3.1243 \(\int \cos ^6(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=152 \[ -\frac {\left (a^2+8 b^2\right ) \cos ^7(c+d x)}{252 d}-\frac {\cos ^7(c+d x) (a+b \sin (c+d x))^2}{9 d}-\frac {a \cos ^7(c+d x) (a+b \sin (c+d x))}{36 d}+\frac {a b \sin (c+d x) \cos ^5(c+d x)}{24 d}+\frac {5 a b \sin (c+d x) \cos ^3(c+d x)}{96 d}+\frac {5 a b \sin (c+d x) \cos (c+d x)}{64 d}+\frac {5 a b x}{64} \]

[Out]

5/64*a*b*x-1/252*(a^2+8*b^2)*cos(d*x+c)^7/d+5/64*a*b*cos(d*x+c)*sin(d*x+c)/d+5/96*a*b*cos(d*x+c)^3*sin(d*x+c)/
d+1/24*a*b*cos(d*x+c)^5*sin(d*x+c)/d-1/36*a*cos(d*x+c)^7*(a+b*sin(d*x+c))/d-1/9*cos(d*x+c)^7*(a+b*sin(d*x+c))^
2/d

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Rubi [A]  time = 0.21, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2862, 2669, 2635, 8} \[ -\frac {\left (a^2+8 b^2\right ) \cos ^7(c+d x)}{252 d}-\frac {\cos ^7(c+d x) (a+b \sin (c+d x))^2}{9 d}-\frac {a \cos ^7(c+d x) (a+b \sin (c+d x))}{36 d}+\frac {a b \sin (c+d x) \cos ^5(c+d x)}{24 d}+\frac {5 a b \sin (c+d x) \cos ^3(c+d x)}{96 d}+\frac {5 a b \sin (c+d x) \cos (c+d x)}{64 d}+\frac {5 a b x}{64} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*Sin[c + d*x]*(a + b*Sin[c + d*x])^2,x]

[Out]

(5*a*b*x)/64 - ((a^2 + 8*b^2)*Cos[c + d*x]^7)/(252*d) + (5*a*b*Cos[c + d*x]*Sin[c + d*x])/(64*d) + (5*a*b*Cos[
c + d*x]^3*Sin[c + d*x])/(96*d) + (a*b*Cos[c + d*x]^5*Sin[c + d*x])/(24*d) - (a*Cos[c + d*x]^7*(a + b*Sin[c +
d*x]))/(36*d) - (Cos[c + d*x]^7*(a + b*Sin[c + d*x])^2)/(9*d)

Rule 8

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

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 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
e + f*x])^(p + 1))/(f*g*(p + 1)), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x]
&& (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 2862

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[1/(m + p + 1), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1)*Simp[a*c*(m + p + 1) + b*d*m + (a*d*
m + b*c*(m + p + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && Gt
Q[m, 0] &&  !LtQ[p, -1] && IntegerQ[2*m] &&  !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && SimplerQ[c + d*x, a + b*x])

Rubi steps

\begin {align*} \int \cos ^6(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx &=-\frac {\cos ^7(c+d x) (a+b \sin (c+d x))^2}{9 d}+\frac {1}{9} \int \cos ^6(c+d x) (2 b+2 a \sin (c+d x)) (a+b \sin (c+d x)) \, dx\\ &=-\frac {a \cos ^7(c+d x) (a+b \sin (c+d x))}{36 d}-\frac {\cos ^7(c+d x) (a+b \sin (c+d x))^2}{9 d}+\frac {1}{72} \int \cos ^6(c+d x) \left (18 a b+2 \left (a^2+8 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=-\frac {\left (a^2+8 b^2\right ) \cos ^7(c+d x)}{252 d}-\frac {a \cos ^7(c+d x) (a+b \sin (c+d x))}{36 d}-\frac {\cos ^7(c+d x) (a+b \sin (c+d x))^2}{9 d}+\frac {1}{4} (a b) \int \cos ^6(c+d x) \, dx\\ &=-\frac {\left (a^2+8 b^2\right ) \cos ^7(c+d x)}{252 d}+\frac {a b \cos ^5(c+d x) \sin (c+d x)}{24 d}-\frac {a \cos ^7(c+d x) (a+b \sin (c+d x))}{36 d}-\frac {\cos ^7(c+d x) (a+b \sin (c+d x))^2}{9 d}+\frac {1}{24} (5 a b) \int \cos ^4(c+d x) \, dx\\ &=-\frac {\left (a^2+8 b^2\right ) \cos ^7(c+d x)}{252 d}+\frac {5 a b \cos ^3(c+d x) \sin (c+d x)}{96 d}+\frac {a b \cos ^5(c+d x) \sin (c+d x)}{24 d}-\frac {a \cos ^7(c+d x) (a+b \sin (c+d x))}{36 d}-\frac {\cos ^7(c+d x) (a+b \sin (c+d x))^2}{9 d}+\frac {1}{32} (5 a b) \int \cos ^2(c+d x) \, dx\\ &=-\frac {\left (a^2+8 b^2\right ) \cos ^7(c+d x)}{252 d}+\frac {5 a b \cos (c+d x) \sin (c+d x)}{64 d}+\frac {5 a b \cos ^3(c+d x) \sin (c+d x)}{96 d}+\frac {a b \cos ^5(c+d x) \sin (c+d x)}{24 d}-\frac {a \cos ^7(c+d x) (a+b \sin (c+d x))}{36 d}-\frac {\cos ^7(c+d x) (a+b \sin (c+d x))^2}{9 d}+\frac {1}{64} (5 a b) \int 1 \, dx\\ &=\frac {5 a b x}{64}-\frac {\left (a^2+8 b^2\right ) \cos ^7(c+d x)}{252 d}+\frac {5 a b \cos (c+d x) \sin (c+d x)}{64 d}+\frac {5 a b \cos ^3(c+d x) \sin (c+d x)}{96 d}+\frac {a b \cos ^5(c+d x) \sin (c+d x)}{24 d}-\frac {a \cos ^7(c+d x) (a+b \sin (c+d x))}{36 d}-\frac {\cos ^7(c+d x) (a+b \sin (c+d x))^2}{9 d}\\ \end {align*}

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Mathematica [A]  time = 1.02, size = 161, normalized size = 1.06 \[ -\frac {126 \left (10 a^2+3 b^2\right ) \cos (c+d x)+84 \left (9 a^2+2 b^2\right ) \cos (3 (c+d x))+252 a^2 \cos (5 (c+d x))+36 a^2 \cos (7 (c+d x))-504 a b \sin (2 (c+d x))+252 a b \sin (4 (c+d x))+168 a b \sin (6 (c+d x))+\frac {63}{2} a b \sin (8 (c+d x))-1260 a b c-1260 a b d x-27 b^2 \cos (7 (c+d x))-7 b^2 \cos (9 (c+d x))}{16128 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*Sin[c + d*x]*(a + b*Sin[c + d*x])^2,x]

[Out]

-1/16128*(-1260*a*b*c - 1260*a*b*d*x + 126*(10*a^2 + 3*b^2)*Cos[c + d*x] + 84*(9*a^2 + 2*b^2)*Cos[3*(c + d*x)]
+ 252*a^2*Cos[5*(c + d*x)] + 36*a^2*Cos[7*(c + d*x)] - 27*b^2*Cos[7*(c + d*x)] - 7*b^2*Cos[9*(c + d*x)] - 504
*a*b*Sin[2*(c + d*x)] + 252*a*b*Sin[4*(c + d*x)] + 168*a*b*Sin[6*(c + d*x)] + (63*a*b*Sin[8*(c + d*x)])/2)/d

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fricas [A]  time = 0.73, size = 97, normalized size = 0.64 \[ \frac {448 \, b^{2} \cos \left (d x + c\right )^{9} - 576 \, {\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{7} + 315 \, a b d x - 21 \, {\left (48 \, a b \cos \left (d x + c\right )^{7} - 8 \, a b \cos \left (d x + c\right )^{5} - 10 \, a b \cos \left (d x + c\right )^{3} - 15 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4032 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*sin(d*x+c)*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/4032*(448*b^2*cos(d*x + c)^9 - 576*(a^2 + b^2)*cos(d*x + c)^7 + 315*a*b*d*x - 21*(48*a*b*cos(d*x + c)^7 - 8*
a*b*cos(d*x + c)^5 - 10*a*b*cos(d*x + c)^3 - 15*a*b*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.28, size = 176, normalized size = 1.16 \[ \frac {5}{64} \, a b x + \frac {b^{2} \cos \left (9 \, d x + 9 \, c\right )}{2304 \, d} - \frac {a^{2} \cos \left (5 \, d x + 5 \, c\right )}{64 \, d} - \frac {a b \sin \left (8 \, d x + 8 \, c\right )}{512 \, d} - \frac {a b \sin \left (6 \, d x + 6 \, c\right )}{96 \, d} - \frac {a b \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {a b \sin \left (2 \, d x + 2 \, c\right )}{32 \, d} - \frac {{\left (4 \, a^{2} - 3 \, b^{2}\right )} \cos \left (7 \, d x + 7 \, c\right )}{1792 \, d} - \frac {{\left (9 \, a^{2} + 2 \, b^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{192 \, d} - \frac {{\left (10 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )}{128 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*sin(d*x+c)*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

5/64*a*b*x + 1/2304*b^2*cos(9*d*x + 9*c)/d - 1/64*a^2*cos(5*d*x + 5*c)/d - 1/512*a*b*sin(8*d*x + 8*c)/d - 1/96
*a*b*sin(6*d*x + 6*c)/d - 1/64*a*b*sin(4*d*x + 4*c)/d + 1/32*a*b*sin(2*d*x + 2*c)/d - 1/1792*(4*a^2 - 3*b^2)*c
os(7*d*x + 7*c)/d - 1/192*(9*a^2 + 2*b^2)*cos(3*d*x + 3*c)/d - 1/128*(10*a^2 + 3*b^2)*cos(d*x + c)/d

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maple [A]  time = 0.32, size = 115, normalized size = 0.76 \[ \frac {-\frac {a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{7}+2 a b \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{7}\left (d x +c \right )\right )}{8}+\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{48}+\frac {5 d x}{128}+\frac {5 c}{128}\right )+b^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{7}\left (d x +c \right )\right )}{9}-\frac {2 \left (\cos ^{7}\left (d x +c \right )\right )}{63}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*sin(d*x+c)*(a+b*sin(d*x+c))^2,x)

[Out]

1/d*(-1/7*a^2*cos(d*x+c)^7+2*a*b*(-1/8*sin(d*x+c)*cos(d*x+c)^7+1/48*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*
x+c))*sin(d*x+c)+5/128*d*x+5/128*c)+b^2*(-1/9*sin(d*x+c)^2*cos(d*x+c)^7-2/63*cos(d*x+c)^7))

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maxima [A]  time = 0.99, size = 92, normalized size = 0.61 \[ -\frac {4608 \, a^{2} \cos \left (d x + c\right )^{7} - 21 \, {\left (64 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 120 \, d x + 120 \, c - 3 \, \sin \left (8 \, d x + 8 \, c\right ) - 24 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a b - 512 \, {\left (7 \, \cos \left (d x + c\right )^{9} - 9 \, \cos \left (d x + c\right )^{7}\right )} b^{2}}{32256 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*sin(d*x+c)*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/32256*(4608*a^2*cos(d*x + c)^7 - 21*(64*sin(2*d*x + 2*c)^3 + 120*d*x + 120*c - 3*sin(8*d*x + 8*c) - 24*sin(
4*d*x + 4*c))*a*b - 512*(7*cos(d*x + c)^9 - 9*cos(d*x + c)^7)*b^2)/d

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mupad [B]  time = 15.05, size = 332, normalized size = 2.18 \[ \frac {5\,a\,b\,x}{64}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}\,\left (4\,a^2+4\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {4\,a^2}{7}+\frac {4\,b^2}{7}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (12\,a^2+12\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (16\,a^2-12\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (12\,a^2-\frac {20\,b^2}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (20\,a^2+20\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {44\,a^2}{7}-\frac {12\,b^2}{7}\right )+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}+\frac {2\,a^2}{7}+\frac {4\,b^2}{63}-\frac {191\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{48}+\frac {83\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{16}-\frac {145\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{16}+\frac {145\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{16}-\frac {83\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{16}+\frac {191\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{48}-\frac {5\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}}{32}+\frac {5\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{32}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6*sin(c + d*x)*(a + b*sin(c + d*x))^2,x)

[Out]

(5*a*b*x)/64 - (tan(c/2 + (d*x)/2)^14*(4*a^2 + 4*b^2) + tan(c/2 + (d*x)/2)^2*((4*a^2)/7 + (4*b^2)/7) + tan(c/2
+ (d*x)/2)^6*(12*a^2 + 12*b^2) + tan(c/2 + (d*x)/2)^8*(16*a^2 - 12*b^2) + tan(c/2 + (d*x)/2)^12*(12*a^2 - (20
*b^2)/3) + tan(c/2 + (d*x)/2)^10*(20*a^2 + 20*b^2) + tan(c/2 + (d*x)/2)^4*((44*a^2)/7 - (12*b^2)/7) + 2*a^2*ta
n(c/2 + (d*x)/2)^16 + (2*a^2)/7 + (4*b^2)/63 - (191*a*b*tan(c/2 + (d*x)/2)^3)/48 + (83*a*b*tan(c/2 + (d*x)/2)^
5)/16 - (145*a*b*tan(c/2 + (d*x)/2)^7)/16 + (145*a*b*tan(c/2 + (d*x)/2)^11)/16 - (83*a*b*tan(c/2 + (d*x)/2)^13
)/16 + (191*a*b*tan(c/2 + (d*x)/2)^15)/48 - (5*a*b*tan(c/2 + (d*x)/2)^17)/32 + (5*a*b*tan(c/2 + (d*x)/2))/32)/
(d*(tan(c/2 + (d*x)/2)^2 + 1)^9)

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sympy [A]  time = 14.94, size = 282, normalized size = 1.86 \[ \begin {cases} - \frac {a^{2} \cos ^{7}{\left (c + d x \right )}}{7 d} + \frac {5 a b x \sin ^{8}{\left (c + d x \right )}}{64} + \frac {5 a b x \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {15 a b x \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{32} + \frac {5 a b x \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{16} + \frac {5 a b x \cos ^{8}{\left (c + d x \right )}}{64} + \frac {5 a b \sin ^{7}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{64 d} + \frac {55 a b \sin ^{5}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{192 d} + \frac {73 a b \sin ^{3}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{192 d} - \frac {5 a b \sin {\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{64 d} - \frac {b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{7 d} - \frac {2 b^{2} \cos ^{9}{\left (c + d x \right )}}{63 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\relax (c )}\right )^{2} \sin {\relax (c )} \cos ^{6}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6*sin(d*x+c)*(a+b*sin(d*x+c))**2,x)

[Out]

Piecewise((-a**2*cos(c + d*x)**7/(7*d) + 5*a*b*x*sin(c + d*x)**8/64 + 5*a*b*x*sin(c + d*x)**6*cos(c + d*x)**2/
16 + 15*a*b*x*sin(c + d*x)**4*cos(c + d*x)**4/32 + 5*a*b*x*sin(c + d*x)**2*cos(c + d*x)**6/16 + 5*a*b*x*cos(c
+ d*x)**8/64 + 5*a*b*sin(c + d*x)**7*cos(c + d*x)/(64*d) + 55*a*b*sin(c + d*x)**5*cos(c + d*x)**3/(192*d) + 73
*a*b*sin(c + d*x)**3*cos(c + d*x)**5/(192*d) - 5*a*b*sin(c + d*x)*cos(c + d*x)**7/(64*d) - b**2*sin(c + d*x)**
2*cos(c + d*x)**7/(7*d) - 2*b**2*cos(c + d*x)**9/(63*d), Ne(d, 0)), (x*(a + b*sin(c))**2*sin(c)*cos(c)**6, Tru
e))

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