3.173 \(\int \frac{\sec ^5(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=84 \[ -\frac{2 i \sec ^3(c+d x)}{3 a^2 d}+\frac{5 \tanh ^{-1}(\sin (c+d x))}{8 a^2 d}-\frac{\tan (c+d x) \sec ^3(c+d x)}{4 a^2 d}+\frac{5 \tan (c+d x) \sec (c+d x)}{8 a^2 d} \]

[Out]

(5*ArcTanh[Sin[c + d*x]])/(8*a^2*d) - (((2*I)/3)*Sec[c + d*x]^3)/(a^2*d) + (5*Sec[c + d*x]*Tan[c + d*x])/(8*a^
2*d) - (Sec[c + d*x]^3*Tan[c + d*x])/(4*a^2*d)

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Rubi [A]  time = 0.192213, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {3092, 3090, 3768, 3770, 2606, 30, 2611} \[ -\frac{2 i \sec ^3(c+d x)}{3 a^2 d}+\frac{5 \tanh ^{-1}(\sin (c+d x))}{8 a^2 d}-\frac{\tan (c+d x) \sec ^3(c+d x)}{4 a^2 d}+\frac{5 \tan (c+d x) \sec (c+d x)}{8 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^2,x]

[Out]

(5*ArcTanh[Sin[c + d*x]])/(8*a^2*d) - (((2*I)/3)*Sec[c + d*x]^3)/(a^2*d) + (5*Sec[c + d*x]*Tan[c + d*x])/(8*a^
2*d) - (Sec[c + d*x]^3*Tan[c + d*x])/(4*a^2*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]

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rubi steps

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

Mathematica [B]  time = 0.94825, size = 215, normalized size = 2.56 \[ -\frac{\sec ^4(c+d x) \left (18 \sin (c+d x)-30 \sin (3 (c+d x))+128 i \cos (c+d x)+45 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+60 \cos (2 (c+d x)) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+15 \cos (4 (c+d x)) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-45 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{192 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^2,x]

[Out]

-(Sec[c + d*x]^4*((128*I)*Cos[c + d*x] + 45*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 60*Cos[2*(c + d*x)]*(Lo
g[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 15*Cos[4*(c + d*x)]*(Log[
Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 45*Log[Cos[(c + d*x)/2] + S
in[(c + d*x)/2]] + 18*Sin[c + d*x] - 30*Sin[3*(c + d*x)]))/(192*a^2*d)

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Maple [B]  time = 0.197, size = 342, normalized size = 4.1 \begin{align*}{\frac{3}{8\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{{\frac{2\,i}{3}}}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}-{\frac{1}{8\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{i}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{1}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{i}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{1}{4\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-4}}+{\frac{5}{8\,d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{3}{8\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{{\frac{2\,i}{3}}}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}+{\frac{1}{8\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{i}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}+{\frac{i}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{1}{4\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-4}}-{\frac{5}{8\,d{a}^{2}}\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)^5/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x)

[Out]

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

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Maxima [B]  time = 1.09991, size = 398, normalized size = 4.74 \begin{align*} \frac{\frac{2 \,{\left (\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{16 i \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{33 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{48 i \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{33 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{48 i \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{9 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - 16 i\right )}}{a^{2} - \frac{4 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{6 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{4 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} + \frac{15 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} - \frac{15 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/24*(2*(9*sin(d*x + c)/(cos(d*x + c) + 1) + 16*I*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 33*sin(d*x + c)^3/(cos
(d*x + c) + 1)^3 - 48*I*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 33*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 48*I*si
n(d*x + c)^6/(cos(d*x + c) + 1)^6 + 9*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 16*I)/(a^2 - 4*a^2*sin(d*x + c)^2/
(cos(d*x + c) + 1)^2 + 6*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 4*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 +
a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8) + 15*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 - 15*log(sin(d*x +
c)/(cos(d*x + c) + 1) - 1)/a^2)/d

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Fricas [B]  time = 0.487309, size = 667, normalized size = 7.94 \begin{align*} \frac{15 \,{\left (e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, 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 (8 i \, d x + 8 i \, c\right )} + 4 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, 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 (7 i \, d x + 7 i \, c\right )} - 110 i \, e^{\left (5 i \, d x + 5 i \, c\right )} - 146 i \, e^{\left (3 i \, d x + 3 i \, c\right )} + 30 i \, e^{\left (i \, d x + i \, c\right )}}{24 \,{\left (a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/24*(15*(e^(8*I*d*x + 8*I*c) + 4*e^(6*I*d*x + 6*I*c) + 6*e^(4*I*d*x + 4*I*c) + 4*e^(2*I*d*x + 2*I*c) + 1)*log
(e^(I*d*x + I*c) + I) - 15*(e^(8*I*d*x + 8*I*c) + 4*e^(6*I*d*x + 6*I*c) + 6*e^(4*I*d*x + 4*I*c) + 4*e^(2*I*d*x
+ 2*I*c) + 1)*log(e^(I*d*x + I*c) - I) - 30*I*e^(7*I*d*x + 7*I*c) - 110*I*e^(5*I*d*x + 5*I*c) - 146*I*e^(3*I*
d*x + 3*I*c) + 30*I*e^(I*d*x + I*c))/(a^2*d*e^(8*I*d*x + 8*I*c) + 4*a^2*d*e^(6*I*d*x + 6*I*c) + 6*a^2*d*e^(4*I
*d*x + 4*I*c) + 4*a^2*d*e^(2*I*d*x + 2*I*c) + a^2*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)**5/(a*cos(d*x+c)+I*a*sin(d*x+c))**2,x)

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.20561, size = 207, normalized size = 2.46 \begin{align*} \frac{\frac{15 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac{15 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac{2 \,{\left (9 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 48 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 33 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 48 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 33 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 16 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 9 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 16 i\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{4} a^{2}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(15*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 15*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 + 2*(9*tan(1/2*d*x
+ 1/2*c)^7 + 48*I*tan(1/2*d*x + 1/2*c)^6 - 33*tan(1/2*d*x + 1/2*c)^5 - 48*I*tan(1/2*d*x + 1/2*c)^4 - 33*tan(1
/2*d*x + 1/2*c)^3 + 16*I*tan(1/2*d*x + 1/2*c)^2 + 9*tan(1/2*d*x + 1/2*c) - 16*I)/((tan(1/2*d*x + 1/2*c)^2 - 1)
^4*a^2))/d