3.1.17 \(\int \cos ^2(c+d x) (a+a \sin (c+d x))^2 \, dx\) [17]

Optimal. Leaf size=78 \[ \frac {5 a^2 x}{8}-\frac {5 a^2 \cos ^3(c+d x)}{12 d}+\frac {5 a^2 \cos (c+d x) \sin (c+d x)}{8 d}-\frac {\cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{4 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.05, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2757, 2748, 2715, 8} \begin {gather*} -\frac {5 a^2 \cos ^3(c+d x)}{12 d}-\frac {\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{4 d}+\frac {5 a^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {5 a^2 x}{8} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^2*x)/8 - (5*a^2*Cos[c + d*x]^3)/(12*d) + (5*a^2*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (Cos[c + d*x]^3*(a^2 +
a^2*Sin[c + d*x]))/(4*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 2757

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

________________________________________________________________________________________

Mathematica [A]
time = 0.20, size = 131, normalized size = 1.68 \begin {gather*} -\frac {a^2 \cos ^3(c+d x) \left (30 \sin ^{-1}\left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right ) \sqrt {1-\sin (c+d x)}+\sqrt {1+\sin (c+d x)} \left (16-25 \sin (c+d x)-7 \sin ^2(c+d x)+10 \sin ^3(c+d x)+6 \sin ^4(c+d x)\right )\right )}{24 d (-1+\sin (c+d x))^2 (1+\sin (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*(a^2*Cos[c + d*x]^3*(30*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sqrt[1 - Sin[c + d*x]] + Sqrt[1 + Sin[c +
d*x]]*(16 - 25*Sin[c + d*x] - 7*Sin[c + d*x]^2 + 10*Sin[c + d*x]^3 + 6*Sin[c + d*x]^4)))/(d*(-1 + Sin[c + d*x
])^2*(1 + Sin[c + d*x])^(3/2))

________________________________________________________________________________________

Maple [A]
time = 0.13, size = 87, normalized size = 1.12

method	result	size
risch	\(\frac {5 a^{2} x}{8}-\frac {a^{2} \cos \left (d x +c \right )}{2 d}-\frac {a^{2} \sin \left (4 d x +4 c \right )}{32 d}-\frac {a^{2} \cos \left (3 d x +3 c \right )}{6 d}+\frac {a^{2} \sin \left (2 d x +2 c \right )}{4 d}\)	\(73\)
derivativedivides	\(\frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right ) a^{2}}{3}+a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(87\)
default	\(\frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right ) a^{2}}{3}+a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(87\)
norman	\(\frac {\frac {5 a^{2} x}{8}-\frac {4 a^{2}}{3 d}+\frac {3 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {11 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {11 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {3 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {5 a^{2} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {15 a^{2} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {5 a^{2} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {5 a^{2} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {4 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {4 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\)	\(231\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.31, size = 65, normalized size = 0.83 \begin {gather*} -\frac {64 \, a^{2} \cos \left (d x + c\right )^{3} - 3 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{96 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(64*a^2*cos(d*x + c)^3 - 3*(4*d*x + 4*c - sin(4*d*x + 4*c))*a^2 - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^
2)/d

________________________________________________________________________________________

Fricas [A]
time = 0.35, size = 59, normalized size = 0.76 \begin {gather*} -\frac {16 \, a^{2} \cos \left (d x + c\right )^{3} - 15 \, a^{2} d x + 3 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{3} - 5 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(16*a^2*cos(d*x + c)^3 - 15*a^2*d*x + 3*(2*a^2*cos(d*x + c)^3 - 5*a^2*cos(d*x + c))*sin(d*x + c))/d

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (71) = 142\).
time = 0.19, size = 180, normalized size = 2.31 \begin {gather*} \begin {cases} \frac {a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {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 ^{3}{\left (c + d x \right )}}{8 d} + \frac {a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} - \frac {2 a^{2} \cos ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{2} \cos ^{2}{\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)**2*(a+a*sin(d*x+c))**2,x)

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 5.86, size = 72, normalized size = 0.92 \begin {gather*} \frac {5}{8} \, a^{2} x - \frac {a^{2} \cos \left (3 \, d x + 3 \, c\right )}{6 \, d} - \frac {a^{2} \cos \left (d x + c\right )}{2 \, d} - \frac {a^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {a^{2} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/8*a^2*x - 1/6*a^2*cos(3*d*x + 3*c)/d - 1/2*a^2*cos(d*x + c)/d - 1/32*a^2*sin(4*d*x + 4*c)/d + 1/4*a^2*sin(2*
d*x + 2*c)/d

________________________________________________________________________________________

Mupad [B]
time = 6.71, size = 237, normalized size = 3.04 \begin {gather*} \frac {5\,a^2\,x}{8}-\frac {\frac {11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {a^2\,\left (15\,c+15\,d\,x\right )}{24}-\frac {a^2\,\left (15\,c+15\,d\,x-32\right )}{24}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^2\,\left (15\,c+15\,d\,x\right )}{6}-\frac {a^2\,\left (60\,c+60\,d\,x-32\right )}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {a^2\,\left (15\,c+15\,d\,x\right )}{6}-\frac {a^2\,\left (60\,c+60\,d\,x-96\right )}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a^2\,\left (15\,c+15\,d\,x\right )}{4}-\frac {a^2\,\left (90\,c+90\,d\,x-96\right )}{24}\right )-\frac {3\,a^2\,\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 )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*a^2*x)/8 - ((11*a^2*tan(c/2 + (d*x)/2)^5)/4 - (11*a^2*tan(c/2 + (d*x)/2)^3)/4 + (3*a^2*tan(c/2 + (d*x)/2)^7
)/4 + (a^2*(15*c + 15*d*x))/24 - (a^2*(15*c + 15*d*x - 32))/24 + tan(c/2 + (d*x)/2)^2*((a^2*(15*c + 15*d*x))/6
- (a^2*(60*c + 60*d*x - 32))/24) + tan(c/2 + (d*x)/2)^6*((a^2*(15*c + 15*d*x))/6 - (a^2*(60*c + 60*d*x - 96))
/24) + tan(c/2 + (d*x)/2)^4*((a^2*(15*c + 15*d*x))/4 - (a^2*(90*c + 90*d*x - 96))/24) - (3*a^2*tan(c/2 + (d*x)
/2))/4)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^4)

________________________________________________________________________________________