3.171 \(\int \frac{\sec ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=56 \[ -\frac{2 i \sec (c+d x)}{a^2 d}+\frac{3 \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}-\frac{\tan (c+d x) \sec (c+d x)}{2 a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.145636, antiderivative size = 56, 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, 3770, 2606, 8, 2611} \[ -\frac{2 i \sec (c+d x)}{a^2 d}+\frac{3 \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}-\frac{\tan (c+d x) \sec (c+d x)}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^2,x]

[Out]

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

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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

Mathematica [B]  time = 0.394743, size = 146, normalized size = 2.61 \[ -\frac{\sec ^2(c+d x) \left (2 \sin (c+d x)+8 i \cos (c+d x)+3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+3 \cos (2 (c+d x)) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{4 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^2,x]

[Out]

-(Sec[c + d*x]^2*((8*I)*Cos[c + d*x] + 3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 3*Cos[2*(c + d*x)]*(Log[Co
s[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 3*Log[Cos[(c + d*x)/2] + Sin[
(c + d*x)/2]] + 2*Sin[c + d*x]))/(4*a^2*d)

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Maple [B]  time = 0.184, size = 170, normalized size = 3. \begin{align*} -{\frac{1}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{2\,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 ) ^{-2}}+{\frac{3}{2\,d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{1}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{2\,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 ) ^{-2}}-{\frac{3}{2\,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)^3/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x)

[Out]

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

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Maxima [B]  time = 1.05135, size = 225, normalized size = 4.02 \begin{align*} -\frac{\frac{2 \,{\left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{4 i \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + 4 i\right )}}{a^{2} - \frac{2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac{3 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac{3 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/2*(2*(sin(d*x + c)/(cos(d*x + c) + 1) - 4*I*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 + 4*I)/(a^2 - 2*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)
- 3*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 3*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2)/d

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Fricas [B]  time = 0.480987, size = 375, normalized size = 6.7 \begin{align*} \frac{3 \,{\left (e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 3 \,{\left (e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 6 i \, e^{\left (3 i \, d x + 3 i \, c\right )} - 10 i \, e^{\left (i \, d x + i \, c\right )}}{2 \,{\left (a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, 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)^3/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/2*(3*(e^(4*I*d*x + 4*I*c) + 2*e^(2*I*d*x + 2*I*c) + 1)*log(e^(I*d*x + I*c) + I) - 3*(e^(4*I*d*x + 4*I*c) + 2
*e^(2*I*d*x + 2*I*c) + 1)*log(e^(I*d*x + I*c) - I) - 6*I*e^(3*I*d*x + 3*I*c) - 10*I*e^(I*d*x + I*c))/(a^2*d*e^
(4*I*d*x + 4*I*c) + 2*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)**3/(a*cos(d*x+c)+I*a*sin(d*x+c))**2,x)

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.19903, size = 131, normalized size = 2.34 \begin{align*} \frac{\frac{3 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac{3 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac{2 \,{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 i\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/2*(3*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 3*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 - 2*(tan(1/2*d*x + 1/
2*c)^3 - 4*I*tan(1/2*d*x + 1/2*c)^2 + tan(1/2*d*x + 1/2*c) + 4*I)/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^2))/d