\(\int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx\) [1192]

Optimal result

Integrand size = 29, antiderivative size = 487 \[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {\left (3 b^4 (5+n)+2 a^4 \left (6+5 n+n^2\right )-2 a^2 b^2 \left (40+13 n+n^2\right )\right ) \cos (c+d x) \sin ^{1+n}(c+d x)}{b^2 d (2+n) (4+n) (5+n) (6+n)}+\frac {3 \left (b^2 (1+n)+a^2 (6+n)\right ) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right ) \sin ^{1+n}(c+d x)}{d (1+n) (2+n) (4+n) (6+n) \sqrt {\cos ^2(c+d x)}}-\frac {2 a \left (a^2 \left (6+5 n+n^2\right )-b^2 \left (39+13 n+n^2\right )\right ) \cos (c+d x) \sin ^{2+n}(c+d x)}{b d (3+n) (4+n) (5+n) (6+n)}+\frac {6 a b \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right ) \sin ^{2+n}(c+d x)}{d (2+n) (3+n) (5+n) \sqrt {\cos ^2(c+d x)}}-\frac {\left (a^2 (2+n) (3+n)-b^2 (5+n) (7+n)\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^2}{b^2 d (4+n) (5+n) (6+n)}+\frac {a (3+n) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (5+n) (6+n)}-\frac {\cos (c+d x) \sin ^{2+n}(c+d x) (a+b \sin (c+d x))^3}{b d (6+n)} \] Output:

-(3*b^4*(5+n)+2*a^4*(n^2+5*n+6)-2*a^2*b^2*(n^2+13*n+40))*cos(d*x+c)*sin(d* 
x+c)^(1+n)/b^2/d/(2+n)/(4+n)/(5+n)/(6+n)+3*(b^2*(1+n)+a^2*(6+n))*cos(d*x+c 
)*hypergeom([1/2, 1/2+1/2*n],[3/2+1/2*n],sin(d*x+c)^2)*sin(d*x+c)^(1+n)/d/ 
(1+n)/(2+n)/(4+n)/(6+n)/(cos(d*x+c)^2)^(1/2)-2*a*(a^2*(n^2+5*n+6)-b^2*(n^2 
+13*n+39))*cos(d*x+c)*sin(d*x+c)^(2+n)/b/d/(3+n)/(4+n)/(5+n)/(6+n)+6*a*b*c 
os(d*x+c)*hypergeom([1/2, 1+1/2*n],[2+1/2*n],sin(d*x+c)^2)*sin(d*x+c)^(2+n 
)/d/(2+n)/(3+n)/(5+n)/(cos(d*x+c)^2)^(1/2)-(a^2*(2+n)*(3+n)-b^2*(5+n)*(7+n 
))*cos(d*x+c)*sin(d*x+c)^(1+n)*(a+b*sin(d*x+c))^2/b^2/d/(4+n)/(5+n)/(6+n)+ 
a*(3+n)*cos(d*x+c)*sin(d*x+c)^(1+n)*(a+b*sin(d*x+c))^3/b^2/d/(5+n)/(6+n)-c 
os(d*x+c)*sin(d*x+c)^(2+n)*(a+b*sin(d*x+c))^3/b/d/(6+n)
 

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.34 \[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\sqrt {\cos ^2(c+d x)} \sec (c+d x) \sin ^{1+n}(c+d x) \left (a^2 \left (6+5 n+n^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right )+b (1+n) \sin (c+d x) \left (2 a (3+n) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right )+b (2+n) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3+n}{2},\frac {5+n}{2},\sin ^2(c+d x)\right ) \sin (c+d x)\right )\right )}{d (1+n) (2+n) (3+n)} \] Input:

Integrate[Cos[c + d*x]^4*Sin[c + d*x]^n*(a + b*Sin[c + d*x])^2,x]
 

Output:

(Sqrt[Cos[c + d*x]^2]*Sec[c + d*x]*Sin[c + d*x]^(1 + n)*(a^2*(6 + 5*n + n^ 
2)*Hypergeometric2F1[-3/2, (1 + n)/2, (3 + n)/2, Sin[c + d*x]^2] + b*(1 + 
n)*Sin[c + d*x]*(2*a*(3 + n)*Hypergeometric2F1[-3/2, (2 + n)/2, (4 + n)/2, 
 Sin[c + d*x]^2] + b*(2 + n)*Hypergeometric2F1[-3/2, (3 + n)/2, (5 + n)/2, 
 Sin[c + d*x]^2]*Sin[c + d*x])))/(d*(1 + n)*(2 + n)*(3 + n))
 

Rubi [A] (verified)

Time = 2.14 (sec) , antiderivative size = 466, normalized size of antiderivative = 0.96, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.448, Rules used = {3042, 3374, 3042, 3528, 3042, 3512, 3042, 3502, 27, 3042, 3227, 3042, 3122}

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.

\(\displaystyle \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (c+d x)^4 \sin (c+d x)^n (a+b \sin (c+d x))^2dx\)

\(\Big \downarrow \) 3374

\(\displaystyle -\frac {\int \sin ^n(c+d x) (a+b \sin (c+d x))^2 \left ((n+1) (n+3) a^2+2 b \sin (c+d x) a-\left (a^2 (n+2) (n+3)-b^2 (n+5) (n+7)\right ) \sin ^2(c+d x)-b^2 (n+5) (n+6)\right )dx}{b^2 (n+5) (n+6)}+\frac {a (n+3) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (n+5) (n+6)}-\frac {\cos (c+d x) \sin ^{n+2}(c+d x) (a+b \sin (c+d x))^3}{b d (n+6)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \sin (c+d x)^n (a+b \sin (c+d x))^2 \left ((n+1) (n+3) a^2+2 b \sin (c+d x) a-\left (a^2 (n+2) (n+3)-b^2 (n+5) (n+7)\right ) \sin (c+d x)^2-b^2 (n+5) (n+6)\right )dx}{b^2 (n+5) (n+6)}+\frac {a (n+3) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (n+5) (n+6)}-\frac {\cos (c+d x) \sin ^{n+2}(c+d x) (a+b \sin (c+d x))^3}{b d (n+6)}\)

\(\Big \downarrow \) 3528

\(\displaystyle -\frac {\frac {\int \sin ^n(c+d x) (a+b \sin (c+d x)) \left (-2 a \left (a^2 \left (n^2+5 n+6\right )-b^2 \left (n^2+13 n+39\right )\right ) \sin ^2(c+d x)+b \left (2 a^2-3 b^2 (n+5)\right ) \sin (c+d x)+a \left (2 a^2 \left (n^2+4 n+3\right )-b^2 \left (2 n^2+27 n+85\right )\right )\right )dx}{n+4}+\frac {\left (a^2 (n+2) (n+3)-b^2 (n+5) (n+7)\right ) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^2}{d (n+4)}}{b^2 (n+5) (n+6)}+\frac {a (n+3) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (n+5) (n+6)}-\frac {\cos (c+d x) \sin ^{n+2}(c+d x) (a+b \sin (c+d x))^3}{b d (n+6)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\int \sin (c+d x)^n (a+b \sin (c+d x)) \left (-2 a \left (a^2 \left (n^2+5 n+6\right )-b^2 \left (n^2+13 n+39\right )\right ) \sin (c+d x)^2+b \left (2 a^2-3 b^2 (n+5)\right ) \sin (c+d x)+a \left (2 a^2 \left (n^2+4 n+3\right )-b^2 \left (2 n^2+27 n+85\right )\right )\right )dx}{n+4}+\frac {\left (a^2 (n+2) (n+3)-b^2 (n+5) (n+7)\right ) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^2}{d (n+4)}}{b^2 (n+5) (n+6)}+\frac {a (n+3) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (n+5) (n+6)}-\frac {\cos (c+d x) \sin ^{n+2}(c+d x) (a+b \sin (c+d x))^3}{b d (n+6)}\)

\(\Big \downarrow \) 3512

\(\displaystyle -\frac {\frac {\frac {\int \sin ^n(c+d x) \left (-6 a (n+4) (n+6) \sin (c+d x) b^3-(n+3) \left (2 \left (n^2+5 n+6\right ) a^4-2 b^2 \left (n^2+13 n+40\right ) a^2+3 b^4 (n+5)\right ) \sin ^2(c+d x)+a^2 (n+3) \left (2 a^2 \left (n^2+4 n+3\right )-b^2 \left (2 n^2+27 n+85\right )\right )\right )dx}{n+3}+\frac {2 a b \left (a^2 \left (n^2+5 n+6\right )-b^2 \left (n^2+13 n+39\right )\right ) \cos (c+d x) \sin ^{n+2}(c+d x)}{d (n+3)}}{n+4}+\frac {\left (a^2 (n+2) (n+3)-b^2 (n+5) (n+7)\right ) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^2}{d (n+4)}}{b^2 (n+5) (n+6)}+\frac {a (n+3) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (n+5) (n+6)}-\frac {\cos (c+d x) \sin ^{n+2}(c+d x) (a+b \sin (c+d x))^3}{b d (n+6)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\frac {\int \sin (c+d x)^n \left (-6 a (n+4) (n+6) \sin (c+d x) b^3-(n+3) \left (2 \left (n^2+5 n+6\right ) a^4-2 b^2 \left (n^2+13 n+40\right ) a^2+3 b^4 (n+5)\right ) \sin (c+d x)^2+a^2 (n+3) \left (2 a^2 \left (n^2+4 n+3\right )-b^2 \left (2 n^2+27 n+85\right )\right )\right )dx}{n+3}+\frac {2 a b \left (a^2 \left (n^2+5 n+6\right )-b^2 \left (n^2+13 n+39\right )\right ) \cos (c+d x) \sin ^{n+2}(c+d x)}{d (n+3)}}{n+4}+\frac {\left (a^2 (n+2) (n+3)-b^2 (n+5) (n+7)\right ) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^2}{d (n+4)}}{b^2 (n+5) (n+6)}+\frac {a (n+3) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (n+5) (n+6)}-\frac {\cos (c+d x) \sin ^{n+2}(c+d x) (a+b \sin (c+d x))^3}{b d (n+6)}\)

\(\Big \downarrow \) 3502

\(\displaystyle -\frac {\frac {\frac {\frac {\int -3 \sin ^n(c+d x) \left (2 a (n+2) (n+4) (n+6) \sin (c+d x) b^3+\left (n^2+8 n+15\right ) \left ((n+6) a^2+b^2 (n+1)\right ) b^2\right )dx}{n+2}+\frac {(n+3) \left (2 a^4 \left (n^2+5 n+6\right )-2 a^2 b^2 \left (n^2+13 n+40\right )+3 b^4 (n+5)\right ) \cos (c+d x) \sin ^{n+1}(c+d x)}{d (n+2)}}{n+3}+\frac {2 a b \left (a^2 \left (n^2+5 n+6\right )-b^2 \left (n^2+13 n+39\right )\right ) \cos (c+d x) \sin ^{n+2}(c+d x)}{d (n+3)}}{n+4}+\frac {\left (a^2 (n+2) (n+3)-b^2 (n+5) (n+7)\right ) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^2}{d (n+4)}}{b^2 (n+5) (n+6)}+\frac {a (n+3) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (n+5) (n+6)}-\frac {\cos (c+d x) \sin ^{n+2}(c+d x) (a+b \sin (c+d x))^3}{b d (n+6)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\frac {\frac {(n+3) \left (2 a^4 \left (n^2+5 n+6\right )-2 a^2 b^2 \left (n^2+13 n+40\right )+3 b^4 (n+5)\right ) \cos (c+d x) \sin ^{n+1}(c+d x)}{d (n+2)}-\frac {3 \int \sin ^n(c+d x) \left (2 a (n+2) (n+4) (n+6) \sin (c+d x) b^3+\left (n^2+8 n+15\right ) \left ((n+6) a^2+b^2 (n+1)\right ) b^2\right )dx}{n+2}}{n+3}+\frac {2 a b \left (a^2 \left (n^2+5 n+6\right )-b^2 \left (n^2+13 n+39\right )\right ) \cos (c+d x) \sin ^{n+2}(c+d x)}{d (n+3)}}{n+4}+\frac {\left (a^2 (n+2) (n+3)-b^2 (n+5) (n+7)\right ) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^2}{d (n+4)}}{b^2 (n+5) (n+6)}+\frac {a (n+3) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (n+5) (n+6)}-\frac {\cos (c+d x) \sin ^{n+2}(c+d x) (a+b \sin (c+d x))^3}{b d (n+6)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\frac {\frac {(n+3) \left (2 a^4 \left (n^2+5 n+6\right )-2 a^2 b^2 \left (n^2+13 n+40\right )+3 b^4 (n+5)\right ) \cos (c+d x) \sin ^{n+1}(c+d x)}{d (n+2)}-\frac {3 \int \sin (c+d x)^n \left (2 a (n+2) (n+4) (n+6) \sin (c+d x) b^3+\left (n^2+8 n+15\right ) \left ((n+6) a^2+b^2 (n+1)\right ) b^2\right )dx}{n+2}}{n+3}+\frac {2 a b \left (a^2 \left (n^2+5 n+6\right )-b^2 \left (n^2+13 n+39\right )\right ) \cos (c+d x) \sin ^{n+2}(c+d x)}{d (n+3)}}{n+4}+\frac {\left (a^2 (n+2) (n+3)-b^2 (n+5) (n+7)\right ) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^2}{d (n+4)}}{b^2 (n+5) (n+6)}+\frac {a (n+3) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (n+5) (n+6)}-\frac {\cos (c+d x) \sin ^{n+2}(c+d x) (a+b \sin (c+d x))^3}{b d (n+6)}\)

\(\Big \downarrow \) 3227

\(\displaystyle -\frac {\frac {\frac {\frac {(n+3) \left (2 a^4 \left (n^2+5 n+6\right )-2 a^2 b^2 \left (n^2+13 n+40\right )+3 b^4 (n+5)\right ) \cos (c+d x) \sin ^{n+1}(c+d x)}{d (n+2)}-\frac {3 \left (b^2 \left (n^2+8 n+15\right ) \left (a^2 (n+6)+b^2 (n+1)\right ) \int \sin ^n(c+d x)dx+2 a b^3 (n+2) (n+4) (n+6) \int \sin ^{n+1}(c+d x)dx\right )}{n+2}}{n+3}+\frac {2 a b \left (a^2 \left (n^2+5 n+6\right )-b^2 \left (n^2+13 n+39\right )\right ) \cos (c+d x) \sin ^{n+2}(c+d x)}{d (n+3)}}{n+4}+\frac {\left (a^2 (n+2) (n+3)-b^2 (n+5) (n+7)\right ) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^2}{d (n+4)}}{b^2 (n+5) (n+6)}+\frac {a (n+3) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (n+5) (n+6)}-\frac {\cos (c+d x) \sin ^{n+2}(c+d x) (a+b \sin (c+d x))^3}{b d (n+6)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\frac {\frac {(n+3) \left (2 a^4 \left (n^2+5 n+6\right )-2 a^2 b^2 \left (n^2+13 n+40\right )+3 b^4 (n+5)\right ) \cos (c+d x) \sin ^{n+1}(c+d x)}{d (n+2)}-\frac {3 \left (b^2 \left (n^2+8 n+15\right ) \left (a^2 (n+6)+b^2 (n+1)\right ) \int \sin (c+d x)^ndx+2 a b^3 (n+2) (n+4) (n+6) \int \sin (c+d x)^{n+1}dx\right )}{n+2}}{n+3}+\frac {2 a b \left (a^2 \left (n^2+5 n+6\right )-b^2 \left (n^2+13 n+39\right )\right ) \cos (c+d x) \sin ^{n+2}(c+d x)}{d (n+3)}}{n+4}+\frac {\left (a^2 (n+2) (n+3)-b^2 (n+5) (n+7)\right ) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^2}{d (n+4)}}{b^2 (n+5) (n+6)}+\frac {a (n+3) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (n+5) (n+6)}-\frac {\cos (c+d x) \sin ^{n+2}(c+d x) (a+b \sin (c+d x))^3}{b d (n+6)}\)

\(\Big \downarrow \) 3122

\(\displaystyle -\frac {\frac {\left (a^2 (n+2) (n+3)-b^2 (n+5) (n+7)\right ) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^2}{d (n+4)}+\frac {\frac {2 a b \left (a^2 \left (n^2+5 n+6\right )-b^2 \left (n^2+13 n+39\right )\right ) \cos (c+d x) \sin ^{n+2}(c+d x)}{d (n+3)}+\frac {\frac {(n+3) \left (2 a^4 \left (n^2+5 n+6\right )-2 a^2 b^2 \left (n^2+13 n+40\right )+3 b^4 (n+5)\right ) \cos (c+d x) \sin ^{n+1}(c+d x)}{d (n+2)}-\frac {3 \left (\frac {b^2 \left (n^2+8 n+15\right ) \left (a^2 (n+6)+b^2 (n+1)\right ) \cos (c+d x) \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+1}{2},\frac {n+3}{2},\sin ^2(c+d x)\right )}{d (n+1) \sqrt {\cos ^2(c+d x)}}+\frac {2 a b^3 (n+4) (n+6) \cos (c+d x) \sin ^{n+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+2}{2},\frac {n+4}{2},\sin ^2(c+d x)\right )}{d \sqrt {\cos ^2(c+d x)}}\right )}{n+2}}{n+3}}{n+4}}{b^2 (n+5) (n+6)}+\frac {a (n+3) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (n+5) (n+6)}-\frac {\cos (c+d x) \sin ^{n+2}(c+d x) (a+b \sin (c+d x))^3}{b d (n+6)}\)

Input:

Int[Cos[c + d*x]^4*Sin[c + d*x]^n*(a + b*Sin[c + d*x])^2,x]
 

Output:

(a*(3 + n)*Cos[c + d*x]*Sin[c + d*x]^(1 + n)*(a + b*Sin[c + d*x])^3)/(b^2* 
d*(5 + n)*(6 + n)) - (Cos[c + d*x]*Sin[c + d*x]^(2 + n)*(a + b*Sin[c + d*x 
])^3)/(b*d*(6 + n)) - (((a^2*(2 + n)*(3 + n) - b^2*(5 + n)*(7 + n))*Cos[c 
+ d*x]*Sin[c + d*x]^(1 + n)*(a + b*Sin[c + d*x])^2)/(d*(4 + n)) + ((2*a*b* 
(a^2*(6 + 5*n + n^2) - b^2*(39 + 13*n + n^2))*Cos[c + d*x]*Sin[c + d*x]^(2 
 + n))/(d*(3 + n)) + (((3 + n)*(3*b^4*(5 + n) + 2*a^4*(6 + 5*n + n^2) - 2* 
a^2*b^2*(40 + 13*n + n^2))*Cos[c + d*x]*Sin[c + d*x]^(1 + n))/(d*(2 + n)) 
- (3*((b^2*(15 + 8*n + n^2)*(b^2*(1 + n) + a^2*(6 + n))*Cos[c + d*x]*Hyper 
geometric2F1[1/2, (1 + n)/2, (3 + n)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(1 + 
n))/(d*(1 + n)*Sqrt[Cos[c + d*x]^2]) + (2*a*b^3*(4 + n)*(6 + n)*Cos[c + d* 
x]*Hypergeometric2F1[1/2, (2 + n)/2, (4 + n)/2, Sin[c + d*x]^2]*Sin[c + d* 
x]^(2 + n))/(d*Sqrt[Cos[c + d*x]^2])))/(2 + n))/(3 + n))/(4 + n))/(b^2*(5 
+ n)*(6 + n))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

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

rule 3122

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 
F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] 
 &&  !IntegerQ[2*n]
 

rule 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3374

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + 
(b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[a*(n + 3)*Cos[e + f* 
x]*(d*Sin[e + f*x])^(n + 1)*((a + b*Sin[e + f*x])^(m + 1)/(b^2*d*f*(m + n + 
 3)*(m + n + 4))), x] + (-Simp[Cos[e + f*x]*(d*Sin[e + f*x])^(n + 2)*((a + 
b*Sin[e + f*x])^(m + 1)/(b*d^2*f*(m + n + 4))), x] - Simp[1/(b^2*(m + n + 3 
)*(m + n + 4))   Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^m*Simp[a^2*(n 
+ 1)*(n + 3) - b^2*(m + n + 3)*(m + n + 4) + a*b*m*Sin[e + f*x] - (a^2*(n + 
 2)*(n + 3) - b^2*(m + n + 3)*(m + n + 5))*Sin[e + f*x]^2, x], x], x]) /; F 
reeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || Integ 
ersQ[2*m, 2*n]) &&  !m < -1 &&  !LtQ[n, -1] && NeQ[m + n + 3, 0] && NeQ[m + 
 n + 4, 0]
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[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[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

rule 3512

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si 
n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3))   Int[(a + b*Si 
n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + 
A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 
0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]
 

rule 3528

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ 
.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x 
])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + 
n + 2))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* 
d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a 
*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + 
 n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} 
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ 
m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
 

Maple [F]

\[\int \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )^{n} \left (a +b \sin \left (d x +c \right )\right )^{2}d x\]

Input:

int(cos(d*x+c)^4*sin(d*x+c)^n*(a+b*sin(d*x+c))^2,x)
 

Output:

int(cos(d*x+c)^4*sin(d*x+c)^n*(a+b*sin(d*x+c))^2,x)
 

Fricas [F]

\[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{2} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4} \,d x } \] Input:

integrate(cos(d*x+c)^4*sin(d*x+c)^n*(a+b*sin(d*x+c))^2,x, algorithm="frica 
s")
 

Output:

integral(-(b^2*cos(d*x + c)^6 - 2*a*b*cos(d*x + c)^4*sin(d*x + c) - (a^2 + 
 b^2)*cos(d*x + c)^4)*sin(d*x + c)^n, x)
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**4*sin(d*x+c)**n*(a+b*sin(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{2} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4} \,d x } \] Input:

integrate(cos(d*x+c)^4*sin(d*x+c)^n*(a+b*sin(d*x+c))^2,x, algorithm="maxim 
a")
 

Output:

integrate((b*sin(d*x + c) + a)^2*sin(d*x + c)^n*cos(d*x + c)^4, x)
                                                                                    
                                                                                    
 

Giac [F]

\[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{2} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4} \,d x } \] Input:

integrate(cos(d*x+c)^4*sin(d*x+c)^n*(a+b*sin(d*x+c))^2,x, algorithm="giac" 
)
 

Output:

integrate((b*sin(d*x + c) + a)^2*sin(d*x + c)^n*cos(d*x + c)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=\int {\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^n\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^2 \,d x \] Input:

int(cos(c + d*x)^4*sin(c + d*x)^n*(a + b*sin(c + d*x))^2,x)
 

Output:

int(cos(c + d*x)^4*sin(c + d*x)^n*(a + b*sin(c + d*x))^2, x)
 

Reduce [F]

\[ \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^2 \, dx=\left (\int \sin \left (d x +c \right )^{n} \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )^{2}d x \right ) b^{2}+2 \left (\int \sin \left (d x +c \right )^{n} \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )d x \right ) a b +\left (\int \sin \left (d x +c \right )^{n} \cos \left (d x +c \right )^{4}d x \right ) a^{2} \] Input:

int(cos(d*x+c)^4*sin(d*x+c)^n*(a+b*sin(d*x+c))^2,x)
 

Output:

int(sin(c + d*x)**n*cos(c + d*x)**4*sin(c + d*x)**2,x)*b**2 + 2*int(sin(c 
+ d*x)**n*cos(c + d*x)**4*sin(c + d*x),x)*a*b + int(sin(c + d*x)**n*cos(c 
+ d*x)**4,x)*a**2