3.31 \(\int \cos ^2(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=106 \[ -\frac {7 a^3 \cos ^3(c+d x)}{12 d}-\frac {7 \cos ^3(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{20 d}+\frac {7 a^3 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {7 a^3 x}{8}-\frac {a \cos ^3(c+d x) (a \sin (c+d x)+a)^2}{5 d} \]

[Out]

7/8*a^3*x-7/12*a^3*cos(d*x+c)^3/d+7/8*a^3*cos(d*x+c)*sin(d*x+c)/d-1/5*a*cos(d*x+c)^3*(a+a*sin(d*x+c))^2/d-7/20
*cos(d*x+c)^3*(a^3+a^3*sin(d*x+c))/d

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Rubi [A]  time = 0.12, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2678, 2669, 2635, 8} \[ -\frac {7 a^3 \cos ^3(c+d x)}{12 d}-\frac {7 \cos ^3(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{20 d}+\frac {7 a^3 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {7 a^3 x}{8}-\frac {a \cos ^3(c+d x) (a \sin (c+d x)+a)^2}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]

[Out]

(7*a^3*x)/8 - (7*a^3*Cos[c + d*x]^3)/(12*d) + (7*a^3*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (a*Cos[c + d*x]^3*(a +
a*Sin[c + d*x])^2)/(5*d) - (7*Cos[c + d*x]^3*(a^3 + a^3*Sin[c + d*x]))/(20*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 ^2(c+d x) (a+a \sin (c+d x))^3 \, dx &=-\frac {a \cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}+\frac {1}{5} (7 a) \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 \, dx\\ &=-\frac {a \cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}-\frac {7 \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{20 d}+\frac {1}{4} \left (7 a^2\right ) \int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac {7 a^3 \cos ^3(c+d x)}{12 d}-\frac {a \cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}-\frac {7 \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{20 d}+\frac {1}{4} \left (7 a^3\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac {7 a^3 \cos ^3(c+d x)}{12 d}+\frac {7 a^3 \cos (c+d x) \sin (c+d x)}{8 d}-\frac {a \cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}-\frac {7 \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{20 d}+\frac {1}{8} \left (7 a^3\right ) \int 1 \, dx\\ &=\frac {7 a^3 x}{8}-\frac {7 a^3 \cos ^3(c+d x)}{12 d}+\frac {7 a^3 \cos (c+d x) \sin (c+d x)}{8 d}-\frac {a \cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}-\frac {7 \cos ^3(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{20 d}\\ \end {align*}

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Mathematica [A]  time = 0.41, size = 141, normalized size = 1.33 \[ -\frac {a^3 \left (210 \sqrt {1-\sin (c+d x)} \sin ^{-1}\left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right )+\sqrt {\sin (c+d x)+1} \left (24 \sin ^5(c+d x)+66 \sin ^4(c+d x)+22 \sin ^3(c+d x)-97 \sin ^2(c+d x)-151 \sin (c+d x)+136\right )\right ) \cos ^3(c+d x)}{120 d (\sin (c+d x)-1)^2 (\sin (c+d x)+1)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]

[Out]

-1/120*(a^3*Cos[c + d*x]^3*(210*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sqrt[1 - Sin[c + d*x]] + Sqrt[1 + Sin[c
+ d*x]]*(136 - 151*Sin[c + d*x] - 97*Sin[c + d*x]^2 + 22*Sin[c + d*x]^3 + 66*Sin[c + d*x]^4 + 24*Sin[c + d*x]
^5)))/(d*(-1 + Sin[c + d*x])^2*(1 + Sin[c + d*x])^(3/2))

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fricas [A]  time = 0.85, size = 72, normalized size = 0.68 \[ \frac {24 \, a^{3} \cos \left (d x + c\right )^{5} - 160 \, a^{3} \cos \left (d x + c\right )^{3} + 105 \, a^{3} d x - 15 \, {\left (6 \, a^{3} \cos \left (d x + c\right )^{3} - 7 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/120*(24*a^3*cos(d*x + c)^5 - 160*a^3*cos(d*x + c)^3 + 105*a^3*d*x - 15*(6*a^3*cos(d*x + c)^3 - 7*a^3*cos(d*x
+ c))*sin(d*x + c))/d

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giac [A]  time = 1.07, size = 89, normalized size = 0.84 \[ \frac {7}{8} \, a^{3} x + \frac {a^{3} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {13 \, a^{3} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {7 \, a^{3} \cos \left (d x + c\right )}{8 \, d} - \frac {3 \, a^{3} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {a^{3} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

7/8*a^3*x + 1/80*a^3*cos(5*d*x + 5*c)/d - 13/48*a^3*cos(3*d*x + 3*c)/d - 7/8*a^3*cos(d*x + c)/d - 3/32*a^3*sin
(4*d*x + 4*c)/d + 1/4*a^3*sin(2*d*x + 2*c)/d

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maple [A]  time = 0.12, size = 121, normalized size = 1.14 \[ \frac {a^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )+3 a^{3} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\left (\cos ^{3}\left (d x +c \right )\right ) a^{3}+a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+a*sin(d*x+c))^3,x)

[Out]

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

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maxima [A]  time = 0.54, size = 91, normalized size = 0.86 \[ -\frac {480 \, a^{3} \cos \left (d x + c\right )^{3} - 32 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{3} - 45 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3} - 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/480*(480*a^3*cos(d*x + c)^3 - 32*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a^3 - 45*(4*d*x + 4*c - sin(4*d*x +
4*c))*a^3 - 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^3)/d

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mupad [B]  time = 6.53, size = 277, normalized size = 2.61 \[ \frac {7\,a^3\,x}{8}-\frac {\frac {13\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}-\frac {13\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}+\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+\frac {a^3\,\left (105\,c+105\,d\,x\right )}{120}-\frac {a^3\,\left (105\,c+105\,d\,x-272\right )}{120}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^3\,\left (105\,c+105\,d\,x\right )}{24}-\frac {a^3\,\left (525\,c+525\,d\,x-640\right )}{120}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {a^3\,\left (105\,c+105\,d\,x\right )}{24}-\frac {a^3\,\left (525\,c+525\,d\,x-720\right )}{120}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a^3\,\left (105\,c+105\,d\,x\right )}{12}-\frac {a^3\,\left (1050\,c+1050\,d\,x-800\right )}{120}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {a^3\,\left (105\,c+105\,d\,x\right )}{12}-\frac {a^3\,\left (1050\,c+1050\,d\,x-1920\right )}{120}\right )-\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2*(a + a*sin(c + d*x))^3,x)

[Out]

(7*a^3*x)/8 - ((13*a^3*tan(c/2 + (d*x)/2)^7)/2 - (13*a^3*tan(c/2 + (d*x)/2)^3)/2 + (a^3*tan(c/2 + (d*x)/2)^9)/
4 + (a^3*(105*c + 105*d*x))/120 - (a^3*(105*c + 105*d*x - 272))/120 + tan(c/2 + (d*x)/2)^2*((a^3*(105*c + 105*
d*x))/24 - (a^3*(525*c + 525*d*x - 640))/120) + tan(c/2 + (d*x)/2)^8*((a^3*(105*c + 105*d*x))/24 - (a^3*(525*c
+ 525*d*x - 720))/120) + tan(c/2 + (d*x)/2)^4*((a^3*(105*c + 105*d*x))/12 - (a^3*(1050*c + 1050*d*x - 800))/1
20) + tan(c/2 + (d*x)/2)^6*((a^3*(105*c + 105*d*x))/12 - (a^3*(1050*c + 1050*d*x - 1920))/120) - (a^3*tan(c/2
+ (d*x)/2))/4)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^5)

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sympy [A]  time = 3.72, size = 226, normalized size = 2.13 \[ \begin {cases} \frac {3 a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {3 a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} - \frac {2 a^{3} \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac {a^{3} \cos ^{3}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{3} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(a+a*sin(d*x+c))**3,x)

[Out]

Piecewise((3*a**3*x*sin(c + d*x)**4/8 + 3*a**3*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + a**3*x*sin(c + d*x)**2/2
+ 3*a**3*x*cos(c + d*x)**4/8 + a**3*x*cos(c + d*x)**2/2 + 3*a**3*sin(c + d*x)**3*cos(c + d*x)/(8*d) - a**3*sin
(c + d*x)**2*cos(c + d*x)**3/(3*d) - 3*a**3*sin(c + d*x)*cos(c + d*x)**3/(8*d) + a**3*sin(c + d*x)*cos(c + d*x
)/(2*d) - 2*a**3*cos(c + d*x)**5/(15*d) - a**3*cos(c + d*x)**3/d, Ne(d, 0)), (x*(a*sin(c) + a)**3*cos(c)**2, T
rue))

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