3.949 \(\int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec (c+d x) \, dx\)

Optimal. Leaf size=134 \[ \frac{\sin (c+d x) \left (2 a^2 C+6 a b B+3 A b^2+2 b^2 C\right )}{3 d}+\frac{1}{2} x \left (2 a^2 B+2 a b (2 A+C)+b^2 B\right )+\frac{a^2 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b (2 a C+3 b B) \sin (c+d x) \cos (c+d x)}{6 d}+\frac{C \sin (c+d x) (a+b \cos (c+d x))^2}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.331725, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.128, Rules used = {3049, 3033, 3023, 2735, 3770} \[ \frac{\sin (c+d x) \left (2 a^2 C+6 a b B+3 A b^2+2 b^2 C\right )}{3 d}+\frac{1}{2} x \left (2 a^2 B+2 a b (2 A+C)+b^2 B\right )+\frac{a^2 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b (2 a C+3 b B) \sin (c+d x) \cos (c+d x)}{6 d}+\frac{C \sin (c+d x) (a+b \cos (c+d x))^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3049

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] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !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))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx &=\frac{C (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 A+(3 A b+3 a B+2 b C) \cos (c+d x)+(3 b B+2 a C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{b (3 b B+2 a C) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{C (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{1}{6} \int \left (6 a^2 A+3 \left (2 a^2 B+b^2 B+2 a b (2 A+C)\right ) \cos (c+d x)+2 \left (3 A b^2+6 a b B+2 a^2 C+2 b^2 C\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{\left (3 A b^2+6 a b B+2 a^2 C+2 b^2 C\right ) \sin (c+d x)}{3 d}+\frac{b (3 b B+2 a C) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{C (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{1}{6} \int \left (6 a^2 A+3 \left (2 a^2 B+b^2 B+2 a b (2 A+C)\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{1}{2} \left (2 a^2 B+b^2 B+2 a b (2 A+C)\right ) x+\frac{\left (3 A b^2+6 a b B+2 a^2 C+2 b^2 C\right ) \sin (c+d x)}{3 d}+\frac{b (3 b B+2 a C) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{C (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\left (a^2 A\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} \left (2 a^2 B+b^2 B+2 a b (2 A+C)\right ) x+\frac{a^2 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{\left (3 A b^2+6 a b B+2 a^2 C+2 b^2 C\right ) \sin (c+d x)}{3 d}+\frac{b (3 b B+2 a C) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{C (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.499848, size = 158, normalized size = 1.18 \[ \frac{6 (c+d x) \left (2 a^2 B+2 a b (2 A+C)+b^2 B\right )+3 \sin (c+d x) \left (4 a^2 C+8 a b B+4 A b^2+3 b^2 C\right )-12 a^2 A \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 a^2 A \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+3 b (2 a C+b B) \sin (2 (c+d x))+b^2 C \sin (3 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*(2*a^2*B + b^2*B + 2*a*b*(2*A + C))*(c + d*x) - 12*a^2*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*a^2*
A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 3*(4*A*b^2 + 8*a*b*B + 4*a^2*C + 3*b^2*C)*Sin[c + d*x] + 3*b*(b*B
+ 2*a*C)*Sin[2*(c + d*x)] + b^2*C*Sin[3*(c + d*x)])/(12*d)

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Maple [A]  time = 0.053, size = 204, normalized size = 1.5 \begin{align*}{\frac{A{b}^{2}\sin \left ( dx+c \right ) }{d}}+{\frac{{b}^{2}B\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{{b}^{2}Bx}{2}}+{\frac{{b}^{2}Bc}{2\,d}}+{\frac{C\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{b}^{2}}{3\,d}}+{\frac{2\,{b}^{2}C\sin \left ( dx+c \right ) }{3\,d}}+2\,aAbx+2\,{\frac{Aabc}{d}}+2\,{\frac{abB\sin \left ( dx+c \right ) }{d}}+{\frac{C\cos \left ( dx+c \right ) ab\sin \left ( dx+c \right ) }{d}}+abCx+{\frac{abCc}{d}}+{\frac{A{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{a}^{2}Bx+{\frac{B{a}^{2}c}{d}}+{\frac{{a}^{2}C\sin \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.988509, size = 203, normalized size = 1.51 \begin{align*} \frac{12 \,{\left (d x + c\right )} B a^{2} + 24 \,{\left (d x + c\right )} A a b + 6 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b + 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{2} - 4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C b^{2} + 12 \, A a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 12 \, C a^{2} \sin \left (d x + c\right ) + 24 \, B a b \sin \left (d x + c\right ) + 12 \, 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))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c),x, algorithm="maxima")

[Out]

1/12*(12*(d*x + c)*B*a^2 + 24*(d*x + c)*A*a*b + 6*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a*b + 3*(2*d*x + 2*c + si
n(2*d*x + 2*c))*B*b^2 - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*b^2 + 12*A*a^2*log(sec(d*x + c) + tan(d*x + c))
+ 12*C*a^2*sin(d*x + c) + 24*B*a*b*sin(d*x + c) + 12*A*b^2*sin(d*x + c))/d

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Fricas [A]  time = 1.83762, size = 313, normalized size = 2.34 \begin{align*} \frac{3 \, A a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, A a^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (2 \, B a^{2} + 2 \,{\left (2 \, A + C\right )} a b + B b^{2}\right )} d x +{\left (2 \, C b^{2} \cos \left (d x + c\right )^{2} + 6 \, C a^{2} + 12 \, B a b + 2 \,{\left (3 \, A + 2 \, C\right )} b^{2} + 3 \,{\left (2 \, C a b + B b^{2}\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))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c),x, algorithm="fricas")

[Out]

1/6*(3*A*a^2*log(sin(d*x + c) + 1) - 3*A*a^2*log(-sin(d*x + c) + 1) + 3*(2*B*a^2 + 2*(2*A + C)*a*b + B*b^2)*d*
x + (2*C*b^2*cos(d*x + c)^2 + 6*C*a^2 + 12*B*a*b + 2*(3*A + 2*C)*b^2 + 3*(2*C*a*b + B*b^2)*cos(d*x + c))*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+b*cos(d*x+c))**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c),x)

[Out]

Timed out

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Giac [B]  time = 1.2862, size = 467, normalized size = 3.49 \begin{align*} \frac{6 \, A a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 6 \, A a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + 3 \,{\left (2 \, B a^{2} + 4 \, A a b + 2 \, C a b + B b^{2}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (6 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 24 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C b^{2} \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))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c),x, algorithm="giac")

[Out]

1/6*(6*A*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 6*A*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 3*(2*B*a^2 + 4*
A*a*b + 2*C*a*b + B*b^2)*(d*x + c) + 2*(6*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 12*B*a*b*tan(1/2*d*x + 1/2*c)^5 - 6*C
*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^2*tan(1/2*d*x + 1/2*c)^5 - 3*B*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*C*b^2*tan(1/
2*d*x + 1/2*c)^5 + 12*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 24*B*a*b*tan(1/2*d*x + 1/2*c)^3 + 12*A*b^2*tan(1/2*d*x +
1/2*c)^3 + 4*C*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*C*a^2*tan(1/2*d*x + 1/2*c) + 12*B*a*b*tan(1/2*d*x + 1/2*c) + 6*C
*a*b*tan(1/2*d*x + 1/2*c) + 6*A*b^2*tan(1/2*d*x + 1/2*c) + 3*B*b^2*tan(1/2*d*x + 1/2*c) + 6*C*b^2*tan(1/2*d*x
+ 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d