3.1105 \(\int \cos ^4(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=301 \[ \frac{\left (9 a^2+4 b^2\right ) \cos ^3(c+d x)}{315 d}-\frac{\left (9 a^2+4 b^2\right ) \cos (c+d x)}{105 d}-\frac{a \left (10 a^2-29 b^2\right ) \sin ^5(c+d x) \cos (c+d x)}{504 b d}-\frac{5 \left (3 a^2-8 b^2\right ) \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{252 b^2 d}-\frac{\left (-44 a^2 b^2+15 a^4+6 b^4\right ) \sin ^4(c+d x) \cos (c+d x)}{630 b^2 d}+\frac{a \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac{\sin ^5(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{9 b d}-\frac{a b \sin ^3(c+d x) \cos (c+d x)}{32 d}-\frac{3 a b \sin (c+d x) \cos (c+d x)}{64 d}+\frac{3 a b x}{64} \]

[Out]

(3*a*b*x)/64 - ((9*a^2 + 4*b^2)*Cos[c + d*x])/(105*d) + ((9*a^2 + 4*b^2)*Cos[c + d*x]^3)/(315*d) - (3*a*b*Cos[
c + d*x]*Sin[c + d*x])/(64*d) - (a*b*Cos[c + d*x]*Sin[c + d*x]^3)/(32*d) - ((15*a^4 - 44*a^2*b^2 + 6*b^4)*Cos[
c + d*x]*Sin[c + d*x]^4)/(630*b^2*d) - (a*(10*a^2 - 29*b^2)*Cos[c + d*x]*Sin[c + d*x]^5)/(504*b*d) - (5*(3*a^2
- 8*b^2)*Cos[c + d*x]*Sin[c + d*x]^4*(a + b*Sin[c + d*x])^2)/(252*b^2*d) + (a*Cos[c + d*x]*Sin[c + d*x]^4*(a
+ b*Sin[c + d*x])^3)/(12*b^2*d) - (Cos[c + d*x]*Sin[c + d*x]^5*(a + b*Sin[c + d*x])^3)/(9*b*d)

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Rubi [A]  time = 0.653838, antiderivative size = 301, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {2895, 3049, 3033, 3023, 2748, 2633, 2635, 8} \[ \frac{\left (9 a^2+4 b^2\right ) \cos ^3(c+d x)}{315 d}-\frac{\left (9 a^2+4 b^2\right ) \cos (c+d x)}{105 d}-\frac{a \left (10 a^2-29 b^2\right ) \sin ^5(c+d x) \cos (c+d x)}{504 b d}-\frac{5 \left (3 a^2-8 b^2\right ) \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^2}{252 b^2 d}-\frac{\left (-44 a^2 b^2+15 a^4+6 b^4\right ) \sin ^4(c+d x) \cos (c+d x)}{630 b^2 d}+\frac{a \sin ^4(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac{\sin ^5(c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{9 b d}-\frac{a b \sin ^3(c+d x) \cos (c+d x)}{32 d}-\frac{3 a b \sin (c+d x) \cos (c+d x)}{64 d}+\frac{3 a b x}{64} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*Sin[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]

[Out]

(3*a*b*x)/64 - ((9*a^2 + 4*b^2)*Cos[c + d*x])/(105*d) + ((9*a^2 + 4*b^2)*Cos[c + d*x]^3)/(315*d) - (3*a*b*Cos[
c + d*x]*Sin[c + d*x])/(64*d) - (a*b*Cos[c + d*x]*Sin[c + d*x]^3)/(32*d) - ((15*a^4 - 44*a^2*b^2 + 6*b^4)*Cos[
c + d*x]*Sin[c + d*x]^4)/(630*b^2*d) - (a*(10*a^2 - 29*b^2)*Cos[c + d*x]*Sin[c + d*x]^5)/(504*b*d) - (5*(3*a^2
- 8*b^2)*Cos[c + d*x]*Sin[c + d*x]^4*(a + b*Sin[c + d*x])^2)/(252*b^2*d) + (a*Cos[c + d*x]*Sin[c + d*x]^4*(a
+ b*Sin[c + d*x])^3)/(12*b^2*d) - (Cos[c + d*x]*Sin[c + d*x]^5*(a + b*Sin[c + d*x])^3)/(9*b*d)

Rule 2895

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(a*(n + 3)*Cos[e + f*x]*(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 1))/(b^2*d*f*(m
+ n + 3)*(m + n + 4)), x] + (-Dist[1/(b^2*(m + n + 3)*(m + n + 4)), Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x
])^m*Simp[a^2*(n + 1)*(n + 3) - b^2*(m + n + 3)*(m + n + 4) + a*b*m*Sin[e + f*x] - (a^2*(n + 2)*(n + 3) - b^2*
(m + n + 3)*(m + n + 5))*Sin[e + f*x]^2, x], x], x] - Simp[(Cos[e + f*x]*(d*Sin[e + f*x])^(n + 2)*(a + b*Sin[e
+ f*x])^(m + 1))/(b*d^2*f*(m + n + 4)), x]) /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[
m, 0] || IntegersQ[2*m, 2*n]) &&  !m < -1 &&  !LtQ[n, -1] && NeQ[m + n + 3, 0] && NeQ[m + n + 4, 0]

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 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 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 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 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 8

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

Rubi steps

\begin{align*} \int \cos ^4(c+d x) \sin ^3(c+d x) (a+b \sin (c+d x))^2 \, dx &=\frac{a \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac{\cos (c+d x) \sin ^5(c+d x) (a+b \sin (c+d x))^3}{9 b d}-\frac{\int \sin ^3(c+d x) (a+b \sin (c+d x))^2 \left (24 \left (a^2-3 b^2\right )+2 a b \sin (c+d x)-10 \left (3 a^2-8 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{72 b^2}\\ &=-\frac{5 \left (3 a^2-8 b^2\right ) \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2}{252 b^2 d}+\frac{a \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac{\cos (c+d x) \sin ^5(c+d x) (a+b \sin (c+d x))^3}{9 b d}-\frac{\int \sin ^3(c+d x) (a+b \sin (c+d x)) \left (8 a \left (6 a^2-23 b^2\right )+2 b \left (a^2-12 b^2\right ) \sin (c+d x)-6 a \left (10 a^2-29 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{504 b^2}\\ &=-\frac{a \left (10 a^2-29 b^2\right ) \cos (c+d x) \sin ^5(c+d x)}{504 b d}-\frac{5 \left (3 a^2-8 b^2\right ) \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2}{252 b^2 d}+\frac{a \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac{\cos (c+d x) \sin ^5(c+d x) (a+b \sin (c+d x))^3}{9 b d}-\frac{\int \sin ^3(c+d x) \left (48 a^2 \left (6 a^2-23 b^2\right )-378 a b^3 \sin (c+d x)-24 \left (15 a^4-44 a^2 b^2+6 b^4\right ) \sin ^2(c+d x)\right ) \, dx}{3024 b^2}\\ &=-\frac{\left (15 a^4-44 a^2 b^2+6 b^4\right ) \cos (c+d x) \sin ^4(c+d x)}{630 b^2 d}-\frac{a \left (10 a^2-29 b^2\right ) \cos (c+d x) \sin ^5(c+d x)}{504 b d}-\frac{5 \left (3 a^2-8 b^2\right ) \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2}{252 b^2 d}+\frac{a \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac{\cos (c+d x) \sin ^5(c+d x) (a+b \sin (c+d x))^3}{9 b d}-\frac{\int \sin ^3(c+d x) \left (-144 b^2 \left (9 a^2+4 b^2\right )-1890 a b^3 \sin (c+d x)\right ) \, dx}{15120 b^2}\\ &=-\frac{\left (15 a^4-44 a^2 b^2+6 b^4\right ) \cos (c+d x) \sin ^4(c+d x)}{630 b^2 d}-\frac{a \left (10 a^2-29 b^2\right ) \cos (c+d x) \sin ^5(c+d x)}{504 b d}-\frac{5 \left (3 a^2-8 b^2\right ) \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2}{252 b^2 d}+\frac{a \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac{\cos (c+d x) \sin ^5(c+d x) (a+b \sin (c+d x))^3}{9 b d}+\frac{1}{8} (a b) \int \sin ^4(c+d x) \, dx-\frac{1}{105} \left (-9 a^2-4 b^2\right ) \int \sin ^3(c+d x) \, dx\\ &=-\frac{a b \cos (c+d x) \sin ^3(c+d x)}{32 d}-\frac{\left (15 a^4-44 a^2 b^2+6 b^4\right ) \cos (c+d x) \sin ^4(c+d x)}{630 b^2 d}-\frac{a \left (10 a^2-29 b^2\right ) \cos (c+d x) \sin ^5(c+d x)}{504 b d}-\frac{5 \left (3 a^2-8 b^2\right ) \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2}{252 b^2 d}+\frac{a \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac{\cos (c+d x) \sin ^5(c+d x) (a+b \sin (c+d x))^3}{9 b d}+\frac{1}{32} (3 a b) \int \sin ^2(c+d x) \, dx-\frac{\left (9 a^2+4 b^2\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{105 d}\\ &=-\frac{\left (9 a^2+4 b^2\right ) \cos (c+d x)}{105 d}+\frac{\left (9 a^2+4 b^2\right ) \cos ^3(c+d x)}{315 d}-\frac{3 a b \cos (c+d x) \sin (c+d x)}{64 d}-\frac{a b \cos (c+d x) \sin ^3(c+d x)}{32 d}-\frac{\left (15 a^4-44 a^2 b^2+6 b^4\right ) \cos (c+d x) \sin ^4(c+d x)}{630 b^2 d}-\frac{a \left (10 a^2-29 b^2\right ) \cos (c+d x) \sin ^5(c+d x)}{504 b d}-\frac{5 \left (3 a^2-8 b^2\right ) \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2}{252 b^2 d}+\frac{a \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac{\cos (c+d x) \sin ^5(c+d x) (a+b \sin (c+d x))^3}{9 b d}+\frac{1}{64} (3 a b) \int 1 \, dx\\ &=\frac{3 a b x}{64}-\frac{\left (9 a^2+4 b^2\right ) \cos (c+d x)}{105 d}+\frac{\left (9 a^2+4 b^2\right ) \cos ^3(c+d x)}{315 d}-\frac{3 a b \cos (c+d x) \sin (c+d x)}{64 d}-\frac{a b \cos (c+d x) \sin ^3(c+d x)}{32 d}-\frac{\left (15 a^4-44 a^2 b^2+6 b^4\right ) \cos (c+d x) \sin ^4(c+d x)}{630 b^2 d}-\frac{a \left (10 a^2-29 b^2\right ) \cos (c+d x) \sin ^5(c+d x)}{504 b d}-\frac{5 \left (3 a^2-8 b^2\right ) \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^2}{252 b^2 d}+\frac{a \cos (c+d x) \sin ^4(c+d x) (a+b \sin (c+d x))^3}{12 b^2 d}-\frac{\cos (c+d x) \sin ^5(c+d x) (a+b \sin (c+d x))^3}{9 b d}\\ \end{align*}

Mathematica [A]  time = 0.956657, size = 144, normalized size = 0.48 \[ \frac{-3780 \left (2 a^2+b^2\right ) \cos (c+d x)-840 \left (3 a^2+b^2\right ) \cos (3 (c+d x))+504 a^2 \cos (5 (c+d x))+360 a^2 \cos (7 (c+d x))-2520 a b \sin (4 (c+d x))+315 a b \sin (8 (c+d x))+7560 a b c+7560 a b d x+504 b^2 \cos (5 (c+d x))+90 b^2 \cos (7 (c+d x))-70 b^2 \cos (9 (c+d x))}{161280 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*Sin[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]

[Out]

(7560*a*b*c + 7560*a*b*d*x - 3780*(2*a^2 + b^2)*Cos[c + d*x] - 840*(3*a^2 + b^2)*Cos[3*(c + d*x)] + 504*a^2*Co
s[5*(c + d*x)] + 504*b^2*Cos[5*(c + d*x)] + 360*a^2*Cos[7*(c + d*x)] + 90*b^2*Cos[7*(c + d*x)] - 70*b^2*Cos[9*
(c + d*x)] - 2520*a*b*Sin[4*(c + d*x)] + 315*a*b*Sin[8*(c + d*x)])/(161280*d)

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Maple [A]  time = 0.043, size = 161, normalized size = 0.5 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{7}}-{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{35}} \right ) +2\,ab \left ( -1/8\, \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{5}-1/16\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{ \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) }{64}}+{\frac{3\,dx}{128}}+{\frac{3\,c}{128}} \right ) +{b}^{2} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{9}}-{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{63}}-{\frac{8\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{315}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*sin(d*x+c)^3*(a+b*sin(d*x+c))^2,x)

[Out]

1/d*(a^2*(-1/7*sin(d*x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5)+2*a*b*(-1/8*sin(d*x+c)^3*cos(d*x+c)^5-1/16*sin(d*x
+c)*cos(d*x+c)^5+1/64*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/128*d*x+3/128*c)+b^2*(-1/9*sin(d*x+c)^4*cos(d
*x+c)^5-4/63*sin(d*x+c)^2*cos(d*x+c)^5-8/315*cos(d*x+c)^5))

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Maxima [A]  time = 0.998958, size = 135, normalized size = 0.45 \begin{align*} \frac{4608 \,{\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} a^{2} + 315 \,{\left (24 \, d x + 24 \, c + \sin \left (8 \, d x + 8 \, c\right ) - 8 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a b - 512 \,{\left (35 \, \cos \left (d x + c\right )^{9} - 90 \, \cos \left (d x + c\right )^{7} + 63 \, \cos \left (d x + c\right )^{5}\right )} b^{2}}{161280 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/161280*(4608*(5*cos(d*x + c)^7 - 7*cos(d*x + c)^5)*a^2 + 315*(24*d*x + 24*c + sin(8*d*x + 8*c) - 8*sin(4*d*x
+ 4*c))*a*b - 512*(35*cos(d*x + c)^9 - 90*cos(d*x + c)^7 + 63*cos(d*x + c)^5)*b^2)/d

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Fricas [A]  time = 1.86421, size = 315, normalized size = 1.05 \begin{align*} -\frac{2240 \, b^{2} \cos \left (d x + c\right )^{9} - 2880 \,{\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{7} + 4032 \,{\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{5} - 945 \, a b d x - 315 \,{\left (16 \, a b \cos \left (d x + c\right )^{7} - 24 \, a b \cos \left (d x + c\right )^{5} + 2 \, a b \cos \left (d x + c\right )^{3} + 3 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{20160 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/20160*(2240*b^2*cos(d*x + c)^9 - 2880*(a^2 + 2*b^2)*cos(d*x + c)^7 + 4032*(a^2 + b^2)*cos(d*x + c)^5 - 945*
a*b*d*x - 315*(16*a*b*cos(d*x + c)^7 - 24*a*b*cos(d*x + c)^5 + 2*a*b*cos(d*x + c)^3 + 3*a*b*cos(d*x + c))*sin(
d*x + c))/d

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Sympy [A]  time = 23.89, size = 335, normalized size = 1.11 \begin{align*} \begin{cases} - \frac{a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac{2 a^{2} \cos ^{7}{\left (c + d x \right )}}{35 d} + \frac{3 a b x \sin ^{8}{\left (c + d x \right )}}{64} + \frac{3 a b x \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{9 a b x \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{32} + \frac{3 a b x \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{16} + \frac{3 a b x \cos ^{8}{\left (c + d x \right )}}{64} + \frac{3 a b \sin ^{7}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{64 d} + \frac{11 a b \sin ^{5}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{64 d} - \frac{11 a b \sin ^{3}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{64 d} - \frac{3 a b \sin{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{64 d} - \frac{b^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac{4 b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{35 d} - \frac{8 b^{2} \cos ^{9}{\left (c + d x \right )}}{315 d} & \text{for}\: d \neq 0 \\x \left (a + b \sin{\left (c \right )}\right )^{2} \sin ^{3}{\left (c \right )} \cos ^{4}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*sin(d*x+c)**3*(a+b*sin(d*x+c))**2,x)

[Out]

Piecewise((-a**2*sin(c + d*x)**2*cos(c + d*x)**5/(5*d) - 2*a**2*cos(c + d*x)**7/(35*d) + 3*a*b*x*sin(c + d*x)*
*8/64 + 3*a*b*x*sin(c + d*x)**6*cos(c + d*x)**2/16 + 9*a*b*x*sin(c + d*x)**4*cos(c + d*x)**4/32 + 3*a*b*x*sin(
c + d*x)**2*cos(c + d*x)**6/16 + 3*a*b*x*cos(c + d*x)**8/64 + 3*a*b*sin(c + d*x)**7*cos(c + d*x)/(64*d) + 11*a
*b*sin(c + d*x)**5*cos(c + d*x)**3/(64*d) - 11*a*b*sin(c + d*x)**3*cos(c + d*x)**5/(64*d) - 3*a*b*sin(c + d*x)
*cos(c + d*x)**7/(64*d) - b**2*sin(c + d*x)**4*cos(c + d*x)**5/(5*d) - 4*b**2*sin(c + d*x)**2*cos(c + d*x)**7/
(35*d) - 8*b**2*cos(c + d*x)**9/(315*d), Ne(d, 0)), (x*(a + b*sin(c))**2*sin(c)**3*cos(c)**4, True))

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Giac [A]  time = 1.35921, size = 192, normalized size = 0.64 \begin{align*} \frac{3}{64} \, a b x - \frac{b^{2} \cos \left (9 \, d x + 9 \, c\right )}{2304 \, d} + \frac{a b \sin \left (8 \, d x + 8 \, c\right )}{512 \, d} - \frac{a b \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac{{\left (4 \, a^{2} + b^{2}\right )} \cos \left (7 \, d x + 7 \, c\right )}{1792 \, d} + \frac{{\left (a^{2} + b^{2}\right )} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac{{\left (3 \, a^{2} + b^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{192 \, d} - \frac{3 \,{\left (2 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )}{128 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

3/64*a*b*x - 1/2304*b^2*cos(9*d*x + 9*c)/d + 1/512*a*b*sin(8*d*x + 8*c)/d - 1/64*a*b*sin(4*d*x + 4*c)/d + 1/17
92*(4*a^2 + b^2)*cos(7*d*x + 7*c)/d + 1/320*(a^2 + b^2)*cos(5*d*x + 5*c)/d - 1/192*(3*a^2 + b^2)*cos(3*d*x + 3
*c)/d - 3/128*(2*a^2 + b^2)*cos(d*x + c)/d