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

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

[Out]

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

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Rubi [A]  time = 0.121883, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.135, Rules used = {4076, 4047, 8, 4045, 3770} \[ x (a B+A b)+\frac{a A \sin (c+d x)}{d}+\frac{(a C+b B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b C \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4076

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[(b*C*Csc[e + f*x]*Cot[e + f*x]*(d*Csc[e + f*x
])^n)/(f*(n + 2)), x] + Dist[1/(n + 2), Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 2) + (B*a*(n + 2) + b*(C*(n + 1)
+ A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*(n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C,
n}, x] &&  !LtQ[n, -1]

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

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

Mathematica [A]  time = 0.0219333, size = 71, normalized size = 1.37 \[ \frac{a A \sin (c) \cos (d x)}{d}+\frac{a A \cos (c) \sin (d x)}{d}+a B x+\frac{a C \tanh ^{-1}(\sin (c+d x))}{d}+A b x+\frac{b B \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b C \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.057, size = 88, normalized size = 1.7 \begin{align*} Abx+aBx+{\frac{A\sin \left ( dx+c \right ) a}{d}}+{\frac{Abc}{d}}+{\frac{Bb\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{Bac}{d}}+{\frac{aC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{Cb\tan \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.985311, size = 124, normalized size = 2.38 \begin{align*} \frac{2 \,{\left (d x + c\right )} B a + 2 \,{\left (d x + c\right )} A b + C a{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + B b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A a \sin \left (d x + c\right ) + 2 \, C b \tan \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.551275, size = 265, normalized size = 5.1 \begin{align*} \frac{2 \,{\left (B a + A b\right )} d x \cos \left (d x + c\right ) +{\left (C a + B b\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (C a + B b\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (A a \cos \left (d x + c\right ) + C b\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(cos(d*x+c)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sec{\left (c + d x \right )}\right ) \left (A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.17684, size = 181, normalized size = 3.48 \begin{align*} \frac{{\left (B a + A b\right )}{\left (d x + c\right )} +{\left (C a + B b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (C a + 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} - 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(cos(d*x+c)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

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