\(\int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx\) [616]

Optimal result

Integrand size = 25, antiderivative size = 193 \[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\frac {d (d-2 c (2+m)) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac {2^{\frac {1}{2}+m} \left (2 c d m (2+m)+d^2 \left (1+m+m^2\right )+c^2 \left (2+3 m+m^2\right )\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac {d^2 \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)} \] Output:

d*(d-2*c*(2+m))*cos(f*x+e)*(a+a*sin(f*x+e))^m/f/(1+m)/(2+m)-2^(1/2+m)*(2*c 
*d*m*(2+m)+d^2*(m^2+m+1)+c^2*(m^2+3*m+2))*cos(f*x+e)*hypergeom([1/2, 1/2-m 
],[3/2],1/2-1/2*sin(f*x+e))*(1+sin(f*x+e))^(-1/2-m)*(a+a*sin(f*x+e))^m/f/( 
1+m)/(2+m)-d^2*cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)/a/f/(2+m)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 21.69 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.61 \[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=-\frac {(a (1+\sin (e+f x)))^m (\cos (e+f x)+i (1+\sin (e+f x))) \left (-\frac {2 \left (2 c^2+d^2\right ) \operatorname {Hypergeometric2F1}(1,1+m,1-m,i \cos (e+f x)-\sin (e+f x))}{m}-\frac {4 i c d \operatorname {Hypergeometric2F1}(1,m,-m,i \cos (e+f x)-\sin (e+f x)) (\cos (e+f x)-i \sin (e+f x))}{1+m}+\frac {4 i c d \operatorname {Hypergeometric2F1}(1,2+m,2-m,i \cos (e+f x)-\sin (e+f x)) (\cos (e+f x)+i \sin (e+f x))}{-1+m}+\frac {d^2 \operatorname {Hypergeometric2F1}(1,-1+m,-1-m,i \cos (e+f x)-\sin (e+f x)) (\cos (2 (e+f x))-i \sin (2 (e+f x)))}{2+m}+\frac {d^2 \operatorname {Hypergeometric2F1}(1,3+m,3-m,i \cos (e+f x)-\sin (e+f x)) (\cos (2 (e+f x))+i \sin (2 (e+f x)))}{-2+m}\right )}{4 f} \] Input:

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

Output:

-1/4*((a*(1 + Sin[e + f*x]))^m*(Cos[e + f*x] + I*(1 + Sin[e + f*x]))*((-2* 
(2*c^2 + d^2)*Hypergeometric2F1[1, 1 + m, 1 - m, I*Cos[e + f*x] - Sin[e + 
f*x]])/m - ((4*I)*c*d*Hypergeometric2F1[1, m, -m, I*Cos[e + f*x] - Sin[e + 
 f*x]]*(Cos[e + f*x] - I*Sin[e + f*x]))/(1 + m) + ((4*I)*c*d*Hypergeometri 
c2F1[1, 2 + m, 2 - m, I*Cos[e + f*x] - Sin[e + f*x]]*(Cos[e + f*x] + I*Sin 
[e + f*x]))/(-1 + m) + (d^2*Hypergeometric2F1[1, -1 + m, -1 - m, I*Cos[e + 
 f*x] - Sin[e + f*x]]*(Cos[2*(e + f*x)] - I*Sin[2*(e + f*x)]))/(2 + m) + ( 
d^2*Hypergeometric2F1[1, 3 + m, 3 - m, I*Cos[e + f*x] - Sin[e + f*x]]*(Cos 
[2*(e + f*x)] + I*Sin[2*(e + f*x)]))/(-2 + m)))/f
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3240, 3042, 3230, 3042, 3131, 3042, 3130}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3240

\(\displaystyle \frac {\int (\sin (e+f x) a+a)^m \left (a \left ((m+2) c^2+d^2 (m+1)\right )-a d (d-2 c (m+2)) \sin (e+f x)\right )dx}{a (m+2)}-\frac {d^2 \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (\sin (e+f x) a+a)^m \left (a \left ((m+2) c^2+d^2 (m+1)\right )-a d (d-2 c (m+2)) \sin (e+f x)\right )dx}{a (m+2)}-\frac {d^2 \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)}\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {\frac {a \left (c^2 \left (m^2+3 m+2\right )+2 c d m (m+2)+d^2 \left (m^2+m+1\right )\right ) \int (\sin (e+f x) a+a)^mdx}{m+1}+\frac {a d (d-2 c (m+2)) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (m+1)}}{a (m+2)}-\frac {d^2 \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a \left (c^2 \left (m^2+3 m+2\right )+2 c d m (m+2)+d^2 \left (m^2+m+1\right )\right ) \int (\sin (e+f x) a+a)^mdx}{m+1}+\frac {a d (d-2 c (m+2)) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (m+1)}}{a (m+2)}-\frac {d^2 \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)}\)

\(\Big \downarrow \) 3131

\(\displaystyle \frac {\frac {a \left (c^2 \left (m^2+3 m+2\right )+2 c d m (m+2)+d^2 \left (m^2+m+1\right )\right ) (\sin (e+f x)+1)^{-m} (a \sin (e+f x)+a)^m \int (\sin (e+f x)+1)^mdx}{m+1}+\frac {a d (d-2 c (m+2)) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (m+1)}}{a (m+2)}-\frac {d^2 \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a \left (c^2 \left (m^2+3 m+2\right )+2 c d m (m+2)+d^2 \left (m^2+m+1\right )\right ) (\sin (e+f x)+1)^{-m} (a \sin (e+f x)+a)^m \int (\sin (e+f x)+1)^mdx}{m+1}+\frac {a d (d-2 c (m+2)) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (m+1)}}{a (m+2)}-\frac {d^2 \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)}\)

\(\Big \downarrow \) 3130

\(\displaystyle \frac {\frac {a d (d-2 c (m+2)) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (m+1)}-\frac {a 2^{m+\frac {1}{2}} \left (c^2 \left (m^2+3 m+2\right )+2 c d m (m+2)+d^2 \left (m^2+m+1\right )\right ) \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{f (m+1)}}{a (m+2)}-\frac {d^2 \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)}\)

Input:

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

Output:

-((d^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(a*f*(2 + m))) + ((a*d*( 
d - 2*c*(2 + m))*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(1 + m)) - (2^(1/ 
2 + m)*a*(2*c*d*m*(2 + m) + d^2*(1 + m + m^2) + c^2*(2 + 3*m + m^2))*Cos[e 
 + f*x]*Hypergeometric2F1[1/2, 1/2 - m, 3/2, (1 - Sin[e + f*x])/2]*(1 + Si 
n[e + f*x])^(-1/2 - m)*(a + a*Sin[e + f*x])^m)/(f*(1 + m)))/(a*(2 + m))
 

Defintions of rubi rules used

rule 3042

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

rule 3130

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 
 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeome 
tric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; FreeQ[{a, 
 b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]
 

rule 3131

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPar 
t[n]*((a + b*Sin[c + d*x])^FracPart[n]/(1 + (b/a)*Sin[c + d*x])^FracPart[n] 
)   Int[(1 + (b/a)*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && 
EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]
 

rule 3230

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

rule 3240

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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