\(\int \sec (c+d x) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [39]

Optimal result

Integrand size = 26, antiderivative size = 47 \[ \int \sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {C \text {arctanh}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x)}{d}+\frac {C \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

1/2*C*arctanh(sin(d*x+c))/d+B*tan(d*x+c)/d+1/2*C*sec(d*x+c)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {4132, 3852, 8, 12, 3853, 3855} \[ \int \sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {C \text {arctanh}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x)}{d}+\frac {C \tan (c+d x) \sec (c+d x)}{2 d} \]

[In]

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

[Out]

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

Rule 8

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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]

Rubi steps \begin{align*} \text {integral}& = B \int \sec ^2(c+d x) \, dx+\int C \sec ^3(c+d x) \, dx \\ & = C \int \sec ^3(c+d x) \, dx-\frac {B \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d} \\ & = \frac {B \tan (c+d x)}{d}+\frac {C \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} C \int \sec (c+d x) \, dx \\ & = \frac {C \text {arctanh}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x)}{d}+\frac {C \sec (c+d x) \tan (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {C \text {arctanh}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x)}{d}+\frac {C \sec (c+d x) \tan (c+d x)}{2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {B \tan \left (d x +c \right )+C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(47\)
default	\(\frac {B \tan \left (d x +c \right )+C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(47\)
parts	\(\frac {B \tan \left (d x +c \right )}{d}+\frac {C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(49\)
parallelrisch	\(\frac {-C \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+C \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 B \sin \left (2 d x +2 c \right )+2 C \sin \left (d x +c \right )}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(92\)
norman	\(\frac {\frac {\left (2 B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {\left (2 B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(96\)
risch	\(-\frac {i \left (C \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B \,{\mathrm e}^{2 i \left (d x +c \right )}-C \,{\mathrm e}^{i \left (d x +c \right )}-2 B \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}\)	\(98\)

[In]

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

[Out]

1/d*(B*tan(d*x+c)+C*(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 = 0.27 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.57 \[ \int \sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {C \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - C \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, B \cos \left (d x + c\right ) + C\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral((B + C*sec(c + d*x))*sec(c + d*x)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.23 \[ \int \sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {C {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, B \tan \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

-1/4*(C*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 4*B*tan(d*x +
c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (43) = 86\).

Time = 0.33 (sec) , antiderivative size = 107, normalized size of antiderivative = 2.28 \[ \int \sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

1/2*(C*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - C*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(2*B*tan(1/2*d*x + 1/2*c)
^3 - C*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c) - C*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 = 15.82 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.81 \[ \int \sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {C\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,B-C\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,B+C\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

[In]

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

[Out]

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