3.1060 \(\int \cos ^2(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=163 \[ \frac {\left (2 a^2-b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac {\left (2 a^2+b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} x \left (2 a^2+b^2\right )+\frac {2 a b \cos ^3(c+d x)}{15 d}-\frac {2 a b \cos (c+d x)}{5 d}+\frac {a b \sin ^4(c+d x) \cos (c+d x)}{15 d}+\frac {\sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{6 d} \]

[Out]

1/16*(2*a^2+b^2)*x-2/5*a*b*cos(d*x+c)/d+2/15*a*b*cos(d*x+c)^3/d-1/16*(2*a^2+b^2)*cos(d*x+c)*sin(d*x+c)/d+1/24*
(2*a^2-b^2)*cos(d*x+c)*sin(d*x+c)^3/d+1/15*a*b*cos(d*x+c)*sin(d*x+c)^4/d+1/6*cos(d*x+c)*sin(d*x+c)^3*(a+b*sin(
d*x+c))^2/d

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Rubi [A]  time = 0.38, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2889, 3050, 3033, 3023, 2748, 2635, 8, 2633} \[ \frac {\left (2 a^2-b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac {\left (2 a^2+b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} x \left (2 a^2+b^2\right )+\frac {2 a b \cos ^3(c+d x)}{15 d}-\frac {2 a b \cos (c+d x)}{5 d}+\frac {a b \sin ^4(c+d x) \cos (c+d x)}{15 d}+\frac {\sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^2 + b^2)*x)/16 - (2*a*b*Cos[c + d*x])/(5*d) + (2*a*b*Cos[c + d*x]^3)/(15*d) - ((2*a^2 + b^2)*Cos[c + d*x
]*Sin[c + d*x])/(16*d) + ((2*a^2 - b^2)*Cos[c + d*x]*Sin[c + d*x]^3)/(24*d) + (a*b*Cos[c + d*x]*Sin[c + d*x]^4
)/(15*d) + (Cos[c + d*x]*Sin[c + d*x]^3*(a + b*Sin[c + d*x])^2)/(6*d)

Rule 8

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

Rule 2633

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

Rule 2635

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] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2748

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

Rule 2889

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f,
m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((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 + 1))/(b*f*(m + 2)), x] + Dist[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 3033

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
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[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 3050

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (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] + Dist[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)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*
x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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])))

Rubi steps

\begin {align*} \int \cos ^2(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx &=\int \sin ^2(c+d x) (a+b \sin (c+d x))^2 \left (1-\sin ^2(c+d x)\right ) \, dx\\ &=\frac {\cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{6 d}+\frac {1}{6} \int \sin ^2(c+d x) (a+b \sin (c+d x)) \left (3 a+b \sin (c+d x)-2 a \sin ^2(c+d x)\right ) \, dx\\ &=\frac {a b \cos (c+d x) \sin ^4(c+d x)}{15 d}+\frac {\cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{6 d}+\frac {1}{30} \int \sin ^2(c+d x) \left (15 a^2+12 a b \sin (c+d x)-5 \left (2 a^2-b^2\right ) \sin ^2(c+d x)\right ) \, dx\\ &=\frac {\left (2 a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac {a b \cos (c+d x) \sin ^4(c+d x)}{15 d}+\frac {\cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{6 d}+\frac {1}{120} \int \sin ^2(c+d x) \left (15 \left (2 a^2+b^2\right )+48 a b \sin (c+d x)\right ) \, dx\\ &=\frac {\left (2 a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac {a b \cos (c+d x) \sin ^4(c+d x)}{15 d}+\frac {\cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{6 d}+\frac {1}{5} (2 a b) \int \sin ^3(c+d x) \, dx+\frac {1}{8} \left (2 a^2+b^2\right ) \int \sin ^2(c+d x) \, dx\\ &=-\frac {\left (2 a^2+b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {\left (2 a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac {a b \cos (c+d x) \sin ^4(c+d x)}{15 d}+\frac {\cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{6 d}+\frac {1}{16} \left (2 a^2+b^2\right ) \int 1 \, dx-\frac {(2 a b) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{5 d}\\ &=\frac {1}{16} \left (2 a^2+b^2\right ) x-\frac {2 a b \cos (c+d x)}{5 d}+\frac {2 a b \cos ^3(c+d x)}{15 d}-\frac {\left (2 a^2+b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {\left (2 a^2-b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac {a b \cos (c+d x) \sin ^4(c+d x)}{15 d}+\frac {\cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{6 d}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 120, normalized size = 0.74 \[ \frac {-30 a^2 \sin (4 (c+d x))+120 a^2 c+120 a^2 d x-240 a b \cos (c+d x)-40 a b \cos (3 (c+d x))+24 a b \cos (5 (c+d x))-15 b^2 \sin (2 (c+d x))-15 b^2 \sin (4 (c+d x))+5 b^2 \sin (6 (c+d x))+60 b^2 c+60 b^2 d x}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(120*a^2*c + 60*b^2*c + 120*a^2*d*x + 60*b^2*d*x - 240*a*b*Cos[c + d*x] - 40*a*b*Cos[3*(c + d*x)] + 24*a*b*Cos
[5*(c + d*x)] - 15*b^2*Sin[2*(c + d*x)] - 30*a^2*Sin[4*(c + d*x)] - 15*b^2*Sin[4*(c + d*x)] + 5*b^2*Sin[6*(c +
d*x)])/(960*d)

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fricas [A]  time = 0.64, size = 103, normalized size = 0.63 \[ \frac {96 \, a b \cos \left (d x + c\right )^{5} - 160 \, a b \cos \left (d x + c\right )^{3} + 15 \, {\left (2 \, a^{2} + b^{2}\right )} d x + 5 \, {\left (8 \, b^{2} \cos \left (d x + c\right )^{5} - 2 \, {\left (6 \, a^{2} + 7 \, b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(96*a*b*cos(d*x + c)^5 - 160*a*b*cos(d*x + c)^3 + 15*(2*a^2 + b^2)*d*x + 5*(8*b^2*cos(d*x + c)^5 - 2*(6*
a^2 + 7*b^2)*cos(d*x + c)^3 + 3*(2*a^2 + b^2)*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.24, size = 115, normalized size = 0.71 \[ \frac {1}{16} \, {\left (2 \, a^{2} + b^{2}\right )} x + \frac {a b \cos \left (5 \, d x + 5 \, c\right )}{40 \, d} - \frac {a b \cos \left (3 \, d x + 3 \, c\right )}{24 \, d} - \frac {a b \cos \left (d x + c\right )}{4 \, d} + \frac {b^{2} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {b^{2} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} - \frac {{\left (2 \, a^{2} + b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*a^2 + b^2)*x + 1/40*a*b*cos(5*d*x + 5*c)/d - 1/24*a*b*cos(3*d*x + 3*c)/d - 1/4*a*b*cos(d*x + c)/d + 1/
192*b^2*sin(6*d*x + 6*c)/d - 1/64*b^2*sin(2*d*x + 2*c)/d - 1/64*(2*a^2 + b^2)*sin(4*d*x + 4*c)/d

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maple [A]  time = 0.22, size = 141, normalized size = 0.87 \[ \frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )+2 a b \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )+b^{2} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{6}-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{8}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^2*(-1/4*cos(d*x+c)^3*sin(d*x+c)+1/8*cos(d*x+c)*sin(d*x+c)+1/8*d*x+1/8*c)+2*a*b*(-1/5*sin(d*x+c)^2*cos(d
*x+c)^3-2/15*cos(d*x+c)^3)+b^2*(-1/6*sin(d*x+c)^3*cos(d*x+c)^3-1/8*cos(d*x+c)^3*sin(d*x+c)+1/16*cos(d*x+c)*sin
(d*x+c)+1/16*d*x+1/16*c))

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maxima [A]  time = 0.39, size = 92, normalized size = 0.56 \[ \frac {30 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} + 128 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a b - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 12 \, d x - 12 \, c + 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} b^{2}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/960*(30*(4*d*x + 4*c - sin(4*d*x + 4*c))*a^2 + 128*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a*b - 5*(4*sin(2*d*
x + 2*c)^3 - 12*d*x - 12*c + 3*sin(4*d*x + 4*c))*b^2)/d

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mupad [B]  time = 9.62, size = 112, normalized size = 0.69 \[ -\frac {\frac {15\,a^2\,\sin \left (4\,c+4\,d\,x\right )}{2}+\frac {15\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {15\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{4}-\frac {5\,b^2\,\sin \left (6\,c+6\,d\,x\right )}{4}+60\,a\,b\,\cos \left (c+d\,x\right )+10\,a\,b\,\cos \left (3\,c+3\,d\,x\right )-6\,a\,b\,\cos \left (5\,c+5\,d\,x\right )-30\,a^2\,d\,x-15\,b^2\,d\,x}{240\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((15*a^2*sin(4*c + 4*d*x))/2 + (15*b^2*sin(2*c + 2*d*x))/4 + (15*b^2*sin(4*c + 4*d*x))/4 - (5*b^2*sin(6*c + 6
*d*x))/4 + 60*a*b*cos(c + d*x) + 10*a*b*cos(3*c + 3*d*x) - 6*a*b*cos(5*c + 5*d*x) - 30*a^2*d*x - 15*b^2*d*x)/(
240*d)

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sympy [A]  time = 3.57, size = 309, normalized size = 1.90 \[ \begin {cases} \frac {a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {2 a b \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {4 a b \cos ^{5}{\left (c + d x \right )}}{15 d} + \frac {b^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {3 b^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {b^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {b^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} - \frac {b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac {b^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\relax (c )}\right )^{2} \sin ^{2}{\relax (c )} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*x*sin(c + d*x)**4/8 + a**2*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + a**2*x*cos(c + d*x)**4/8 + a*
*2*sin(c + d*x)**3*cos(c + d*x)/(8*d) - a**2*sin(c + d*x)*cos(c + d*x)**3/(8*d) - 2*a*b*sin(c + d*x)**2*cos(c
+ d*x)**3/(3*d) - 4*a*b*cos(c + d*x)**5/(15*d) + b**2*x*sin(c + d*x)**6/16 + 3*b**2*x*sin(c + d*x)**4*cos(c +
d*x)**2/16 + 3*b**2*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + b**2*x*cos(c + d*x)**6/16 + b**2*sin(c + d*x)**5*co
s(c + d*x)/(16*d) - b**2*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) - b**2*sin(c + d*x)*cos(c + d*x)**5/(16*d), Ne(
d, 0)), (x*(a + b*sin(c))**2*sin(c)**2*cos(c)**2, True))

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