3.3 \(\int \sec ^6(c+d x) (a+i a \tan (c+d x)) \, dx\)

Optimal. Leaf size=62 \[ \frac{a \tan ^5(c+d x)}{5 d}+\frac{2 a \tan ^3(c+d x)}{3 d}+\frac{a \tan (c+d x)}{d}+\frac{i a \sec ^6(c+d x)}{6 d} \]

[Out]

((I/6)*a*Sec[c + d*x]^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 = 0.0373904, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {3486, 3767} \[ \frac{a \tan ^5(c+d x)}{5 d}+\frac{2 a \tan ^3(c+d x)}{3 d}+\frac{a \tan (c+d x)}{d}+\frac{i a \sec ^6(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/6)*a*Sec[c + d*x]^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+i a \tan (c+d x)) \, dx &=\frac{i a \sec ^6(c+d x)}{6 d}+a \int \sec ^6(c+d x) \, dx\\ &=\frac{i a \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{i a \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 = 0.10323, size = 55, normalized size = 0.89 \[ \frac{a \left (\frac{1}{5} \tan ^5(c+d x)+\frac{2}{3} \tan ^3(c+d x)+\tan (c+d x)\right )}{d}+\frac{i a \sec ^6(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/6)*a*Sec[c + d*x]^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 = 0.082, size = 49, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({\frac{{\frac{i}{6}}a}{ \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+I*a*tan(d*x+c)),x)

[Out]

1/d*(1/6*I*a/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 = 1.09728, size = 95, normalized size = 1.53 \begin{align*} \frac{5 i \, a \tan \left (d x + c\right )^{6} + 6 \, a \tan \left (d x + c\right )^{5} + 15 i \, a \tan \left (d x + c\right )^{4} + 20 \, a \tan \left (d x + c\right )^{3} + 15 i \, a \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+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.11756, size = 363, normalized size = 5.85 \begin{align*} \frac{320 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} + 240 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 96 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 16 i \, a}{15 \,{\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(320*I*a*e^(6*I*d*x + 6*I*c) + 240*I*a*e^(4*I*d*x + 4*I*c) + 96*I*a*e^(2*I*d*x + 2*I*c) + 16*I*a)/(d*e^(1
2*I*d*x + 12*I*c) + 6*d*e^(10*I*d*x + 10*I*c) + 15*d*e^(8*I*d*x + 8*I*c) + 20*d*e^(6*I*d*x + 6*I*c) + 15*d*e^(
4*I*d*x + 4*I*c) + 6*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [A]  time = 18.4573, size = 60, normalized size = 0.97 \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{i a \sec ^{6}{\left (c + d x \right )}}{6}}{d} & \text{for}\: d \neq 0 \\x \left (i a \tan{\left (c \right )} + a\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+I*a*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)) + I*a*sec(c + d*x)**6/6)/d, Ne(d, 0)),
(x*(I*a*tan(c) + a)*sec(c)**6, True))

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Giac [A]  time = 1.13942, size = 95, normalized size = 1.53 \begin{align*} -\frac{-5 i \, a \tan \left (d x + c\right )^{6} - 6 \, a \tan \left (d x + c\right )^{5} - 15 i \, a \tan \left (d x + c\right )^{4} - 20 \, a \tan \left (d x + c\right )^{3} - 15 i \, a \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+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

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