3.15 \(\int \cos ^4(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=102 \[ -\frac{7 a^2 \cos ^5(c+d x)}{30 d}-\frac{\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{6 d}+\frac{7 a^2 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{7 a^2 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{7 a^2 x}{16} \]

[Out]

(7*a^2*x)/16 - (7*a^2*Cos[c + d*x]^5)/(30*d) + (7*a^2*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (7*a^2*Cos[c + d*x]^
3*Sin[c + d*x])/(24*d) - (Cos[c + d*x]^5*(a^2 + a^2*Sin[c + d*x]))/(6*d)

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Rubi [A]  time = 0.0925856, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2678, 2669, 2635, 8} \[ -\frac{7 a^2 \cos ^5(c+d x)}{30 d}-\frac{\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{6 d}+\frac{7 a^2 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{7 a^2 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{7 a^2 x}{16} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]

[Out]

(7*a^2*x)/16 - (7*a^2*Cos[c + d*x]^5)/(30*d) + (7*a^2*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (7*a^2*Cos[c + d*x]^
3*Sin[c + d*x])/(24*d) - (Cos[c + d*x]^5*(a^2 + a^2*Sin[c + d*x]))/(6*d)

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]

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 ^4(c+d x) (a+a \sin (c+d x))^2 \, dx &=-\frac{\cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{6 d}+\frac{1}{6} (7 a) \int \cos ^4(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac{7 a^2 \cos ^5(c+d x)}{30 d}-\frac{\cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{6 d}+\frac{1}{6} \left (7 a^2\right ) \int \cos ^4(c+d x) \, dx\\ &=-\frac{7 a^2 \cos ^5(c+d x)}{30 d}+\frac{7 a^2 \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{\cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{6 d}+\frac{1}{8} \left (7 a^2\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac{7 a^2 \cos ^5(c+d x)}{30 d}+\frac{7 a^2 \cos (c+d x) \sin (c+d x)}{16 d}+\frac{7 a^2 \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{\cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{6 d}+\frac{1}{16} \left (7 a^2\right ) \int 1 \, dx\\ &=\frac{7 a^2 x}{16}-\frac{7 a^2 \cos ^5(c+d x)}{30 d}+\frac{7 a^2 \cos (c+d x) \sin (c+d x)}{16 d}+\frac{7 a^2 \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{\cos ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.561608, size = 151, normalized size = 1.48 \[ -\frac{a^2 \left (\sqrt{\sin (c+d x)+1} \left (40 \sin ^6(c+d x)+56 \sin ^5(c+d x)-106 \sin ^4(c+d x)-182 \sin ^3(c+d x)+57 \sin ^2(c+d x)+231 \sin (c+d x)-96\right )-210 \sin ^{-1}\left (\frac{\sqrt{1-\sin (c+d x)}}{\sqrt{2}}\right ) \sqrt{1-\sin (c+d x)}\right ) \cos ^5(c+d x)}{240 d (\sin (c+d x)-1)^3 (\sin (c+d x)+1)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]

[Out]

-(a^2*Cos[c + d*x]^5*(-210*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sqrt[1 - Sin[c + d*x]] + Sqrt[1 + Sin[c + d*
x]]*(-96 + 231*Sin[c + d*x] + 57*Sin[c + d*x]^2 - 182*Sin[c + d*x]^3 - 106*Sin[c + d*x]^4 + 56*Sin[c + d*x]^5
+ 40*Sin[c + d*x]^6)))/(240*d*(-1 + Sin[c + d*x])^3*(1 + Sin[c + d*x])^(5/2))

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Maple [A]  time = 0.036, size = 109, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( -{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{6}}+{\frac{\sin \left ( dx+c \right ) }{24} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{dx}{16}}+{\frac{c}{16}} \right ) -{\frac{2\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5}}+{a}^{2} \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+a*sin(d*x+c))^2,x)

[Out]

1/d*(a^2*(-1/6*sin(d*x+c)*cos(d*x+c)^5+1/24*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+1/16*d*x+1/16*c)-2/5*a^2*
cos(d*x+c)^5+a^2*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c))

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Maxima [A]  time = 0.957817, size = 120, normalized size = 1.18 \begin{align*} -\frac{384 \, a^{2} \cos \left (d x + c\right )^{5} - 5 \,{\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} - 30 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/960*(384*a^2*cos(d*x + c)^5 - 5*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*a^2 - 30*(12*d*
x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^2)/d

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Fricas [A]  time = 1.6799, size = 181, normalized size = 1.77 \begin{align*} -\frac{96 \, a^{2} \cos \left (d x + c\right )^{5} - 105 \, a^{2} d x + 5 \,{\left (8 \, a^{2} \cos \left (d x + c\right )^{5} - 14 \, a^{2} \cos \left (d x + c\right )^{3} - 21 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/240*(96*a^2*cos(d*x + c)^5 - 105*a^2*d*x + 5*(8*a^2*cos(d*x + c)^5 - 14*a^2*cos(d*x + c)^3 - 21*a^2*cos(d*x
+ c))*sin(d*x + c))/d

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Sympy [A]  time = 5.30316, size = 287, normalized size = 2.81 \begin{align*} \begin{cases} \frac{a^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{3 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{3 a^{2} x \sin ^{4}{\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 )}}{16} + \frac{3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{a^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{3 a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{a^{2} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} + \frac{a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac{3 a^{2} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} - \frac{a^{2} \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac{5 a^{2} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac{2 a^{2} \cos ^{5}{\left (c + d x \right )}}{5 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\right )^{2} \cos ^{4}{\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)**4*(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((a**2*x*sin(c + d*x)**6/16 + 3*a**2*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 3*a**2*x*sin(c + d*x)**4/
8 + 3*a**2*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 3*a**2*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + a**2*x*cos(c +
d*x)**6/16 + 3*a**2*x*cos(c + d*x)**4/8 + a**2*sin(c + d*x)**5*cos(c + d*x)/(16*d) + a**2*sin(c + d*x)**3*cos(
c + d*x)**3/(6*d) + 3*a**2*sin(c + d*x)**3*cos(c + d*x)/(8*d) - a**2*sin(c + d*x)*cos(c + d*x)**5/(16*d) + 5*a
**2*sin(c + d*x)*cos(c + d*x)**3/(8*d) - 2*a**2*cos(c + d*x)**5/(5*d), Ne(d, 0)), (x*(a*sin(c) + a)**2*cos(c)*
*4, True))

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Giac [A]  time = 1.15864, size = 143, normalized size = 1.4 \begin{align*} \frac{7}{16} \, a^{2} x - \frac{a^{2} \cos \left (5 \, d x + 5 \, c\right )}{40 \, d} - \frac{a^{2} \cos \left (3 \, d x + 3 \, c\right )}{8 \, d} - \frac{a^{2} \cos \left (d x + c\right )}{4 \, d} - \frac{a^{2} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac{a^{2} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac{17 \, a^{2} \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)^4*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

7/16*a^2*x - 1/40*a^2*cos(5*d*x + 5*c)/d - 1/8*a^2*cos(3*d*x + 3*c)/d - 1/4*a^2*cos(d*x + c)/d - 1/192*a^2*sin
(6*d*x + 6*c)/d + 1/64*a^2*sin(4*d*x + 4*c)/d + 17/64*a^2*sin(2*d*x + 2*c)/d