\(\int \sec (c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx\) [279]

Optimal result

Integrand size = 27, antiderivative size = 61 \[ \int \sec (c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {(2 a A+b B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {(A b+a B) \tan (c+d x)}{d}+\frac {b B \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

1/2*(2*A*a+B*b)*arctanh(sin(d*x+c))/d+(A*b+B*a)*tan(d*x+c)/d+1/2*b*B*sec(d*x+c)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {4082, 3872, 3855, 3852, 8} \[ \int \sec (c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {(2 a A+b B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {(a B+A b) \tan (c+d x)}{d}+\frac {b B \tan (c+d x) \sec (c+d x)}{2 d} \]

[In]

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

[Out]

((2*a*A + b*B)*ArcTanh[Sin[c + d*x]])/(2*d) + ((A*b + a*B)*Tan[c + d*x])/d + (b*B*Sec[c + d*x]*Tan[c + d*x])/(
2*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 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4082

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

Rubi steps \begin{align*} \text {integral}& = \frac {b B \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int \sec (c+d x) (2 a A+b B+2 (A b+a B) \sec (c+d x)) \, dx \\ & = \frac {b B \sec (c+d x) \tan (c+d x)}{2 d}+(A b+a B) \int \sec ^2(c+d x) \, dx+\frac {1}{2} (2 a A+b B) \int \sec (c+d x) \, dx \\ & = \frac {(2 a A+b B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {b B \sec (c+d x) \tan (c+d x)}{2 d}-\frac {(A b+a B) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d} \\ & = \frac {(2 a A+b B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {(A b+a B) \tan (c+d x)}{d}+\frac {b B \sec (c+d x) \tan (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.23 \[ \int \sec (c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a A \text {arctanh}(\sin (c+d x))}{d}+\frac {b B \text {arctanh}(\sin (c+d x))}{2 d}+\frac {A b \tan (c+d x)}{d}+\frac {a B \tan (c+d x)}{d}+\frac {b B \sec (c+d x) \tan (c+d x)}{2 d} \]

[In]

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

[Out]

(a*A*ArcTanh[Sin[c + d*x]])/d + (b*B*ArcTanh[Sin[c + d*x]])/(2*d) + (A*b*Tan[c + d*x])/d + (a*B*Tan[c + d*x])/
d + (b*B*Sec[c + d*x]*Tan[c + d*x])/(2*d)

Maple [A] (verified)

Time = 2.55 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.23

method	result	size
derivativedivides	\(\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \tan \left (d x +c \right ) a +A \tan \left (d x +c \right ) b +B b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(75\)
default	\(\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \tan \left (d x +c \right ) a +A \tan \left (d x +c \right ) b +B b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(75\)
parts	\(\frac {\left (A b +B a \right ) \tan \left (d x +c \right )}{d}+\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {B b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(76\)
parallelrisch	\(\frac {-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (a A +\frac {B b}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1+\cos \left (2 d x +2 c \right )\right ) \left (a A +\frac {B b}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (A b +B a \right ) \sin \left (2 d x +2 c \right )+B \sin \left (d x +c \right ) b}{d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(110\)
norman	\(\frac {\frac {\left (2 A b +2 B a +B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {\left (2 A b +2 B a -B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {\left (2 a A +B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (2 a A +B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(123\)
risch	\(-\frac {i \left (B b \,{\mathrm e}^{3 i \left (d x +c \right )}-2 A b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-B b \,{\mathrm e}^{i \left (d x +c \right )}-2 A b -2 B a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a A}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a A}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{2 d}\)	\(160\)

[In]

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

[Out]

1/d*(a*A*ln(sec(d*x+c)+tan(d*x+c))+B*tan(d*x+c)*a+A*tan(d*x+c)*b+B*b*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x
+c)+tan(d*x+c))))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.57 \[ \int \sec (c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {{\left (2 \, A a + B b\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, A a + B b\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (B b + 2 \, {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

1/4*((2*A*a + B*b)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (2*A*a + B*b)*cos(d*x + c)^2*log(-sin(d*x + c) + 1)
+ 2*(B*b + 2*(B*a + A*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2)

Sympy [F]

\[ \int \sec (c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\int \left (A + B \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.44 \[ \int \sec (c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=-\frac {B b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, A a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 4 \, B a \tan \left (d x + c\right ) - 4 \, A b \tan \left (d x + c\right )}{4 \, d} \]

[In]

integrate(sec(d*x+c)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

-1/4*(B*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 4*A*a*log(se
c(d*x + c) + tan(d*x + c)) - 4*B*a*tan(d*x + c) - 4*A*b*tan(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (57) = 114\).

Time = 0.31 (sec) , antiderivative size = 153, normalized size of antiderivative = 2.51 \[ \int \sec (c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {{\left (2 \, A a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, A a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 15.90 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.70 \[ \int \sec (c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,b+2\,B\,a+B\,b\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,A\,b+2\,B\,a-B\,b\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,A\,a+B\,b\right )}{d} \]

[In]

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

[Out]

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