\(\int \frac {(e+f x)^m \cos ^2(c+d x)}{a+a \sin (c+d x)} \, dx\) [289]

Optimal result

Integrand size = 28, antiderivative size = 154 \[ \int \frac {(e+f x)^m \cos ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {(e+f x)^{1+m}}{a f (1+m)}+\frac {e^{i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (-\frac {i d (e+f x)}{f}\right )^{-m} \Gamma \left (1+m,-\frac {i d (e+f x)}{f}\right )}{2 a d}+\frac {e^{-i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (\frac {i d (e+f x)}{f}\right )^{-m} \Gamma \left (1+m,\frac {i d (e+f x)}{f}\right )}{2 a d} \] Output:

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

Mathematica [A] (verified)

Time = 1.34 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.43 \[ \int \frac {(e+f x)^m \cos ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {e^{i \left (c-\frac {d e}{f}\right )} (e+f x)^m \left (\frac {d^2 (e+f x)^2}{f^2}\right )^{-m} \left (2 d e^{-i \left (c-\frac {d e}{f}\right )} (e+f x) \left (\frac {d^2 (e+f x)^2}{f^2}\right )^m+f (1+m) \left (\frac {i d (e+f x)}{f}\right )^m \Gamma \left (1+m,-\frac {i d (e+f x)}{f}\right )+e^{-2 i \left (c-\frac {d e}{f}\right )} f (1+m) \left (-\frac {i d (e+f x)}{f}\right )^m \Gamma \left (1+m,\frac {i d (e+f x)}{f}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}{2 a d f (1+m) (1+\sin (c+d x))} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {5034, 17, 3042, 3789, 2612}

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.

\(\displaystyle \int \frac {\cos ^2(c+d x) (e+f x)^m}{a \sin (c+d x)+a} \, dx\)

\(\Big \downarrow \) 5034

\(\displaystyle \frac {\int (e+f x)^mdx}{a}-\frac {\int (e+f x)^m \sin (c+d x)dx}{a}\)

\(\Big \downarrow \) 17

\(\displaystyle \frac {(e+f x)^{m+1}}{a f (m+1)}-\frac {\int (e+f x)^m \sin (c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(e+f x)^{m+1}}{a f (m+1)}-\frac {\int (e+f x)^m \sin (c+d x)dx}{a}\)

\(\Big \downarrow \) 3789

\(\displaystyle \frac {(e+f x)^{m+1}}{a f (m+1)}-\frac {\frac {1}{2} i \int e^{-i (c+d x)} (e+f x)^mdx-\frac {1}{2} i \int e^{i (c+d x)} (e+f x)^mdx}{a}\)

\(\Big \downarrow \) 2612

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 17

Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 
)/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
 

rule 2612

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 3042

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

rule 3789

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

rule 5034

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

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.84 \[ \int \frac {(e+f x)^m \cos ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {{\left (f m + f\right )} e^{\left (-\frac {f m \log \left (\frac {i \, d}{f}\right ) - i \, d e + i \, c f}{f}\right )} \Gamma \left (m + 1, \frac {i \, d f x + i \, d e}{f}\right ) + {\left (f m + f\right )} e^{\left (-\frac {f m \log \left (-\frac {i \, d}{f}\right ) + i \, d e - i \, c f}{f}\right )} \Gamma \left (m + 1, \frac {-i \, d f x - i \, d e}{f}\right ) + 2 \, {\left (d f x + d e\right )} {\left (f x + e\right )}^{m}}{2 \, {\left (a d f m + a d f\right )}} \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {(e+f x)^m \cos ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\left (e + f x\right )^{m} \cos ^{2}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:

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

Output:

Integral((e + f*x)**m*cos(c + d*x)**2/(sin(c + d*x) + 1), x)/a
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(e+f x)^m \cos ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\left (f x +e \right )^{m} \cos \left (d x +c \right ) f m +\left (f x +e \right )^{m} \cos \left (d x +c \right ) f +\left (f x +e \right )^{m} d e +\left (f x +e \right )^{m} d f x -\left (\int \frac {\left (f x +e \right )^{m} \cos \left (d x +c \right )}{f x +e}d x \right ) f^{2} m^{2}-\left (\int \frac {\left (f x +e \right )^{m} \cos \left (d x +c \right )}{f x +e}d x \right ) f^{2} m}{a d f \left (m +1\right )} \] Input:

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

Output:

((e + f*x)**m*cos(c + d*x)*f*m + (e + f*x)**m*cos(c + d*x)*f + (e + f*x)** 
m*d*e + (e + f*x)**m*d*f*x - int(((e + f*x)**m*cos(c + d*x))/(e + f*x),x)* 
f**2*m**2 - int(((e + f*x)**m*cos(c + d*x))/(e + f*x),x)*f**2*m)/(a*d*f*(m 
 + 1))