3.29 \(\int \sec ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=94 \[ \frac{5 i a^2 \sec ^3(c+d x)}{12 d}+\frac{5 a^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{i \sec ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\frac{5 a^2 \tan (c+d x) \sec (c+d x)}{8 d} \]

[Out]

(5*a^2*ArcTanh[Sin[c + d*x]])/(8*d) + (((5*I)/12)*a^2*Sec[c + d*x]^3)/d + (5*a^2*Sec[c + d*x]*Tan[c + d*x])/(8
*d) + ((I/4)*Sec[c + d*x]^3*(a^2 + I*a^2*Tan[c + d*x]))/d

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Rubi [A]  time = 0.0766581, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3498, 3486, 3768, 3770} \[ \frac{5 i a^2 \sec ^3(c+d x)}{12 d}+\frac{5 a^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{i \sec ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\frac{5 a^2 \tan (c+d x) \sec (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^2*ArcTanh[Sin[c + d*x]])/(8*d) + (((5*I)/12)*a^2*Sec[c + d*x]^3)/d + (5*a^2*Sec[c + d*x]*Tan[c + d*x])/(8
*d) + ((I/4)*Sec[c + d*x]^3*(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 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 ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx &=\frac{i \sec ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\frac{1}{4} (5 a) \int \sec ^3(c+d x) (a+i a \tan (c+d x)) \, dx\\ &=\frac{5 i a^2 \sec ^3(c+d x)}{12 d}+\frac{i \sec ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\frac{1}{4} \left (5 a^2\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac{5 i a^2 \sec ^3(c+d x)}{12 d}+\frac{5 a^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{i \sec ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\frac{1}{8} \left (5 a^2\right ) \int \sec (c+d x) \, dx\\ &=\frac{5 a^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{5 i a^2 \sec ^3(c+d x)}{12 d}+\frac{5 a^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{i \sec ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}\\ \end{align*}

Mathematica [B]  time = 1.051, size = 215, normalized size = 2.29 \[ \frac{a^2 \sec ^4(c+d x) \left (-18 \sin (c+d x)+30 \sin (3 (c+d x))+128 i \cos (c+d x)-45 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-60 \cos (2 (c+d x)) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-15 \cos (4 (c+d x)) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+45 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{192 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*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)]*
(Log[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)]*(L
og[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]
+ Sin[(c + d*x)/2]] - 18*Sin[c + d*x] + 30*Sin[3*(c + d*x)]))/(192*d)

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Maple [A]  time = 0.052, size = 123, normalized size = 1.3 \begin{align*} -{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{a}^{2}\sin \left ( dx+c \right ) }{8\,d}}+{\frac{5\,{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{{\frac{2\,i}{3}}{a}^{2}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.06383, size = 176, normalized size = 1.87 \begin{align*} -\frac{3 \, a^{2}{\left (\frac{2 \,{\left (\sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \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 )} + 12 \, 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 )} - \frac{32 i \, a^{2}}{\cos \left (d x + c\right )^{3}}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(3*a^2*(2*(sin(d*x + c)^3 + sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - log(sin(d*x + c) + 1
) + log(sin(d*x + c) - 1)) + 12*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)) - 32*I*a^2/cos(d*x + c)^3)/d

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Fricas [B]  time = 1.2719, size = 711, normalized size = 7.56 \begin{align*} \frac{-30 i \, a^{2} e^{\left (7 i \, d x + 7 i \, c\right )} + 146 i \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + 110 i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 30 i \, a^{2} e^{\left (i \, d x + i \, c\right )} + 15 \,{\left (a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, 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 ) - 15 \,{\left (a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, 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 )}{24 \,{\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, 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)^3*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/24*(-30*I*a^2*e^(7*I*d*x + 7*I*c) + 146*I*a^2*e^(5*I*d*x + 5*I*c) + 110*I*a^2*e^(3*I*d*x + 3*I*c) + 30*I*a^2
*e^(I*d*x + I*c) + 15*(a^2*e^(8*I*d*x + 8*I*c) + 4*a^2*e^(6*I*d*x + 6*I*c) + 6*a^2*e^(4*I*d*x + 4*I*c) + 4*a^2
*e^(2*I*d*x + 2*I*c) + a^2)*log(e^(I*d*x + I*c) + I) - 15*(a^2*e^(8*I*d*x + 8*I*c) + 4*a^2*e^(6*I*d*x + 6*I*c)
+ 6*a^2*e^(4*I*d*x + 4*I*c) + 4*a^2*e^(2*I*d*x + 2*I*c) + a^2)*log(e^(I*d*x + I*c) - I))/(d*e^(8*I*d*x + 8*I*
c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*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 ^{3}{\left (c + d x \right )}\, dx + \int 2 i \tan{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \sec ^{3}{\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)**3*(a+I*a*tan(d*x+c))**2,x)

[Out]

a**2*(Integral(-tan(c + d*x)**2*sec(c + d*x)**3, x) + Integral(2*I*tan(c + d*x)*sec(c + d*x)**3, x) + Integral
(sec(c + d*x)**3, x))

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Giac [B]  time = 1.2133, size = 236, normalized size = 2.51 \begin{align*} \frac{15 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (9 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 48 i \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 33 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 48 i \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 33 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 16 i \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 9 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 16 i \, a^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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