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

Optimal. Leaf size=32 \[ \frac {a B \sin (c+d x)}{d}+a x (B+C)+\frac {a C \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {4072, 3996, 3770} \[ \frac {a B \sin (c+d x)}{d}+a x (B+C)+\frac {a C \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3770

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

Rule 3996

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.)
+ (A_)), x_Symbol] :> Simp[(A*a*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*n), x] + Dist[1/(d*n), Int[(d*Csc[e + f*x
])^(n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B},
x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]

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

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Mathematica [A]  time = 0.03, size = 46, normalized size = 1.44 \[ \frac {a B \sin (c) \cos (d x)}{d}+\frac {a B \cos (c) \sin (d x)}{d}+a B x+\frac {a C \tanh ^{-1}(\sin (c+d x))}{d}+a C x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.49, size = 51, normalized size = 1.59 \[ \frac {2 \, {\left (B + C\right )} a d x + C a \log \left (\sin \left (d x + c\right ) + 1\right ) - C a \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, B a \sin \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.27, size = 79, normalized size = 2.47 \[ \frac {C a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - C a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (B a + C a\right )} {\left (d x + c\right )} + \frac {2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(C*a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - C*a*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + (B*a + C*a)*(d*x + c) + 2*B
*a*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 = 56, normalized size = 1.75 \[ a B x +a C x +\frac {a B \sin \left (d x +c \right )}{d}+\frac {B a c}{d}+\frac {a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {C a c}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.35, size = 58, normalized size = 1.81 \[ \frac {2 \, {\left (d x + c\right )} B a + 2 \, {\left (d x + c\right )} C a + C a {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, B a \sin \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 3.02, size = 100, normalized size = 3.12 \[ \frac {B\,a\,\sin \left (c+d\,x\right )}{d}+\frac {2\,B\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*a*sin(c + d*x))/d + (2*B*a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*C*a*atan(sin(c/2 + (d*x)/2)/
cos(c/2 + (d*x)/2)))/d + (2*C*a*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int B \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(Integral(B*cos(c + d*x)**2*sec(c + d*x), x) + Integral(B*cos(c + d*x)**2*sec(c + d*x)**2, x) + Integral(C*c
os(c + d*x)**2*sec(c + d*x)**2, x) + Integral(C*cos(c + d*x)**2*sec(c + d*x)**3, x))

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