\(\int \sec ^2(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx\) [278]

Optimal result

Integrand size = 29, antiderivative size = 93 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {(A b+a B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {(3 a A+2 b B) \tan (c+d x)}{3 d}+\frac {(A b+a B) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {b B \sec ^2(c+d x) \tan (c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 93, 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 = {4082, 3872, 3852, 8, 3853, 3855} \[ \int \sec ^2(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {(a B+A b) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {(3 a A+2 b B) \tan (c+d x)}{3 d}+\frac {(a B+A b) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {b B \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

[In]

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

[Out]

((A*b + a*B)*ArcTanh[Sin[c + d*x]])/(2*d) + ((3*a*A + 2*b*B)*Tan[c + d*x])/(3*d) + ((A*b + a*B)*Sec[c + d*x]*T
an[c + d*x])/(2*d) + (b*B*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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.72 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 (A b+a B) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (6 a A+6 b B+3 (A b+a B) \sec (c+d x)+2 b B \tan ^2(c+d x)\right )}{6 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.81 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.87

method	result	size
parts	\(\frac {\left (A b +B a \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {a A \tan \left (d x +c \right )}{d}-\frac {B b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(81\)
derivativedivides	\(\frac {a A \tan \left (d x +c \right )+B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(105\)
default	\(\frac {a A \tan \left (d x +c \right )+B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(105\)
norman	\(\frac {\frac {4 \left (3 a A +B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {\left (2 a A -A b -B a +2 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {\left (2 a A +A b +B a +2 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-\frac {\left (A b +B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (A b +B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(153\)
parallelrisch	\(\frac {-9 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A b +B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+9 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A b +B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (6 A b +6 B a \right ) \sin \left (2 d x +2 c \right )+\left (6 a A +4 B b \right ) \sin \left (3 d x +3 c \right )+6 \sin \left (d x +c \right ) \left (a A +2 B b \right )}{6 d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(159\)
risch	\(-\frac {i \left (3 A b \,{\mathrm e}^{5 i \left (d x +c \right )}+3 B a \,{\mathrm e}^{5 i \left (d x +c \right )}-6 A a \,{\mathrm e}^{4 i \left (d x +c \right )}-12 a A \,{\mathrm e}^{2 i \left (d x +c \right )}-12 B b \,{\mathrm e}^{2 i \left (d x +c \right )}-3 A b \,{\mathrm e}^{i \left (d x +c \right )}-3 B a \,{\mathrm e}^{i \left (d x +c \right )}-6 a A -4 B b \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A b}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A b}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}\)	\(201\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.24 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (B a + A b\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (B a + A b\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (3 \, A a + 2 \, B b\right )} \cos \left (d x + c\right )^{2} + 2 \, B b + 3 \, {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate(sec(d*x+c)**2*(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)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.37 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B b - 3 \, B a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, A b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a \tan \left (d x + c\right )}{12 \, d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 210 vs. \(2 (85) = 170\).

Time = 0.31 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.26 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (B a + A b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (B a + A b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

[In]

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

[Out]

1/6*(3*(B*a + A*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(B*a + A*b)*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*B*a*tan(1/2*d*x + 1/2*c)^5 - 3*A*b*tan(1/2*d*x + 1/2*c)^5 + 6*B*b*tan(1/2*d*x
+ 1/2*c)^5 - 12*A*a*tan(1/2*d*x + 1/2*c)^3 - 4*B*b*tan(1/2*d*x + 1/2*c)^3 + 6*A*a*tan(1/2*d*x + 1/2*c) + 3*B*
a*tan(1/2*d*x + 1/2*c) + 3*A*b*tan(1/2*d*x + 1/2*c) + 6*B*b*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 = 16.99 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.56 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A\,b+B\,a\right )}{d}-\frac {\left (2\,A\,a-A\,b-B\,a+2\,B\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-4\,A\,a-\frac {4\,B\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a+A\,b+B\,a+2\,B\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[In]

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

[Out]

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