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

Optimal. Leaf size=126 \[ \frac{1}{2} x \left (2 a^2 C+4 a b B+2 A b^2+b^2 C\right )-\frac{b \sin (c+d x) (2 a A-2 a C-b B)}{d}+\frac{a (a B+2 A b) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{A \tan (c+d x) (a+b \cos (c+d x))^2}{d}-\frac{b^2 (2 A-C) \sin (c+d x) \cos (c+d x)}{2 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.322606, antiderivative size = 126, 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, 3033, 3023, 2735, 3770} \[ \frac{1}{2} x \left (2 a^2 C+4 a b B+2 A b^2+b^2 C\right )-\frac{b \sin (c+d x) (2 a A-2 a C-b B)}{d}+\frac{a (a B+2 A b) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{A \tan (c+d x) (a+b \cos (c+d x))^2}{d}-\frac{b^2 (2 A-C) \sin (c+d x) \cos (c+d x)}{2 d} \]

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]^2,x]

[Out]

((2*A*b^2 + 4*a*b*B + 2*a^2*C + b^2*C)*x)/2 + (a*(2*A*b + a*B)*ArcTanh[Sin[c + d*x]])/d - (b*(2*a*A - b*B - 2*
a*C)*Sin[c + d*x])/d - (b^2*(2*A - C)*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (A*(a + b*Cos[c + d*x])^2*Tan[c + d*x
])/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 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 ^2(c+d x) \, dx &=\frac{A (a+b \cos (c+d x))^2 \tan (c+d x)}{d}+\int (a+b \cos (c+d x)) \left (2 A b+a B+(b B+a C) \cos (c+d x)-b (2 A-C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b^2 (2 A-C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac{A (a+b \cos (c+d x))^2 \tan (c+d x)}{d}+\frac{1}{2} \int \left (2 a (2 A b+a B)+\left (2 A b^2+4 a b B+2 a^2 C+b^2 C\right ) \cos (c+d x)-2 b (2 a A-b B-2 a C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b (2 a A-b B-2 a C) \sin (c+d x)}{d}-\frac{b^2 (2 A-C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac{A (a+b \cos (c+d x))^2 \tan (c+d x)}{d}+\frac{1}{2} \int \left (2 a (2 A b+a B)+\left (2 A b^2+4 a b B+2 a^2 C+b^2 C\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{1}{2} \left (2 A b^2+4 a b B+2 a^2 C+b^2 C\right ) x-\frac{b (2 a A-b B-2 a C) \sin (c+d x)}{d}-\frac{b^2 (2 A-C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac{A (a+b \cos (c+d x))^2 \tan (c+d x)}{d}+(a (2 A b+a B)) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} \left (2 A b^2+4 a b B+2 a^2 C+b^2 C\right ) x+\frac{a (2 A b+a B) \tanh ^{-1}(\sin (c+d x))}{d}-\frac{b (2 a A-b B-2 a C) \sin (c+d x)}{d}-\frac{b^2 (2 A-C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac{A (a+b \cos (c+d x))^2 \tan (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 1.02564, size = 155, normalized size = 1.23 \[ \frac{2 (c+d x) \left (2 a^2 C+4 a b B+2 A b^2+b^2 C\right )+\tan (c+d x) \left (4 a^2 A+4 b (2 a C+b B) \cos (c+d x)+b^2 C \cos (2 (c+d x))+b^2 C\right )-4 a (a B+2 A b) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 a (a B+2 A b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{4 d} \]

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]^2,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.063, size = 171, normalized size = 1.4 \begin{align*} A{b}^{2}x+{\frac{A{b}^{2}c}{d}}+{\frac{{b}^{2}B\sin \left ( dx+c \right ) }{d}}+{\frac{{b}^{2}C\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{{b}^{2}Cx}{2}}+{\frac{{b}^{2}Cc}{2\,d}}+2\,{\frac{aAb\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,abBx+2\,{\frac{Babc}{d}}+2\,{\frac{abC\sin \left ( dx+c \right ) }{d}}+{\frac{A{a}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{a}^{2}Cx+{\frac{C{a}^{2}c}{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)^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.02231, size = 200, normalized size = 1.59 \begin{align*} \frac{4 \,{\left (d x + c\right )} C a^{2} + 8 \,{\left (d x + c\right )} B a b + 4 \,{\left (d x + c\right )} A b^{2} +{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{2} + 2 \, B a^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, A a b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, C a b \sin \left (d x + c\right ) + 4 \, B b^{2} \sin \left (d x + c\right ) + 4 \, A a^{2} \tan \left (d x + c\right )}{4 \, 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)^2,x, algorithm="maxima")

[Out]

1/4*(4*(d*x + c)*C*a^2 + 8*(d*x + c)*B*a*b + 4*(d*x + c)*A*b^2 + (2*d*x + 2*c + sin(2*d*x + 2*c))*C*b^2 + 2*B*
a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*A*a*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1))
+ 8*C*a*b*sin(d*x + c) + 4*B*b^2*sin(d*x + c) + 4*A*a^2*tan(d*x + c))/d

________________________________________________________________________________________

Fricas [A]  time = 1.81429, size = 366, normalized size = 2.9 \begin{align*} \frac{{\left (2 \, C a^{2} + 4 \, B a b +{\left (2 \, A + C\right )} b^{2}\right )} d x \cos \left (d x + c\right ) +{\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (C b^{2} \cos \left (d x + c\right )^{2} + 2 \, A a^{2} + 2 \,{\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \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)^2,x, algorithm="fricas")

[Out]

1/2*((2*C*a^2 + 4*B*a*b + (2*A + C)*b^2)*d*x*cos(d*x + c) + (B*a^2 + 2*A*a*b)*cos(d*x + c)*log(sin(d*x + c) +
1) - (B*a^2 + 2*A*a*b)*cos(d*x + c)*log(-sin(d*x + c) + 1) + (C*b^2*cos(d*x + c)^2 + 2*A*a^2 + 2*(2*C*a*b + B*
b^2)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))

________________________________________________________________________________________

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)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.24385, size = 309, normalized size = 2.45 \begin{align*} -\frac{\frac{4 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1} -{\left (2 \, C a^{2} + 4 \, B a b + 2 \, A b^{2} + C b^{2}\right )}{\left (d x + c\right )} - 2 \,{\left (B a^{2} + 2 \, A a b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) + 2 \,{\left (B a^{2} + 2 \, A a b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (4 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 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 )}^{2}}}{2 \, 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)^2,x, algorithm="giac")

[Out]

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