\(\int (c+d x)^m (a+b \sin (e+f x))^2 \, dx\) [175]

Optimal result

Integrand size = 20, antiderivative size = 318 \[ \int (c+d x)^m (a+b \sin (e+f x))^2 \, dx=\frac {a^2 (c+d x)^{1+m}}{d (1+m)}+\frac {b^2 (c+d x)^{1+m}}{2 d (1+m)}-\frac {a b e^{i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i f (c+d x)}{d}\right )}{f}-\frac {a b e^{-i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i f (c+d x)}{d}\right )}{f}+\frac {i 2^{-3-m} b^2 e^{2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 i f (c+d x)}{d}\right )}{f}-\frac {i 2^{-3-m} b^2 e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right )}{f} \] Output:

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

Mathematica [A] (verified)

Time = 10.26 (sec) , antiderivative size = 268, normalized size of antiderivative = 0.84 \[ \int (c+d x)^m (a+b \sin (e+f x))^2 \, dx=-\frac {(c+d x)^m \left (-\frac {4 \left (2 a^2+b^2\right ) f (c+d x)}{d (1+m)}+8 a b e^{i \left (e-\frac {c f}{d}\right )} \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i f (c+d x)}{d}\right )+8 a b e^{-i \left (e-\frac {c f}{d}\right )} \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i f (c+d x)}{d}\right )-i 2^{-m} b^2 e^{2 i \left (e-\frac {c f}{d}\right )} \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 i f (c+d x)}{d}\right )+i 2^{-m} b^2 e^{-2 i \left (e-\frac {c f}{d}\right )} \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right )\right )}{8 f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3042, 3798, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int (c+d x)^m (a+b \sin (e+f x))^2dx\)

\(\Big \downarrow \) 3798

\(\displaystyle \int \left (a^2 (c+d x)^m+2 a b (c+d x)^m \sin (e+f x)+b^2 (c+d x)^m \sin ^2(e+f x)\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(a^2*(c + d*x)^(1 + m))/(d*(1 + m)) + (b^2*(c + d*x)^(1 + m))/(2*d*(1 + m) 
) - (a*b*E^(I*(e - (c*f)/d))*(c + d*x)^m*Gamma[1 + m, ((-I)*f*(c + d*x))/d 
])/(f*(((-I)*f*(c + d*x))/d)^m) - (a*b*(c + d*x)^m*Gamma[1 + m, (I*f*(c + 
d*x))/d])/(E^(I*(e - (c*f)/d))*f*((I*f*(c + d*x))/d)^m) + (I*2^(-3 - m)*b^ 
2*E^((2*I)*(e - (c*f)/d))*(c + d*x)^m*Gamma[1 + m, ((-2*I)*f*(c + d*x))/d] 
)/(f*(((-I)*f*(c + d*x))/d)^m) - (I*2^(-3 - m)*b^2*(c + d*x)^m*Gamma[1 + m 
, ((2*I)*f*(c + d*x))/d])/(E^((2*I)*(e - (c*f)/d))*f*((I*f*(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 3798

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], 
 x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] || IGtQ[ 
m, 0] || NeQ[a^2 - b^2, 0])
 

Maple [F]

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

Input:

int((d*x+c)^m*(a+b*sin(f*x+e))^2,x)
 

Output:

int((d*x+c)^m*(a+b*sin(f*x+e))^2,x)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 278, normalized size of antiderivative = 0.87 \[ \int (c+d x)^m (a+b \sin (e+f x))^2 \, dx=-\frac {8 \, {\left (a b d m + a b d\right )} e^{\left (-\frac {d m \log \left (\frac {i \, f}{d}\right ) + i \, d e - i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {i \, d f x + i \, c f}{d}\right ) - {\left (i \, b^{2} d m + i \, b^{2} d\right )} e^{\left (-\frac {d m \log \left (-\frac {2 i \, f}{d}\right ) - 2 i \, d e + 2 i \, c f}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (i \, d f x + i \, c f\right )}}{d}\right ) + 8 \, {\left (a b d m + a b d\right )} e^{\left (-\frac {d m \log \left (-\frac {i \, f}{d}\right ) - i \, d e + i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {-i \, d f x - i \, c f}{d}\right ) - {\left (-i \, b^{2} d m - i \, b^{2} d\right )} e^{\left (-\frac {d m \log \left (\frac {2 i \, f}{d}\right ) + 2 i \, d e - 2 i \, c f}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (-i \, d f x - i \, c f\right )}}{d}\right ) - 4 \, {\left ({\left (2 \, a^{2} + b^{2}\right )} d f x + {\left (2 \, a^{2} + b^{2}\right )} c f\right )} {\left (d x + c\right )}^{m}}{8 \, {\left (d f m + d f\right )}} \] Input:

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

Output:

-1/8*(8*(a*b*d*m + a*b*d)*e^(-(d*m*log(I*f/d) + I*d*e - I*c*f)/d)*gamma(m 
+ 1, (I*d*f*x + I*c*f)/d) - (I*b^2*d*m + I*b^2*d)*e^(-(d*m*log(-2*I*f/d) - 
 2*I*d*e + 2*I*c*f)/d)*gamma(m + 1, -2*(I*d*f*x + I*c*f)/d) + 8*(a*b*d*m + 
 a*b*d)*e^(-(d*m*log(-I*f/d) - I*d*e + I*c*f)/d)*gamma(m + 1, (-I*d*f*x - 
I*c*f)/d) - (-I*b^2*d*m - I*b^2*d)*e^(-(d*m*log(2*I*f/d) + 2*I*d*e - 2*I*c 
*f)/d)*gamma(m + 1, -2*(-I*d*f*x - I*c*f)/d) - 4*((2*a^2 + b^2)*d*f*x + (2 
*a^2 + b^2)*c*f)*(d*x + c)^m)/(d*f*m + d*f)
 

Sympy [F]

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

integrate((d*x+c)**m*(a+b*sin(f*x+e))**2,x)
 

Output:

Integral((a + b*sin(e + f*x))**2*(c + d*x)**m, x)
 

Maxima [F]

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

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

Output:

(d*x + c)^(m + 1)*a^2/(d*(m + 1)) + 1/2*(b^2*e^(m*log(d*x + c) + log(d*x + 
 c)) - (b^2*d*m + b^2*d)*integrate((d*x + c)^m*cos(2*f*x + 2*e), x) + 4*(a 
*b*d*m + a*b*d)*integrate((d*x + c)^m*sin(f*x + e), x))/(d*m + d)
 

Giac [F]

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

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

Output:

integrate((b*sin(f*x + e) + a)^2*(d*x + c)^m, x)
 

Mupad [F(-1)]

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

int((a + b*sin(e + f*x))^2*(c + d*x)^m,x)
 

Output:

int((a + b*sin(e + f*x))^2*(c + d*x)^m, x)
 

Reduce [F]

\[ \int (c+d x)^m (a+b \sin (e+f x))^2 \, dx=\text {too large to display} \] Input:

int((d*x+c)^m*(a+b*sin(f*x+e))^2,x)
 

Output:

( - 2*(c + d*x)**m*cos(e + f*x)*sin(e + f*x)*tan((e + f*x)/2)**4*b**2*c*d* 
f*m - 2*(c + d*x)**m*cos(e + f*x)*sin(e + f*x)*tan((e + f*x)/2)**4*b**2*c* 
d*f - 4*(c + d*x)**m*cos(e + f*x)*sin(e + f*x)*tan((e + f*x)/2)**2*b**2*c* 
d*f*m - 4*(c + d*x)**m*cos(e + f*x)*sin(e + f*x)*tan((e + f*x)/2)**2*b**2* 
c*d*f - 2*(c + d*x)**m*cos(e + f*x)*sin(e + f*x)*b**2*c*d*f*m - 2*(c + d*x 
)**m*cos(e + f*x)*sin(e + f*x)*b**2*c*d*f - 6*(c + d*x)**m*cos(e + f*x)*ta 
n((e + f*x)/2)**4*a*b*c*d*f*m - 6*(c + d*x)**m*cos(e + f*x)*tan((e + f*x)/ 
2)**4*a*b*c*d*f - 12*(c + d*x)**m*cos(e + f*x)*tan((e + f*x)/2)**2*a*b*c*d 
*f*m - 12*(c + d*x)**m*cos(e + f*x)*tan((e + f*x)/2)**2*a*b*c*d*f - 6*(c + 
 d*x)**m*cos(e + f*x)*a*b*c*d*f*m - 6*(c + d*x)**m*cos(e + f*x)*a*b*c*d*f 
- 2*(c + d*x)**m*sin(e + f*x)*tan((e + f*x)/2)**4*b**2*c*d*f*m - 2*(c + d* 
x)**m*sin(e + f*x)*tan((e + f*x)/2)**4*b**2*c*d*f - 4*(c + d*x)**m*sin(e + 
 f*x)*tan((e + f*x)/2)**2*b**2*c*d*f*m - 4*(c + d*x)**m*sin(e + f*x)*tan(( 
e + f*x)/2)**2*b**2*c*d*f - 2*(c + d*x)**m*sin(e + f*x)*b**2*c*d*f*m - 2*( 
c + d*x)**m*sin(e + f*x)*b**2*c*d*f + 3*(c + d*x)**m*tan((e + f*x)/2)**4*a 
**2*c**2*f**2 + 3*(c + d*x)**m*tan((e + f*x)/2)**4*a**2*c*d*f**2*x - 6*(c 
+ d*x)**m*tan((e + f*x)/2)**4*a*b*c*d*f*m - 6*(c + d*x)**m*tan((e + f*x)/2 
)**4*a*b*c*d*f + 6*(c + d*x)**m*tan((e + f*x)/2)**2*a**2*c**2*f**2 + 6*(c 
+ d*x)**m*tan((e + f*x)/2)**2*a**2*c*d*f**2*x - 12*(c + d*x)**m*tan((e + f 
*x)/2)**2*a*b*c*d*f*m - 12*(c + d*x)**m*tan((e + f*x)/2)**2*a*b*c*d*f +...