3.12 \(\int \sec ^5(c+d x) (a+i a \tan (c+d x)) \, dx\)

Optimal. Leaf size=76 \[ \frac{i a \sec ^5(c+d x)}{5 d}+\frac{3 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{3 a \tan (c+d x) \sec (c+d x)}{8 d} \]

[Out]

(3*a*ArcTanh[Sin[c + d*x]])/(8*d) + ((I/5)*a*Sec[c + d*x]^5)/d + (3*a*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (a*Se
c[c + d*x]^3*Tan[c + d*x])/(4*d)

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Rubi [A]  time = 0.0483808, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {3486, 3768, 3770} \[ \frac{i a \sec ^5(c+d x)}{5 d}+\frac{3 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{3 a \tan (c+d x) \sec (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*(a + I*a*Tan[c + d*x]),x]

[Out]

(3*a*ArcTanh[Sin[c + d*x]])/(8*d) + ((I/5)*a*Sec[c + d*x]^5)/d + (3*a*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (a*Se
c[c + d*x]^3*Tan[c + d*x])/(4*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 3768

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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \sec ^5(c+d x) (a+i a \tan (c+d x)) \, dx &=\frac{i a \sec ^5(c+d x)}{5 d}+a \int \sec ^5(c+d x) \, dx\\ &=\frac{i a \sec ^5(c+d x)}{5 d}+\frac{a \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{4} (3 a) \int \sec ^3(c+d x) \, dx\\ &=\frac{i a \sec ^5(c+d x)}{5 d}+\frac{3 a \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{8} (3 a) \int \sec (c+d x) \, dx\\ &=\frac{3 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{i a \sec ^5(c+d x)}{5 d}+\frac{3 a \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.151544, size = 70, normalized size = 0.92 \[ \frac{i a \sec ^5(c+d x)}{5 d}+\frac{a \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{3 a \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5*(a + I*a*Tan[c + d*x]),x]

[Out]

((I/5)*a*Sec[c + d*x]^5)/d + (a*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (3*a*(ArcTanh[Sin[c + d*x]] + Sec[c + d*x
]*Tan[c + d*x]))/(8*d)

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Maple [A]  time = 0.085, size = 75, normalized size = 1. \begin{align*}{\frac{{\frac{i}{5}}a}{d \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{a \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,a\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,a\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(a+I*a*tan(d*x+c)),x)

[Out]

1/5*I/d*a/cos(d*x+c)^5+1/4*a*sec(d*x+c)^3*tan(d*x+c)/d+3/8*a*sec(d*x+c)*tan(d*x+c)/d+3/8/d*a*ln(sec(d*x+c)+tan
(d*x+c))

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Maxima [A]  time = 1.11564, size = 116, normalized size = 1.53 \begin{align*} -\frac{5 \, a{\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 )} - \frac{16 i \, a}{\cos \left (d x + c\right )^{5}}}{80 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/80*(5*a*(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)) - 16*I*a/cos(d*x + c)^5)/d

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Fricas [B]  time = 1.2512, size = 836, normalized size = 11. \begin{align*} \frac{-30 i \, a e^{\left (9 i \, d x + 9 i \, c\right )} - 140 i \, a e^{\left (7 i \, d x + 7 i \, c\right )} + 256 i \, a e^{\left (5 i \, d x + 5 i \, c\right )} + 140 i \, a e^{\left (3 i \, d x + 3 i \, c\right )} + 30 i \, a e^{\left (i \, d x + i \, c\right )} + 15 \,{\left (a e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 15 \,{\left (a e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{40 \,{\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, 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)^5*(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/40*(-30*I*a*e^(9*I*d*x + 9*I*c) - 140*I*a*e^(7*I*d*x + 7*I*c) + 256*I*a*e^(5*I*d*x + 5*I*c) + 140*I*a*e^(3*I
*d*x + 3*I*c) + 30*I*a*e^(I*d*x + I*c) + 15*(a*e^(10*I*d*x + 10*I*c) + 5*a*e^(8*I*d*x + 8*I*c) + 10*a*e^(6*I*d
*x + 6*I*c) + 10*a*e^(4*I*d*x + 4*I*c) + 5*a*e^(2*I*d*x + 2*I*c) + a)*log(e^(I*d*x + I*c) + I) - 15*(a*e^(10*I
*d*x + 10*I*c) + 5*a*e^(8*I*d*x + 8*I*c) + 10*a*e^(6*I*d*x + 6*I*c) + 10*a*e^(4*I*d*x + 4*I*c) + 5*a*e^(2*I*d*
x + 2*I*c) + a)*log(e^(I*d*x + I*c) - I))/(d*e^(10*I*d*x + 10*I*c) + 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x
+ 6*I*c) + 10*d*e^(4*I*d*x + 4*I*c) + 5*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int i \tan{\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx + \int \sec ^{5}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5*(a+I*a*tan(d*x+c)),x)

[Out]

a*(Integral(I*tan(c + d*x)*sec(c + d*x)**5, x) + Integral(sec(c + d*x)**5, x))

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Giac [B]  time = 1.19293, size = 190, normalized size = 2.5 \begin{align*} \frac{15 \, a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (25 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 40 i \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} - 10 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 80 i \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 10 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 25 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 8 i \, a\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{5}}}{40 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

1/40*(15*a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*a*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(25*a*tan(1/2*d*x
+ 1/2*c)^9 - 40*I*a*tan(1/2*d*x + 1/2*c)^8 - 10*a*tan(1/2*d*x + 1/2*c)^7 - 80*I*a*tan(1/2*d*x + 1/2*c)^4 + 10*
a*tan(1/2*d*x + 1/2*c)^3 - 25*a*tan(1/2*d*x + 1/2*c) - 8*I*a)/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d