3.2.90 \(\int \cos ^2(a+b x) \sin ^3(a+b x) \sin ^2(2 a+2 b x) \, dx\) [190]

Optimal. Leaf size=46 \[ -\frac {4 \cos ^5(a+b x)}{5 b}+\frac {8 \cos ^7(a+b x)}{7 b}-\frac {4 \cos ^9(a+b x)}{9 b} \]

[Out]

-4/5*cos(b*x+a)^5/b+8/7*cos(b*x+a)^7/b-4/9*cos(b*x+a)^9/b

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {4397, 2645, 276} \begin {gather*} -\frac {4 \cos ^9(a+b x)}{9 b}+\frac {8 \cos ^7(a+b x)}{7 b}-\frac {4 \cos ^5(a+b x)}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2*Sin[a + b*x]^3*Sin[2*a + 2*b*x]^2,x]

[Out]

(-4*Cos[a + b*x]^5)/(5*b) + (8*Cos[a + b*x]^7)/(7*b) - (4*Cos[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 2645

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

Rule 4397

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*((f_.)*sin[(a_.) + (b_.)*(x_)])^(n_.)*sin[(c_.) + (d_.)*(x_)]^(p_.),
x_Symbol] :> Dist[2^p/(e^p*f^p), Int[(e*Cos[a + b*x])^(m + p)*(f*Sin[a + b*x])^(n + p), x], x] /; FreeQ[{a, b
, c, d, e, f, m, n}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && IntegerQ[p]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.18, size = 37, normalized size = 0.80 \begin {gather*} \frac {\cos ^5(a+b x) (-249+220 \cos (2 (a+b x))-35 \cos (4 (a+b x)))}{630 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^2*Sin[a + b*x]^3*Sin[2*a + 2*b*x]^2,x]

[Out]

(Cos[a + b*x]^5*(-249 + 220*Cos[2*(a + b*x)] - 35*Cos[4*(a + b*x)]))/(630*b)

________________________________________________________________________________________

Maple [A]
time = 0.22, size = 69, normalized size = 1.50

method	result	size
default	\(-\frac {3 \cos \left (b x +a \right )}{32 b}-\frac {\cos \left (3 b x +3 a \right )}{48 b}+\frac {\cos \left (5 b x +5 a \right )}{80 b}+\frac {\cos \left (7 b x +7 a \right )}{448 b}-\frac {\cos \left (9 b x +9 a \right )}{576 b}\)	\(69\)
risch	\(-\frac {3 \cos \left (b x +a \right )}{32 b}-\frac {\cos \left (3 b x +3 a \right )}{48 b}+\frac {\cos \left (5 b x +5 a \right )}{80 b}+\frac {\cos \left (7 b x +7 a \right )}{448 b}-\frac {\cos \left (9 b x +9 a \right )}{576 b}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^2*sin(b*x+a)^3*sin(2*b*x+2*a)^2,x,method=_RETURNVERBOSE)

[Out]

-3/32*cos(b*x+a)/b-1/48*cos(3*b*x+3*a)/b+1/80/b*cos(5*b*x+5*a)+1/448/b*cos(7*b*x+7*a)-1/576/b*cos(9*b*x+9*a)

________________________________________________________________________________________

Maxima [A]
time = 0.27, size = 58, normalized size = 1.26 \begin {gather*} -\frac {35 \, \cos \left (9 \, b x + 9 \, a\right ) - 45 \, \cos \left (7 \, b x + 7 \, a\right ) - 252 \, \cos \left (5 \, b x + 5 \, a\right ) + 420 \, \cos \left (3 \, b x + 3 \, a\right ) + 1890 \, \cos \left (b x + a\right )}{20160 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2*sin(b*x+a)^3*sin(2*b*x+2*a)^2,x, algorithm="maxima")

[Out]

-1/20160*(35*cos(9*b*x + 9*a) - 45*cos(7*b*x + 7*a) - 252*cos(5*b*x + 5*a) + 420*cos(3*b*x + 3*a) + 1890*cos(b
*x + a))/b

________________________________________________________________________________________

Fricas [A]
time = 2.70, size = 36, normalized size = 0.78 \begin {gather*} -\frac {4 \, {\left (35 \, \cos \left (b x + a\right )^{9} - 90 \, \cos \left (b x + a\right )^{7} + 63 \, \cos \left (b x + a\right )^{5}\right )}}{315 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2*sin(b*x+a)^3*sin(2*b*x+2*a)^2,x, algorithm="fricas")

[Out]

-4/315*(35*cos(b*x + a)^9 - 90*cos(b*x + a)^7 + 63*cos(b*x + a)^5)/b

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 318 vs. \(2 (39) = 78\).
time = 14.95, size = 318, normalized size = 6.91 \begin {gather*} \begin {cases} - \frac {8 \sin ^{5}{\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos {\left (2 a + 2 b x \right )}}{315 b} + \frac {16 \sin ^{4}{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )}}{315 b} - \frac {16 \sin ^{4}{\left (a + b x \right )} \cos {\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{315 b} + \frac {44 \sin ^{3}{\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{315 b} - \frac {113 \sin ^{2}{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{3}{\left (a + b x \right )}}{315 b} + \frac {8 \sin ^{2}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{315 b} - \frac {88 \sin {\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos ^{4}{\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{315 b} - \frac {2 \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{5}{\left (a + b x \right )}}{63 b} - \frac {32 \cos ^{5}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{315 b} & \text {for}\: b \neq 0 \\x \sin ^{3}{\left (a \right )} \sin ^{2}{\left (2 a \right )} \cos ^{2}{\left (a \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-8*sin(a + b*x)**5*sin(2*a + 2*b*x)*cos(2*a + 2*b*x)/(315*b) + 16*sin(a + b*x)**4*sin(2*a + 2*b*x)*
*2*cos(a + b*x)/(315*b) - 16*sin(a + b*x)**4*cos(a + b*x)*cos(2*a + 2*b*x)**2/(315*b) + 44*sin(a + b*x)**3*sin
(2*a + 2*b*x)*cos(a + b*x)**2*cos(2*a + 2*b*x)/(315*b) - 113*sin(a + b*x)**2*sin(2*a + 2*b*x)**2*cos(a + b*x)*
*3/(315*b) + 8*sin(a + b*x)**2*cos(a + b*x)**3*cos(2*a + 2*b*x)**2/(315*b) - 88*sin(a + b*x)*sin(2*a + 2*b*x)*
cos(a + b*x)**4*cos(2*a + 2*b*x)/(315*b) - 2*sin(2*a + 2*b*x)**2*cos(a + b*x)**5/(63*b) - 32*cos(a + b*x)**5*c
os(2*a + 2*b*x)**2/(315*b), Ne(b, 0)), (x*sin(a)**3*sin(2*a)**2*cos(a)**2, True))

________________________________________________________________________________________

Giac [A]
time = 0.46, size = 36, normalized size = 0.78 \begin {gather*} -\frac {4 \, {\left (35 \, \cos \left (b x + a\right )^{9} - 90 \, \cos \left (b x + a\right )^{7} + 63 \, \cos \left (b x + a\right )^{5}\right )}}{315 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/315*(35*cos(b*x + a)^9 - 90*cos(b*x + a)^7 + 63*cos(b*x + a)^5)/b

________________________________________________________________________________________

Mupad [B]
time = 0.20, size = 36, normalized size = 0.78 \begin {gather*} -\frac {4\,\left (35\,{\cos \left (a+b\,x\right )}^9-90\,{\cos \left (a+b\,x\right )}^7+63\,{\cos \left (a+b\,x\right )}^5\right )}{315\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*(63*cos(a + b*x)^5 - 90*cos(a + b*x)^7 + 35*cos(a + b*x)^9))/(315*b)

________________________________________________________________________________________