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

Optimal. Leaf size=137 \[ -\frac {7 a^4 \cos ^3(c+d x)}{8 d}-\frac {21 \cos ^3(c+d x) \left (a^4 \sin (c+d x)+a^4\right )}{40 d}+\frac {21 a^4 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {21 a^4 x}{16}-\frac {3 \cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )^2}{10 d}-\frac {a \cos ^3(c+d x) (a \sin (c+d x)+a)^3}{6 d} \]

[Out]

21/16*a^4*x-7/8*a^4*cos(d*x+c)^3/d+21/16*a^4*cos(d*x+c)*sin(d*x+c)/d-1/6*a*cos(d*x+c)^3*(a+a*sin(d*x+c))^3/d-3
/10*cos(d*x+c)^3*(a^2+a^2*sin(d*x+c))^2/d-21/40*cos(d*x+c)^3*(a^4+a^4*sin(d*x+c))/d

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Rubi [A]  time = 0.16, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2678, 2669, 2635, 8} \[ -\frac {7 a^4 \cos ^3(c+d x)}{8 d}-\frac {3 \cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )^2}{10 d}-\frac {21 \cos ^3(c+d x) \left (a^4 \sin (c+d x)+a^4\right )}{40 d}+\frac {21 a^4 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {21 a^4 x}{16}-\frac {a \cos ^3(c+d x) (a \sin (c+d x)+a)^3}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(21*a^4*x)/16 - (7*a^4*Cos[c + d*x]^3)/(8*d) + (21*a^4*Cos[c + d*x]*Sin[c + d*x])/(16*d) - (a*Cos[c + d*x]^3*(
a + a*Sin[c + d*x])^3)/(6*d) - (3*Cos[c + d*x]^3*(a^2 + a^2*Sin[c + d*x])^2)/(10*d) - (21*Cos[c + d*x]^3*(a^4
+ a^4*Sin[c + d*x]))/(40*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))^4 \, dx &=-\frac {a \cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}+\frac {1}{2} (3 a) \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))^3}{6 d}-\frac {3 \cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )^2}{10 d}+\frac {1}{10} \left (21 a^2\right ) \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))^3}{6 d}-\frac {3 \cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )^2}{10 d}-\frac {21 \cos ^3(c+d x) \left (a^4+a^4 \sin (c+d x)\right )}{40 d}+\frac {1}{8} \left (21 a^3\right ) \int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac {7 a^4 \cos ^3(c+d x)}{8 d}-\frac {a \cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}-\frac {3 \cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )^2}{10 d}-\frac {21 \cos ^3(c+d x) \left (a^4+a^4 \sin (c+d x)\right )}{40 d}+\frac {1}{8} \left (21 a^4\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac {7 a^4 \cos ^3(c+d x)}{8 d}+\frac {21 a^4 \cos (c+d x) \sin (c+d x)}{16 d}-\frac {a \cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}-\frac {3 \cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )^2}{10 d}-\frac {21 \cos ^3(c+d x) \left (a^4+a^4 \sin (c+d x)\right )}{40 d}+\frac {1}{16} \left (21 a^4\right ) \int 1 \, dx\\ &=\frac {21 a^4 x}{16}-\frac {7 a^4 \cos ^3(c+d x)}{8 d}+\frac {21 a^4 \cos (c+d x) \sin (c+d x)}{16 d}-\frac {a \cos ^3(c+d x) (a+a \sin (c+d x))^3}{6 d}-\frac {3 \cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )^2}{10 d}-\frac {21 \cos ^3(c+d x) \left (a^4+a^4 \sin (c+d x)\right )}{40 d}\\ \end {align*}

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Mathematica [A]  time = 0.44, size = 151, normalized size = 1.10 \[ -\frac {a^4 \left (630 \sqrt {1-\sin (c+d x)} \sin ^{-1}\left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right )+\sqrt {\sin (c+d x)+1} \left (40 \sin ^6(c+d x)+152 \sin ^5(c+d x)+158 \sin ^4(c+d x)-94 \sin ^3(c+d x)-331 \sin ^2(c+d x)-373 \sin (c+d x)+448\right )\right ) \cos ^3(c+d x)}{240 d (\sin (c+d x)-1)^2 (\sin (c+d x)+1)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/240*(a^4*Cos[c + d*x]^3*(630*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sqrt[1 - Sin[c + d*x]] + Sqrt[1 + Sin[c
+ d*x]]*(448 - 373*Sin[c + d*x] - 331*Sin[c + d*x]^2 - 94*Sin[c + d*x]^3 + 158*Sin[c + d*x]^4 + 152*Sin[c + d
*x]^5 + 40*Sin[c + d*x]^6)))/(d*(-1 + Sin[c + d*x])^2*(1 + Sin[c + d*x])^(3/2))

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fricas [A]  time = 0.49, size = 85, normalized size = 0.62 \[ \frac {192 \, a^{4} \cos \left (d x + c\right )^{5} - 640 \, a^{4} \cos \left (d x + c\right )^{3} + 315 \, a^{4} d x + 5 \, {\left (8 \, a^{4} \cos \left (d x + c\right )^{5} - 86 \, a^{4} \cos \left (d x + c\right )^{3} + 63 \, a^{4} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(192*a^4*cos(d*x + c)^5 - 640*a^4*cos(d*x + c)^3 + 315*a^4*d*x + 5*(8*a^4*cos(d*x + c)^5 - 86*a^4*cos(d*
x + c)^3 + 63*a^4*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.21, size = 106, normalized size = 0.77 \[ \frac {21}{16} \, a^{4} x + \frac {a^{4} \cos \left (5 \, d x + 5 \, c\right )}{20 \, d} - \frac {5 \, a^{4} \cos \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac {3 \, a^{4} \cos \left (d x + c\right )}{2 \, d} + \frac {a^{4} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {13 \, a^{4} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {15 \, a^{4} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

21/16*a^4*x + 1/20*a^4*cos(5*d*x + 5*c)/d - 5/12*a^4*cos(3*d*x + 3*c)/d - 3/2*a^4*cos(d*x + c)/d + 1/192*a^4*s
in(6*d*x + 6*c)/d - 13/64*a^4*sin(4*d*x + 4*c)/d + 15/64*a^4*sin(2*d*x + 2*c)/d

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maple [A]  time = 0.32, size = 182, normalized size = 1.33 \[ \frac {a^{4} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{6}-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{8}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )+4 a^{4} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )+6 a^{4} \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {4 a^{4} \left (\cos ^{3}\left (d x +c \right )\right )}{3}+a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^4*(-1/6*sin(d*x+c)^3*cos(d*x+c)^3-1/8*cos(d*x+c)^3*sin(d*x+c)+1/16*cos(d*x+c)*sin(d*x+c)+1/16*d*x+1/16*
c)+4*a^4*(-1/5*sin(d*x+c)^2*cos(d*x+c)^3-2/15*cos(d*x+c)^3)+6*a^4*(-1/4*cos(d*x+c)^3*sin(d*x+c)+1/8*cos(d*x+c)
*sin(d*x+c)+1/8*d*x+1/8*c)-4/3*a^4*cos(d*x+c)^3+a^4*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))

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maxima [A]  time = 0.32, size = 128, normalized size = 0.93 \[ -\frac {1280 \, a^{4} \cos \left (d x + c\right )^{3} - 256 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{4} + 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 12 \, d x - 12 \, c + 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{4} - 180 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{4} - 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/960*(1280*a^4*cos(d*x + c)^3 - 256*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a^4 + 5*(4*sin(2*d*x + 2*c)^3 - 12
*d*x - 12*c + 3*sin(4*d*x + 4*c))*a^4 - 180*(4*d*x + 4*c - sin(4*d*x + 4*c))*a^4 - 240*(2*d*x + 2*c + sin(2*d*
x + 2*c))*a^4)/d

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mupad [B]  time = 10.73, size = 349, normalized size = 2.55 \[ \frac {21\,a^4\,x}{16}-\frac {\frac {63\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}-\frac {63\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {235\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}+\frac {235\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}-\frac {5\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}+\frac {a^4\,\left (315\,c+315\,d\,x\right )}{240}-\frac {a^4\,\left (315\,c+315\,d\,x-896\right )}{240}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (\frac {a^4\,\left (315\,c+315\,d\,x\right )}{40}-\frac {a^4\,\left (1890\,c+1890\,d\,x-1920\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^4\,\left (315\,c+315\,d\,x\right )}{40}-\frac {a^4\,\left (1890\,c+1890\,d\,x-3456\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a^4\,\left (315\,c+315\,d\,x\right )}{16}-\frac {a^4\,\left (4725\,c+4725\,d\,x-3840\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {a^4\,\left (315\,c+315\,d\,x\right )}{16}-\frac {a^4\,\left (4725\,c+4725\,d\,x-9600\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {a^4\,\left (315\,c+315\,d\,x\right )}{12}-\frac {a^4\,\left (6300\,c+6300\,d\,x-8960\right )}{240}\right )+\frac {5\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(21*a^4*x)/16 - ((63*a^4*tan(c/2 + (d*x)/2)^7)/4 - (63*a^4*tan(c/2 + (d*x)/2)^5)/4 - (235*a^4*tan(c/2 + (d*x)/
2)^3)/24 + (235*a^4*tan(c/2 + (d*x)/2)^9)/24 - (5*a^4*tan(c/2 + (d*x)/2)^11)/8 + (a^4*(315*c + 315*d*x))/240 -
(a^4*(315*c + 315*d*x - 896))/240 + tan(c/2 + (d*x)/2)^10*((a^4*(315*c + 315*d*x))/40 - (a^4*(1890*c + 1890*d
*x - 1920))/240) + tan(c/2 + (d*x)/2)^2*((a^4*(315*c + 315*d*x))/40 - (a^4*(1890*c + 1890*d*x - 3456))/240) +
tan(c/2 + (d*x)/2)^4*((a^4*(315*c + 315*d*x))/16 - (a^4*(4725*c + 4725*d*x - 3840))/240) + tan(c/2 + (d*x)/2)^
8*((a^4*(315*c + 315*d*x))/16 - (a^4*(4725*c + 4725*d*x - 9600))/240) + tan(c/2 + (d*x)/2)^6*((a^4*(315*c + 31
5*d*x))/12 - (a^4*(6300*c + 6300*d*x - 8960))/240) + (5*a^4*tan(c/2 + (d*x)/2))/8)/(d*(tan(c/2 + (d*x)/2)^2 +
1)^6)

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sympy [A]  time = 5.09, size = 381, normalized size = 2.78 \[ \begin {cases} \frac {a^{4} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {3 a^{4} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 a^{4} x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {3 a^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {3 a^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {a^{4} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {a^{4} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {3 a^{4} x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {a^{4} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {a^{4} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} - \frac {a^{4} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {3 a^{4} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} - \frac {4 a^{4} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {a^{4} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {3 a^{4} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac {a^{4} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} - \frac {8 a^{4} \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac {4 a^{4} \cos ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{4} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**4*x*sin(c + d*x)**6/16 + 3*a**4*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 3*a**4*x*sin(c + d*x)**4/
4 + 3*a**4*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 3*a**4*x*sin(c + d*x)**2*cos(c + d*x)**2/2 + a**4*x*sin(c +
d*x)**2/2 + a**4*x*cos(c + d*x)**6/16 + 3*a**4*x*cos(c + d*x)**4/4 + a**4*x*cos(c + d*x)**2/2 + a**4*sin(c + d
*x)**5*cos(c + d*x)/(16*d) - a**4*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) + 3*a**4*sin(c + d*x)**3*cos(c + d*x)/
(4*d) - 4*a**4*sin(c + d*x)**2*cos(c + d*x)**3/(3*d) - a**4*sin(c + d*x)*cos(c + d*x)**5/(16*d) - 3*a**4*sin(c
+ d*x)*cos(c + d*x)**3/(4*d) + a**4*sin(c + d*x)*cos(c + d*x)/(2*d) - 8*a**4*cos(c + d*x)**5/(15*d) - 4*a**4*
cos(c + d*x)**3/(3*d), Ne(d, 0)), (x*(a*sin(c) + a)**4*cos(c)**2, True))

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