\(\int \sec ^6(c+d x) (a+b \tan (c+d x)) \, dx\) [507]

Optimal result

Integrand size = 19, antiderivative size = 60 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x)) \, dx=\frac {b \sec ^6(c+d x)}{6 d}+\frac {a \tan (c+d x)}{d}+\frac {2 a \tan ^3(c+d x)}{3 d}+\frac {a \tan ^5(c+d x)}{5 d} \]

[Out]

1/6*b*sec(d*x+c)^6/d+a*tan(d*x+c)/d+2/3*a*tan(d*x+c)^3/d+1/5*a*tan(d*x+c)^5/d

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3567, 3852} \[ \int \sec ^6(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a \tan ^5(c+d x)}{5 d}+\frac {2 a \tan ^3(c+d x)}{3 d}+\frac {a \tan (c+d x)}{d}+\frac {b \sec ^6(c+d x)}{6 d} \]

[In]

Int[Sec[c + d*x]^6*(a + b*Tan[c + d*x]),x]

[Out]

(b*Sec[c + d*x]^6)/(6*d) + (a*Tan[c + d*x])/d + (2*a*Tan[c + d*x]^3)/(3*d) + (a*Tan[c + d*x]^5)/(5*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 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]

Rubi steps \begin{align*} \text {integral}& = \frac {b \sec ^6(c+d x)}{6 d}+a \int \sec ^6(c+d x) \, dx \\ & = \frac {b \sec ^6(c+d x)}{6 d}-\frac {a \text {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (c+d x)\right )}{d} \\ & = \frac {b \sec ^6(c+d x)}{6 d}+\frac {a \tan (c+d x)}{d}+\frac {2 a \tan ^3(c+d x)}{3 d}+\frac {a \tan ^5(c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.88 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x)) \, dx=\frac {b \sec ^6(c+d x)}{6 d}+\frac {a \left (\tan (c+d x)+\frac {2}{3} \tan ^3(c+d x)+\frac {1}{5} \tan ^5(c+d x)\right )}{d} \]

[In]

Integrate[Sec[c + d*x]^6*(a + b*Tan[c + d*x]),x]

[Out]

(b*Sec[c + d*x]^6)/(6*d) + (a*(Tan[c + d*x] + (2*Tan[c + d*x]^3)/3 + Tan[c + d*x]^5/5))/d

Maple [A] (verified)

Time = 14.75 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.15

method	result	size
derivativedivides	\(\frac {\frac {b \left (\tan ^{6}\left (d x +c \right )\right )}{6}+\frac {a \left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\tan ^{4}\left (d x +c \right )\right ) b}{2}+\frac {2 a \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {b \left (\tan ^{2}\left (d x +c \right )\right )}{2}+a \tan \left (d x +c \right )}{d}\)	\(69\)
default	\(\frac {\frac {b \left (\tan ^{6}\left (d x +c \right )\right )}{6}+\frac {a \left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\tan ^{4}\left (d x +c \right )\right ) b}{2}+\frac {2 a \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {b \left (\tan ^{2}\left (d x +c \right )\right )}{2}+a \tan \left (d x +c \right )}{d}\)	\(69\)
risch	\(\frac {\frac {32 i a \,{\mathrm e}^{6 i \left (d x +c \right )}}{3}+\frac {32 b \,{\mathrm e}^{6 i \left (d x +c \right )}}{3}+16 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+\frac {32 i a \,{\mathrm e}^{2 i \left (d x +c \right )}}{5}+\frac {16 i a}{15}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}\)	\(75\)

[In]

int(sec(d*x+c)^6*(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.95 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x)) \, dx=\frac {2 \, {\left (8 \, a \cos \left (d x + c\right )^{5} + 4 \, a \cos \left (d x + c\right )^{3} + 3 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 5 \, b}{30 \, d \cos \left (d x + c\right )^{6}} \]

[In]

integrate(sec(d*x+c)^6*(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/30*(2*(8*a*cos(d*x + c)^5 + 4*a*cos(d*x + c)^3 + 3*a*cos(d*x + c))*sin(d*x + c) + 5*b)/(d*cos(d*x + c)^6)

Sympy [A] (verification not implemented)

Time = 1.48 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.93 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x)) \, dx=\begin {cases} \frac {a \left (\frac {\tan ^{5}{\left (c + d x \right )}}{5} + \frac {2 \tan ^{3}{\left (c + d x \right )}}{3} + \tan {\left (c + d x \right )}\right ) + \frac {b \sec ^{6}{\left (c + d x \right )}}{6}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\left (c \right )}\right ) \sec ^{6}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(sec(d*x+c)**6*(a+b*tan(d*x+c)),x)

[Out]

Piecewise(((a*(tan(c + d*x)**5/5 + 2*tan(c + d*x)**3/3 + tan(c + d*x)) + b*sec(c + d*x)**6/6)/d, Ne(d, 0)), (x
*(a + b*tan(c))*sec(c)**6, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.17 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x)) \, dx=\frac {5 \, b \tan \left (d x + c\right )^{6} + 6 \, a \tan \left (d x + c\right )^{5} + 15 \, b \tan \left (d x + c\right )^{4} + 20 \, a \tan \left (d x + c\right )^{3} + 15 \, b \tan \left (d x + c\right )^{2} + 30 \, a \tan \left (d x + c\right )}{30 \, d} \]

[In]

integrate(sec(d*x+c)^6*(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/30*(5*b*tan(d*x + c)^6 + 6*a*tan(d*x + c)^5 + 15*b*tan(d*x + c)^4 + 20*a*tan(d*x + c)^3 + 15*b*tan(d*x + c)^
2 + 30*a*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.17 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x)) \, dx=\frac {5 \, b \tan \left (d x + c\right )^{6} + 6 \, a \tan \left (d x + c\right )^{5} + 15 \, b \tan \left (d x + c\right )^{4} + 20 \, a \tan \left (d x + c\right )^{3} + 15 \, b \tan \left (d x + c\right )^{2} + 30 \, a \tan \left (d x + c\right )}{30 \, d} \]

[In]

integrate(sec(d*x+c)^6*(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

1/30*(5*b*tan(d*x + c)^6 + 6*a*tan(d*x + c)^5 + 15*b*tan(d*x + c)^4 + 20*a*tan(d*x + c)^3 + 15*b*tan(d*x + c)^
2 + 30*a*tan(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 4.10 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.13 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x)) \, dx=\frac {\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^6}{6}+\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5}+\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^4}{2}+\frac {2\,a\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3}+\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}+a\,\mathrm {tan}\left (c+d\,x\right )}{d} \]

[In]

int((a + b*tan(c + d*x))/cos(c + d*x)^6,x)

[Out]

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