3.1069 \(\int \cos ^2(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=232 \[ \frac {b \left (21 a^2+4 b^2\right ) \cos ^3(c+d x)}{105 d}-\frac {b \left (21 a^2+4 b^2\right ) \cos (c+d x)}{35 d}+\frac {b \left (a^2-b^2\right ) \sin ^4(c+d x) \cos (c+d x)}{35 d}+\frac {a \left (2 a^2-7 b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{56 d}-\frac {a \left (2 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} a x \left (2 a^2+3 b^2\right )+\frac {\sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{7 d}+\frac {a \sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{14 d} \]

[Out]

1/16*a*(2*a^2+3*b^2)*x-1/35*b*(21*a^2+4*b^2)*cos(d*x+c)/d+1/105*b*(21*a^2+4*b^2)*cos(d*x+c)^3/d-1/16*a*(2*a^2+
3*b^2)*cos(d*x+c)*sin(d*x+c)/d+1/56*a*(2*a^2-7*b^2)*cos(d*x+c)*sin(d*x+c)^3/d+1/35*b*(a^2-b^2)*cos(d*x+c)*sin(
d*x+c)^4/d+1/14*a*cos(d*x+c)*sin(d*x+c)^3*(a+b*sin(d*x+c))^2/d+1/7*cos(d*x+c)*sin(d*x+c)^3*(a+b*sin(d*x+c))^3/
d

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Rubi [A]  time = 0.57, antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {2889, 3050, 3049, 3033, 3023, 2748, 2635, 8, 2633} \[ \frac {b \left (21 a^2+4 b^2\right ) \cos ^3(c+d x)}{105 d}-\frac {b \left (21 a^2+4 b^2\right ) \cos (c+d x)}{35 d}+\frac {b \left (a^2-b^2\right ) \sin ^4(c+d x) \cos (c+d x)}{35 d}+\frac {a \left (2 a^2-7 b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{56 d}-\frac {a \left (2 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} a x \left (2 a^2+3 b^2\right )+\frac {\sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{7 d}+\frac {a \sin ^3(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{14 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(2*a^2 + 3*b^2)*x)/16 - (b*(21*a^2 + 4*b^2)*Cos[c + d*x])/(35*d) + (b*(21*a^2 + 4*b^2)*Cos[c + d*x]^3)/(105
*d) - (a*(2*a^2 + 3*b^2)*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (a*(2*a^2 - 7*b^2)*Cos[c + d*x]*Sin[c + d*x]^3)/(
56*d) + (b*(a^2 - b^2)*Cos[c + d*x]*Sin[c + d*x]^4)/(35*d) + (a*Cos[c + d*x]*Sin[c + d*x]^3*(a + b*Sin[c + d*x
])^2)/(14*d) + (Cos[c + d*x]*Sin[c + d*x]^3*(a + b*Sin[c + d*x])^3)/(7*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 3049

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] + 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)) + (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])))

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))^3 \, dx &=\int \sin ^2(c+d x) (a+b \sin (c+d x))^3 \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))^3}{7 d}+\frac {1}{7} \int \sin ^2(c+d x) (a+b \sin (c+d x))^2 \left (4 a+b \sin (c+d x)-3 a \sin ^2(c+d x)\right ) \, dx\\ &=\frac {a \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{14 d}+\frac {\cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^3}{7 d}+\frac {1}{42} \int \sin ^2(c+d x) (a+b \sin (c+d x)) \left (15 a^2+15 a b \sin (c+d x)-6 \left (a^2-b^2\right ) \sin ^2(c+d x)\right ) \, dx\\ &=\frac {b \left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{35 d}+\frac {a \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{14 d}+\frac {\cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^3}{7 d}+\frac {1}{210} \int \sin ^2(c+d x) \left (75 a^3+6 b \left (21 a^2+4 b^2\right ) \sin (c+d x)-15 a \left (2 a^2-7 b^2\right ) \sin ^2(c+d x)\right ) \, dx\\ &=\frac {a \left (2 a^2-7 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{56 d}+\frac {b \left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{35 d}+\frac {a \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{14 d}+\frac {\cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^3}{7 d}+\frac {1}{840} \int \sin ^2(c+d x) \left (105 a \left (2 a^2+3 b^2\right )+24 b \left (21 a^2+4 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=\frac {a \left (2 a^2-7 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{56 d}+\frac {b \left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{35 d}+\frac {a \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{14 d}+\frac {\cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^3}{7 d}+\frac {1}{8} \left (a \left (2 a^2+3 b^2\right )\right ) \int \sin ^2(c+d x) \, dx+\frac {1}{35} \left (b \left (21 a^2+4 b^2\right )\right ) \int \sin ^3(c+d x) \, dx\\ &=-\frac {a \left (2 a^2+3 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a \left (2 a^2-7 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{56 d}+\frac {b \left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{35 d}+\frac {a \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{14 d}+\frac {\cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^3}{7 d}+\frac {1}{16} \left (a \left (2 a^2+3 b^2\right )\right ) \int 1 \, dx-\frac {\left (b \left (21 a^2+4 b^2\right )\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{35 d}\\ &=\frac {1}{16} a \left (2 a^2+3 b^2\right ) x-\frac {b \left (21 a^2+4 b^2\right ) \cos (c+d x)}{35 d}+\frac {b \left (21 a^2+4 b^2\right ) \cos ^3(c+d x)}{105 d}-\frac {a \left (2 a^2+3 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a \left (2 a^2-7 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{56 d}+\frac {b \left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{35 d}+\frac {a \cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2}{14 d}+\frac {\cos (c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^3}{7 d}\\ \end {align*}

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Mathematica [A]  time = 0.82, size = 157, normalized size = 0.68 \[ \frac {-35 \left (12 a^2 b+b^3\right ) \cos (3 (c+d x))+63 \left (4 a^2 b+b^3\right ) \cos (5 (c+d x))+105 a \left (-\left (2 a^2+3 b^2\right ) \sin (4 (c+d x))+8 a^2 c+8 a^2 d x-3 b^2 \sin (2 (c+d x))+b^2 \sin (6 (c+d x))+12 b^2 c+12 b^2 d x\right )-105 b \left (24 a^2+5 b^2\right ) \cos (c+d x)-15 b^3 \cos (7 (c+d x))}{6720 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-105*b*(24*a^2 + 5*b^2)*Cos[c + d*x] - 35*(12*a^2*b + b^3)*Cos[3*(c + d*x)] + 63*(4*a^2*b + b^3)*Cos[5*(c + d
*x)] - 15*b^3*Cos[7*(c + d*x)] + 105*a*(8*a^2*c + 12*b^2*c + 8*a^2*d*x + 12*b^2*d*x - 3*b^2*Sin[2*(c + d*x)] -
(2*a^2 + 3*b^2)*Sin[4*(c + d*x)] + b^2*Sin[6*(c + d*x)]))/(6720*d)

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fricas [A]  time = 0.72, size = 141, normalized size = 0.61 \[ -\frac {240 \, b^{3} \cos \left (d x + c\right )^{7} - 336 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{5} + 560 \, {\left (3 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{3} - 105 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} d x - 105 \, {\left (8 \, a b^{2} \cos \left (d x + c\right )^{5} - 2 \, {\left (2 \, a^{3} + 7 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{1680 \, d} \]

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))^3,x, algorithm="fricas")

[Out]

-1/1680*(240*b^3*cos(d*x + c)^7 - 336*(3*a^2*b + 2*b^3)*cos(d*x + c)^5 + 560*(3*a^2*b + b^3)*cos(d*x + c)^3 -
105*(2*a^3 + 3*a*b^2)*d*x - 105*(8*a*b^2*cos(d*x + c)^5 - 2*(2*a^3 + 7*a*b^2)*cos(d*x + c)^3 + (2*a^3 + 3*a*b^
2)*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.28, size = 166, normalized size = 0.72 \[ -\frac {b^{3} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {a b^{2} \sin \left (6 \, d x + 6 \, c\right )}{64 \, d} - \frac {3 \, a b^{2} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {1}{16} \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} x + \frac {3 \, {\left (4 \, a^{2} b + b^{3}\right )} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {{\left (12 \, a^{2} b + b^{3}\right )} \cos \left (3 \, d x + 3 \, c\right )}{192 \, d} - \frac {{\left (24 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (d x + c\right )}{64 \, d} - \frac {{\left (2 \, a^{3} + 3 \, a b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} \]

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))^3,x, algorithm="giac")

[Out]

-1/448*b^3*cos(7*d*x + 7*c)/d + 1/64*a*b^2*sin(6*d*x + 6*c)/d - 3/64*a*b^2*sin(2*d*x + 2*c)/d + 1/16*(2*a^3 +
3*a*b^2)*x + 3/320*(4*a^2*b + b^3)*cos(5*d*x + 5*c)/d - 1/192*(12*a^2*b + b^3)*cos(3*d*x + 3*c)/d - 1/64*(24*a
^2*b + 5*b^3)*cos(d*x + c)/d - 1/64*(2*a^3 + 3*a*b^2)*sin(4*d*x + 4*c)/d

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maple [A]  time = 0.22, size = 196, normalized size = 0.84 \[ \frac {a^{3} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )+3 a^{2} b \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )+3 a \,b^{2} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{6}-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{8}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )+b^{3} \left (-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{7}-\frac {4 \left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{35}-\frac {8 \left (\cos ^{3}\left (d x +c \right )\right )}{105}\right )}{d} \]

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))^3,x)

[Out]

1/d*(a^3*(-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)+3*a^2*b*(-1/5*sin(d*x+c)^2*cos
(d*x+c)^3-2/15*cos(d*x+c)^3)+3*a*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)+b^3*(-1/7*sin(d*x+c)^4*cos(d*x+c)^3-4/35*sin(d*x+c)^2*cos(d*x+c)^3-8/105*cos(d*
x+c)^3))

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maxima [A]  time = 0.33, size = 131, normalized size = 0.56 \[ \frac {210 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3} + 1344 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{2} b - 105 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 12 \, d x - 12 \, c + 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a b^{2} - 64 \, {\left (15 \, \cos \left (d x + c\right )^{7} - 42 \, \cos \left (d x + c\right )^{5} + 35 \, \cos \left (d x + c\right )^{3}\right )} b^{3}}{6720 \, d} \]

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))^3,x, algorithm="maxima")

[Out]

1/6720*(210*(4*d*x + 4*c - sin(4*d*x + 4*c))*a^3 + 1344*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a^2*b - 105*(4*s
in(2*d*x + 2*c)^3 - 12*d*x - 12*c + 3*sin(4*d*x + 4*c))*a*b^2 - 64*(15*cos(d*x + c)^7 - 42*cos(d*x + c)^5 + 35
*cos(d*x + c)^3)*b^3)/d

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mupad [B]  time = 10.66, size = 455, normalized size = 1.96 \[ \frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2+3\,b^2\right )}{8\,\left (\frac {a^3}{4}+\frac {3\,a\,b^2}{8}\right )}\right )\,\left (2\,a^2+3\,b^2\right )}{8\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {a^3}{4}+\frac {3\,a\,b^2}{8}\right )+\frac {4\,a^2\,b}{5}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {5\,a\,b^2}{2}-a^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (\frac {5\,a\,b^2}{2}-a^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}\,\left (\frac {a^3}{4}+\frac {3\,a\,b^2}{8}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {11\,a^3}{4}+\frac {97\,a\,b^2}{8}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {11\,a^3}{4}+\frac {97\,a\,b^2}{8}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (8\,a^2\,b-\frac {16\,b^3}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {24\,a^2\,b}{5}+\frac {16\,b^3}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (20\,a^2\,b+\frac {32\,b^3}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {28\,a^2\,b}{5}+\frac {16\,b^3}{15}\right )+\frac {16\,b^3}{105}+12\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {a\,\left (2\,a^2+3\,b^2\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{8\,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))^3,x)

[Out]

(a*atan((a*tan(c/2 + (d*x)/2)*(2*a^2 + 3*b^2))/(8*((3*a*b^2)/8 + a^3/4)))*(2*a^2 + 3*b^2))/(8*d) - (tan(c/2 +
(d*x)/2)*((3*a*b^2)/8 + a^3/4) + (4*a^2*b)/5 + tan(c/2 + (d*x)/2)^3*((5*a*b^2)/2 - a^3) - tan(c/2 + (d*x)/2)^1
1*((5*a*b^2)/2 - a^3) - tan(c/2 + (d*x)/2)^13*((3*a*b^2)/8 + a^3/4) - tan(c/2 + (d*x)/2)^5*((97*a*b^2)/8 + (11
*a^3)/4) + tan(c/2 + (d*x)/2)^9*((97*a*b^2)/8 + (11*a^3)/4) + tan(c/2 + (d*x)/2)^6*(8*a^2*b - (16*b^3)/3) + ta
n(c/2 + (d*x)/2)^4*((24*a^2*b)/5 + (16*b^3)/5) + tan(c/2 + (d*x)/2)^8*(20*a^2*b + (32*b^3)/3) + tan(c/2 + (d*x
)/2)^2*((28*a^2*b)/5 + (16*b^3)/15) + (16*b^3)/105 + 12*a^2*b*tan(c/2 + (d*x)/2)^10)/(d*(7*tan(c/2 + (d*x)/2)^
2 + 21*tan(c/2 + (d*x)/2)^4 + 35*tan(c/2 + (d*x)/2)^6 + 35*tan(c/2 + (d*x)/2)^8 + 21*tan(c/2 + (d*x)/2)^10 + 7
*tan(c/2 + (d*x)/2)^12 + tan(c/2 + (d*x)/2)^14 + 1)) - (a*(2*a^2 + 3*b^2)*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2)
)/(8*d)

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sympy [A]  time = 6.29, size = 394, normalized size = 1.70 \[ \begin {cases} \frac {a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {a^{2} b \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{d} - \frac {2 a^{2} b \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac {3 a b^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {9 a b^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {9 a b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {3 a b^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {3 a b^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} - \frac {a b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{2 d} - \frac {3 a b^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {b^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {4 b^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac {8 b^{3} \cos ^{7}{\left (c + d x \right )}}{105 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\relax (c )}\right )^{3} \sin ^{2}{\relax (c )} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

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))**3,x)

[Out]

Piecewise((a**3*x*sin(c + d*x)**4/8 + a**3*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + a**3*x*cos(c + d*x)**4/8 + a*
*3*sin(c + d*x)**3*cos(c + d*x)/(8*d) - a**3*sin(c + d*x)*cos(c + d*x)**3/(8*d) - a**2*b*sin(c + d*x)**2*cos(c
+ d*x)**3/d - 2*a**2*b*cos(c + d*x)**5/(5*d) + 3*a*b**2*x*sin(c + d*x)**6/16 + 9*a*b**2*x*sin(c + d*x)**4*cos
(c + d*x)**2/16 + 9*a*b**2*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 3*a*b**2*x*cos(c + d*x)**6/16 + 3*a*b**2*sin
(c + d*x)**5*cos(c + d*x)/(16*d) - a*b**2*sin(c + d*x)**3*cos(c + d*x)**3/(2*d) - 3*a*b**2*sin(c + d*x)*cos(c
+ d*x)**5/(16*d) - b**3*sin(c + d*x)**4*cos(c + d*x)**3/(3*d) - 4*b**3*sin(c + d*x)**2*cos(c + d*x)**5/(15*d)
- 8*b**3*cos(c + d*x)**7/(105*d), Ne(d, 0)), (x*(a + b*sin(c))**3*sin(c)**2*cos(c)**2, True))

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