3.246 \(\int (a+a \cos (c+d x))^3 (B \cos (c+d x)+C \cos ^2(c+d x)) \sec (c+d x) \, dx\)

Optimal. Leaf size=116 \[ -\frac{a^3 (4 B+3 C) \sin ^3(c+d x)}{12 d}+\frac{a^3 (4 B+3 C) \sin (c+d x)}{d}+\frac{3 a^3 (4 B+3 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{5}{8} a^3 x (4 B+3 C)+\frac{C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.166009, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.184, Rules used = {3029, 2751, 2645, 2637, 2635, 8, 2633} \[ -\frac{a^3 (4 B+3 C) \sin ^3(c+d x)}{12 d}+\frac{a^3 (4 B+3 C) \sin (c+d x)}{d}+\frac{3 a^3 (4 B+3 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{5}{8} a^3 x (4 B+3 C)+\frac{C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3029

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 2751

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

Rule 2645

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandTrig[(a + b*sin[c + d*x])^n, x], x] /;
FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, 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 8

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

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

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

Mathematica [A]  time = 0.301944, size = 86, normalized size = 0.74 \[ \frac{a^3 (24 (15 B+13 C) \sin (c+d x)+24 (3 B+4 C) \sin (2 (c+d x))+8 B \sin (3 (c+d x))+240 B d x+24 C \sin (3 (c+d x))+3 C \sin (4 (c+d x))+180 C d x)}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(240*B*d*x + 180*C*d*x + 24*(15*B + 13*C)*Sin[c + d*x] + 24*(3*B + 4*C)*Sin[2*(c + d*x)] + 8*B*Sin[3*(c +
d*x)] + 24*C*Sin[3*(c + d*x)] + 3*C*Sin[4*(c + d*x)]))/(96*d)

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Maple [A]  time = 0.055, size = 176, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ({a}^{3}C \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +{\frac{{a}^{3}B \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{a}^{3}C \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +3\,{a}^{3}B \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +3\,{a}^{3}C \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +3\,{a}^{3}B\sin \left ( dx+c \right ) +{a}^{3}C\sin \left ( dx+c \right ) +{a}^{3}B \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c),x)

[Out]

1/d*(a^3*C*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+1/3*a^3*B*(2+cos(d*x+c)^2)*sin(d*x+c)+
a^3*C*(2+cos(d*x+c)^2)*sin(d*x+c)+3*a^3*B*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+3*a^3*C*(1/2*cos(d*x+c)*si
n(d*x+c)+1/2*d*x+1/2*c)+3*a^3*B*sin(d*x+c)+a^3*C*sin(d*x+c)+a^3*B*(d*x+c))

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Maxima [A]  time = 1.06128, size = 225, normalized size = 1.94 \begin{align*} -\frac{32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} - 72 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 96 \,{\left (d x + c\right )} B a^{3} + 96 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} - 3 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} - 72 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} - 288 \, B a^{3} \sin \left (d x + c\right ) - 96 \, C a^{3} \sin \left (d x + c\right )}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^3 - 72*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3 - 96*(d*x + c)*B
*a^3 + 96*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^3 - 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*
C*a^3 - 72*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3 - 288*B*a^3*sin(d*x + c) - 96*C*a^3*sin(d*x + c))/d

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Fricas [A]  time = 1.68229, size = 216, normalized size = 1.86 \begin{align*} \frac{15 \,{\left (4 \, B + 3 \, C\right )} a^{3} d x +{\left (6 \, C a^{3} \cos \left (d x + c\right )^{3} + 8 \,{\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 9 \,{\left (4 \, B + 5 \, C\right )} a^{3} \cos \left (d x + c\right ) + 8 \,{\left (11 \, B + 9 \, C\right )} a^{3}\right )} \sin \left (d x + c\right )}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(15*(4*B + 3*C)*a^3*d*x + (6*C*a^3*cos(d*x + c)^3 + 8*(B + 3*C)*a^3*cos(d*x + c)^2 + 9*(4*B + 5*C)*a^3*co
s(d*x + c) + 8*(11*B + 9*C)*a^3)*sin(d*x + c))/d

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))**3*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c),x)

[Out]

Timed out

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Giac [A]  time = 1.49682, size = 238, normalized size = 2.05 \begin{align*} \frac{15 \,{\left (4 \, B a^{3} + 3 \, C a^{3}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (60 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 45 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 220 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 165 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 292 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 219 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 132 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 147 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(15*(4*B*a^3 + 3*C*a^3)*(d*x + c) + 2*(60*B*a^3*tan(1/2*d*x + 1/2*c)^7 + 45*C*a^3*tan(1/2*d*x + 1/2*c)^7
+ 220*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 165*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 292*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 219
*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 132*B*a^3*tan(1/2*d*x + 1/2*c) + 147*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x
+ 1/2*c)^2 + 1)^4)/d