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

Optimal. Leaf size=60 \[ \frac{a \tan ^5(c+d x)}{5 d}+\frac{2 a \tan ^3(c+d x)}{3 d}+\frac{a \tan (c+d x)}{d}+\frac{b \sec ^6(c+d x)}{6 d} \]

[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)

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Rubi [A]  time = 0.0369258, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3486, 3767} \[ \frac{a \tan ^5(c+d x)}{5 d}+\frac{2 a \tan ^3(c+d x)}{3 d}+\frac{a \tan (c+d x)}{d}+\frac{b \sec ^6(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[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 3486

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 3767

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*} \int \sec ^6(c+d x) (a+b \tan (c+d x)) \, dx &=\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 \operatorname{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]  time = 0.160658, size = 53, normalized size = 0.88 \[ \frac{a \left (\frac{1}{5} \tan ^5(c+d x)+\frac{2}{3} \tan ^3(c+d x)+\tan (c+d x)\right )}{d}+\frac{b \sec ^6(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[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

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Maple [A]  time = 0.042, size = 48, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({\frac{b}{6\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}-a \left ( -{\frac{8}{15}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15}} \right ) \tan \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/6*b/cos(d*x+c)^6-a*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c))

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Maxima [A]  time = 1.47607, size = 95, normalized size = 1.58 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.8321, size = 147, normalized size = 2.45 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 5.74066, size = 56, normalized size = 0.93 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [A]  time = 1.26476, size = 95, normalized size = 1.58 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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