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

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

[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)

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Rubi [A]  time = 0.0610808, antiderivative size = 87, normalized size of antiderivative = 1., 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 = {2669, 2635, 8} \[ -\frac{a \cos ^7(c+d x)}{7 d}+\frac{a \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{5 a \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{5 a \sin (c+d x) \cos (c+d x)}{16 d}+\frac{5 a x}{16} \]

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 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 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 8

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

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*}

Mathematica [A]  time = 0.160345, size = 57, normalized size = 0.66 \[ \frac{a \left (7 (45 \sin (2 (c+d x))+9 \sin (4 (c+d x))+\sin (6 (c+d x))+60 c+60 d x)-192 \cos ^7(c+d x)\right )}{1344 d} \]

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 = 0.026, size = 62, normalized size = 0.7 \begin{align*}{\frac{1}{d} \left ( -{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{7}}+a \left ({\frac{\sin \left ( dx+c \right ) }{6} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\cos \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.973284, size = 85, normalized size = 0.98 \begin{align*} -\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{align*}

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 = 1.73268, size = 167, normalized size = 1.92 \begin{align*} -\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{align*}

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 [A]  time = 8.81304, size = 172, normalized size = 1.98 \begin{align*} \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{align*}

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 = 1.18363, size = 144, normalized size = 1.66 \begin{align*} \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{align*}

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