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

Optimal. Leaf size=137 \[ \frac{b \left (8 a^2 B+9 a A b+2 b^2 B\right ) \sin (c+d x)}{3 d}+\frac{1}{2} x \left (6 a^2 A b+2 a^3 B+3 a b^2 B+A b^3\right )+\frac{a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b^2 (5 a B+3 A b) \sin (c+d x) \cos (c+d x)}{6 d}+\frac{b B \sin (c+d x) (a+b \cos (c+d x))^2}{3 d} \]

[Out]

((6*a^2*A*b + A*b^3 + 2*a^3*B + 3*a*b^2*B)*x)/2 + (a^3*A*ArcTanh[Sin[c + d*x]])/d + (b*(9*a*A*b + 8*a^2*B + 2*
b^2*B)*Sin[c + d*x])/(3*d) + (b^2*(3*A*b + 5*a*B)*Cos[c + d*x]*Sin[c + d*x])/(6*d) + (b*B*(a + b*Cos[c + d*x])
^2*Sin[c + d*x])/(3*d)

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Rubi [A]  time = 0.323917, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {2990, 3033, 3023, 2735, 3770} \[ \frac{b \left (8 a^2 B+9 a A b+2 b^2 B\right ) \sin (c+d x)}{3 d}+\frac{1}{2} x \left (6 a^2 A b+2 a^3 B+3 a b^2 B+A b^3\right )+\frac{a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b^2 (5 a B+3 A b) \sin (c+d x) \cos (c+d x)}{6 d}+\frac{b B \sin (c+d x) (a+b \cos (c+d x))^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*a^2*A*b + A*b^3 + 2*a^3*B + 3*a*b^2*B)*x)/2 + (a^3*A*ArcTanh[Sin[c + d*x]])/d + (b*(9*a*A*b + 8*a^2*B + 2*
b^2*B)*Sin[c + d*x])/(3*d) + (b^2*(3*A*b + 5*a*B)*Cos[c + d*x]*Sin[c + d*x])/(6*d) + (b*B*(a + b*Cos[c + d*x])
^2*Sin[c + d*x])/(3*d)

Rule 2990

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 - 2)*(c + d*Sin[e + f*x
])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*(m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c -
b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m,
1] &&  !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3033

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

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+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec (c+d x) \, dx &=\frac{b B (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{1}{3} \int (a+b \cos (c+d x)) \left (3 a^2 A+\left (6 a A b+3 a^2 B+2 b^2 B\right ) \cos (c+d x)+b (3 A b+5 a B) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{b^2 (3 A b+5 a B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{b B (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{1}{6} \int \left (6 a^3 A+3 \left (6 a^2 A b+A b^3+2 a^3 B+3 a b^2 B\right ) \cos (c+d x)+2 b \left (9 a A b+8 a^2 B+2 b^2 B\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{b \left (9 a A b+8 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 d}+\frac{b^2 (3 A b+5 a B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{b B (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{1}{6} \int \left (6 a^3 A+3 \left (6 a^2 A b+A b^3+2 a^3 B+3 a b^2 B\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{1}{2} \left (6 a^2 A b+A b^3+2 a^3 B+3 a b^2 B\right ) x+\frac{b \left (9 a A b+8 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 d}+\frac{b^2 (3 A b+5 a B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{b B (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\left (a^3 A\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} \left (6 a^2 A b+A b^3+2 a^3 B+3 a b^2 B\right ) x+\frac{a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b \left (9 a A b+8 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 d}+\frac{b^2 (3 A b+5 a B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{b B (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.370471, size = 159, normalized size = 1.16 \[ \frac{6 (c+d x) \left (6 a^2 A b+2 a^3 B+3 a b^2 B+A b^3\right )+9 b \left (4 a^2 B+4 a A b+b^2 B\right ) \sin (c+d x)-12 a^3 A \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 a^3 A \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+3 b^2 (3 a B+A b) \sin (2 (c+d x))+b^3 B \sin (3 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*(6*a^2*A*b + A*b^3 + 2*a^3*B + 3*a*b^2*B)*(c + d*x) - 12*a^3*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 1
2*a^3*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 9*b*(4*a*A*b + 4*a^2*B + b^2*B)*Sin[c + d*x] + 3*b^2*(A*b +
3*a*B)*Sin[2*(c + d*x)] + b^3*B*Sin[3*(c + d*x)])/(12*d)

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Maple [A]  time = 0.077, size = 207, normalized size = 1.5 \begin{align*}{\frac{A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{a}^{3}Bx+{\frac{B{a}^{3}c}{d}}+3\,A{a}^{2}bx+3\,{\frac{A{a}^{2}bc}{d}}+3\,{\frac{{a}^{2}bB\sin \left ( dx+c \right ) }{d}}+3\,{\frac{Aa{b}^{2}\sin \left ( dx+c \right ) }{d}}+{\frac{3\,Ba{b}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{3\,Ba{b}^{2}x}{2}}+{\frac{3\,Ba{b}^{2}c}{2\,d}}+{\frac{A{b}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{A{b}^{3}x}{2}}+{\frac{A{b}^{3}c}{2\,d}}+{\frac{B\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{b}^{3}}{3\,d}}+{\frac{2\,B{b}^{3}\sin \left ( dx+c \right ) }{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.13425, size = 196, normalized size = 1.43 \begin{align*} \frac{12 \,{\left (d x + c\right )} B a^{3} + 36 \,{\left (d x + c\right )} A a^{2} b + 9 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a b^{2} + 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{3} - 4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B b^{3} + 12 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 36 \, B a^{2} b \sin \left (d x + c\right ) + 36 \, A a b^{2} \sin \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.46861, size = 317, normalized size = 2.31 \begin{align*} \frac{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 ) + 3 \,{\left (2 \, B a^{3} + 6 \, A a^{2} b + 3 \, B a b^{2} + A b^{3}\right )} d x +{\left (2 \, B b^{3} \cos \left (d x + c\right )^{2} + 18 \, B a^{2} b + 18 \, A a b^{2} + 4 \, B b^{3} + 3 \,{\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*A*a^3*log(sin(d*x + c) + 1) - 3*A*a^3*log(-sin(d*x + c) + 1) + 3*(2*B*a^3 + 6*A*a^2*b + 3*B*a*b^2 + A*b
^3)*d*x + (2*B*b^3*cos(d*x + c)^2 + 18*B*a^2*b + 18*A*a*b^2 + 4*B*b^3 + 3*(3*B*a*b^2 + A*b^3)*cos(d*x + c))*si
n(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+b*cos(d*x+c))**3*(A+B*cos(d*x+c))*sec(d*x+c),x)

[Out]

Timed out

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Giac [B]  time = 1.60186, size = 424, normalized size = 3.09 \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 (2 \, B a^{3} + 6 \, A a^{2} b + 3 \, B a b^{2} + A b^{3}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (18 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 18 \, A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, A b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, B b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 36 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 36 \, A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, B b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 18 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, A b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, B b^{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+b*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*(2*B*a^3 + 6*
A*a^2*b + 3*B*a*b^2 + A*b^3)*(d*x + c) + 2*(18*B*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*A*a*b^2*tan(1/2*d*x + 1/2*c
)^5 - 9*B*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 3*A*b^3*tan(1/2*d*x + 1/2*c)^5 + 6*B*b^3*tan(1/2*d*x + 1/2*c)^5 + 36*
B*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 36*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 4*B*b^3*tan(1/2*d*x + 1/2*c)^3 + 18*B*a^2
*b*tan(1/2*d*x + 1/2*c) + 18*A*a*b^2*tan(1/2*d*x + 1/2*c) + 9*B*a*b^2*tan(1/2*d*x + 1/2*c) + 3*A*b^3*tan(1/2*d
*x + 1/2*c) + 6*B*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d