\(\int \cos ^7(a+b x) \sin ^2(a+b x) \, dx\) [49]

Optimal result

Integrand size = 17, antiderivative size = 61 \[ \int \cos ^7(a+b x) \sin ^2(a+b x) \, dx=\frac {\sin ^3(a+b x)}{3 b}-\frac {3 \sin ^5(a+b x)}{5 b}+\frac {3 \sin ^7(a+b x)}{7 b}-\frac {\sin ^9(a+b x)}{9 b} \]

[Out]

1/3*sin(b*x+a)^3/b-3/5*sin(b*x+a)^5/b+3/7*sin(b*x+a)^7/b-1/9*sin(b*x+a)^9/b

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2644, 276} \[ \int \cos ^7(a+b x) \sin ^2(a+b x) \, dx=-\frac {\sin ^9(a+b x)}{9 b}+\frac {3 \sin ^7(a+b x)}{7 b}-\frac {3 \sin ^5(a+b x)}{5 b}+\frac {\sin ^3(a+b x)}{3 b} \]

[In]

Int[Cos[a + b*x]^7*Sin[a + b*x]^2,x]

[Out]

Sin[a + b*x]^3/(3*b) - (3*Sin[a + b*x]^5)/(5*b) + (3*Sin[a + b*x]^7)/(7*b) - Sin[a + b*x]^9/(9*b)

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int x^2 \left (1-x^2\right )^3 \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {\text {Subst}\left (\int \left (x^2-3 x^4+3 x^6-x^8\right ) \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {\sin ^3(a+b x)}{3 b}-\frac {3 \sin ^5(a+b x)}{5 b}+\frac {3 \sin ^7(a+b x)}{7 b}-\frac {\sin ^9(a+b x)}{9 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.77 \[ \int \cos ^7(a+b x) \sin ^2(a+b x) \, dx=\frac {(1606+1389 \cos (2 (a+b x))+330 \cos (4 (a+b x))+35 \cos (6 (a+b x))) \sin ^3(a+b x)}{10080 b} \]

[In]

Integrate[Cos[a + b*x]^7*Sin[a + b*x]^2,x]

[Out]

((1606 + 1389*Cos[2*(a + b*x)] + 330*Cos[4*(a + b*x)] + 35*Cos[6*(a + b*x)])*Sin[a + b*x]^3)/(10080*b)

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(-\frac {\frac {\left (\sin ^{9}\left (b x +a \right )\right )}{9}-\frac {3 \left (\sin ^{7}\left (b x +a \right )\right )}{7}+\frac {3 \left (\sin ^{5}\left (b x +a \right )\right )}{5}-\frac {\left (\sin ^{3}\left (b x +a \right )\right )}{3}}{b}\)	\(47\)
default	\(-\frac {\frac {\left (\sin ^{9}\left (b x +a \right )\right )}{9}-\frac {3 \left (\sin ^{7}\left (b x +a \right )\right )}{7}+\frac {3 \left (\sin ^{5}\left (b x +a \right )\right )}{5}-\frac {\left (\sin ^{3}\left (b x +a \right )\right )}{3}}{b}\)	\(47\)
risch	\(\frac {7 \sin \left (b x +a \right )}{128 b}-\frac {\sin \left (9 b x +9 a \right )}{2304 b}-\frac {5 \sin \left (7 b x +7 a \right )}{1792 b}-\frac {\sin \left (5 b x +5 a \right )}{160 b}\)	\(55\)
parallelrisch	\(-\frac {\left (\sin \left (\frac {3 b x}{2}+\frac {3 a}{2}\right )-3 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (35 \cos \left (6 b x +6 a \right )+1389 \cos \left (2 b x +2 a \right )+330 \cos \left (4 b x +4 a \right )+1606\right ) \left (\cos \left (\frac {3 b x}{2}+\frac {3 a}{2}\right )+3 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{20160 b}\)	\(83\)
norman	\(\frac {\frac {8 \left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}-\frac {16 \left (\tan ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{5 b}+\frac {632 \left (\tan ^{7}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{35 b}-\frac {2848 \left (\tan ^{9}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{315 b}+\frac {632 \left (\tan ^{11}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{35 b}-\frac {16 \left (\tan ^{13}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{5 b}+\frac {8 \left (\tan ^{15}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{9}}\)	\(130\)

[In]

int(cos(b*x+a)^7*sin(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

-1/b*(1/9*sin(b*x+a)^9-3/7*sin(b*x+a)^7+3/5*sin(b*x+a)^5-1/3*sin(b*x+a)^3)

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87 \[ \int \cos ^7(a+b x) \sin ^2(a+b x) \, dx=-\frac {{\left (35 \, \cos \left (b x + a\right )^{8} - 5 \, \cos \left (b x + a\right )^{6} - 6 \, \cos \left (b x + a\right )^{4} - 8 \, \cos \left (b x + a\right )^{2} - 16\right )} \sin \left (b x + a\right )}{315 \, b} \]

[In]

integrate(cos(b*x+a)^7*sin(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/315*(35*cos(b*x + a)^8 - 5*cos(b*x + a)^6 - 6*cos(b*x + a)^4 - 8*cos(b*x + a)^2 - 16)*sin(b*x + a)/b

Sympy [A] (verification not implemented)

Time = 0.91 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.44 \[ \int \cos ^7(a+b x) \sin ^2(a+b x) \, dx=\begin {cases} \frac {16 \sin ^{9}{\left (a + b x \right )}}{315 b} + \frac {8 \sin ^{7}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{35 b} + \frac {2 \sin ^{5}{\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{5 b} + \frac {\sin ^{3}{\left (a + b x \right )} \cos ^{6}{\left (a + b x \right )}}{3 b} & \text {for}\: b \neq 0 \\x \sin ^{2}{\left (a \right )} \cos ^{7}{\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(b*x+a)**7*sin(b*x+a)**2,x)

[Out]

Piecewise((16*sin(a + b*x)**9/(315*b) + 8*sin(a + b*x)**7*cos(a + b*x)**2/(35*b) + 2*sin(a + b*x)**5*cos(a + b
*x)**4/(5*b) + sin(a + b*x)**3*cos(a + b*x)**6/(3*b), Ne(b, 0)), (x*sin(a)**2*cos(a)**7, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.75 \[ \int \cos ^7(a+b x) \sin ^2(a+b x) \, dx=-\frac {35 \, \sin \left (b x + a\right )^{9} - 135 \, \sin \left (b x + a\right )^{7} + 189 \, \sin \left (b x + a\right )^{5} - 105 \, \sin \left (b x + a\right )^{3}}{315 \, b} \]

[In]

integrate(cos(b*x+a)^7*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

-1/315*(35*sin(b*x + a)^9 - 135*sin(b*x + a)^7 + 189*sin(b*x + a)^5 - 105*sin(b*x + a)^3)/b

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89 \[ \int \cos ^7(a+b x) \sin ^2(a+b x) \, dx=-\frac {\sin \left (9 \, b x + 9 \, a\right )}{2304 \, b} - \frac {5 \, \sin \left (7 \, b x + 7 \, a\right )}{1792 \, b} - \frac {\sin \left (5 \, b x + 5 \, a\right )}{160 \, b} + \frac {7 \, \sin \left (b x + a\right )}{128 \, b} \]

[In]

integrate(cos(b*x+a)^7*sin(b*x+a)^2,x, algorithm="giac")

[Out]

-1/2304*sin(9*b*x + 9*a)/b - 5/1792*sin(7*b*x + 7*a)/b - 1/160*sin(5*b*x + 5*a)/b + 7/128*sin(b*x + a)/b

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.74 \[ \int \cos ^7(a+b x) \sin ^2(a+b x) \, dx=\frac {-\frac {{\sin \left (a+b\,x\right )}^9}{9}+\frac {3\,{\sin \left (a+b\,x\right )}^7}{7}-\frac {3\,{\sin \left (a+b\,x\right )}^5}{5}+\frac {{\sin \left (a+b\,x\right )}^3}{3}}{b} \]

[In]

int(cos(a + b*x)^7*sin(a + b*x)^2,x)

[Out]

(sin(a + b*x)^3/3 - (3*sin(a + b*x)^5)/5 + (3*sin(a + b*x)^7)/7 - sin(a + b*x)^9/9)/b