3.161 \(\int \frac{\sec ^6(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx\)

Optimal. Leaf size=84 \[ -\frac{i \sec ^5(c+d x)}{5 a d}+\frac{3 \tanh ^{-1}(\sin (c+d x))}{8 a d}+\frac{\tan (c+d x) \sec ^3(c+d x)}{4 a d}+\frac{3 \tan (c+d x) \sec (c+d x)}{8 a d} \]

[Out]

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

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Rubi [A]  time = 0.13277, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3092, 3090, 3768, 3770, 2606, 30} \[ -\frac{i \sec ^5(c+d x)}{5 a d}+\frac{3 \tanh ^{-1}(\sin (c+d x))}{8 a d}+\frac{\tan (c+d x) \sec ^3(c+d x)}{4 a d}+\frac{3 \tan (c+d x) \sec (c+d x)}{8 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3092

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> Dist[a^n*b^n, Int[Cos[c + d*x]^m/(b*Cos[c + d*x] + a*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, m},
x] && EqQ[a^2 + b^2, 0] && ILtQ[n, 0]

Rule 3090

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[cos[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
IntegerQ[m] && IGtQ[n, 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]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\sec ^6(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx &=-\frac{i \int \sec ^6(c+d x) (i a \cos (c+d x)+a \sin (c+d x)) \, dx}{a^2}\\ &=-\frac{i \int \left (i a \sec ^5(c+d x)+a \sec ^5(c+d x) \tan (c+d x)\right ) \, dx}{a^2}\\ &=-\frac{i \int \sec ^5(c+d x) \tan (c+d x) \, dx}{a}+\frac{\int \sec ^5(c+d x) \, dx}{a}\\ &=\frac{\sec ^3(c+d x) \tan (c+d x)}{4 a d}+\frac{3 \int \sec ^3(c+d x) \, dx}{4 a}-\frac{i \operatorname{Subst}\left (\int x^4 \, dx,x,\sec (c+d x)\right )}{a d}\\ &=-\frac{i \sec ^5(c+d x)}{5 a d}+\frac{3 \sec (c+d x) \tan (c+d x)}{8 a d}+\frac{\sec ^3(c+d x) \tan (c+d x)}{4 a d}+\frac{3 \int \sec (c+d x) \, dx}{8 a}\\ &=\frac{3 \tanh ^{-1}(\sin (c+d x))}{8 a d}-\frac{i \sec ^5(c+d x)}{5 a d}+\frac{3 \sec (c+d x) \tan (c+d x)}{8 a d}+\frac{\sec ^3(c+d x) \tan (c+d x)}{4 a d}\\ \end{align*}

Mathematica [A]  time = 0.477167, size = 66, normalized size = 0.79 \[ -\frac{i \left ((70 i \sin (2 (c+d x))+15 i \sin (4 (c+d x))+64) \sec ^5(c+d x)+240 i \tanh ^{-1}\left (\cos (c) \tan \left (\frac{d x}{2}\right )+\sin (c)\right )\right )}{320 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/320)*((240*I)*ArcTanh[Sin[c] + Cos[c]*Tan[(d*x)/2]] + Sec[c + d*x]^5*(64 + (70*I)*Sin[2*(c + d*x)] + (15*
I)*Sin[4*(c + d*x)])))/(a*d)

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Maple [B]  time = 0.169, size = 430, normalized size = 5.1 \begin{align*}{\frac{{\frac{5\,i}{8}}}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{5}{8\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{{\frac{3\,i}{8}}}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{1}{2\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{{\frac{5\,i}{8}}}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{1}{4\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-4}}+{\frac{{\frac{i}{5}}}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-5}}-{\frac{7}{8\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{{\frac{3\,i}{4}}}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{3}{8\,ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{{\frac{i}{2}}}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-4}}+{\frac{7}{8\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{{\frac{3\,i}{8}}}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{5}{8\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{{\frac{3\,i}{4}}}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}+{\frac{1}{2\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}+{\frac{{\frac{i}{2}}}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-4}}+{\frac{1}{4\,ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-4}}-{\frac{{\frac{i}{5}}}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-5}}-{\frac{3}{8\,ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6/(a*cos(d*x+c)+I*a*sin(d*x+c)),x)

[Out]

5/8*I/d/a/(tan(1/2*d*x+1/2*c)-1)^2+5/8/d/a/(tan(1/2*d*x+1/2*c)+1)+3/8*I/d/a/(tan(1/2*d*x+1/2*c)-1)+1/2/d/a/(ta
n(1/2*d*x+1/2*c)+1)^3+5/8*I/d/a/(tan(1/2*d*x+1/2*c)+1)^2-1/4/d/a/(tan(1/2*d*x+1/2*c)+1)^4+1/5*I/d/a/(tan(1/2*d
*x+1/2*c)-1)^5-7/8/d/a/(tan(1/2*d*x+1/2*c)+1)^2-3/4*I/d/a/(tan(1/2*d*x+1/2*c)+1)^3+3/8/d/a*ln(tan(1/2*d*x+1/2*
c)+1)+1/2*I/d/a/(tan(1/2*d*x+1/2*c)+1)^4+7/8/d/a/(tan(1/2*d*x+1/2*c)-1)^2-3/8*I/d/a/(tan(1/2*d*x+1/2*c)+1)+5/8
/d/a/(tan(1/2*d*x+1/2*c)-1)+3/4*I/d/a/(tan(1/2*d*x+1/2*c)-1)^3+1/2/d/a/(tan(1/2*d*x+1/2*c)-1)^3+1/2*I/d/a/(tan
(1/2*d*x+1/2*c)-1)^4+1/4/d/a/(tan(1/2*d*x+1/2*c)-1)^4-1/5*I/d/a/(tan(1/2*d*x+1/2*c)+1)^5-3/8/d/a*ln(tan(1/2*d*
x+1/2*c)-1)

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Maxima [B]  time = 1.27656, size = 390, normalized size = 4.64 \begin{align*} \frac{\frac{16 \,{\left (-\frac{75 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{30 i \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{240 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{30 i \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac{120 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac{75 i \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - 24\right )}}{-120 i \, a + \frac{600 i \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{1200 i \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{1200 i \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{600 i \, a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac{120 i \, a \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} + \frac{3 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac{3 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a*cos(d*x+c)+I*a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/8*(16*(-75*I*sin(d*x + c)/(cos(d*x + c) + 1) + 30*I*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 240*sin(d*x + c)^4
/(cos(d*x + c) + 1)^4 - 30*I*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 120*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 7
5*I*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 24)/(-120*I*a + 600*I*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1200*I
*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1200*I*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 600*I*a*sin(d*x + c)^8
/(cos(d*x + c) + 1)^8 + 120*I*a*sin(d*x + c)^10/(cos(d*x + c) + 1)^10) + 3*log(sin(d*x + c)/(cos(d*x + c) + 1)
+ 1)/a - 3*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a)/d

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Fricas [B]  time = 0.496565, size = 810, normalized size = 9.64 \begin{align*} \frac{15 \,{\left (e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 15 \,{\left (e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 30 i \, e^{\left (9 i \, d x + 9 i \, c\right )} - 140 i \, e^{\left (7 i \, d x + 7 i \, c\right )} - 256 i \, e^{\left (5 i \, d x + 5 i \, c\right )} + 140 i \, e^{\left (3 i \, d x + 3 i \, c\right )} + 30 i \, e^{\left (i \, d x + i \, c\right )}}{40 \,{\left (a d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**6/(a*cos(d*x+c)+I*a*sin(d*x+c)),x)

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.1778, size = 189, normalized size = 2.25 \begin{align*} \frac{\frac{15 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac{15 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a} + \frac{2 \,{\left (25 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 40 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} - 10 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 80 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 10 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 25 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 8 i\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{5} a}}{40 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a*cos(d*x+c)+I*a*sin(d*x+c)),x, algorithm="giac")

[Out]

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