3.4.94 \(\int \cos ^4(c+d x) (a+b \sin (c+d x))^2 \, dx\) [394]

Optimal. Leaf size=116 \[ \frac {1}{16} \left (6 a^2+b^2\right ) x-\frac {7 a b \cos ^5(c+d x)}{30 d}+\frac {\left (6 a^2+b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {\left (6 a^2+b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {b \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d} \]

[Out]

1/16*(6*a^2+b^2)*x-7/30*a*b*cos(d*x+c)^5/d+1/16*(6*a^2+b^2)*cos(d*x+c)*sin(d*x+c)/d+1/24*(6*a^2+b^2)*cos(d*x+c
)^3*sin(d*x+c)/d-1/6*b*cos(d*x+c)^5*(a+b*sin(d*x+c))/d

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Rubi [A]
time = 0.08, antiderivative size = 116, 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 = {2771, 2748, 2715, 8} \begin {gather*} \frac {\left (6 a^2+b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {\left (6 a^2+b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} x \left (6 a^2+b^2\right )-\frac {7 a b \cos ^5(c+d x)}{30 d}-\frac {b \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 2715

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] && Integ
erQ[2*n]

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[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 2771

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[1/(m + p), Int[(g*Cos[e + f*x]
)^p*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(m + p) + a*b*(2*m + p - 1)*Sin[e + f*x]), x], x] /; FreeQ
[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + p, 0] && (IntegersQ[2*m, 2*p] || IntegerQ
[m])

Rubi steps

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

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Mathematica [A]
time = 0.21, size = 133, normalized size = 1.15 \begin {gather*} \frac {360 a^2 c+60 b^2 c+360 a^2 d x+60 b^2 d x-240 a b \cos (c+d x)-120 a b \cos (3 (c+d x))-24 a b \cos (5 (c+d x))+240 a^2 \sin (2 (c+d x))+15 b^2 \sin (2 (c+d x))+30 a^2 \sin (4 (c+d x))-15 b^2 \sin (4 (c+d x))-5 b^2 \sin (6 (c+d x))}{960 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(360*a^2*c + 60*b^2*c + 360*a^2*d*x + 60*b^2*d*x - 240*a*b*Cos[c + d*x] - 120*a*b*Cos[3*(c + d*x)] - 24*a*b*Co
s[5*(c + d*x)] + 240*a^2*Sin[2*(c + d*x)] + 15*b^2*Sin[2*(c + d*x)] + 30*a^2*Sin[4*(c + d*x)] - 15*b^2*Sin[4*(
c + d*x)] - 5*b^2*Sin[6*(c + d*x)])/(960*d)

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Maple [A]
time = 0.42, size = 108, normalized size = 0.93

method	result	size
derivativedivides	\(\frac {b^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {2 a b \left (\cos ^{5}\left (d x +c \right )\right )}{5}+a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(108\)
default	\(\frac {b^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {2 a b \left (\cos ^{5}\left (d x +c \right )\right )}{5}+a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(108\)
risch	\(\frac {3 a^{2} x}{8}+\frac {b^{2} x}{16}-\frac {a b \cos \left (d x +c \right )}{4 d}-\frac {\sin \left (6 d x +6 c \right ) b^{2}}{192 d}-\frac {a b \cos \left (5 d x +5 c \right )}{40 d}+\frac {a^{2} \sin \left (4 d x +4 c \right )}{32 d}-\frac {\sin \left (4 d x +4 c \right ) b^{2}}{64 d}-\frac {a b \cos \left (3 d x +3 c \right )}{8 d}+\frac {a^{2} \sin \left (2 d x +2 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) b^{2}}{64 d}\)	\(144\)
norman	\(\frac {\left (\frac {3 a^{2}}{8}+\frac {b^{2}}{16}\right ) x +\left (\frac {3 a^{2}}{8}+\frac {b^{2}}{16}\right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {9 a^{2}}{4}+\frac {3 b^{2}}{8}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {9 a^{2}}{4}+\frac {3 b^{2}}{8}\right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {15 a^{2}}{2}+\frac {5 b^{2}}{4}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {45 a^{2}}{8}+\frac {15 b^{2}}{16}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {45 a^{2}}{8}+\frac {15 b^{2}}{16}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4 a b}{5 d}+\frac {\left (2 a^{2}-13 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {\left (2 a^{2}-13 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (10 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}-\frac {\left (10 a^{2}-b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {\left (42 a^{2}+47 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}-\frac {\left (42 a^{2}+47 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}-\frac {4 a b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}-\frac {8 a b \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a b \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a b \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a b \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\)	\(432\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(b^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*b*
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 = 0.30, size = 88, normalized size = 0.76 \begin {gather*} -\frac {384 \, a b \cos \left (d x + c\right )^{5} - 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} - 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 )} b^{2}}{960 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 89, normalized size = 0.77 \begin {gather*} -\frac {96 \, a b \cos \left (d x + c\right )^{5} - 15 \, {\left (6 \, a^{2} + b^{2}\right )} d x + 5 \, {\left (8 \, b^{2} \cos \left (d x + c\right )^{5} - 2 \, {\left (6 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (6 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(96*a*b*cos(d*x + c)^5 - 15*(6*a^2 + b^2)*d*x + 5*(8*b^2*cos(d*x + c)^5 - 2*(6*a^2 + b^2)*cos(d*x + c)^
3 - 3*(6*a^2 + b^2)*cos(d*x + c))*sin(d*x + c))/d

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (107) = 214\).
time = 0.44, size = 287, normalized size = 2.47 \begin {gather*} \begin {cases} \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 ^{2}{\left (c + d x \right )}}{4} + \frac {3 a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {2 a b \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac {b^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {3 b^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {b^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {b^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac {b^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right )^{2} \cos ^{4}{\left (c \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 4.98, size = 123, normalized size = 1.06 \begin {gather*} \frac {1}{16} \, {\left (6 \, a^{2} + b^{2}\right )} x - \frac {a b \cos \left (5 \, d x + 5 \, c\right )}{40 \, d} - \frac {a b \cos \left (3 \, d x + 3 \, c\right )}{8 \, d} - \frac {a b \cos \left (d x + c\right )}{4 \, d} - \frac {b^{2} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {{\left (2 \, a^{2} - b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {{\left (16 \, a^{2} + b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 5.31, size = 134, normalized size = 1.16 \begin {gather*} \frac {3\,a^2\,x}{8}+\frac {b^2\,x}{16}+\frac {a^2\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {b^2\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{24\,d}-\frac {b^2\,{\cos \left (c+d\,x\right )}^5\,\sin \left (c+d\,x\right )}{6\,d}-\frac {2\,a\,b\,{\cos \left (c+d\,x\right )}^5}{5\,d}+\frac {3\,a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{8\,d}+\frac {b^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{16\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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