3.943 \(\int (a+b \cos (c+d x)) (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x) \, dx\)

Optimal. Leaf size=69 \[ \frac{(a (A+2 C)+2 b B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(a B+A b) \tan (c+d x)}{d}+\frac{a A \tan (c+d x) \sec (c+d x)}{2 d}+b C x \]

[Out]

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

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Rubi [A]  time = 0.168074, antiderivative size = 69, 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, 3021, 2735, 3770} \[ \frac{(a (A+2 C)+2 b B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(a B+A b) \tan (c+d x)}{d}+\frac{a A \tan (c+d x) \sec (c+d x)}{2 d}+b C x \]

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]^3,x]

[Out]

b*C*x + ((2*b*B + a*(A + 2*C))*ArcTanh[Sin[c + d*x]])/(2*d) + ((A*b + a*B)*Tan[c + d*x])/d + (a*A*Sec[c + d*x]
*Tan[c + d*x])/(2*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 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)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx &=\frac{a A \sec (c+d x) \tan (c+d x)}{2 d}-\frac{1}{2} \int \left (-2 (A b+a B)-(2 b B+a (A+2 C)) \cos (c+d x)-2 b C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{(A b+a B) \tan (c+d x)}{d}+\frac{a A \sec (c+d x) \tan (c+d x)}{2 d}-\frac{1}{2} \int (-2 b B-a (A+2 C)-2 b C \cos (c+d x)) \sec (c+d x) \, dx\\ &=b C x+\frac{(A b+a B) \tan (c+d x)}{d}+\frac{a A \sec (c+d x) \tan (c+d x)}{2 d}-\frac{1}{2} (-2 b B-a (A+2 C)) \int \sec (c+d x) \, dx\\ &=b C x+\frac{(2 b B+a (A+2 C)) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(A b+a B) \tan (c+d x)}{d}+\frac{a A \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.0275851, size = 92, normalized size = 1.33 \[ \frac{a A \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a A \tan (c+d x) \sec (c+d x)}{2 d}+\frac{a B \tan (c+d x)}{d}+\frac{a C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{A b \tan (c+d x)}{d}+\frac{b B \tanh ^{-1}(\sin (c+d x))}{d}+b C x \]

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]^3,x]

[Out]

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

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Maple [A]  time = 0.053, size = 117, normalized size = 1.7 \begin{align*}{\frac{Ab\tan \left ( dx+c \right ) }{d}}+{\frac{bB\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+bCx+{\frac{Cbc}{d}}+{\frac{aA\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{aA\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{Ba\tan \left ( dx+c \right ) }{d}}+{\frac{aC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{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)^3,x)

[Out]

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

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Maxima [A]  time = 1.01096, size = 176, normalized size = 2.55 \begin{align*} \frac{4 \,{\left (d x + c\right )} C b - A a{\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 )} + 2 \, C a{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, B b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a \tan \left (d x + c\right ) + 4 \, A b \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))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.8409, size = 305, normalized size = 4.42 \begin{align*} \frac{4 \, C b d x \cos \left (d x + c\right )^{2} +{\left ({\left (A + 2 \, C\right )} a + 2 \, B b\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left ({\left (A + 2 \, C\right )} a + 2 \, B b\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (A a + 2 \,{\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \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)^3,x, algorithm="fricas")

[Out]

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

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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))*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.24343, size = 227, normalized size = 3.29 \begin{align*} \frac{2 \,{\left (d x + c\right )} C b +{\left (A a + 2 \, C a + 2 \, B b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (A a + 2 \, C a + 2 \, B 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} - 2 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, A 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 ) + 2 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, A b \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))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="giac")

[Out]

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