3.30 \(\int \sec (c+d x) (a+i a \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=68 \[ \frac{3 i a^2 \sec (c+d x)}{2 d}+\frac{3 a^2 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0399934, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {3498, 3486, 3770} \[ \frac{3 i a^2 \sec (c+d x)}{2 d}+\frac{3 a^2 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

Rule 3498

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

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

Mathematica [B]  time = 0.801456, size = 146, normalized size = 2.15 \[ -\frac{a^2 \sec ^2(c+d x) \left (2 \sin (c+d x)-8 i \cos (c+d x)+3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+3 \cos (2 (c+d x)) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]*(a + I*a*Tan[c + d*x])^2,x]

[Out]

-(a^2*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)]*(L
og[Cos[(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*d)

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Maple [A]  time = 0.021, size = 79, normalized size = 1.2 \begin{align*} -{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{a}^{2}\sin \left ( dx+c \right ) }{2\,d}}+{\frac{3\,{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,i{a}^{2}}{d\cos \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(a+I*a*tan(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.12143, size = 112, normalized size = 1.65 \begin{align*} \frac{a^{2}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + \frac{8 i \, a^{2}}{\cos \left (d x + c\right )}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

1/4*(a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*a^2*log(sec
(d*x + c) + tan(d*x + c)) + 8*I*a^2/cos(d*x + c))/d

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Fricas [B]  time = 1.26849, size = 397, normalized size = 5.84 \begin{align*} \frac{10 i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 6 i \, a^{2} e^{\left (i \, d x + i \, c\right )} + 3 \,{\left (a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 3 \,{\left (a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{2 \,{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, 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)*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/2*(10*I*a^2*e^(3*I*d*x + 3*I*c) + 6*I*a^2*e^(I*d*x + I*c) + 3*(a^2*e^(4*I*d*x + 4*I*c) + 2*a^2*e^(2*I*d*x +
2*I*c) + a^2)*log(e^(I*d*x + I*c) + I) - 3*(a^2*e^(4*I*d*x + 4*I*c) + 2*a^2*e^(2*I*d*x + 2*I*c) + a^2)*log(e^(
I*d*x + I*c) - I))/(d*e^(4*I*d*x + 4*I*c) + 2*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^{2} \left (\int - \tan ^{2}{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int 2 i \tan{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int \sec{\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)*(a+I*a*tan(d*x+c))**2,x)

[Out]

a**2*(Integral(-tan(c + d*x)**2*sec(c + d*x), x) + Integral(2*I*tan(c + d*x)*sec(c + d*x), x) + Integral(sec(c
+ d*x), x))

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Giac [A]  time = 1.19868, size = 147, normalized size = 2.16 \begin{align*} \frac{3 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 i \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 i \, a^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

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