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

Optimal. Leaf size=52 \[ \frac{(a B+A b) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a A \tan (c+d x)}{d}+x (a C+b B)+\frac{b C \sin (c+d x)}{d} \]

[Out]

(b*B + a*C)*x + ((A*b + a*B)*ArcTanh[Sin[c + d*x]])/d + (b*C*Sin[c + d*x])/d + (a*A*Tan[c + d*x])/d

________________________________________________________________________________________

Rubi [A]  time = 0.13825, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {3031, 3023, 2735, 3770} \[ \frac{(a B+A b) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a A \tan (c+d x)}{d}+x (a C+b B)+\frac{b C \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*B + a*C)*x + ((A*b + a*B)*ArcTanh[Sin[c + d*x]])/d + (b*C*Sin[c + d*x])/d + (a*A*Tan[c + d*x])/d

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

Mathematica [A]  time = 0.0202367, size = 71, normalized size = 1.37 \[ \frac{a A \tan (c+d x)}{d}+\frac{a B \tanh ^{-1}(\sin (c+d x))}{d}+a C x+\frac{A b \tanh ^{-1}(\sin (c+d x))}{d}+b B x+\frac{b C \sin (c) \cos (d x)}{d}+\frac{b C \cos (c) \sin (d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.049, size = 88, normalized size = 1.7 \begin{align*} bBx+aCx+{\frac{Ab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{A\tan \left ( dx+c \right ) a}{d}}+{\frac{Ba\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{Bbc}{d}}+{\frac{Cb\sin \left ( dx+c \right ) }{d}}+{\frac{Cac}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.999385, size = 124, normalized size = 2.38 \begin{align*} \frac{2 \,{\left (d x + c\right )} C a + 2 \,{\left (d x + c\right )} B b + B a{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + A b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C b \sin \left (d x + c\right ) + 2 \, A a \tan \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.7503, size = 265, normalized size = 5.1 \begin{align*} \frac{2 \,{\left (C a + B b\right )} d x \cos \left (d x + c\right ) +{\left (B a + A b\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (B a + A b\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (C b \cos \left (d x + c\right ) + A a\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))*(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 + B*b)*d*x*cos(d*x + c) + (B*a + A*b)*cos(d*x + c)*log(sin(d*x + c) + 1) - (B*a + A*b)*cos(d*x + c
)*log(-sin(d*x + c) + 1) + 2*(C*b*cos(d*x + c) + A*a)*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))*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.19432, size = 178, normalized size = 3.42 \begin{align*} \frac{{\left (C a + B b\right )}{\left (d x + c\right )} +{\left (B a + A b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (B 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 (A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - C b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + C b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 1}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm="giac")

[Out]

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