3.22 \(\int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec (c+d x) \, dx\)

Optimal. Leaf size=111 \[ \frac{5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac{(3 A+5 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{6 d}+\frac{1}{2} a^3 x (7 A+5 B)+\frac{a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a B \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

[Out]

(a^3*(7*A + 5*B)*x)/2 + (a^3*A*ArcTanh[Sin[c + d*x]])/d + (5*a^3*(A + B)*Sin[c + d*x])/(2*d) + (a*B*(a + a*Cos
[c + d*x])^2*Sin[c + d*x])/(3*d) + ((3*A + 5*B)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(6*d)

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Rubi [A]  time = 0.30435, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {2976, 2968, 3023, 2735, 3770} \[ \frac{5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac{(3 A+5 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{6 d}+\frac{1}{2} a^3 x (7 A+5 B)+\frac{a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a B \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(7*A + 5*B)*x)/2 + (a^3*A*ArcTanh[Sin[c + d*x]])/d + (5*a^3*(A + B)*Sin[c + d*x])/(2*d) + (a*B*(a + a*Cos
[c + d*x])^2*Sin[c + d*x])/(3*d) + ((3*A + 5*B)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(6*d)

Rule 2976

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

Rule 2968

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

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2735

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec (c+d x) \, dx &=\frac{a B (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{1}{3} \int (a+a \cos (c+d x))^2 (3 a A+a (3 A+5 B) \cos (c+d x)) \sec (c+d x) \, dx\\ &=\frac{a B (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{(3 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{1}{6} \int (a+a \cos (c+d x)) \left (6 a^2 A+15 a^2 (A+B) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{a B (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{(3 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{1}{6} \int \left (6 a^3 A+\left (6 a^3 A+15 a^3 (A+B)\right ) \cos (c+d x)+15 a^3 (A+B) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac{a B (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{(3 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{1}{6} \int \left (6 a^3 A+3 a^3 (7 A+5 B) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{1}{2} a^3 (7 A+5 B) x+\frac{5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac{a B (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{(3 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\left (a^3 A\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} a^3 (7 A+5 B) x+\frac{a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac{a B (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{(3 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.24898, size = 113, normalized size = 1.02 \[ \frac{a^3 \left (9 (4 A+5 B) \sin (c+d x)+3 (A+3 B) \sin (2 (c+d x))-12 A \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 A \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+42 A d x+B \sin (3 (c+d x))+30 B d x\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(42*A*d*x + 30*B*d*x - 12*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*A*Log[Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2]] + 9*(4*A + 5*B)*Sin[c + d*x] + 3*(A + 3*B)*Sin[2*(c + d*x)] + B*Sin[3*(c + d*x)]))/(12*d)

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Maple [A]  time = 0.111, size = 153, normalized size = 1.4 \begin{align*}{\frac{A{a}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{7\,A{a}^{3}x}{2}}+{\frac{7\,A{a}^{3}c}{2\,d}}+{\frac{B\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{3}}{3\,d}}+{\frac{11\,{a}^{3}B\sin \left ( dx+c \right ) }{3\,d}}+3\,{\frac{A{a}^{3}\sin \left ( dx+c \right ) }{d}}+{\frac{3\,{a}^{3}B\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{5\,{a}^{3}Bx}{2}}+{\frac{5\,{a}^{3}Bc}{2\,d}}+{\frac{A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*A*a^3*cos(d*x+c)*sin(d*x+c)+7/2*A*a^3*x+7/2/d*A*a^3*c+1/3/d*B*sin(d*x+c)*cos(d*x+c)^2*a^3+11/3*a^3*B*sin
(d*x+c)/d+3*a^3*A*sin(d*x+c)/d+3/2/d*a^3*B*cos(d*x+c)*sin(d*x+c)+5/2*a^3*B*x+5/2/d*a^3*B*c+1/d*A*a^3*ln(sec(d*
x+c)+tan(d*x+c))

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Maxima [A]  time = 1.00012, size = 190, normalized size = 1.71 \begin{align*} \frac{3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 36 \,{\left (d x + c\right )} A a^{3} - 4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} + 9 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 12 \,{\left (d x + c\right )} B a^{3} + 12 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 36 \, A a^{3} \sin \left (d x + c\right ) + 36 \, B a^{3} \sin \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^3 + 36*(d*x + c)*A*a^3 - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^
3 + 9*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3 + 12*(d*x + c)*B*a^3 + 12*A*a^3*log(sec(d*x + c) + tan(d*x + c))
+ 36*A*a^3*sin(d*x + c) + 36*B*a^3*sin(d*x + c))/d

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Fricas [A]  time = 1.49719, size = 254, normalized size = 2.29 \begin{align*} \frac{3 \,{\left (7 \, A + 5 \, B\right )} a^{3} d x + 3 \, A a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, A a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (2 \, B a^{3} \cos \left (d x + c\right )^{2} + 3 \,{\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right ) + 2 \,{\left (9 \, A + 11 \, B\right )} a^{3}\right )} \sin \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(7*A + 5*B)*a^3*d*x + 3*A*a^3*log(sin(d*x + c) + 1) - 3*A*a^3*log(-sin(d*x + c) + 1) + (2*B*a^3*cos(d*x
+ c)^2 + 3*(A + 3*B)*a^3*cos(d*x + c) + 2*(9*A + 11*B)*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*(A+B*cos(d*x+c))*sec(d*x+c),x)

[Out]

Timed out

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Giac [A]  time = 1.28875, size = 243, normalized size = 2.19 \begin{align*} \frac{6 \, A a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 6 \, A a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + 3 \,{\left (7 \, A a^{3} + 5 \, B a^{3}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (15 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 15 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 36 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 40 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 21 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 33 \, B 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 )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*A*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 6*A*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 3*(7*A*a^3 + 5*
B*a^3)*(d*x + c) + 2*(15*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 15*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 36*A*a^3*tan(1/2*d*x
+ 1/2*c)^3 + 40*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 21*A*a^3*tan(1/2*d*x + 1/2*c) + 33*B*a^3*tan(1/2*d*x + 1/2*c))
/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d