\(\int \cos ^4(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2 \, dx\) [1101]

Optimal result

Integrand size = 29, antiderivative size = 301 \[ \int \cos ^4(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {3 a b x}{64}-\frac {\left (9 a^2+4 b^2\right ) \cos (c+d x)}{105 d}+\frac {\left (9 a^2+4 b^2\right ) \cos ^3(c+d x)}{315 d}-\frac {3 a b \cos (c+d x) \sin (c+d x)}{64 d}-\frac {a b \cos (c+d x) \sin ^3(c+d x)}{32 d}-\frac {\left (15 a^4-44 a^2 b^2+6 b^4\right ) \cos (c+d x) \sin ^4(c+d x)}{630 b^2 d}-\frac {a \left (10 a^2-29 b^2\right ) \cos (c+d x) \sin ^5(c+d x)}{504 b d}-\frac {5 \left (3 a^2-8 b^2\right ) \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2}{252 b^2 d}+\frac {a \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac {\cos (c+d x) \sin ^5(c+d x) (a+b \sin (c+d x))^3}{9 b d} \] Output:

3/64*a*b*x-1/105*(9*a^2+4*b^2)*cos(d*x+c)/d+1/315*(9*a^2+4*b^2)*cos(d*x+c) 
^3/d-3/64*a*b*cos(d*x+c)*sin(d*x+c)/d-1/32*a*b*cos(d*x+c)*sin(d*x+c)^3/d-1 
/630*(15*a^4-44*a^2*b^2+6*b^4)*cos(d*x+c)*sin(d*x+c)^4/b^2/d-1/504*a*(10*a 
^2-29*b^2)*cos(d*x+c)*sin(d*x+c)^5/b/d-5/252*(3*a^2-8*b^2)*cos(d*x+c)*sin( 
d*x+c)^4*(a+b*sin(d*x+c))^2/b^2/d+1/12*a*cos(d*x+c)*sin(d*x+c)^4*(a+b*sin( 
d*x+c))^3/b^2/d-1/9*cos(d*x+c)*sin(d*x+c)^5*(a+b*sin(d*x+c))^3/b/d
 

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.48 \[ \int \cos ^4(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {7560 a b c+7560 a b d x-3780 \left (2 a^2+b^2\right ) \cos (c+d x)-840 \left (3 a^2+b^2\right ) \cos (3 (c+d x))+504 a^2 \cos (5 (c+d x))+504 b^2 \cos (5 (c+d x))+360 a^2 \cos (7 (c+d x))+90 b^2 \cos (7 (c+d x))-70 b^2 \cos (9 (c+d x))-2520 a b \sin (4 (c+d x))+315 a b \sin (8 (c+d x))}{161280 d} \] Input:

Integrate[Cos[c + d*x]^4*Sin[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]
 

Output:

(7560*a*b*c + 7560*a*b*d*x - 3780*(2*a^2 + b^2)*Cos[c + d*x] - 840*(3*a^2 
+ b^2)*Cos[3*(c + d*x)] + 504*a^2*Cos[5*(c + d*x)] + 504*b^2*Cos[5*(c + d* 
x)] + 360*a^2*Cos[7*(c + d*x)] + 90*b^2*Cos[7*(c + d*x)] - 70*b^2*Cos[9*(c 
 + d*x)] - 2520*a*b*Sin[4*(c + d*x)] + 315*a*b*Sin[8*(c + d*x)])/(161280*d 
)
 

Rubi [A] (verified)

Time = 1.85 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.03, number of steps used = 21, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.690, Rules used = {3042, 3374, 27, 3042, 3528, 3042, 3512, 27, 3042, 3502, 27, 3042, 3227, 3042, 3113, 2009, 3115, 3042, 3115, 24}

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.

\(\displaystyle \int \sin ^3(c+d x) \cos ^4(c+d x) (a+b \sin (c+d x))^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (c+d x)^3 \cos (c+d x)^4 (a+b \sin (c+d x))^2dx\)

\(\Big \downarrow \) 3374

\(\displaystyle -\frac {\int 2 \sin ^3(c+d x) (a+b \sin (c+d x))^2 \left (-5 \left (3 a^2-8 b^2\right ) \sin ^2(c+d x)+a b \sin (c+d x)+12 \left (a^2-3 b^2\right )\right )dx}{72 b^2}+\frac {a \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \sin ^3(c+d x) (a+b \sin (c+d x))^2 \left (-5 \left (3 a^2-8 b^2\right ) \sin ^2(c+d x)+a b \sin (c+d x)+12 \left (a^2-3 b^2\right )\right )dx}{36 b^2}+\frac {a \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \sin (c+d x)^3 (a+b \sin (c+d x))^2 \left (-5 \left (3 a^2-8 b^2\right ) \sin (c+d x)^2+a b \sin (c+d x)+12 \left (a^2-3 b^2\right )\right )dx}{36 b^2}+\frac {a \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{9 b d}\)

\(\Big \downarrow \) 3528

\(\displaystyle -\frac {\frac {1}{7} \int \sin ^3(c+d x) (a+b \sin (c+d x)) \left (-3 a \left (10 a^2-29 b^2\right ) \sin ^2(c+d x)+b \left (a^2-12 b^2\right ) \sin (c+d x)+4 a \left (6 a^2-23 b^2\right )\right )dx+\frac {5 \left (3 a^2-8 b^2\right ) \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}}{36 b^2}+\frac {a \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {1}{7} \int \sin (c+d x)^3 (a+b \sin (c+d x)) \left (-3 a \left (10 a^2-29 b^2\right ) \sin (c+d x)^2+b \left (a^2-12 b^2\right ) \sin (c+d x)+4 a \left (6 a^2-23 b^2\right )\right )dx+\frac {5 \left (3 a^2-8 b^2\right ) \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}}{36 b^2}+\frac {a \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{9 b d}\)

\(\Big \downarrow \) 3512

\(\displaystyle -\frac {\frac {1}{7} \left (\frac {1}{6} \int 3 \sin ^3(c+d x) \left (-63 a \sin (c+d x) b^3-4 \left (15 a^4-44 b^2 a^2+6 b^4\right ) \sin ^2(c+d x)+8 a^2 \left (6 a^2-23 b^2\right )\right )dx+\frac {a b \left (10 a^2-29 b^2\right ) \sin ^5(c+d x) \cos (c+d x)}{2 d}\right )+\frac {5 \left (3 a^2-8 b^2\right ) \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}}{36 b^2}+\frac {a \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {1}{7} \left (\frac {1}{2} \int \sin ^3(c+d x) \left (-63 a \sin (c+d x) b^3-4 \left (15 a^4-44 b^2 a^2+6 b^4\right ) \sin ^2(c+d x)+8 a^2 \left (6 a^2-23 b^2\right )\right )dx+\frac {a b \left (10 a^2-29 b^2\right ) \sin ^5(c+d x) \cos (c+d x)}{2 d}\right )+\frac {5 \left (3 a^2-8 b^2\right ) \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}}{36 b^2}+\frac {a \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {1}{7} \left (\frac {1}{2} \int \sin (c+d x)^3 \left (-63 a \sin (c+d x) b^3-4 \left (15 a^4-44 b^2 a^2+6 b^4\right ) \sin (c+d x)^2+8 a^2 \left (6 a^2-23 b^2\right )\right )dx+\frac {a b \left (10 a^2-29 b^2\right ) \sin ^5(c+d x) \cos (c+d x)}{2 d}\right )+\frac {5 \left (3 a^2-8 b^2\right ) \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}}{36 b^2}+\frac {a \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{9 b d}\)

\(\Big \downarrow \) 3502

\(\displaystyle -\frac {\frac {1}{7} \left (\frac {1}{2} \left (\frac {1}{5} \int -3 \sin ^3(c+d x) \left (105 a \sin (c+d x) b^3+8 \left (9 a^2+4 b^2\right ) b^2\right )dx+\frac {4 \left (15 a^4-44 a^2 b^2+6 b^4\right ) \sin ^4(c+d x) \cos (c+d x)}{5 d}\right )+\frac {a b \left (10 a^2-29 b^2\right ) \sin ^5(c+d x) \cos (c+d x)}{2 d}\right )+\frac {5 \left (3 a^2-8 b^2\right ) \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}}{36 b^2}+\frac {a \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {1}{7} \left (\frac {1}{2} \left (\frac {4 \left (15 a^4-44 a^2 b^2+6 b^4\right ) \sin ^4(c+d x) \cos (c+d x)}{5 d}-\frac {3}{5} \int \sin ^3(c+d x) \left (105 a \sin (c+d x) b^3+8 \left (9 a^2+4 b^2\right ) b^2\right )dx\right )+\frac {a b \left (10 a^2-29 b^2\right ) \sin ^5(c+d x) \cos (c+d x)}{2 d}\right )+\frac {5 \left (3 a^2-8 b^2\right ) \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}}{36 b^2}+\frac {a \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {1}{7} \left (\frac {1}{2} \left (\frac {4 \left (15 a^4-44 a^2 b^2+6 b^4\right ) \sin ^4(c+d x) \cos (c+d x)}{5 d}-\frac {3}{5} \int \sin (c+d x)^3 \left (105 a \sin (c+d x) b^3+8 \left (9 a^2+4 b^2\right ) b^2\right )dx\right )+\frac {a b \left (10 a^2-29 b^2\right ) \sin ^5(c+d x) \cos (c+d x)}{2 d}\right )+\frac {5 \left (3 a^2-8 b^2\right ) \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}}{36 b^2}+\frac {a \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{9 b d}\)

\(\Big \downarrow \) 3227

\(\displaystyle -\frac {\frac {1}{7} \left (\frac {1}{2} \left (\frac {4 \left (15 a^4-44 a^2 b^2+6 b^4\right ) \sin ^4(c+d x) \cos (c+d x)}{5 d}-\frac {3}{5} \left (8 b^2 \left (9 a^2+4 b^2\right ) \int \sin ^3(c+d x)dx+105 a b^3 \int \sin ^4(c+d x)dx\right )\right )+\frac {a b \left (10 a^2-29 b^2\right ) \sin ^5(c+d x) \cos (c+d x)}{2 d}\right )+\frac {5 \left (3 a^2-8 b^2\right ) \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}}{36 b^2}+\frac {a \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {1}{7} \left (\frac {1}{2} \left (\frac {4 \left (15 a^4-44 a^2 b^2+6 b^4\right ) \sin ^4(c+d x) \cos (c+d x)}{5 d}-\frac {3}{5} \left (8 b^2 \left (9 a^2+4 b^2\right ) \int \sin (c+d x)^3dx+105 a b^3 \int \sin (c+d x)^4dx\right )\right )+\frac {a b \left (10 a^2-29 b^2\right ) \sin ^5(c+d x) \cos (c+d x)}{2 d}\right )+\frac {5 \left (3 a^2-8 b^2\right ) \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}}{36 b^2}+\frac {a \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{9 b d}\)

\(\Big \downarrow \) 3113

\(\displaystyle -\frac {\frac {1}{7} \left (\frac {1}{2} \left (\frac {4 \left (15 a^4-44 a^2 b^2+6 b^4\right ) \sin ^4(c+d x) \cos (c+d x)}{5 d}-\frac {3}{5} \left (105 a b^3 \int \sin (c+d x)^4dx-\frac {8 b^2 \left (9 a^2+4 b^2\right ) \int \left (1-\cos ^2(c+d x)\right )d\cos (c+d x)}{d}\right )\right )+\frac {a b \left (10 a^2-29 b^2\right ) \sin ^5(c+d x) \cos (c+d x)}{2 d}\right )+\frac {5 \left (3 a^2-8 b^2\right ) \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}}{36 b^2}+\frac {a \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{9 b d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\frac {1}{7} \left (\frac {1}{2} \left (\frac {4 \left (15 a^4-44 a^2 b^2+6 b^4\right ) \sin ^4(c+d x) \cos (c+d x)}{5 d}-\frac {3}{5} \left (105 a b^3 \int \sin (c+d x)^4dx-\frac {8 b^2 \left (9 a^2+4 b^2\right ) \left (\cos (c+d x)-\frac {1}{3} \cos ^3(c+d x)\right )}{d}\right )\right )+\frac {a b \left (10 a^2-29 b^2\right ) \sin ^5(c+d x) \cos (c+d x)}{2 d}\right )+\frac {5 \left (3 a^2-8 b^2\right ) \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}}{36 b^2}+\frac {a \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{9 b d}\)

\(\Big \downarrow \) 3115

\(\displaystyle -\frac {\frac {1}{7} \left (\frac {1}{2} \left (\frac {4 \left (15 a^4-44 a^2 b^2+6 b^4\right ) \sin ^4(c+d x) \cos (c+d x)}{5 d}-\frac {3}{5} \left (105 a b^3 \left (\frac {3}{4} \int \sin ^2(c+d x)dx-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {8 b^2 \left (9 a^2+4 b^2\right ) \left (\cos (c+d x)-\frac {1}{3} \cos ^3(c+d x)\right )}{d}\right )\right )+\frac {a b \left (10 a^2-29 b^2\right ) \sin ^5(c+d x) \cos (c+d x)}{2 d}\right )+\frac {5 \left (3 a^2-8 b^2\right ) \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}}{36 b^2}+\frac {a \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {1}{7} \left (\frac {1}{2} \left (\frac {4 \left (15 a^4-44 a^2 b^2+6 b^4\right ) \sin ^4(c+d x) \cos (c+d x)}{5 d}-\frac {3}{5} \left (105 a b^3 \left (\frac {3}{4} \int \sin (c+d x)^2dx-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {8 b^2 \left (9 a^2+4 b^2\right ) \left (\cos (c+d x)-\frac {1}{3} \cos ^3(c+d x)\right )}{d}\right )\right )+\frac {a b \left (10 a^2-29 b^2\right ) \sin ^5(c+d x) \cos (c+d x)}{2 d}\right )+\frac {5 \left (3 a^2-8 b^2\right ) \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}}{36 b^2}+\frac {a \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{9 b d}\)

\(\Big \downarrow \) 3115

\(\displaystyle -\frac {\frac {1}{7} \left (\frac {1}{2} \left (\frac {4 \left (15 a^4-44 a^2 b^2+6 b^4\right ) \sin ^4(c+d x) \cos (c+d x)}{5 d}-\frac {3}{5} \left (105 a b^3 \left (\frac {3}{4} \left (\frac {\int 1dx}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {8 b^2 \left (9 a^2+4 b^2\right ) \left (\cos (c+d x)-\frac {1}{3} \cos ^3(c+d x)\right )}{d}\right )\right )+\frac {a b \left (10 a^2-29 b^2\right ) \sin ^5(c+d x) \cos (c+d x)}{2 d}\right )+\frac {5 \left (3 a^2-8 b^2\right ) \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}}{36 b^2}+\frac {a \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{9 b d}\)

\(\Big \downarrow \) 24

\(\displaystyle -\frac {\frac {5 \left (3 a^2-8 b^2\right ) \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac {1}{7} \left (\frac {a b \left (10 a^2-29 b^2\right ) \sin ^5(c+d x) \cos (c+d x)}{2 d}+\frac {1}{2} \left (\frac {4 \left (15 a^4-44 a^2 b^2+6 b^4\right ) \sin ^4(c+d x) \cos (c+d x)}{5 d}-\frac {3}{5} \left (105 a b^3 \left (\frac {3}{4} \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {8 b^2 \left (9 a^2+4 b^2\right ) \left (\cos (c+d x)-\frac {1}{3} \cos ^3(c+d x)\right )}{d}\right )\right )\right )}{36 b^2}+\frac {a \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{9 b d}\)

Input:

Int[Cos[c + d*x]^4*Sin[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]
 

Output:

(a*Cos[c + d*x]*Sin[c + d*x]^4*(a + b*Sin[c + d*x])^3)/(12*b^2*d) - (Cos[c 
 + d*x]*Sin[c + d*x]^5*(a + b*Sin[c + d*x])^3)/(9*b*d) - ((5*(3*a^2 - 8*b^ 
2)*Cos[c + d*x]*Sin[c + d*x]^4*(a + b*Sin[c + d*x])^2)/(7*d) + ((a*b*(10*a 
^2 - 29*b^2)*Cos[c + d*x]*Sin[c + d*x]^5)/(2*d) + ((4*(15*a^4 - 44*a^2*b^2 
 + 6*b^4)*Cos[c + d*x]*Sin[c + d*x]^4)/(5*d) - (3*((-8*b^2*(9*a^2 + 4*b^2) 
*(Cos[c + d*x] - Cos[c + d*x]^3/3))/d + 105*a*b^3*(-1/4*(Cos[c + d*x]*Sin[ 
c + d*x]^3)/d + (3*(x/2 - (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4)))/5)/2)/7 
)/(36*b^2)
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

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 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 3113

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] 
 && IGtQ[(n - 1)/2, 0]
 

rule 3115

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && 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 [A] (verified)

Time = 0.38 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.53

\[\frac {a^{2} \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{5}}{7}-\frac {2 \cos \left (d x +c \right )^{5}}{35}\right )+2 a b \left (-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{5}}{8}-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{5}}{16}+\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{64}+\frac {3 d x}{128}+\frac {3 c}{128}\right )+b^{2} \left (-\frac {\sin \left (d x +c \right )^{4} \cos \left (d x +c \right )^{5}}{9}-\frac {4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{5}}{63}-\frac {8 \cos \left (d x +c \right )^{5}}{315}\right )}{d}\]

Input:

int(cos(d*x+c)^4*sin(d*x+c)^3*(a+b*sin(d*x+c))^2,x)
 

Output:

1/d*(a^2*(-1/7*sin(d*x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5)+2*a*b*(-1/8*si 
n(d*x+c)^3*cos(d*x+c)^5-1/16*sin(d*x+c)*cos(d*x+c)^5+1/64*(cos(d*x+c)^3+3/ 
2*cos(d*x+c))*sin(d*x+c)+3/128*d*x+3/128*c)+b^2*(-1/9*sin(d*x+c)^4*cos(d*x 
+c)^5-4/63*sin(d*x+c)^2*cos(d*x+c)^5-8/315*cos(d*x+c)^5))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.39 \[ \int \cos ^4(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {2240 \, b^{2} \cos \left (d x + c\right )^{9} - 2880 \, {\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{7} + 4032 \, {\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{5} - 945 \, a b d x - 315 \, {\left (16 \, a b \cos \left (d x + c\right )^{7} - 24 \, a b \cos \left (d x + c\right )^{5} + 2 \, a b \cos \left (d x + c\right )^{3} + 3 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{20160 \, d} \] Input:

integrate(cos(d*x+c)^4*sin(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="frica 
s")
 

Output:

-1/20160*(2240*b^2*cos(d*x + c)^9 - 2880*(a^2 + 2*b^2)*cos(d*x + c)^7 + 40 
32*(a^2 + b^2)*cos(d*x + c)^5 - 945*a*b*d*x - 315*(16*a*b*cos(d*x + c)^7 - 
 24*a*b*cos(d*x + c)^5 + 2*a*b*cos(d*x + c)^3 + 3*a*b*cos(d*x + c))*sin(d* 
x + c))/d
 

Sympy [A] (verification not implemented)

Time = 0.95 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.11 \[ \int \cos ^4(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\begin {cases} - \frac {a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {2 a^{2} \cos ^{7}{\left (c + d x \right )}}{35 d} + \frac {3 a b x \sin ^{8}{\left (c + d x \right )}}{64} + \frac {3 a b x \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {9 a b x \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{32} + \frac {3 a b x \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{16} + \frac {3 a b x \cos ^{8}{\left (c + d x \right )}}{64} + \frac {3 a b \sin ^{7}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{64 d} + \frac {11 a b \sin ^{5}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{64 d} - \frac {11 a b \sin ^{3}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{64 d} - \frac {3 a b \sin {\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{64 d} - \frac {b^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {4 b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{35 d} - \frac {8 b^{2} \cos ^{9}{\left (c + d x \right )}}{315 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right )^{2} \sin ^{3}{\left (c \right )} \cos ^{4}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:

integrate(cos(d*x+c)**4*sin(d*x+c)**3*(a+b*sin(d*x+c))**2,x)
 

Output:

Piecewise((-a**2*sin(c + d*x)**2*cos(c + d*x)**5/(5*d) - 2*a**2*cos(c + d* 
x)**7/(35*d) + 3*a*b*x*sin(c + d*x)**8/64 + 3*a*b*x*sin(c + d*x)**6*cos(c 
+ d*x)**2/16 + 9*a*b*x*sin(c + d*x)**4*cos(c + d*x)**4/32 + 3*a*b*x*sin(c 
+ d*x)**2*cos(c + d*x)**6/16 + 3*a*b*x*cos(c + d*x)**8/64 + 3*a*b*sin(c + 
d*x)**7*cos(c + d*x)/(64*d) + 11*a*b*sin(c + d*x)**5*cos(c + d*x)**3/(64*d 
) - 11*a*b*sin(c + d*x)**3*cos(c + d*x)**5/(64*d) - 3*a*b*sin(c + d*x)*cos 
(c + d*x)**7/(64*d) - b**2*sin(c + d*x)**4*cos(c + d*x)**5/(5*d) - 4*b**2* 
sin(c + d*x)**2*cos(c + d*x)**7/(35*d) - 8*b**2*cos(c + d*x)**9/(315*d), N 
e(d, 0)), (x*(a + b*sin(c))**2*sin(c)**3*cos(c)**4, True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.33 \[ \int \cos ^4(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {4608 \, {\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} a^{2} + 315 \, {\left (24 \, d x + 24 \, c + \sin \left (8 \, d x + 8 \, c\right ) - 8 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a b - 512 \, {\left (35 \, \cos \left (d x + c\right )^{9} - 90 \, \cos \left (d x + c\right )^{7} + 63 \, \cos \left (d x + c\right )^{5}\right )} b^{2}}{161280 \, d} \] Input:

integrate(cos(d*x+c)^4*sin(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="maxim 
a")
 

Output:

1/161280*(4608*(5*cos(d*x + c)^7 - 7*cos(d*x + c)^5)*a^2 + 315*(24*d*x + 2 
4*c + sin(8*d*x + 8*c) - 8*sin(4*d*x + 4*c))*a*b - 512*(35*cos(d*x + c)^9 
- 90*cos(d*x + c)^7 + 63*cos(d*x + c)^5)*b^2)/d
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.47 \[ \int \cos ^4(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {3}{64} \, a b x - \frac {b^{2} \cos \left (9 \, d x + 9 \, c\right )}{2304 \, d} + \frac {a b \sin \left (8 \, d x + 8 \, c\right )}{512 \, d} - \frac {a b \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {{\left (4 \, a^{2} + b^{2}\right )} \cos \left (7 \, d x + 7 \, c\right )}{1792 \, d} + \frac {{\left (a^{2} + b^{2}\right )} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {{\left (3 \, a^{2} + b^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{192 \, d} - \frac {3 \, {\left (2 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )}{128 \, d} \] Input:

integrate(cos(d*x+c)^4*sin(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="giac" 
)
 

Output:

3/64*a*b*x - 1/2304*b^2*cos(9*d*x + 9*c)/d + 1/512*a*b*sin(8*d*x + 8*c)/d 
- 1/64*a*b*sin(4*d*x + 4*c)/d + 1/1792*(4*a^2 + b^2)*cos(7*d*x + 7*c)/d + 
1/320*(a^2 + b^2)*cos(5*d*x + 5*c)/d - 1/192*(3*a^2 + b^2)*cos(3*d*x + 3*c 
)/d - 3/128*(2*a^2 + b^2)*cos(d*x + c)/d
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 38.22 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.03 \[ \int \cos ^4(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {3\,a\,b\,x}{64}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (4\,a^2-16\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (4\,a^2+\frac {32\,b^2}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {28\,a^2}{5}-\frac {32\,b^2}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {36\,a^2}{35}+\frac {16\,b^2}{35}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {4\,a^2}{35}+\frac {64\,b^2}{35}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {52\,a^2}{5}+\frac {112\,b^2}{5}\right )+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+\frac {4\,a^2}{35}+\frac {16\,b^2}{315}+\frac {13\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{16}-\frac {155\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{16}+\frac {169\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{16}-\frac {169\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{16}+\frac {155\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{16}-\frac {13\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{16}-\frac {3\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}}{32}+\frac {3\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{32}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^9} \] Input:

int(cos(c + d*x)^4*sin(c + d*x)^3*(a + b*sin(c + d*x))^2,x)
 

Output:

(3*a*b*x)/64 - (tan(c/2 + (d*x)/2)^10*(4*a^2 - 16*b^2) + tan(c/2 + (d*x)/2 
)^12*(4*a^2 + (32*b^2)/3) + tan(c/2 + (d*x)/2)^6*((28*a^2)/5 - (32*b^2)/5) 
 + tan(c/2 + (d*x)/2)^2*((36*a^2)/35 + (16*b^2)/35) + tan(c/2 + (d*x)/2)^4 
*((4*a^2)/35 + (64*b^2)/35) + tan(c/2 + (d*x)/2)^8*((52*a^2)/5 + (112*b^2) 
/5) + 4*a^2*tan(c/2 + (d*x)/2)^14 + (4*a^2)/35 + (16*b^2)/315 + (13*a*b*ta 
n(c/2 + (d*x)/2)^3)/16 - (155*a*b*tan(c/2 + (d*x)/2)^5)/16 + (169*a*b*tan( 
c/2 + (d*x)/2)^7)/16 - (169*a*b*tan(c/2 + (d*x)/2)^11)/16 + (155*a*b*tan(c 
/2 + (d*x)/2)^13)/16 - (13*a*b*tan(c/2 + (d*x)/2)^15)/16 - (3*a*b*tan(c/2 
+ (d*x)/2)^17)/32 + (3*a*b*tan(c/2 + (d*x)/2))/32)/(d*(tan(c/2 + (d*x)/2)^ 
2 + 1)^9)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.82 \[ \int \cos ^4(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {-2240 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{8} b^{2}-5040 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{7} a b -2880 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6} a^{2}+3200 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6} b^{2}+7560 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} a b +4608 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a^{2}-192 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} b^{2}-630 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a b -576 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{2}-256 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{2}-945 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b -1152 \cos \left (d x +c \right ) a^{2}-512 \cos \left (d x +c \right ) b^{2}+1152 a^{2}+945 a b d x +512 b^{2}}{20160 d} \] Input:

int(cos(d*x+c)^4*sin(d*x+c)^3*(a+b*sin(d*x+c))^2,x)
 

Output:

( - 2240*cos(c + d*x)*sin(c + d*x)**8*b**2 - 5040*cos(c + d*x)*sin(c + d*x 
)**7*a*b - 2880*cos(c + d*x)*sin(c + d*x)**6*a**2 + 3200*cos(c + d*x)*sin( 
c + d*x)**6*b**2 + 7560*cos(c + d*x)*sin(c + d*x)**5*a*b + 4608*cos(c + d* 
x)*sin(c + d*x)**4*a**2 - 192*cos(c + d*x)*sin(c + d*x)**4*b**2 - 630*cos( 
c + d*x)*sin(c + d*x)**3*a*b - 576*cos(c + d*x)*sin(c + d*x)**2*a**2 - 256 
*cos(c + d*x)*sin(c + d*x)**2*b**2 - 945*cos(c + d*x)*sin(c + d*x)*a*b - 1 
152*cos(c + d*x)*a**2 - 512*cos(c + d*x)*b**2 + 1152*a**2 + 945*a*b*d*x + 
512*b**2)/(20160*d)