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

Optimal. Leaf size=87 \[ \frac {5 a x}{16}-\frac {a \cos ^7(c+d x)}{7 d}+\frac {5 a \cos (c+d x) \sin (c+d x)}{16 d}+\frac {5 a \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a \cos ^5(c+d x) \sin (c+d x)}{6 d} \]

[Out]

5/16*a*x-1/7*a*cos(d*x+c)^7/d+5/16*a*cos(d*x+c)*sin(d*x+c)/d+5/24*a*cos(d*x+c)^3*sin(d*x+c)/d+1/6*a*cos(d*x+c)
^5*sin(d*x+c)/d

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Rubi [A]
time = 0.04, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2748, 2715, 8} \begin {gather*} -\frac {a \cos ^7(c+d x)}{7 d}+\frac {a \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {5 a \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {5 a \sin (c+d x) \cos (c+d x)}{16 d}+\frac {5 a x}{16} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a*x)/16 - (a*Cos[c + d*x]^7)/(7*d) + (5*a*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (5*a*Cos[c + d*x]^3*Sin[c + d
*x])/(24*d) + (a*Cos[c + d*x]^5*Sin[c + d*x])/(6*d)

Rule 8

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

Rule 2715

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] && Integ
erQ[2*n]

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[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])

Rubi steps

\begin {align*} \int \cos ^6(c+d x) (a+a \sin (c+d x)) \, dx &=-\frac {a \cos ^7(c+d x)}{7 d}+a \int \cos ^6(c+d x) \, dx\\ &=-\frac {a \cos ^7(c+d x)}{7 d}+\frac {a \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac {1}{6} (5 a) \int \cos ^4(c+d x) \, dx\\ &=-\frac {a \cos ^7(c+d x)}{7 d}+\frac {5 a \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac {1}{8} (5 a) \int \cos ^2(c+d x) \, dx\\ &=-\frac {a \cos ^7(c+d x)}{7 d}+\frac {5 a \cos (c+d x) \sin (c+d x)}{16 d}+\frac {5 a \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac {1}{16} (5 a) \int 1 \, dx\\ &=\frac {5 a x}{16}-\frac {a \cos ^7(c+d x)}{7 d}+\frac {5 a \cos (c+d x) \sin (c+d x)}{16 d}+\frac {5 a \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a \cos ^5(c+d x) \sin (c+d x)}{6 d}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 57, normalized size = 0.66 \begin {gather*} \frac {a \left (-192 \cos ^7(c+d x)+7 (60 c+60 d x+45 \sin (2 (c+d x))+9 \sin (4 (c+d x))+\sin (6 (c+d x)))\right )}{1344 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(-192*Cos[c + d*x]^7 + 7*(60*c + 60*d*x + 45*Sin[2*(c + d*x)] + 9*Sin[4*(c + d*x)] + Sin[6*(c + d*x)])))/(1
344*d)

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Maple [A]
time = 0.16, size = 62, normalized size = 0.71

method	result	size
derivativedivides	\(\frac {-\frac {a \left (\cos ^{7}\left (d x +c \right )\right )}{7}+a \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}\)	\(62\)
default	\(\frac {-\frac {a \left (\cos ^{7}\left (d x +c \right )\right )}{7}+a \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}\)	\(62\)
risch	\(\frac {5 a x}{16}-\frac {5 a \cos \left (d x +c \right )}{64 d}-\frac {a \cos \left (7 d x +7 c \right )}{448 d}+\frac {a \sin \left (6 d x +6 c \right )}{192 d}-\frac {a \cos \left (5 d x +5 c \right )}{64 d}+\frac {3 a \sin \left (4 d x +4 c \right )}{64 d}-\frac {3 a \cos \left (3 d x +3 c \right )}{64 d}+\frac {15 a \sin \left (2 d x +2 c \right )}{64 d}\)	\(108\)
norman	\(\frac {\frac {5 a x}{16}-\frac {2 a}{7 d}+\frac {11 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {7 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {85 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}-\frac {85 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}-\frac {7 a \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}-\frac {11 a \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {35 a x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {105 a x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {175 a x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {175 a x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {105 a x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {35 a x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {5 a x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}-\frac {6 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {10 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}\)	\(284\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(-1/7*a*cos(d*x+c)^7+a*(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 = 0.30, size = 63, normalized size = 0.72 \begin {gather*} -\frac {192 \, a \cos \left (d x + c\right )^{7} + 7 \, {\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}{1344 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1344*(192*a*cos(d*x + c)^7 + 7*(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)/d

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Fricas [A]
time = 0.36, size = 62, normalized size = 0.71 \begin {gather*} -\frac {48 \, a \cos \left (d x + c\right )^{7} - 105 \, a d x - 7 \, {\left (8 \, a \cos \left (d x + c\right )^{5} + 10 \, a \cos \left (d x + c\right )^{3} + 15 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{336 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/336*(48*a*cos(d*x + c)^7 - 105*a*d*x - 7*(8*a*cos(d*x + c)^5 + 10*a*cos(d*x + c)^3 + 15*a*cos(d*x + c))*sin
(d*x + c))/d

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (82) = 164\).
time = 0.71, size = 172, normalized size = 1.98 \begin {gather*} \begin {cases} \frac {5 a x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 a x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {15 a x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {5 a x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {5 a \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {5 a \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {11 a \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {a \cos ^{7}{\left (c + d x \right )}}{7 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right ) \cos ^{6}{\left (c \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 7.09, size = 107, normalized size = 1.23 \begin {gather*} \frac {5}{16} \, a x - \frac {a \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac {a \cos \left (5 \, d x + 5 \, c\right )}{64 \, d} - \frac {3 \, a \cos \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac {5 \, a \cos \left (d x + c\right )}{64 \, d} + \frac {a \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {3 \, a \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {15 \, a \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/16*a*x - 1/448*a*cos(7*d*x + 7*c)/d - 1/64*a*cos(5*d*x + 5*c)/d - 3/64*a*cos(3*d*x + 3*c)/d - 5/64*a*cos(d*x
+ c)/d + 1/192*a*sin(6*d*x + 6*c)/d + 3/64*a*sin(4*d*x + 4*c)/d + 15/64*a*sin(2*d*x + 2*c)/d

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Mupad [B]
time = 8.23, size = 226, normalized size = 2.60 \begin {gather*} \frac {5\,a\,x}{16}+\frac {-\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{8}+\left (\frac {a\,\left (735\,c+735\,d\,x-672\right )}{336}-\frac {35\,a\,\left (c+d\,x\right )}{16}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-\frac {7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{6}-\frac {85\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+\left (\frac {a\,\left (3675\,c+3675\,d\,x-3360\right )}{336}-\frac {175\,a\,\left (c+d\,x\right )}{16}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {85\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{24}+\left (\frac {a\,\left (2205\,c+2205\,d\,x-2016\right )}{336}-\frac {105\,a\,\left (c+d\,x\right )}{16}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}+\frac {11\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}+\frac {a\,\left (105\,c+105\,d\,x-96\right )}{336}-\frac {5\,a\,\left (c+d\,x\right )}{16}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*a*x)/16 + ((a*(105*c + 105*d*x - 96))/336 + (11*a*tan(c/2 + (d*x)/2))/8 - (5*a*(c + d*x))/16 + tan(c/2 + (d
*x)/2)^12*((a*(735*c + 735*d*x - 672))/336 - (35*a*(c + d*x))/16) + tan(c/2 + (d*x)/2)^4*((a*(2205*c + 2205*d*
x - 2016))/336 - (105*a*(c + d*x))/16) + tan(c/2 + (d*x)/2)^8*((a*(3675*c + 3675*d*x - 3360))/336 - (175*a*(c
+ d*x))/16) + (7*a*tan(c/2 + (d*x)/2)^3)/6 + (85*a*tan(c/2 + (d*x)/2)^5)/24 - (85*a*tan(c/2 + (d*x)/2)^9)/24 -
(7*a*tan(c/2 + (d*x)/2)^11)/6 - (11*a*tan(c/2 + (d*x)/2)^13)/8)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^7)

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