3.9.65 \(\int \cos (c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [865]

Optimal. Leaf size=52 \[ (A b+a B) x+\frac {(b B+a C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a A \sin (c+d x)}{d}+\frac {b C \tan (c+d x)}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 52, 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 = {4161, 4132, 8, 4130, 3855} \begin {gather*} x (a B+A b)+\frac {a A \sin (c+d x)}{d}+\frac {(a C+b B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b C \tan (c+d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(A*b + a*B)*x + ((b*B + a*C)*ArcTanh[Sin[c + d*x]])/d + (a*A*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 3855

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

Rule 4130

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 4132

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 4161

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

________________________________________________________________________________________

Mathematica [A]
time = 0.03, size = 71, normalized size = 1.37 \begin {gather*} A b x+a B x+\frac {b B \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a A \cos (d x) \sin (c)}{d}+\frac {a A \cos (c) \sin (d x)}{d}+\frac {b C \tan (c+d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.08, size = 74, normalized size = 1.42

method	result	size
derivativedivides	\(\frac {A b \left (d x +c \right )+b B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C b \tan \left (d x +c \right )+A \sin \left (d x +c \right ) a +B a \left (d x +c \right )+a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(74\)
default	\(\frac {A b \left (d x +c \right )+b B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C b \tan \left (d x +c \right )+A \sin \left (d x +c \right ) a +B a \left (d x +c \right )+a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(74\)
risch	\(A b x +a B x -\frac {i a A \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {2 i C b}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b B}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b B}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\)	\(143\)
norman	\(\frac {\left (A b +B a \right ) x +\left (-A b -B a \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-A b -B a \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (A b +B a \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {2 \left (a A -C b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (a A +C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {4 a A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\left (b B +a C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {\left (b B +a C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(217\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.29, size = 92, normalized size = 1.77 \begin {gather*} \frac {2 \, {\left (d x + c\right )} B a + 2 \, {\left (d x + c\right )} A b + 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 \, A a \sin \left (d x + c\right ) + 2 \, C b \tan \left (d x + c\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

Fricas [A]
time = 3.10, size = 101, normalized size = 1.94 \begin {gather*} \frac {2 \, {\left (B a + A b\right )} 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 \, {\left (A a \cos \left (d x + c\right ) + C b\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sec {\left (c + d x \right )}\right ) \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos {\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (52) = 104\).
time = 0.49, size = 134, normalized size = 2.58 \begin {gather*} \frac {{\left (B a + A b\right )} {\left (d x + c\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 ) - {\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 \, {\left (A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b \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 b \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 4.26, size = 153, normalized size = 2.94 \begin {gather*} \frac {C\,b\,\mathrm {tan}\left (c+d\,x\right )}{d}+\frac {2\,A\,b\,\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 {2\,B\,b\,\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 {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 {A\,a\,\sin \left (2\,c+2\,d\,x\right )}{2\,d\,\cos \left (c+d\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________