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

Optimal. Leaf size=102 \[ -\frac {7 a^2 \cos ^5(c+d x)}{30 d}-\frac {\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{6 d}+\frac {7 a^2 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {7 a^2 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {7 a^2 x}{16} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 102, 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} \[ -\frac {7 a^2 \cos ^5(c+d x)}{30 d}-\frac {\cos ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{6 d}+\frac {7 a^2 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {7 a^2 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {7 a^2 x}{16} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.54, size = 151, normalized size = 1.48 \[ -\frac {a^2 \left (\sqrt {\sin (c+d x)+1} \left (40 \sin ^6(c+d x)+56 \sin ^5(c+d x)-106 \sin ^4(c+d x)-182 \sin ^3(c+d x)+57 \sin ^2(c+d x)+231 \sin (c+d x)-96\right )-210 \sin ^{-1}\left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right ) \sqrt {1-\sin (c+d x)}\right ) \cos ^5(c+d x)}{240 d (\sin (c+d x)-1)^3 (\sin (c+d x)+1)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/240*(a^2*Cos[c + d*x]^5*(-210*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sqrt[1 - Sin[c + d*x]] + Sqrt[1 + Sin[
c + d*x]]*(-96 + 231*Sin[c + d*x] + 57*Sin[c + d*x]^2 - 182*Sin[c + d*x]^3 - 106*Sin[c + d*x]^4 + 56*Sin[c + d
*x]^5 + 40*Sin[c + d*x]^6)))/(d*(-1 + Sin[c + d*x])^3*(1 + Sin[c + d*x])^(5/2))

________________________________________________________________________________________

fricas [A]  time = 0.98, size = 72, normalized size = 0.71 \[ -\frac {96 \, a^{2} \cos \left (d x + c\right )^{5} - 105 \, a^{2} d x + 5 \, {\left (8 \, a^{2} \cos \left (d x + c\right )^{5} - 14 \, a^{2} \cos \left (d x + c\right )^{3} - 21 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(96*a^2*cos(d*x + c)^5 - 105*a^2*d*x + 5*(8*a^2*cos(d*x + c)^5 - 14*a^2*cos(d*x + c)^3 - 21*a^2*cos(d*x
+ c))*sin(d*x + c))/d

________________________________________________________________________________________

giac [A]  time = 0.61, size = 106, normalized size = 1.04 \[ \frac {7}{16} \, a^{2} x - \frac {a^{2} \cos \left (5 \, d x + 5 \, c\right )}{40 \, d} - \frac {a^{2} \cos \left (3 \, d x + 3 \, c\right )}{8 \, d} - \frac {a^{2} \cos \left (d x + c\right )}{4 \, d} - \frac {a^{2} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {a^{2} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {17 \, a^{2} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/16*a^2*x - 1/40*a^2*cos(5*d*x + 5*c)/d - 1/8*a^2*cos(3*d*x + 3*c)/d - 1/4*a^2*cos(d*x + c)/d - 1/192*a^2*sin
(6*d*x + 6*c)/d + 1/64*a^2*sin(4*d*x + 4*c)/d + 17/64*a^2*sin(2*d*x + 2*c)/d

________________________________________________________________________________________

maple [A]  time = 0.18, size = 109, normalized size = 1.07 \[ \frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {2 \left (\cos ^{5}\left (d x +c \right )\right ) a^{2}}{5}+a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^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*cos(
d*x+c)^5*a^2+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))

________________________________________________________________________________________

maxima [A]  time = 0.32, size = 89, normalized size = 0.87 \[ -\frac {384 \, a^{2} \cos \left (d x + c\right )^{5} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} - 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 6.79, size = 349, normalized size = 3.42 \[ \frac {7\,a^2\,x}{16}-\frac {\frac {11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {89\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}-\frac {11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {89\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+\frac {9\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}+\frac {a^2\,\left (105\,c+105\,d\,x\right )}{240}-\frac {a^2\,\left (105\,c+105\,d\,x-192\right )}{240}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^2\,\left (105\,c+105\,d\,x\right )}{40}-\frac {a^2\,\left (630\,c+630\,d\,x-192\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (\frac {a^2\,\left (105\,c+105\,d\,x\right )}{40}-\frac {a^2\,\left (630\,c+630\,d\,x-960\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {a^2\,\left (105\,c+105\,d\,x\right )}{16}-\frac {a^2\,\left (1575\,c+1575\,d\,x-960\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a^2\,\left (105\,c+105\,d\,x\right )}{16}-\frac {a^2\,\left (1575\,c+1575\,d\,x-1920\right )}{240}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {a^2\,\left (105\,c+105\,d\,x\right )}{12}-\frac {a^2\,\left (2100\,c+2100\,d\,x-1920\right )}{240}\right )-\frac {9\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(7*a^2*x)/16 - ((11*a^2*tan(c/2 + (d*x)/2)^5)/4 - (89*a^2*tan(c/2 + (d*x)/2)^3)/24 - (11*a^2*tan(c/2 + (d*x)/2
)^7)/4 + (89*a^2*tan(c/2 + (d*x)/2)^9)/24 + (9*a^2*tan(c/2 + (d*x)/2)^11)/8 + (a^2*(105*c + 105*d*x))/240 - (a
^2*(105*c + 105*d*x - 192))/240 + tan(c/2 + (d*x)/2)^2*((a^2*(105*c + 105*d*x))/40 - (a^2*(630*c + 630*d*x - 1
92))/240) + tan(c/2 + (d*x)/2)^10*((a^2*(105*c + 105*d*x))/40 - (a^2*(630*c + 630*d*x - 960))/240) + tan(c/2 +
(d*x)/2)^8*((a^2*(105*c + 105*d*x))/16 - (a^2*(1575*c + 1575*d*x - 960))/240) + tan(c/2 + (d*x)/2)^4*((a^2*(1
05*c + 105*d*x))/16 - (a^2*(1575*c + 1575*d*x - 1920))/240) + tan(c/2 + (d*x)/2)^6*((a^2*(105*c + 105*d*x))/12
- (a^2*(2100*c + 2100*d*x - 1920))/240) - (9*a^2*tan(c/2 + (d*x)/2))/8)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^6)

________________________________________________________________________________________

sympy [A]  time = 5.22, size = 287, normalized size = 2.81 \[ \begin {cases} \frac {a^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {3 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {a^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {3 a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {3 a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac {5 a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {2 a^{2} \cos ^{5}{\left (c + d x \right )}}{5 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{2} \cos ^{4}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________