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

Optimal. Leaf size=35 \[ \frac{(a C+b B) \tanh ^{-1}(\sin (c+d x))}{d}+a B x+\frac{b C \tan (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.0736953, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used = {4072, 3914, 3767, 8, 3770} \[ \frac{(a C+b B) \tanh ^{-1}(\sin (c+d x))}{d}+a B x+\frac{b C \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4072

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

Rule 3914

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[a*c*x, x]
+ (Dist[b*d, Int[Csc[e + f*x]^2, x], x] + Dist[b*c + a*d, Int[Csc[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

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

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

Mathematica [A]  time = 0.0135357, size = 43, normalized size = 1.23 \[ a B x+\frac{a C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b B \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b C \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.046, size = 65, normalized size = 1.9 \begin{align*} aBx+{\frac{Bb\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{Bac}{d}}+{\frac{Cb\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(cos(d*x+c)*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

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

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Maxima [B]  time = 0.971548, size = 99, normalized size = 2.83 \begin{align*} \frac{2 \,{\left (d x + c\right )} B a + 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 \, 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))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/2*(2*(d*x + c)*B*a + 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*C*b*tan(d*x + c))/d

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Fricas [B]  time = 0.510689, size = 225, normalized size = 6.43 \begin{align*} \frac{2 \, B a 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 \, C b \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))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/2*(2*B*a*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(-s
in(d*x + c) + 1) + 2*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 (B + C \sec{\left (c + d x \right )}\right ) \left (a + b \sec{\left (c + d x \right )}\right ) \cos{\left (c + d x \right )} \sec{\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))*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

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

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Giac [B]  time = 1.15949, size = 113, normalized size = 3.23 \begin{align*} \frac{{\left (d x + c\right )} B a +{\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 \, C b \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}}{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))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

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