3.952 \(\int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\)

Optimal. Leaf size=141 \[ \frac{\tan (c+d x) \left (a^2 (2 A+3 C)+6 a b B+2 A b^2\right )}{3 d}+\frac{\left (a^2 B+2 a b (A+2 C)+2 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a (3 a B+2 A b) \tan (c+d x) \sec (c+d x)}{6 d}+\frac{A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}+b^2 C x \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.366121, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {3047, 3031, 3021, 2735, 3770} \[ \frac{\tan (c+d x) \left (a^2 (2 A+3 C)+6 a b B+2 A b^2\right )}{3 d}+\frac{\left (a^2 B+2 a b (A+2 C)+2 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a (3 a B+2 A b) \tan (c+d x) \sec (c+d x)}{6 d}+\frac{A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}+b^2 C x \]

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]^4,x]

[Out]

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

Rule 3047

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

Rule 3031

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[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3021

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

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 ^4(c+d x) \, dx &=\frac{A (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{3} \int (a+b \cos (c+d x)) \left (2 A b+3 a B+(2 a A+3 b B+3 a C) \cos (c+d x)+3 b C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{a (2 A b+3 a B) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{1}{6} \int \left (-2 \left (2 A b^2+6 a b B+a^2 (2 A+3 C)\right )-3 \left (a^2 B+2 b^2 B+2 a b (A+2 C)\right ) \cos (c+d x)-6 b^2 C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{\left (2 A b^2+6 a b B+a^2 (2 A+3 C)\right ) \tan (c+d x)}{3 d}+\frac{a (2 A b+3 a B) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{1}{6} \int \left (-3 \left (a^2 B+2 b^2 B+2 a b (A+2 C)\right )-6 b^2 C \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=b^2 C x+\frac{\left (2 A b^2+6 a b B+a^2 (2 A+3 C)\right ) \tan (c+d x)}{3 d}+\frac{a (2 A b+3 a B) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{1}{2} \left (-a^2 B-2 b^2 B-2 a b (A+2 C)\right ) \int \sec (c+d x) \, dx\\ &=b^2 C x+\frac{\left (a^2 B+2 b^2 B+2 a b (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{\left (2 A b^2+6 a b B+a^2 (2 A+3 C)\right ) \tan (c+d x)}{3 d}+\frac{a (2 A b+3 a B) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.615442, size = 104, normalized size = 0.74 \[ \frac{3 \left (a^2 B+2 a b (A+2 C)+2 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))+3 \tan (c+d x) \left (2 a^2 (A+C)+a (a B+2 A b) \sec (c+d x)+4 a b B+2 A b^2\right )+2 a^2 A \tan ^3(c+d x)+6 b^2 C d x}{6 d} \]

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]^4,x]

[Out]

(6*b^2*C*d*x + 3*(a^2*B + 2*b^2*B + 2*a*b*(A + 2*C))*ArcTanh[Sin[c + d*x]] + 3*(2*A*b^2 + 4*a*b*B + 2*a^2*(A +
C) + a*(2*A*b + a*B)*Sec[c + d*x])*Tan[c + d*x] + 2*a^2*A*Tan[c + d*x]^3)/(6*d)

________________________________________________________________________________________

Maple [A]  time = 0.066, size = 225, normalized size = 1.6 \begin{align*}{\frac{A{b}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{{b}^{2}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{b}^{2}Cx+{\frac{C{b}^{2}c}{d}}+{\frac{aAb\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+{\frac{aAb\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{abB\tan \left ( dx+c \right ) }{d}}+2\,{\frac{abC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{2\,A{a}^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{A{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{2}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{2}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{a}^{2}C\tan \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)^4,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.00142, size = 298, normalized size = 2.11 \begin{align*} \frac{4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 12 \,{\left (d x + c\right )} C b^{2} - 3 \, B a^{2}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, A a b{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C a b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B b^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C a^{2} \tan \left (d x + c\right ) + 24 \, B a b \tan \left (d x + c\right ) + 12 \, A b^{2} \tan \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)^4,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.81486, size = 444, normalized size = 3.15 \begin{align*} \frac{12 \, C b^{2} d x \cos \left (d x + c\right )^{3} + 3 \,{\left (B a^{2} + 2 \,{\left (A + 2 \, C\right )} a b + 2 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (B a^{2} + 2 \,{\left (A + 2 \, C\right )} a b + 2 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, A a^{2} + 2 \,{\left ({\left (2 \, A + 3 \, C\right )} a^{2} + 6 \, B a b + 3 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \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)^4,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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)**4,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.23938, size = 491, normalized size = 3.48 \begin{align*} \frac{6 \,{\left (d x + c\right )} C b^{2} + 3 \,{\left (B a^{2} + 2 \, A a b + 4 \, C a b + 2 \, B b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (B a^{2} + 2 \, A a b + 4 \, C a b + 2 \, B b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (6 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, A a b \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 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 4 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 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} + 6 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, A a b \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 \, A 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)^4,x, algorithm="giac")

[Out]

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