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

Optimal. Leaf size=126 \[ -\frac {9 a^2 \cos ^7(c+d x)}{56 d}-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{8 d}+\frac {3 a^2 \sin (c+d x) \cos ^5(c+d x)}{16 d}+\frac {15 a^2 \sin (c+d x) \cos ^3(c+d x)}{64 d}+\frac {45 a^2 \sin (c+d x) \cos (c+d x)}{128 d}+\frac {45 a^2 x}{128} \]

[Out]

45/128*a^2*x-9/56*a^2*cos(d*x+c)^7/d+45/128*a^2*cos(d*x+c)*sin(d*x+c)/d+15/64*a^2*cos(d*x+c)^3*sin(d*x+c)/d+3/
16*a^2*cos(d*x+c)^5*sin(d*x+c)/d-1/8*cos(d*x+c)^7*(a^2+a^2*sin(d*x+c))/d

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Rubi [A]  time = 0.11, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2678, 2669, 2635, 8} \[ -\frac {9 a^2 \cos ^7(c+d x)}{56 d}-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{8 d}+\frac {3 a^2 \sin (c+d x) \cos ^5(c+d x)}{16 d}+\frac {15 a^2 \sin (c+d x) \cos ^3(c+d x)}{64 d}+\frac {45 a^2 \sin (c+d x) \cos (c+d x)}{128 d}+\frac {45 a^2 x}{128} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(45*a^2*x)/128 - (9*a^2*Cos[c + d*x]^7)/(56*d) + (45*a^2*Cos[c + d*x]*Sin[c + d*x])/(128*d) + (15*a^2*Cos[c +
d*x]^3*Sin[c + d*x])/(64*d) + (3*a^2*Cos[c + d*x]^5*Sin[c + d*x])/(16*d) - (Cos[c + d*x]^7*(a^2 + a^2*Sin[c +
d*x]))/(8*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 2678

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

Rubi steps

\begin {align*} \int \cos ^6(c+d x) (a+a \sin (c+d x))^2 \, dx &=-\frac {\cos ^7(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{8 d}+\frac {1}{8} (9 a) \int \cos ^6(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac {9 a^2 \cos ^7(c+d x)}{56 d}-\frac {\cos ^7(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{8 d}+\frac {1}{8} \left (9 a^2\right ) \int \cos ^6(c+d x) \, dx\\ &=-\frac {9 a^2 \cos ^7(c+d x)}{56 d}+\frac {3 a^2 \cos ^5(c+d x) \sin (c+d x)}{16 d}-\frac {\cos ^7(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{8 d}+\frac {1}{16} \left (15 a^2\right ) \int \cos ^4(c+d x) \, dx\\ &=-\frac {9 a^2 \cos ^7(c+d x)}{56 d}+\frac {15 a^2 \cos ^3(c+d x) \sin (c+d x)}{64 d}+\frac {3 a^2 \cos ^5(c+d x) \sin (c+d x)}{16 d}-\frac {\cos ^7(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{8 d}+\frac {1}{64} \left (45 a^2\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac {9 a^2 \cos ^7(c+d x)}{56 d}+\frac {45 a^2 \cos (c+d x) \sin (c+d x)}{128 d}+\frac {15 a^2 \cos ^3(c+d x) \sin (c+d x)}{64 d}+\frac {3 a^2 \cos ^5(c+d x) \sin (c+d x)}{16 d}-\frac {\cos ^7(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{8 d}+\frac {1}{128} \left (45 a^2\right ) \int 1 \, dx\\ &=\frac {45 a^2 x}{128}-\frac {9 a^2 \cos ^7(c+d x)}{56 d}+\frac {45 a^2 \cos (c+d x) \sin (c+d x)}{128 d}+\frac {15 a^2 \cos ^3(c+d x) \sin (c+d x)}{64 d}+\frac {3 a^2 \cos ^5(c+d x) \sin (c+d x)}{16 d}-\frac {\cos ^7(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{8 d}\\ \end {align*}

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Mathematica [A]  time = 1.52, size = 171, normalized size = 1.36 \[ -\frac {a^2 \left (630 \sqrt {1-\sin (c+d x)} \sin ^{-1}\left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right )+\sqrt {\sin (c+d x)+1} \left (112 \sin ^8(c+d x)+144 \sin ^7(c+d x)-424 \sin ^6(c+d x)-600 \sin ^5(c+d x)+558 \sin ^4(c+d x)+978 \sin ^3(c+d x)-187 \sin ^2(c+d x)-837 \sin (c+d x)+256\right )\right ) \cos ^7(c+d x)}{896 d (\sin (c+d x)-1)^4 (\sin (c+d x)+1)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/896*(a^2*Cos[c + d*x]^7*(630*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sqrt[1 - Sin[c + d*x]] + Sqrt[1 + Sin[c
+ d*x]]*(256 - 837*Sin[c + d*x] - 187*Sin[c + d*x]^2 + 978*Sin[c + d*x]^3 + 558*Sin[c + d*x]^4 - 600*Sin[c +
d*x]^5 - 424*Sin[c + d*x]^6 + 144*Sin[c + d*x]^7 + 112*Sin[c + d*x]^8)))/(d*(-1 + Sin[c + d*x])^4*(1 + Sin[c +
d*x])^(7/2))

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fricas [A]  time = 0.96, size = 85, normalized size = 0.67 \[ -\frac {256 \, a^{2} \cos \left (d x + c\right )^{7} - 315 \, a^{2} d x + 7 \, {\left (16 \, a^{2} \cos \left (d x + c\right )^{7} - 24 \, a^{2} \cos \left (d x + c\right )^{5} - 30 \, a^{2} \cos \left (d x + c\right )^{3} - 45 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{896 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/896*(256*a^2*cos(d*x + c)^7 - 315*a^2*d*x + 7*(16*a^2*cos(d*x + c)^7 - 24*a^2*cos(d*x + c)^5 - 30*a^2*cos(d
*x + c)^3 - 45*a^2*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.52, size = 123, normalized size = 0.98 \[ \frac {45}{128} \, a^{2} x - \frac {a^{2} \cos \left (7 \, d x + 7 \, c\right )}{224 \, d} - \frac {a^{2} \cos \left (5 \, d x + 5 \, c\right )}{32 \, d} - \frac {3 \, a^{2} \cos \left (3 \, d x + 3 \, c\right )}{32 \, d} - \frac {5 \, a^{2} \cos \left (d x + c\right )}{32 \, d} - \frac {a^{2} \sin \left (8 \, d x + 8 \, c\right )}{1024 \, d} + \frac {5 \, a^{2} \sin \left (4 \, d x + 4 \, c\right )}{128 \, d} + \frac {a^{2} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

45/128*a^2*x - 1/224*a^2*cos(7*d*x + 7*c)/d - 1/32*a^2*cos(5*d*x + 5*c)/d - 3/32*a^2*cos(3*d*x + 3*c)/d - 5/32
*a^2*cos(d*x + c)/d - 1/1024*a^2*sin(8*d*x + 8*c)/d + 5/128*a^2*sin(4*d*x + 4*c)/d + 1/4*a^2*sin(2*d*x + 2*c)/
d

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maple [A]  time = 0.18, size = 129, normalized size = 1.02 \[ \frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{7}\left (d x +c \right )\right )}{8}+\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{48}+\frac {5 d x}{128}+\frac {5 c}{128}\right )-\frac {2 a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{7}+a^{2} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.61, size = 115, normalized size = 0.91 \[ -\frac {6144 \, a^{2} \cos \left (d x + c\right )^{7} - 7 \, {\left (64 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 120 \, d x + 120 \, c - 3 \, \sin \left (8 \, d x + 8 \, c\right ) - 24 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} + 112 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{21504 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/21504*(6144*a^2*cos(d*x + c)^7 - 7*(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^2 + 112*(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^2)/
d

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mupad [B]  time = 6.92, size = 461, normalized size = 3.66 \[ \frac {45\,a^2\,x}{128}-\frac {\frac {815\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{64}-\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{64}-\frac {815\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{64}-\frac {295\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{64}+\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{64}+\frac {295\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{64}+\frac {83\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{64}+\frac {a^2\,\left (315\,c+315\,d\,x\right )}{896}-\frac {a^2\,\left (315\,c+315\,d\,x-512\right )}{896}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^2\,\left (315\,c+315\,d\,x\right )}{112}-\frac {a^2\,\left (2520\,c+2520\,d\,x-512\right )}{896}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}\,\left (\frac {a^2\,\left (315\,c+315\,d\,x\right )}{112}-\frac {a^2\,\left (2520\,c+2520\,d\,x-3584\right )}{896}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (\frac {a^2\,\left (315\,c+315\,d\,x\right )}{32}-\frac {a^2\,\left (8820\,c+8820\,d\,x-3584\right )}{896}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a^2\,\left (315\,c+315\,d\,x\right )}{32}-\frac {a^2\,\left (8820\,c+8820\,d\,x-10752\right )}{896}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {a^2\,\left (315\,c+315\,d\,x\right )}{16}-\frac {a^2\,\left (17640\,c+17640\,d\,x-10752\right )}{896}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (\frac {a^2\,\left (315\,c+315\,d\,x\right )}{16}-\frac {a^2\,\left (17640\,c+17640\,d\,x-17920\right )}{896}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {5\,a^2\,\left (315\,c+315\,d\,x\right )}{64}-\frac {a^2\,\left (22050\,c+22050\,d\,x-17920\right )}{896}\right )-\frac {83\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(45*a^2*x)/128 - ((815*a^2*tan(c/2 + (d*x)/2)^9)/64 - (3*a^2*tan(c/2 + (d*x)/2)^5)/64 - (815*a^2*tan(c/2 + (d*
x)/2)^7)/64 - (295*a^2*tan(c/2 + (d*x)/2)^3)/64 + (3*a^2*tan(c/2 + (d*x)/2)^11)/64 + (295*a^2*tan(c/2 + (d*x)/
2)^13)/64 + (83*a^2*tan(c/2 + (d*x)/2)^15)/64 + (a^2*(315*c + 315*d*x))/896 - (a^2*(315*c + 315*d*x - 512))/89
6 + tan(c/2 + (d*x)/2)^2*((a^2*(315*c + 315*d*x))/112 - (a^2*(2520*c + 2520*d*x - 512))/896) + tan(c/2 + (d*x)
/2)^14*((a^2*(315*c + 315*d*x))/112 - (a^2*(2520*c + 2520*d*x - 3584))/896) + tan(c/2 + (d*x)/2)^12*((a^2*(315
*c + 315*d*x))/32 - (a^2*(8820*c + 8820*d*x - 3584))/896) + tan(c/2 + (d*x)/2)^4*((a^2*(315*c + 315*d*x))/32 -
(a^2*(8820*c + 8820*d*x - 10752))/896) + tan(c/2 + (d*x)/2)^6*((a^2*(315*c + 315*d*x))/16 - (a^2*(17640*c + 1
7640*d*x - 10752))/896) + tan(c/2 + (d*x)/2)^10*((a^2*(315*c + 315*d*x))/16 - (a^2*(17640*c + 17640*d*x - 1792
0))/896) + tan(c/2 + (d*x)/2)^8*((5*a^2*(315*c + 315*d*x))/64 - (a^2*(22050*c + 22050*d*x - 17920))/896) - (83
*a^2*tan(c/2 + (d*x)/2))/64)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^8)

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sympy [A]  time = 14.95, size = 398, normalized size = 3.16 \[ \begin {cases} \frac {5 a^{2} x \sin ^{8}{\left (c + d x \right )}}{128} + \frac {5 a^{2} x \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{32} + \frac {5 a^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{64} + \frac {15 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {5 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{32} + \frac {15 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {5 a^{2} x \cos ^{8}{\left (c + d x \right )}}{128} + \frac {5 a^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {5 a^{2} \sin ^{7}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{128 d} + \frac {55 a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{384 d} + \frac {5 a^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {73 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{384 d} + \frac {5 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac {5 a^{2} \sin {\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{128 d} + \frac {11 a^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {2 a^{2} \cos ^{7}{\left (c + d x \right )}}{7 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{2} \cos ^{6}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((5*a**2*x*sin(c + d*x)**8/128 + 5*a**2*x*sin(c + d*x)**6*cos(c + d*x)**2/32 + 5*a**2*x*sin(c + d*x)*
*6/16 + 15*a**2*x*sin(c + d*x)**4*cos(c + d*x)**4/64 + 15*a**2*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 5*a**2*x
*sin(c + d*x)**2*cos(c + d*x)**6/32 + 15*a**2*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 5*a**2*x*cos(c + d*x)**8/
128 + 5*a**2*x*cos(c + d*x)**6/16 + 5*a**2*sin(c + d*x)**7*cos(c + d*x)/(128*d) + 55*a**2*sin(c + d*x)**5*cos(
c + d*x)**3/(384*d) + 5*a**2*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 73*a**2*sin(c + d*x)**3*cos(c + d*x)**5/(38
4*d) + 5*a**2*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) - 5*a**2*sin(c + d*x)*cos(c + d*x)**7/(128*d) + 11*a**2*si
n(c + d*x)*cos(c + d*x)**5/(16*d) - 2*a**2*cos(c + d*x)**7/(7*d), Ne(d, 0)), (x*(a*sin(c) + a)**2*cos(c)**6, T
rue))

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