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

Optimal. Leaf size=42 \[ \frac {a A \sin (c+d x)}{d}+a A x+\frac {a C \tan (c+d x)}{d}+\frac {a C \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

a*A*x+a*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 = 0.10, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {4077, 4047, 8, 4045, 3770} \[ \frac {a A \sin (c+d x)}{d}+a A x+\frac {a C \tan (c+d x)}{d}+\frac {a C \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*A*x + (a*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 4077

Int[((A_.) + 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*(C*(n + 1) + A*(n + 2))*Csc[e + f*x] + a*C*(n + 2)*Csc[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] &&  !LtQ[n, -1]

Rubi steps

\begin {align*} \int \cos (c+d x) (a+a \sec (c+d x)) \left (A+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 \sec (c+d x)+a C \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a C \tan (c+d x)}{d}+(a A) \int 1 \, dx+\int \cos (c+d x) \left (a A+a C \sec ^2(c+d x)\right ) \, dx\\ &=a A x+\frac {a A \sin (c+d x)}{d}+\frac {a C \tan (c+d x)}{d}+(a C) \int \sec (c+d x) \, dx\\ &=a A x+\frac {a 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 = 54, normalized size = 1.29 \[ \frac {a A \sin (c) \cos (d x)}{d}+\frac {a A \cos (c) \sin (d x)}{d}+a A x+\frac {a C \tan (c+d x)}{d}+\frac {a C \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*A*x + (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 [B]  time = 0.49, size = 86, normalized size = 2.05 \[ \frac {2 \, A a d x \cos \left (d x + c\right ) + C a \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - C a \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a \cos \left (d x + c\right ) + C a\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*A*a*d*x*cos(d*x + c) + C*a*cos(d*x + c)*log(sin(d*x + c) + 1) - 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 = 0.25, size = 119, normalized size = 2.83 \[ \frac {{\left (d x + c\right )} A a + C a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - C a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d*x + c)*A*a + 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)) + 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 = 0.78, size = 57, normalized size = 1.36 \[ a A x +\frac {a A \sin \left (d x +c \right )}{d}+\frac {A a c}{d}+\frac {a C \tan \left (d x +c \right )}{d}+\frac {a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*A*x+a*A*sin(d*x+c)/d+1/d*A*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 = 0.33, size = 59, normalized size = 1.40 \[ \frac {2 \, {\left (d x + c\right )} A a + C a {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A a \sin \left (d x + c\right ) + 2 \, C a \tan \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(d*x + c)*A*a + 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 = 2.68, size = 91, normalized size = 2.17 \[ \frac {A\,a\,\sin \left (c+d\,x\right )}{d}+\frac {2\,A\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {C\,a\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*a*sin(c + d*x))/d + (2*A*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 + (C*a*sin(c + d*x))/(d*cos(c + d*x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))*(A+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(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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