\(\int \sec ^3(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [37]

Optimal result

Integrand size = 28, antiderivative size = 85 \[ \int \sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 C \text {arctanh}(\sin (c+d x))}{8 d}+\frac {B \tan (c+d x)}{d}+\frac {3 C \sec (c+d x) \tan (c+d x)}{8 d}+\frac {C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {B \tan ^3(c+d x)}{3 d} \]

[Out]

3/8*C*arctanh(sin(d*x+c))/d+B*tan(d*x+c)/d+3/8*C*sec(d*x+c)*tan(d*x+c)/d+1/4*C*sec(d*x+c)^3*tan(d*x+c)/d+1/3*B
*tan(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {4132, 3852, 12, 3853, 3855} \[ \int \sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 C \text {arctanh}(\sin (c+d x))}{8 d}+\frac {B \tan ^3(c+d x)}{3 d}+\frac {B \tan (c+d x)}{d}+\frac {C \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {3 C \tan (c+d x) \sec (c+d x)}{8 d} \]

[In]

Int[Sec[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(3*C*ArcTanh[Sin[c + d*x]])/(8*d) + (B*Tan[c + d*x])/d + (3*C*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (C*Sec[c + d*
x]^3*Tan[c + d*x])/(4*d) + (B*Tan[c + d*x]^3)/(3*d)

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 ^4(c+d x) \, dx+\int C \sec ^5(c+d x) \, dx \\ & = C \int \sec ^5(c+d x) \, dx-\frac {B \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d} \\ & = \frac {B \tan (c+d x)}{d}+\frac {C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {B \tan ^3(c+d x)}{3 d}+\frac {1}{4} (3 C) \int \sec ^3(c+d x) \, dx \\ & = \frac {B \tan (c+d x)}{d}+\frac {3 C \sec (c+d x) \tan (c+d x)}{8 d}+\frac {C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {B \tan ^3(c+d x)}{3 d}+\frac {1}{8} (3 C) \int \sec (c+d x) \, dx \\ & = \frac {3 C \text {arctanh}(\sin (c+d x))}{8 d}+\frac {B \tan (c+d x)}{d}+\frac {3 C \sec (c+d x) \tan (c+d x)}{8 d}+\frac {C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {B \tan ^3(c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.69 \[ \int \sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {9 C \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (9 C \sec (c+d x)+6 C \sec ^3(c+d x)+8 B \left (3+\tan ^2(c+d x)\right )\right )}{24 d} \]

[In]

Integrate[Sec[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(9*C*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(9*C*Sec[c + d*x] + 6*C*Sec[c + d*x]^3 + 8*B*(3 + Tan[c + d*x]^2)))/
(24*d)

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {-B \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(73\)
default	\(\frac {-B \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(73\)
parts	\(-\frac {B \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(75\)
risch	\(-\frac {i \left (9 C \,{\mathrm e}^{7 i \left (d x +c \right )}+33 C \,{\mathrm e}^{5 i \left (d x +c \right )}-48 B \,{\mathrm e}^{4 i \left (d x +c \right )}-33 C \,{\mathrm e}^{3 i \left (d x +c \right )}-64 B \,{\mathrm e}^{2 i \left (d x +c \right )}-9 C \,{\mathrm e}^{i \left (d x +c \right )}-16 B \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}\)	\(135\)
norman	\(\frac {-\frac {\left (8 B -5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {\left (8 B +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (40 B -9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}+\frac {\left (40 B +9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {3 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {3 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(145\)
parallelrisch	\(\frac {-18 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+18 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+8 B \sin \left (4 d x +4 c \right )+32 B \sin \left (2 d x +2 c \right )+9 C \sin \left (3 d x +3 c \right )+33 C \sin \left (d x +c \right )}{12 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(150\)

[In]

int(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/d*(-B*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+C*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)
+tan(d*x+c))))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.16 \[ \int \sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {9 \, C \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, C \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, B \cos \left (d x + c\right )^{3} + 9 \, C \cos \left (d x + c\right )^{2} + 8 \, B \cos \left (d x + c\right ) + 6 \, C\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

[In]

integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/48*(9*C*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 9*C*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(16*B*cos(d*x +
c)^3 + 9*C*cos(d*x + c)^2 + 8*B*cos(d*x + c) + 6*C)*sin(d*x + c))/(d*cos(d*x + c)^4)

Sympy [F]

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

[In]

integrate(sec(d*x+c)**3*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Integral((B + C*sec(c + d*x))*sec(c + d*x)**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.12 \[ \int \sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B - 3 \, C {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \]

[In]

integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B - 3*C*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*
sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (77) = 154\).

Time = 0.31 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.93 \[ \int \sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {9 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 9 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (24 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 40 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

[In]

integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/24*(9*C*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 9*C*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(24*B*tan(1/2*d*x +
1/2*c)^7 - 15*C*tan(1/2*d*x + 1/2*c)^7 - 40*B*tan(1/2*d*x + 1/2*c)^5 - 9*C*tan(1/2*d*x + 1/2*c)^5 + 40*B*tan(1
/2*d*x + 1/2*c)^3 - 9*C*tan(1/2*d*x + 1/2*c)^3 - 24*B*tan(1/2*d*x + 1/2*c) - 15*C*tan(1/2*d*x + 1/2*c))/(tan(1
/2*d*x + 1/2*c)^2 - 1)^4)/d

Mupad [B] (verification not implemented)

Time = 18.03 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.80 \[ \int \sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3\,C\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {\left (2\,B-\frac {5\,C}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (-\frac {10\,B}{3}-\frac {3\,C}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {10\,B}{3}-\frac {3\,C}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-2\,B-\frac {5\,C}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

[In]

int((B/cos(c + d*x) + C/cos(c + d*x)^2)/cos(c + d*x)^3,x)

[Out]

(3*C*atanh(tan(c/2 + (d*x)/2)))/(4*d) - (tan(c/2 + (d*x)/2)^7*(2*B - (5*C)/4) + tan(c/2 + (d*x)/2)^3*((10*B)/3
- (3*C)/4) - tan(c/2 + (d*x)/2)^5*((10*B)/3 + (3*C)/4) - tan(c/2 + (d*x)/2)*(2*B + (5*C)/4))/(d*(6*tan(c/2 +
(d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1))