3.14 \(\int \cos ^5(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=67 \[ \frac{(a \sin (c+d x)+a)^7}{7 a^5 d}-\frac{2 (a \sin (c+d x)+a)^6}{3 a^4 d}+\frac{4 (a \sin (c+d x)+a)^5}{5 a^3 d} \]

[Out]

(4*(a + a*Sin[c + d*x])^5)/(5*a^3*d) - (2*(a + a*Sin[c + d*x])^6)/(3*a^4*d) + (a + a*Sin[c + d*x])^7/(7*a^5*d)

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Rubi [A]  time = 0.0610944, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2667, 43} \[ \frac{(a \sin (c+d x)+a)^7}{7 a^5 d}-\frac{2 (a \sin (c+d x)+a)^6}{3 a^4 d}+\frac{4 (a \sin (c+d x)+a)^5}{5 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + a*Sin[c + d*x])^2,x]

[Out]

(4*(a + a*Sin[c + d*x])^5)/(5*a^3*d) - (2*(a + a*Sin[c + d*x])^6)/(3*a^4*d) + (a + a*Sin[c + d*x])^7/(7*a^5*d)

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \cos ^5(c+d x) (a+a \sin (c+d x))^2 \, dx &=\frac{\operatorname{Subst}\left (\int (a-x)^2 (a+x)^4 \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (4 a^2 (a+x)^4-4 a (a+x)^5+(a+x)^6\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac{4 (a+a \sin (c+d x))^5}{5 a^3 d}-\frac{2 (a+a \sin (c+d x))^6}{3 a^4 d}+\frac{(a+a \sin (c+d x))^7}{7 a^5 d}\\ \end{align*}

Mathematica [A]  time = 0.0795505, size = 58, normalized size = 0.87 \[ -\frac{a^2 (\sin (c+d x)+1)^2 \left (15 \sin ^2(c+d x)-40 \sin (c+d x)+29\right ) \cos ^6(c+d x)}{105 d (\sin (c+d x)-1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + a*Sin[c + d*x])^2,x]

[Out]

-(a^2*Cos[c + d*x]^6*(1 + Sin[c + d*x])^2*(29 - 40*Sin[c + d*x] + 15*Sin[c + d*x]^2))/(105*d*(-1 + Sin[c + d*x
])^3)

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Maple [A]  time = 0.039, size = 99, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( -{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{7}}+{\frac{\sin \left ( dx+c \right ) }{35} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) } \right ) -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{3}}+{\frac{{a}^{2}\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+a*sin(d*x+c))^2,x)

[Out]

1/d*(a^2*(-1/7*sin(d*x+c)*cos(d*x+c)^6+1/35*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))-1/3*a^2*cos(d*x+c)
^6+1/5*a^2*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))

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Maxima [A]  time = 0.950871, size = 128, normalized size = 1.91 \begin{align*} \frac{15 \, a^{2} \sin \left (d x + c\right )^{7} + 35 \, a^{2} \sin \left (d x + c\right )^{6} - 21 \, a^{2} \sin \left (d x + c\right )^{5} - 105 \, a^{2} \sin \left (d x + c\right )^{4} - 35 \, a^{2} \sin \left (d x + c\right )^{3} + 105 \, a^{2} \sin \left (d x + c\right )^{2} + 105 \, a^{2} \sin \left (d x + c\right )}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/105*(15*a^2*sin(d*x + c)^7 + 35*a^2*sin(d*x + c)^6 - 21*a^2*sin(d*x + c)^5 - 105*a^2*sin(d*x + c)^4 - 35*a^2
*sin(d*x + c)^3 + 105*a^2*sin(d*x + c)^2 + 105*a^2*sin(d*x + c))/d

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Fricas [A]  time = 1.75967, size = 176, normalized size = 2.63 \begin{align*} -\frac{35 \, a^{2} \cos \left (d x + c\right )^{6} +{\left (15 \, a^{2} \cos \left (d x + c\right )^{6} - 24 \, a^{2} \cos \left (d x + c\right )^{4} - 32 \, a^{2} \cos \left (d x + c\right )^{2} - 64 \, a^{2}\right )} \sin \left (d x + c\right )}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/105*(35*a^2*cos(d*x + c)^6 + (15*a^2*cos(d*x + c)^6 - 24*a^2*cos(d*x + c)^4 - 32*a^2*cos(d*x + c)^2 - 64*a^
2)*sin(d*x + c))/d

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Sympy [A]  time = 8.32056, size = 202, normalized size = 3.01 \begin{align*} \begin{cases} \frac{8 a^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac{a^{2} \sin ^{6}{\left (c + d x \right )}}{3 d} + \frac{4 a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac{8 a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{a^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} + \frac{4 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac{a^{2} \sin{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\right )^{2} \cos ^{5}{\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)**5*(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((8*a**2*sin(c + d*x)**7/(105*d) + a**2*sin(c + d*x)**6/(3*d) + 4*a**2*sin(c + d*x)**5*cos(c + d*x)**
2/(15*d) + 8*a**2*sin(c + d*x)**5/(15*d) + a**2*sin(c + d*x)**4*cos(c + d*x)**2/d + a**2*sin(c + d*x)**3*cos(c
+ d*x)**4/(3*d) + 4*a**2*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + a**2*sin(c + d*x)**2*cos(c + d*x)**4/d + a**
2*sin(c + d*x)*cos(c + d*x)**4/d, Ne(d, 0)), (x*(a*sin(c) + a)**2*cos(c)**5, True))

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Giac [A]  time = 1.14066, size = 158, normalized size = 2.36 \begin{align*} -\frac{a^{2} \cos \left (6 \, d x + 6 \, c\right )}{96 \, d} - \frac{a^{2} \cos \left (4 \, d x + 4 \, c\right )}{16 \, d} - \frac{5 \, a^{2} \cos \left (2 \, d x + 2 \, c\right )}{32 \, d} - \frac{a^{2} \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac{a^{2} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac{19 \, a^{2} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac{45 \, a^{2} \sin \left (d x + c\right )}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/96*a^2*cos(6*d*x + 6*c)/d - 1/16*a^2*cos(4*d*x + 4*c)/d - 5/32*a^2*cos(2*d*x + 2*c)/d - 1/448*a^2*sin(7*d*x
+ 7*c)/d + 1/320*a^2*sin(5*d*x + 5*c)/d + 19/192*a^2*sin(3*d*x + 3*c)/d + 45/64*a^2*sin(d*x + c)/d