\(\int \sec ^3(c+d x) (a+b \tan (c+d x)) \, dx\) [510]

Optimal result

Integrand size = 19, antiderivative size = 52 \[ \int \sec ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{2 d}+\frac {b \sec ^3(c+d x)}{3 d}+\frac {a \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3567, 3853, 3855} \[ \int \sec ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a \tan (c+d x) \sec (c+d x)}{2 d}+\frac {b \sec ^3(c+d x)}{3 d} \]

[In]

Int[Sec[c + d*x]^3*(a + b*Tan[c + d*x]),x]

[Out]

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

Rule 3567

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*((d*Sec[
e + f*x])^m/(f*m)), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^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]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int \sec ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{2 d}+\frac {b \sec ^3(c+d x)}{3 d}+\frac {a \sec (c+d x) \tan (c+d x)}{2 d} \]

[In]

Integrate[Sec[c + d*x]^3*(a + b*Tan[c + d*x]),x]

[Out]

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

Maple [A] (verified)

Time = 2.25 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {b}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(50\)
default	\(\frac {a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {b}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(50\)
risch	\(\frac {-3 i a \,{\mathrm e}^{5 i \left (d x +c \right )}+8 b \,{\mathrm e}^{3 i \left (d x +c \right )}+3 i a \,{\mathrm e}^{i \left (d x +c \right )}}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\)	\(97\)

[In]

int(sec(d*x+c)^3*(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.42 \[ \int \sec ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {3 \, a \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 6 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 4 \, b}{12 \, d \cos \left (d x + c\right )^{3}} \]

[In]

integrate(sec(d*x+c)^3*(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/12*(3*a*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*a*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 6*a*cos(d*x + c)*
sin(d*x + c) + 4*b)/(d*cos(d*x + c)^3)

Sympy [F]

\[ \int \sec ^3(c+d x) (a+b \tan (c+d x)) \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}\, dx \]

[In]

integrate(sec(d*x+c)**3*(a+b*tan(d*x+c)),x)

[Out]

Integral((a + b*tan(c + d*x))*sec(c + d*x)**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.17 \[ \int \sec ^3(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {3 \, a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - \frac {4 \, b}{\cos \left (d x + c\right )^{3}}}{12 \, d} \]

[In]

integrate(sec(d*x+c)^3*(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/12*(3*a*(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/cos(d*x
+ c)^3)/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (46) = 92\).

Time = 0.38 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.90 \[ \int \sec ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {3 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

[In]

integrate(sec(d*x+c)^3*(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.55 (sec) , antiderivative size = 105, normalized size of antiderivative = 2.02 \[ \int \sec ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {2\,b}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[In]

int((a + b*tan(c + d*x))/cos(c + d*x)^3,x)

[Out]

(a*atanh(tan(c/2 + (d*x)/2)))/d - ((2*b)/3 + a*tan(c/2 + (d*x)/2) - a*tan(c/2 + (d*x)/2)^5 + 2*b*tan(c/2 + (d*
x)/2)^4)/(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))