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+C*a)*arctanh(sin(d*x+c))/d+b*C*tan(d*x+c)/d

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Rubi [A]  time = 0.07, antiderivative size = 35, normalized size of antiderivative = 1.00, 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 8

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

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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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 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]

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*}

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Mathematica [A]  time = 0.01, 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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fricas [B]  time = 0.50, size = 85, normalized size = 2.43 \[ \frac {2 \, B a d x \cos \left (d x + c\right ) + {\left (C a + B b\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (C a + B b\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, C b \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

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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giac [B]  time = 0.23, size = 84, normalized size = 2.40 \[ \frac {{\left (d x + c\right )} B a + {\left (C a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (C a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}}{d} \]

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

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maple [A]  time = 0.72, size = 65, normalized size = 1.86 \[ a B x +\frac {B b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {B a c}{d}+\frac {b C \tan \left (d x +c \right )}{d}+\frac {a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

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 = 0.35, size = 73, normalized size = 2.09 \[ \frac {2 \, {\left (d x + c\right )} B a + C a {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + B b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C b \tan \left (d x + c\right )}{2 \, d} \]

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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mupad [B]  time = 3.87, size = 114, normalized size = 3.26 \[ \frac {2\,B\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {C\,b\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}-\frac {B\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}-\frac {C\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (B + C \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx \]

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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