\(\int \sec (c+d x) (a+a \sec (c+d x)) (A+C \sec ^2(c+d x)) \, dx\) [86]

Optimal result

Integrand size = 29, antiderivative size = 86 \[ \int \sec (c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a (2 A+C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a (3 A+2 C) \tan (c+d x)}{3 d}+\frac {a C \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a C \sec ^2(c+d x) \tan (c+d x)}{3 d} \]

[Out]

1/2*a*(2*A+C)*arctanh(sin(d*x+c))/d+1/3*a*(3*A+2*C)*tan(d*x+c)/d+1/2*a*C*sec(d*x+c)*tan(d*x+c)/d+1/3*a*C*sec(d
*x+c)^2*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {4162, 4132, 3852, 8, 4131, 3855} \[ \int \sec (c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a (2 A+C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a (3 A+2 C) \tan (c+d x)}{3 d}+\frac {a C \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac {a C \tan (c+d x) \sec (c+d x)}{2 d} \]

[In]

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

[Out]

(a*(2*A + C)*ArcTanh[Sin[c + d*x]])/(2*d) + (a*(3*A + 2*C)*Tan[c + d*x])/(3*d) + (a*C*Sec[c + d*x]*Tan[c + d*x
])/(2*d) + (a*C*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

Rule 8

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

Rule 3852

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 3855

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

Rule 4131

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot
[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x
] /; FreeQ[{b, e, f, A, C, m}, 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 4162

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*} \text {integral}& = \frac {a C \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{3} \int \sec (c+d x) \left (3 a A+a (3 A+2 C) \sec (c+d x)+3 a C \sec ^2(c+d x)\right ) \, dx \\ & = \frac {a C \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{3} \int \sec (c+d x) \left (3 a A+3 a C \sec ^2(c+d x)\right ) \, dx+\frac {1}{3} (a (3 A+2 C)) \int \sec ^2(c+d x) \, dx \\ & = \frac {a C \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a C \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{2} (a (2 A+C)) \int \sec (c+d x) \, dx-\frac {(a (3 A+2 C)) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d} \\ & = \frac {a (2 A+C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a (3 A+2 C) \tan (c+d x)}{3 d}+\frac {a C \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a C \sec ^2(c+d x) \tan (c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.65 \[ \int \sec (c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a \left (3 (2 A+C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (6 (A+C)+3 C \sec (c+d x)+2 C \tan ^2(c+d x)\right )\right )}{6 d} \]

[In]

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

[Out]

(a*(3*(2*A + C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(6*(A + C) + 3*C*Sec[c + d*x] + 2*C*Tan[c + d*x]^2)))/(6*
d)

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.02

method	result	size
derivativedivides	\(\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a A \tan \left (d x +c \right )-C a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(88\)
default	\(\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a A \tan \left (d x +c \right )-C a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(88\)
parts	\(\frac {C a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {C a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a A \tan \left (d x +c \right )}{d}\)	\(96\)
norman	\(\frac {-\frac {a \left (2 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {a \left (2 A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 a \left (3 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {a \left (2 A +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a \left (2 A +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(131\)
parallelrisch	\(\frac {a \left (-3 \left (A +\frac {C}{2}\right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \left (A +\frac {C}{2}\right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (A +\frac {2 C}{3}\right ) \sin \left (3 d x +3 c \right )+\sin \left (2 d x +2 c \right ) C +\sin \left (d x +c \right ) \left (A +2 C \right )\right )}{d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(139\)
risch	\(-\frac {i a \left (3 C \,{\mathrm e}^{5 i \left (d x +c \right )}-6 A \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C \,{\mathrm e}^{2 i \left (d x +c \right )}-3 C \,{\mathrm e}^{i \left (d x +c \right )}-6 A -4 C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}\)	\(168\)

[In]

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

[Out]

1/d*(a*A*ln(sec(d*x+c)+tan(d*x+c))+C*a*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+a*A*tan(d*x+c
)-C*a*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.24 \[ \int \sec (c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (2 \, A + C\right )} a \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, A + C\right )} a \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (3 \, A + 2 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \, C a \cos \left (d x + c\right ) + 2 \, C a\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

[In]

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

[Out]

1/12*(3*(2*A + C)*a*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(2*A + C)*a*cos(d*x + c)^3*log(-sin(d*x + c) + 1)
+ 2*(2*(3*A + 2*C)*a*cos(d*x + c)^2 + 3*C*a*cos(d*x + c) + 2*C*a)*sin(d*x + c))/(d*cos(d*x + c)^3)

Sympy [F]

\[ \int \sec (c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=a \left (\int A \sec {\left (c + d x \right )}\, dx + \int A \sec ^{2}{\left (c + d x \right )}\, dx + \int C \sec ^{3}{\left (c + d x \right )}\, dx + \int C \sec ^{4}{\left (c + d x \right )}\, dx\right ) \]

[In]

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

[Out]

a*(Integral(A*sec(c + d*x), x) + Integral(A*sec(c + d*x)**2, x) + Integral(C*sec(c + d*x)**3, x) + Integral(C*
sec(c + d*x)**4, x))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.16 \[ \int \sec (c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a - 3 \, C a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 12 \, A a \tan \left (d x + c\right )}{12 \, d} \]

[In]

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

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a - 3*C*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c)
+ 1) + log(sin(d*x + c) - 1)) + 12*A*a*log(sec(d*x + c) + tan(d*x + c)) + 12*A*a*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.81 \[ \int \sec (c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (2 \, A a + C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (2 \, A a + C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

[In]

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

[Out]

1/6*(3*(2*A*a + C*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*A*a + C*a)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) -
2*(6*A*a*tan(1/2*d*x + 1/2*c)^5 + 3*C*a*tan(1/2*d*x + 1/2*c)^5 - 12*A*a*tan(1/2*d*x + 1/2*c)^3 - 4*C*a*tan(1/
2*d*x + 1/2*c)^3 + 6*A*a*tan(1/2*d*x + 1/2*c) + 9*C*a*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 18.86 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.50 \[ \int \sec (c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,A+C\right )}{d}-\frac {\left (2\,A\,a+C\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-4\,A\,a-\frac {4\,C\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a+3\,C\,a\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[In]

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

[Out]

(a*atanh(tan(c/2 + (d*x)/2))*(2*A + C))/d - (tan(c/2 + (d*x)/2)*(2*A*a + 3*C*a) + tan(c/2 + (d*x)/2)^5*(2*A*a
+ C*a) - tan(c/2 + (d*x)/2)^3*(4*A*a + (4*C*a)/3))/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c
/2 + (d*x)/2)^6 - 1))