Integrand size = 21, antiderivative size = 99 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {a b \cos ^6(c+d x)}{3 d}+\frac {a^2 \sin (c+d x)}{d}-\frac {\left (2 a^2-b^2\right ) \sin ^3(c+d x)}{3 d}+\frac {\left (a^2-2 b^2\right ) \sin ^5(c+d x)}{5 d}+\frac {b^2 \sin ^7(c+d x)}{7 d} \] Output:
-1/3*a*b*cos(d*x+c)^6/d+a^2*sin(d*x+c)/d-1/3*(2*a^2-b^2)*sin(d*x+c)^3/d+1/ 5*(a^2-2*b^2)*sin(d*x+c)^5/d+1/7*b^2*sin(d*x+c)^7/d
Time = 0.24 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.05 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\sin (c+d x) \left (105 a^2+105 a b \sin (c+d x)+35 \left (-2 a^2+b^2\right ) \sin ^2(c+d x)-105 a b \sin ^3(c+d x)+21 \left (a^2-2 b^2\right ) \sin ^4(c+d x)+35 a b \sin ^5(c+d x)+15 b^2 \sin ^6(c+d x)\right )}{105 d} \] Input:
Integrate[Cos[c + d*x]^5*(a + b*Sin[c + d*x])^2,x]
Output:
(Sin[c + d*x]*(105*a^2 + 105*a*b*Sin[c + d*x] + 35*(-2*a^2 + b^2)*Sin[c + d*x]^2 - 105*a*b*Sin[c + d*x]^3 + 21*(a^2 - 2*b^2)*Sin[c + d*x]^4 + 35*a*b *Sin[c + d*x]^5 + 15*b^2*Sin[c + d*x]^6))/(105*d)
Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.11, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3147, 475, 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.
\(\displaystyle \int \cos ^5(c+d x) (a+b \sin (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^5 (a+b \sin (c+d x))^2dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {\int (a+b \sin (c+d x))^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2d(b \sin (c+d x))}{b^5 d}\) |
\(\Big \downarrow \) 475 |
\(\displaystyle \frac {\int \left (b^6 \sin ^6(c+d x)+b^4 \left (a^2-2 b^2\right ) \sin ^4(c+d x)+b^4 \left (b^2-2 a^2\right ) \sin ^2(c+d x)+a^2 b^4\right )d(b \sin (c+d x))-\frac {1}{3} a \left (b^2-b^2 \sin ^2(c+d x)\right )^3}{b^5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 b^5 \sin (c+d x)+\frac {1}{5} b^5 \left (a^2-2 b^2\right ) \sin ^5(c+d x)-\frac {1}{3} b^5 \left (2 a^2-b^2\right ) \sin ^3(c+d x)-\frac {1}{3} a \left (b^2-b^2 \sin ^2(c+d x)\right )^3+\frac {1}{7} b^7 \sin ^7(c+d x)}{b^5 d}\) |
Input:
Int[Cos[c + d*x]^5*(a + b*Sin[c + d*x])^2,x]
Output:
(a^2*b^5*Sin[c + d*x] - (b^5*(2*a^2 - b^2)*Sin[c + d*x]^3)/3 + (b^5*(a^2 - 2*b^2)*Sin[c + d*x]^5)/5 + (b^7*Sin[c + d*x]^7)/7 - (a*(b^2 - b^2*Sin[c + d*x]^2)^3)/3)/(b^5*d)
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp [d*n*c^(n - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Int[ExpandIntegran d[((c + d*x)^n - d*n*c^(n - 1)*x)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 0] && IGtQ[n, 0] && LeQ[n, p]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 30.39 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.03
method | result | size |
derivativedivides | \(\frac {\frac {b^{2} \sin \left (d x +c \right )^{7}}{7}+\frac {a b \sin \left (d x +c \right )^{6}}{3}+\frac {\left (a^{2}-2 b^{2}\right ) \sin \left (d x +c \right )^{5}}{5}-a b \sin \left (d x +c \right )^{4}+\frac {\left (-2 a^{2}+b^{2}\right ) \sin \left (d x +c \right )^{3}}{3}+a b \sin \left (d x +c \right )^{2}+a^{2} \sin \left (d x +c \right )}{d}\) | \(102\) |
default | \(\frac {\frac {b^{2} \sin \left (d x +c \right )^{7}}{7}+\frac {a b \sin \left (d x +c \right )^{6}}{3}+\frac {\left (a^{2}-2 b^{2}\right ) \sin \left (d x +c \right )^{5}}{5}-a b \sin \left (d x +c \right )^{4}+\frac {\left (-2 a^{2}+b^{2}\right ) \sin \left (d x +c \right )^{3}}{3}+a b \sin \left (d x +c \right )^{2}+a^{2} \sin \left (d x +c \right )}{d}\) | \(102\) |
parallelrisch | \(\frac {4200 a^{2} \sin \left (d x +c \right )+700 \sin \left (3 d x +3 c \right ) a^{2}-35 \sin \left (3 d x +3 c \right ) b^{2}+525 \sin \left (d x +c \right ) b^{2}-1050 a b \cos \left (2 d x +2 c \right )+1540 a b -15 b^{2} \sin \left (7 d x +7 c \right )-63 \sin \left (5 d x +5 c \right ) b^{2}-420 a b \cos \left (4 d x +4 c \right )-70 a b \cos \left (6 d x +6 c \right )+84 \sin \left (5 d x +5 c \right ) a^{2}}{6720 d}\) | \(142\) |
risch | \(\frac {5 a^{2} \sin \left (d x +c \right )}{8 d}+\frac {5 b^{2} \sin \left (d x +c \right )}{64 d}-\frac {b^{2} \sin \left (7 d x +7 c \right )}{448 d}-\frac {a b \cos \left (6 d x +6 c \right )}{96 d}+\frac {a^{2} \sin \left (5 d x +5 c \right )}{80 d}-\frac {3 \sin \left (5 d x +5 c \right ) b^{2}}{320 d}-\frac {a b \cos \left (4 d x +4 c \right )}{16 d}+\frac {5 a^{2} \sin \left (3 d x +3 c \right )}{48 d}-\frac {\sin \left (3 d x +3 c \right ) b^{2}}{192 d}-\frac {5 a b \cos \left (2 d x +2 c \right )}{32 d}\) | \(163\) |
norman | \(\frac {\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d}+\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}+\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}+\frac {4 \left (5 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {4 \left (5 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}+\frac {8 \left (91 a^{2}+38 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{35 d}+\frac {2 \left (113 a^{2}-16 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {2 \left (113 a^{2}-16 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{15 d}+\frac {40 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d}+\frac {40 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{7}}\) | \(297\) |
orering | \(\text {Expression too large to display}\) | \(1592\) |
Input:
int(cos(d*x+c)^5*(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(1/7*b^2*sin(d*x+c)^7+1/3*a*b*sin(d*x+c)^6+1/5*(a^2-2*b^2)*sin(d*x+c)^ 5-a*b*sin(d*x+c)^4+1/3*(-2*a^2+b^2)*sin(d*x+c)^3+a*b*sin(d*x+c)^2+a^2*sin( d*x+c))
Time = 0.10 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.88 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {35 \, a b \cos \left (d x + c\right )^{6} + {\left (15 \, b^{2} \cos \left (d x + c\right )^{6} - 3 \, {\left (7 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4} - 4 \, {\left (7 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} - 56 \, a^{2} - 8 \, b^{2}\right )} \sin \left (d x + c\right )}{105 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+b*sin(d*x+c))^2,x, algorithm="fricas")
Output:
-1/105*(35*a*b*cos(d*x + c)^6 + (15*b^2*cos(d*x + c)^6 - 3*(7*a^2 + b^2)*c os(d*x + c)^4 - 4*(7*a^2 + b^2)*cos(d*x + c)^2 - 56*a^2 - 8*b^2)*sin(d*x + c))/d
Time = 0.58 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.60 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\begin {cases} \frac {8 a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {a b \cos ^{6}{\left (c + d x \right )}}{3 d} + \frac {8 b^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {4 b^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right )^{2} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**5*(a+b*sin(d*x+c))**2,x)
Output:
Piecewise((8*a**2*sin(c + d*x)**5/(15*d) + 4*a**2*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + a**2*sin(c + d*x)*cos(c + d*x)**4/d - a*b*cos(c + d*x)**6/ (3*d) + 8*b**2*sin(c + d*x)**7/(105*d) + 4*b**2*sin(c + d*x)**5*cos(c + d* x)**2/(15*d) + b**2*sin(c + d*x)**3*cos(c + d*x)**4/(3*d), Ne(d, 0)), (x*( a + b*sin(c))**2*cos(c)**5, True))
Time = 0.03 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.07 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {15 \, b^{2} \sin \left (d x + c\right )^{7} + 35 \, a b \sin \left (d x + c\right )^{6} - 105 \, a b \sin \left (d x + c\right )^{4} + 21 \, {\left (a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{5} + 105 \, a b \sin \left (d x + c\right )^{2} - 35 \, {\left (2 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{3} + 105 \, a^{2} \sin \left (d x + c\right )}{105 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+b*sin(d*x+c))^2,x, algorithm="maxima")
Output:
1/105*(15*b^2*sin(d*x + c)^7 + 35*a*b*sin(d*x + c)^6 - 105*a*b*sin(d*x + c )^4 + 21*(a^2 - 2*b^2)*sin(d*x + c)^5 + 105*a*b*sin(d*x + c)^2 - 35*(2*a^2 - b^2)*sin(d*x + c)^3 + 105*a^2*sin(d*x + c))/d
Time = 0.14 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.19 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {15 \, b^{2} \sin \left (d x + c\right )^{7} + 35 \, a b \sin \left (d x + c\right )^{6} + 21 \, a^{2} \sin \left (d x + c\right )^{5} - 42 \, b^{2} \sin \left (d x + c\right )^{5} - 105 \, a b \sin \left (d x + c\right )^{4} - 70 \, a^{2} \sin \left (d x + c\right )^{3} + 35 \, b^{2} \sin \left (d x + c\right )^{3} + 105 \, a b \sin \left (d x + c\right )^{2} + 105 \, a^{2} \sin \left (d x + c\right )}{105 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+b*sin(d*x+c))^2,x, algorithm="giac")
Output:
1/105*(15*b^2*sin(d*x + c)^7 + 35*a*b*sin(d*x + c)^6 + 21*a^2*sin(d*x + c) ^5 - 42*b^2*sin(d*x + c)^5 - 105*a*b*sin(d*x + c)^4 - 70*a^2*sin(d*x + c)^ 3 + 35*b^2*sin(d*x + c)^3 + 105*a*b*sin(d*x + c)^2 + 105*a^2*sin(d*x + c)) /d
Time = 15.80 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.05 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a^2\,\sin \left (c+d\,x\right )-{\sin \left (c+d\,x\right )}^3\,\left (\frac {2\,a^2}{3}-\frac {b^2}{3}\right )+{\sin \left (c+d\,x\right )}^5\,\left (\frac {a^2}{5}-\frac {2\,b^2}{5}\right )+\frac {b^2\,{\sin \left (c+d\,x\right )}^7}{7}+a\,b\,{\sin \left (c+d\,x\right )}^2-a\,b\,{\sin \left (c+d\,x\right )}^4+\frac {a\,b\,{\sin \left (c+d\,x\right )}^6}{3}}{d} \] Input:
int(cos(c + d*x)^5*(a + b*sin(c + d*x))^2,x)
Output:
(a^2*sin(c + d*x) - sin(c + d*x)^3*((2*a^2)/3 - b^2/3) + sin(c + d*x)^5*(a ^2/5 - (2*b^2)/5) + (b^2*sin(c + d*x)^7)/7 + a*b*sin(c + d*x)^2 - a*b*sin( c + d*x)^4 + (a*b*sin(c + d*x)^6)/3)/d
Time = 0.17 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.17 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\sin \left (d x +c \right ) \left (15 \sin \left (d x +c \right )^{6} b^{2}+35 \sin \left (d x +c \right )^{5} a b +21 \sin \left (d x +c \right )^{4} a^{2}-42 \sin \left (d x +c \right )^{4} b^{2}-105 \sin \left (d x +c \right )^{3} a b -70 \sin \left (d x +c \right )^{2} a^{2}+35 \sin \left (d x +c \right )^{2} b^{2}+105 \sin \left (d x +c \right ) a b +105 a^{2}\right )}{105 d} \] Input:
int(cos(d*x+c)^5*(a+b*sin(d*x+c))^2,x)
Output:
(sin(c + d*x)*(15*sin(c + d*x)**6*b**2 + 35*sin(c + d*x)**5*a*b + 21*sin(c + d*x)**4*a**2 - 42*sin(c + d*x)**4*b**2 - 105*sin(c + d*x)**3*a*b - 70*s in(c + d*x)**2*a**2 + 35*sin(c + d*x)**2*b**2 + 105*sin(c + d*x)*a*b + 105 *a**2))/(105*d)