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

Optimal. Leaf size=135 \[ -\frac {a^2 \cos ^7(c+d x)}{7 d}+\frac {3 a^2 \cos ^5(c+d x)}{5 d}-\frac {2 a^2 \cos ^3(c+d x)}{3 d}-\frac {a^2 \sin ^3(c+d x) \cos ^3(c+d x)}{3 d}-\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {a^2 x}{8} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2873, 2565, 14, 2568, 2635, 8, 270} \[ -\frac {a^2 \cos ^7(c+d x)}{7 d}+\frac {3 a^2 \cos ^5(c+d x)}{5 d}-\frac {2 a^2 \cos ^3(c+d x)}{3 d}-\frac {a^2 \sin ^3(c+d x) \cos ^3(c+d x)}{3 d}-\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {a^2 x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*x)/8 - (2*a^2*Cos[c + d*x]^3)/(3*d) + (3*a^2*Cos[c + d*x]^5)/(5*d) - (a^2*Cos[c + d*x]^7)/(7*d) + (a^2*Co
s[c + d*x]*Sin[c + d*x])/(8*d) - (a^2*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) - (a^2*Cos[c + d*x]^3*Sin[c + d*x]^3)
/(3*d)

Rule 8

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 270

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 2565

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 2568

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

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 2873

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

Rubi steps

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

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Mathematica [A]  time = 0.50, size = 86, normalized size = 0.64 \[ \frac {a^2 (-210 \sin (2 (c+d x))-210 \sin (4 (c+d x))+70 \sin (6 (c+d x))-1365 \cos (c+d x)-175 \cos (3 (c+d x))+147 \cos (5 (c+d x))-15 \cos (7 (c+d x))+840 c+840 d x)}{6720 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(840*c + 840*d*x - 1365*Cos[c + d*x] - 175*Cos[3*(c + d*x)] + 147*Cos[5*(c + d*x)] - 15*Cos[7*(c + d*x)]
- 210*Sin[2*(c + d*x)] - 210*Sin[4*(c + d*x)] + 70*Sin[6*(c + d*x)]))/(6720*d)

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fricas [A]  time = 0.54, size = 98, normalized size = 0.73 \[ -\frac {120 \, a^{2} \cos \left (d x + c\right )^{7} - 504 \, a^{2} \cos \left (d x + c\right )^{5} + 560 \, a^{2} \cos \left (d x + c\right )^{3} - 105 \, a^{2} d x - 35 \, {\left (8 \, a^{2} \cos \left (d x + c\right )^{5} - 14 \, a^{2} \cos \left (d x + c\right )^{3} + 3 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/840*(120*a^2*cos(d*x + c)^7 - 504*a^2*cos(d*x + c)^5 + 560*a^2*cos(d*x + c)^3 - 105*a^2*d*x - 35*(8*a^2*cos
(d*x + c)^5 - 14*a^2*cos(d*x + c)^3 + 3*a^2*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.22, size = 123, normalized size = 0.91 \[ \frac {1}{8} \, a^{2} x - \frac {a^{2} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {7 \, a^{2} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {5 \, a^{2} \cos \left (3 \, d x + 3 \, c\right )}{192 \, d} - \frac {13 \, a^{2} \cos \left (d x + c\right )}{64 \, d} + \frac {a^{2} \sin \left (6 \, d x + 6 \, c\right )}{96 \, d} - \frac {a^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {a^{2} \sin \left (2 \, d x + 2 \, c\right )}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a^2*x - 1/448*a^2*cos(7*d*x + 7*c)/d + 7/320*a^2*cos(5*d*x + 5*c)/d - 5/192*a^2*cos(3*d*x + 3*c)/d - 13/64
*a^2*cos(d*x + c)/d + 1/96*a^2*sin(6*d*x + 6*c)/d - 1/32*a^2*sin(4*d*x + 4*c)/d - 1/32*a^2*sin(2*d*x + 2*c)/d

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maple [A]  time = 0.17, size = 151, normalized size = 1.12 \[ \frac {a^{2} \left (-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{7}-\frac {4 \left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{35}-\frac {8 \left (\cos ^{3}\left (d x +c \right )\right )}{105}\right )+2 a^{2} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{6}-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{8}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )+a^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^2*(-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)+2*a^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)+a^2*(-1/5*sin(d*x+
c)^2*cos(d*x+c)^3-2/15*cos(d*x+c)^3))

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maxima [A]  time = 0.33, size = 105, normalized size = 0.78 \[ -\frac {32 \, {\left (15 \, \cos \left (d x + c\right )^{7} - 42 \, \cos \left (d x + c\right )^{5} + 35 \, \cos \left (d x + c\right )^{3}\right )} a^{2} - 224 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{2} + 35 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 12 \, d x - 12 \, c + 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2}}{3360 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3360*(32*(15*cos(d*x + c)^7 - 42*cos(d*x + c)^5 + 35*cos(d*x + c)^3)*a^2 - 224*(3*cos(d*x + c)^5 - 5*cos(d*
x + c)^3)*a^2 + 35*(4*sin(2*d*x + 2*c)^3 - 12*d*x - 12*c + 3*sin(4*d*x + 4*c))*a^2)/d

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mupad [B]  time = 12.31, size = 331, normalized size = 2.45 \[ \frac {a^2\,x}{8}-\frac {\frac {a^2\,\left (c+d\,x\right )}{8}+\frac {5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-\frac {97\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}+\frac {97\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{12}-\frac {5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{3}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{4}-\frac {a^2\,\left (105\,c+105\,d\,x-352\right )}{840}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {7\,a^2\,\left (c+d\,x\right )}{8}-\frac {a^2\,\left (735\,c+735\,d\,x-2464\right )}{840}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (\frac {21\,a^2\,\left (c+d\,x\right )}{8}-\frac {a^2\,\left (2205\,c+2205\,d\,x-3360\right )}{840}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {21\,a^2\,\left (c+d\,x\right )}{8}-\frac {a^2\,\left (2205\,c+2205\,d\,x-4032\right )}{840}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {35\,a^2\,\left (c+d\,x\right )}{8}-\frac {a^2\,\left (3675\,c+3675\,d\,x+2240\right )}{840}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {35\,a^2\,\left (c+d\,x\right )}{8}-\frac {a^2\,\left (3675\,c+3675\,d\,x-14560\right )}{840}\right )+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*x)/8 - ((a^2*(c + d*x))/8 + (5*a^2*tan(c/2 + (d*x)/2)^3)/3 - (97*a^2*tan(c/2 + (d*x)/2)^5)/12 + (97*a^2*t
an(c/2 + (d*x)/2)^9)/12 - (5*a^2*tan(c/2 + (d*x)/2)^11)/3 - (a^2*tan(c/2 + (d*x)/2)^13)/4 - (a^2*(105*c + 105*
d*x - 352))/840 + tan(c/2 + (d*x)/2)^2*((7*a^2*(c + d*x))/8 - (a^2*(735*c + 735*d*x - 2464))/840) + tan(c/2 +
(d*x)/2)^10*((21*a^2*(c + d*x))/8 - (a^2*(2205*c + 2205*d*x - 3360))/840) + tan(c/2 + (d*x)/2)^4*((21*a^2*(c +
d*x))/8 - (a^2*(2205*c + 2205*d*x - 4032))/840) + tan(c/2 + (d*x)/2)^6*((35*a^2*(c + d*x))/8 - (a^2*(3675*c +
3675*d*x + 2240))/840) + tan(c/2 + (d*x)/2)^8*((35*a^2*(c + d*x))/8 - (a^2*(3675*c + 3675*d*x - 14560))/840)
+ (a^2*tan(c/2 + (d*x)/2))/4)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^7)

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sympy [A]  time = 7.69, size = 275, normalized size = 2.04 \[ \begin {cases} \frac {a^{2} x \sin ^{6}{\left (c + d x \right )}}{8} + \frac {3 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8} + \frac {3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a^{2} x \cos ^{6}{\left (c + d x \right )}}{8} + \frac {a^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {a^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {4 a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac {a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{8 d} - \frac {8 a^{2} \cos ^{7}{\left (c + d x \right )}}{105 d} - \frac {2 a^{2} \cos ^{5}{\left (c + d x \right )}}{15 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{2} \sin ^{3}{\relax (c )} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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