)**m*cos(a + b*x)*sin(a + b*x)*d - 3*(c + d*x)**m*sin(a + 
 b*x)**3*tan((a + b*x)/2)**6*d*m - 3*(c + d*x)**m*sin(a + b*x)**3*tan((a + 
 b*x)/2)**6*d - 9*(c + d*x)**m*sin(a + b*x)**3*tan((a + b*x)/2)**4*d*m - 9 
*(c + d*x)**m*sin(a + b*x)**3*tan((a + b*x)/2)**4*d - 9*(c + d*x)**m*sin(a 
 + b*x)**3*tan((a + b*x)/2)**2*d*m - 9*(c + d*x)**m*sin(a + b*x)**3*tan((a 
 + b*x)/2)**2*d - 3*(c + d*x)**m*sin(a + b*x)**3*d*m - 3*(c + d*x)**m*sin( 
a + b*x)**3*d - 9*(c + d*x)**m*sin(a + b*x)*tan((a + b*x)/2)**6*d*m - 9*(c 
 + d*x)**m*sin(a + b*x)*tan((a + b*x)/2)**6*d - 27*(c + d*x)**m*sin(a + b* 
x)*tan((a + b*x)/2)**4*d*m - 27*(c + d*x)**m*sin(a + b*x)*tan((a + b*x)/2) 
**4*d - 27*(c + d*x)**m*sin(a + b*x)*tan((a + b*x)/2)**2*d*m - 27*(c + d*x 
)**m*sin(a + b*x)*tan((a + b*x)/2)**2*d - 9*(c + d*x)**m*sin(a + b*x)*d*m 
- 9*(c + d*x)**m*sin(a + b*x)*d - 15*(c + d*x)**m*tan((a + b*x)/2)**6*b*c 
- 15*(c + d*x)**m*tan((a + b*x)/2)**6*b*d*x - 45*(c + d*x)**m*tan((a + b*x 
)/2)**4*b*c - 45*(c + d*x)**m*tan((a + b*x)/2)**4*b*d*x + 160*(c + d*x)...