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

Optimal. Leaf size=161 \[ -\frac {\left (a^2+3 b^2\right ) \cos ^5(c+d x)}{5 d}+\frac {\left (2 a^2+3 b^2\right ) \cos ^3(c+d x)}{3 d}-\frac {\left (a^2+b^2\right ) \cos (c+d x)}{d}-\frac {a b \sin ^5(c+d x) \cos (c+d x)}{3 d}-\frac {5 a b \sin ^3(c+d x) \cos (c+d x)}{12 d}-\frac {5 a b \sin (c+d x) \cos (c+d x)}{8 d}+\frac {5 a b x}{8}+\frac {b^2 \cos ^7(c+d x)}{7 d} \]

[Out]

5/8*a*b*x-(a^2+b^2)*cos(d*x+c)/d+1/3*(2*a^2+3*b^2)*cos(d*x+c)^3/d-1/5*(a^2+3*b^2)*cos(d*x+c)^5/d+1/7*b^2*cos(d
*x+c)^7/d-5/8*a*b*cos(d*x+c)*sin(d*x+c)/d-5/12*a*b*cos(d*x+c)*sin(d*x+c)^3/d-1/3*a*b*cos(d*x+c)*sin(d*x+c)^5/d

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Rubi [A]  time = 0.27, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4393, 2789, 2635, 8, 3013, 373} \[ -\frac {\left (a^2+3 b^2\right ) \cos ^5(c+d x)}{5 d}+\frac {\left (2 a^2+3 b^2\right ) \cos ^3(c+d x)}{3 d}-\frac {\left (a^2+b^2\right ) \cos (c+d x)}{d}-\frac {a b \sin ^5(c+d x) \cos (c+d x)}{3 d}-\frac {5 a b \sin ^3(c+d x) \cos (c+d x)}{12 d}-\frac {5 a b \sin (c+d x) \cos (c+d x)}{8 d}+\frac {5 a b x}{8}+\frac {b^2 \cos ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a*b*x)/8 - ((a^2 + b^2)*Cos[c + d*x])/d + ((2*a^2 + 3*b^2)*Cos[c + d*x]^3)/(3*d) - ((a^2 + 3*b^2)*Cos[c + d
*x]^5)/(5*d) + (b^2*Cos[c + d*x]^7)/(7*d) - (5*a*b*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (5*a*b*Cos[c + d*x]*Sin[
c + d*x]^3)/(12*d) - (a*b*Cos[c + d*x]*Sin[c + d*x]^5)/(3*d)

Rule 8

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

Rule 373

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 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 2789

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Dist[(2*c*d)/b
, Int[(b*Sin[e + f*x])^(m + 1), x], x] + Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c,
d, e, f, m}, x]

Rule 3013

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Dist[f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

Rule 4393

Int[(u_)*((a_)*(F_)[(c_.) + (d_.)*(x_)]^(p_.) + (b_.)*(F_)[(c_.) + (d_.)*(x_)]^(q_.))^(n_.), x_Symbol] :> Int[
ActivateTrig[u*F[c + d*x]^(n*p)*(a + b*F[c + d*x]^(q - p))^n], x] /; FreeQ[{a, b, c, d, p, q}, x] && InertTrig
Q[F] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int \sin (c+d x) \left (a \sin ^2(c+d x)+b \sin ^3(c+d x)\right )^2 \, dx &=\int \sin ^5(c+d x) (a+b \sin (c+d x))^2 \, dx\\ &=(2 a b) \int \sin ^6(c+d x) \, dx+\int \sin ^5(c+d x) \left (a^2+b^2 \sin ^2(c+d x)\right ) \, dx\\ &=-\frac {a b \cos (c+d x) \sin ^5(c+d x)}{3 d}+\frac {1}{3} (5 a b) \int \sin ^4(c+d x) \, dx-\frac {\operatorname {Subst}\left (\int \left (1-x^2\right )^2 \left (a^2+b^2-b^2 x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {5 a b \cos (c+d x) \sin ^3(c+d x)}{12 d}-\frac {a b \cos (c+d x) \sin ^5(c+d x)}{3 d}+\frac {1}{4} (5 a b) \int \sin ^2(c+d x) \, dx-\frac {\operatorname {Subst}\left (\int \left (a^2 \left (1+\frac {b^2}{a^2}\right )-\left (2 a^2+3 b^2\right ) x^2+\left (a^2+3 b^2\right ) x^4-b^2 x^6\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {\left (a^2+b^2\right ) \cos (c+d x)}{d}+\frac {\left (2 a^2+3 b^2\right ) \cos ^3(c+d x)}{3 d}-\frac {\left (a^2+3 b^2\right ) \cos ^5(c+d x)}{5 d}+\frac {b^2 \cos ^7(c+d x)}{7 d}-\frac {5 a b \cos (c+d x) \sin (c+d x)}{8 d}-\frac {5 a b \cos (c+d x) \sin ^3(c+d x)}{12 d}-\frac {a b \cos (c+d x) \sin ^5(c+d x)}{3 d}+\frac {1}{8} (5 a b) \int 1 \, dx\\ &=\frac {5 a b x}{8}-\frac {\left (a^2+b^2\right ) \cos (c+d x)}{d}+\frac {\left (2 a^2+3 b^2\right ) \cos ^3(c+d x)}{3 d}-\frac {\left (a^2+3 b^2\right ) \cos ^5(c+d x)}{5 d}+\frac {b^2 \cos ^7(c+d x)}{7 d}-\frac {5 a b \cos (c+d x) \sin (c+d x)}{8 d}-\frac {5 a b \cos (c+d x) \sin ^3(c+d x)}{12 d}-\frac {a b \cos (c+d x) \sin ^5(c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 134, normalized size = 0.83 \[ \frac {-525 \left (8 a^2+7 b^2\right ) \cos (c+d x)+35 \left (20 a^2+21 b^2\right ) \cos (3 (c+d x))-84 a^2 \cos (5 (c+d x))-3150 a b \sin (2 (c+d x))+630 a b \sin (4 (c+d x))-70 a b \sin (6 (c+d x))+4200 a b c+4200 a b d x-147 b^2 \cos (5 (c+d x))+15 b^2 \cos (7 (c+d x))}{6720 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4200*a*b*c + 4200*a*b*d*x - 525*(8*a^2 + 7*b^2)*Cos[c + d*x] + 35*(20*a^2 + 21*b^2)*Cos[3*(c + d*x)] - 84*a^2
*Cos[5*(c + d*x)] - 147*b^2*Cos[5*(c + d*x)] + 15*b^2*Cos[7*(c + d*x)] - 3150*a*b*Sin[2*(c + d*x)] + 630*a*b*S
in[4*(c + d*x)] - 70*a*b*Sin[6*(c + d*x)])/(6720*d)

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fricas [A]  time = 0.91, size = 123, normalized size = 0.76 \[ \frac {120 \, b^{2} \cos \left (d x + c\right )^{7} - 168 \, {\left (a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{5} + 525 \, a b d x + 280 \, {\left (2 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{3} - 840 \, {\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right ) - 35 \, {\left (8 \, a b \cos \left (d x + c\right )^{5} - 26 \, a b \cos \left (d x + c\right )^{3} + 33 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/840*(120*b^2*cos(d*x + c)^7 - 168*(a^2 + 3*b^2)*cos(d*x + c)^5 + 525*a*b*d*x + 280*(2*a^2 + 3*b^2)*cos(d*x +
c)^3 - 840*(a^2 + b^2)*cos(d*x + c) - 35*(8*a*b*cos(d*x + c)^5 - 26*a*b*cos(d*x + c)^3 + 33*a*b*cos(d*x + c))
*sin(d*x + c))/d

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giac [A]  time = 0.19, size = 143, normalized size = 0.89 \[ \frac {5}{8} \, a b x + \frac {b^{2} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac {a b \sin \left (6 \, d x + 6 \, c\right )}{96 \, d} + \frac {3 \, a b \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {15 \, a b \sin \left (2 \, d x + 2 \, c\right )}{32 \, d} - \frac {{\left (4 \, a^{2} + 7 \, b^{2}\right )} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac {{\left (20 \, a^{2} + 21 \, b^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{192 \, d} - \frac {5 \, {\left (8 \, a^{2} + 7 \, b^{2}\right )} \cos \left (d x + c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/8*a*b*x + 1/448*b^2*cos(7*d*x + 7*c)/d - 1/96*a*b*sin(6*d*x + 6*c)/d + 3/32*a*b*sin(4*d*x + 4*c)/d - 15/32*a
*b*sin(2*d*x + 2*c)/d - 1/320*(4*a^2 + 7*b^2)*cos(5*d*x + 5*c)/d + 1/192*(20*a^2 + 21*b^2)*cos(3*d*x + 3*c)/d
- 5/64*(8*a^2 + 7*b^2)*cos(d*x + c)/d

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maple [A]  time = 0.08, size = 125, normalized size = 0.78 \[ \frac {-\frac {b^{2} \left (\frac {16}{5}+\sin ^{6}\left (d x +c \right )+\frac {6 \left (\sin ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (d x +c \right )\right )}{5}\right ) \cos \left (d x +c \right )}{7}+2 a b \left (-\frac {\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )-\frac {a^{2} \left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )}{5}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-1/7*b^2*(16/5+sin(d*x+c)^6+6/5*sin(d*x+c)^4+8/5*sin(d*x+c)^2)*cos(d*x+c)+2*a*b*(-1/6*(sin(d*x+c)^5+5/4*s
in(d*x+c)^3+15/8*sin(d*x+c))*cos(d*x+c)+5/16*d*x+5/16*c)-1/5*a^2*(8/3+sin(d*x+c)^4+4/3*sin(d*x+c)^2)*cos(d*x+c
))

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maxima [A]  time = 0.33, size = 131, normalized size = 0.81 \[ -\frac {224 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 10 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )} a^{2} - 35 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 60 \, d x + 60 \, c + 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a b - 96 \, {\left (5 \, \cos \left (d x + c\right )^{7} - 21 \, \cos \left (d x + c\right )^{5} + 35 \, \cos \left (d x + c\right )^{3} - 35 \, \cos \left (d x + c\right )\right )} b^{2}}{3360 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3360*(224*(3*cos(d*x + c)^5 - 10*cos(d*x + c)^3 + 15*cos(d*x + c))*a^2 - 35*(4*sin(2*d*x + 2*c)^3 + 60*d*x
+ 60*c + 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*a*b - 96*(5*cos(d*x + c)^7 - 21*cos(d*x + c)^5 + 35*cos(d*x
+ c)^3 - 35*cos(d*x + c))*b^2)/d

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mupad [B]  time = 6.74, size = 210, normalized size = 1.30 \[ \frac {5\,a\,b\,x}{8}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {80\,a^2}{3}+32\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {112\,a^2}{15}+\frac {32\,b^2}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {112\,a^2}{5}+\frac {96\,b^2}{5}\right )+\frac {32\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{3}+\frac {16\,a^2}{15}+\frac {32\,b^2}{35}+\frac {25\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {283\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}-\frac {283\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{12}-\frac {25\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{3}-\frac {5\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{4}+\frac {5\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*a*b*x)/8 - (tan(c/2 + (d*x)/2)^6*((80*a^2)/3 + 32*b^2) + tan(c/2 + (d*x)/2)^2*((112*a^2)/15 + (32*b^2)/5) +
tan(c/2 + (d*x)/2)^4*((112*a^2)/5 + (96*b^2)/5) + (32*a^2*tan(c/2 + (d*x)/2)^8)/3 + (16*a^2)/15 + (32*b^2)/35
+ (25*a*b*tan(c/2 + (d*x)/2)^3)/3 + (283*a*b*tan(c/2 + (d*x)/2)^5)/12 - (283*a*b*tan(c/2 + (d*x)/2)^9)/12 - (
25*a*b*tan(c/2 + (d*x)/2)^11)/3 - (5*a*b*tan(c/2 + (d*x)/2)^13)/4 + (5*a*b*tan(c/2 + (d*x)/2))/4)/(d*(tan(c/2
+ (d*x)/2)^2 + 1)^7)

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sympy [A]  time = 6.01, size = 326, normalized size = 2.02 \[ \begin {cases} - \frac {a^{2} \sin ^{4}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} - \frac {4 a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {8 a^{2} \cos ^{5}{\left (c + d x \right )}}{15 d} + \frac {5 a b x \sin ^{6}{\left (c + d x \right )}}{8} + \frac {15 a b x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8} + \frac {15 a b x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{8} + \frac {5 a b x \cos ^{6}{\left (c + d x \right )}}{8} - \frac {11 a b \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {5 a b \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {5 a b \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{8 d} - \frac {b^{2} \sin ^{6}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} - \frac {2 b^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{d} - \frac {8 b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {16 b^{2} \cos ^{7}{\left (c + d x \right )}}{35 d} & \text {for}\: d \neq 0 \\x \left (a \sin ^{2}{\relax (c )} + b \sin ^{3}{\relax (c )}\right )^{2} \sin {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*sin(c + d*x)**4*cos(c + d*x)/d - 4*a**2*sin(c + d*x)**2*cos(c + d*x)**3/(3*d) - 8*a**2*cos(c
+ d*x)**5/(15*d) + 5*a*b*x*sin(c + d*x)**6/8 + 15*a*b*x*sin(c + d*x)**4*cos(c + d*x)**2/8 + 15*a*b*x*sin(c + d
*x)**2*cos(c + d*x)**4/8 + 5*a*b*x*cos(c + d*x)**6/8 - 11*a*b*sin(c + d*x)**5*cos(c + d*x)/(8*d) - 5*a*b*sin(c
+ d*x)**3*cos(c + d*x)**3/(3*d) - 5*a*b*sin(c + d*x)*cos(c + d*x)**5/(8*d) - b**2*sin(c + d*x)**6*cos(c + d*x
)/d - 2*b**2*sin(c + d*x)**4*cos(c + d*x)**3/d - 8*b**2*sin(c + d*x)**2*cos(c + d*x)**5/(5*d) - 16*b**2*cos(c
+ d*x)**7/(35*d), Ne(d, 0)), (x*(a*sin(c)**2 + b*sin(c)**3)**2*sin(c), True))

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