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

Optimal. Leaf size=46 \[ a x (A+B)+\frac {a A \sin (c+d x)}{d}+\frac {a (B+C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a C \tan (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 46, normalized size of antiderivative = 1.00, 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} \[ a x (A+B)+\frac {a A \sin (c+d x)}{d}+\frac {a (B+C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a C \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 71, normalized size = 1.54 \[ \frac {a A \sin (c) \cos (d x)}{d}+\frac {a A \cos (c) \sin (d x)}{d}+a A x+\frac {a B \tanh ^{-1}(\sin (c+d x))}{d}+a B x+\frac {a C \tan (c+d x)}{d}+\frac {a C \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.49, size = 92, normalized size = 2.00 \[ \frac {2 \, {\left (A + B\right )} a d x \cos \left (d x + c\right ) + {\left (B + C\right )} a \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B + C\right )} a \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a \cos \left (d x + c\right ) + C a\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.22, size = 134, normalized size = 2.91 \[ \frac {{\left (A a + B a\right )} {\left (d x + c\right )} + {\left (B a + C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (B a + C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((A*a + B*a)*(d*x + c) + (B*a + C*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (B*a + C*a)*log(abs(tan(1/2*d*x + 1/
2*c) - 1)) + 2*(A*a*tan(1/2*d*x + 1/2*c)^3 - C*a*tan(1/2*d*x + 1/2*c)^3 - A*a*tan(1/2*d*x + 1/2*c) - C*a*tan(1
/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^4 - 1))/d

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maple [A]  time = 0.81, size = 88, normalized size = 1.91 \[ a A x +a B x +\frac {a A \sin \left (d x +c \right )}{d}+\frac {A a c}{d}+\frac {a B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {B a c}{d}+\frac {a C \tan \left (d x +c \right )}{d}+\frac {a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*A*x+a*B*x+a*A*sin(d*x+c)/d+1/d*A*a*c+1/d*a*B*ln(sec(d*x+c)+tan(d*x+c))+1/d*B*a*c+a*C*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 = 0.40, size = 92, normalized size = 2.00 \[ \frac {2 \, {\left (d x + c\right )} A a + 2 \, {\left (d x + c\right )} B a + B a {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + C a {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A a \sin \left (d x + c\right ) + 2 \, C a \tan \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 3.38, size = 159, normalized size = 3.46 \[ \frac {C\,a\,\mathrm {tan}\left (c+d\,x\right )}{d}+\frac {2\,A\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a\,\sin \left (2\,c+2\,d\,x\right )}{2\,d\,\cos \left (c+d\,x\right )}-\frac {B\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}-\frac {C\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(C*a*tan(c + d*x))/d + (2*A*a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*B*a*atan(sin(c/2 + (d*x)/2)/
cos(c/2 + (d*x)/2)))/d - (B*a*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 + (A*a*sin(2*c + 2*d*x))/(2*d*cos(c + d*x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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