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

Optimal. Leaf size=127 \[ \frac{7 i a^3 \sec ^3(c+d x)}{12 d}+\frac{7 a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{7 i \sec ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\frac{7 a^3 \tan (c+d x) \sec (c+d x)}{8 d}+\frac{i a \sec ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d} \]

[Out]

(7*a^3*ArcTanh[Sin[c + d*x]])/(8*d) + (((7*I)/12)*a^3*Sec[c + d*x]^3)/d + (7*a^3*Sec[c + d*x]*Tan[c + d*x])/(8
*d) + ((I/5)*a*Sec[c + d*x]^3*(a + I*a*Tan[c + d*x])^2)/d + (((7*I)/20)*Sec[c + d*x]^3*(a^3 + I*a^3*Tan[c + d*
x]))/d

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Rubi [A]  time = 0.119956, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3498, 3486, 3768, 3770} \[ \frac{7 i a^3 \sec ^3(c+d x)}{12 d}+\frac{7 a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{7 i \sec ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\frac{7 a^3 \tan (c+d x) \sec (c+d x)}{8 d}+\frac{i a \sec ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*a^3*ArcTanh[Sin[c + d*x]])/(8*d) + (((7*I)/12)*a^3*Sec[c + d*x]^3)/d + (7*a^3*Sec[c + d*x]*Tan[c + d*x])/(8
*d) + ((I/5)*a*Sec[c + d*x]^3*(a + I*a*Tan[c + d*x])^2)/d + (((7*I)/20)*Sec[c + d*x]^3*(a^3 + I*a^3*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))^3 \, dx &=\frac{i a \sec ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d}+\frac{1}{5} (7 a) \int \sec ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx\\ &=\frac{i a \sec ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d}+\frac{7 i \sec ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\frac{1}{4} \left (7 a^2\right ) \int \sec ^3(c+d x) (a+i a \tan (c+d x)) \, dx\\ &=\frac{7 i a^3 \sec ^3(c+d x)}{12 d}+\frac{i a \sec ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d}+\frac{7 i \sec ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\frac{1}{4} \left (7 a^3\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac{7 i a^3 \sec ^3(c+d x)}{12 d}+\frac{7 a^3 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{i a \sec ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d}+\frac{7 i \sec ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\frac{1}{8} \left (7 a^3\right ) \int \sec (c+d x) \, dx\\ &=\frac{7 a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{7 i a^3 \sec ^3(c+d x)}{12 d}+\frac{7 a^3 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{i a \sec ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d}+\frac{7 i \sec ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}\\ \end{align*}

Mathematica [A]  time = 0.652026, size = 102, normalized size = 0.8 \[ \frac{a^3 (\cos (3 d x)+i \sin (3 d x)) \left (1680 \tanh ^{-1}\left (\cos (c) \tan \left (\frac{d x}{2}\right )+\sin (c)\right )+\sec ^5(c+d x) (-150 \sin (2 (c+d x))+105 \sin (4 (c+d x))+640 i \cos (2 (c+d x))+448 i)\right )}{960 d (\cos (d x)+i \sin (d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(Cos[3*d*x] + I*Sin[3*d*x])*(1680*ArcTanh[Sin[c] + Cos[c]*Tan[(d*x)/2]] + Sec[c + d*x]^5*(448*I + (640*I)
*Cos[2*(c + d*x)] - 150*Sin[2*(c + d*x)] + 105*Sin[4*(c + d*x)])))/(960*d*(Cos[d*x] + I*Sin[d*x])^3)

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Maple [B]  time = 0.095, size = 236, normalized size = 1.9 \begin{align*}{\frac{-{\frac{i}{5}}{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}-{\frac{{\frac{i}{15}}{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{{\frac{i}{15}}{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d\cos \left ( dx+c \right ) }}+{\frac{{\frac{i}{15}}{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{2}\cos \left ( dx+c \right ) }{d}}+{\frac{{\frac{2\,i}{15}}{a}^{3}\cos \left ( dx+c \right ) }{d}}-{\frac{3\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{3\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,{a}^{3}\sin \left ( dx+c \right ) }{8\,d}}+{\frac{7\,{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{i{a}^{3}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{{a}^{3}\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))^3,x)

[Out]

-1/5*I/d*a^3*sin(d*x+c)^4/cos(d*x+c)^5-1/15*I/d*a^3*sin(d*x+c)^4/cos(d*x+c)^3+1/15*I/d*a^3*sin(d*x+c)^4/cos(d*
x+c)+1/15*I/d*a^3*sin(d*x+c)^2*cos(d*x+c)+2/15*I/d*a^3*cos(d*x+c)-3/4/d*a^3*sin(d*x+c)^3/cos(d*x+c)^4-3/8/d*a^
3*sin(d*x+c)^3/cos(d*x+c)^2-3/8*a^3*sin(d*x+c)/d+7/8/d*a^3*ln(sec(d*x+c)+tan(d*x+c))+I/d*a^3/cos(d*x+c)^3+1/2*
a^3*sec(d*x+c)*tan(d*x+c)/d

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Maxima [A]  time = 1.14215, size = 209, normalized size = 1.65 \begin{align*} -\frac{45 \, a^{3}{\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 )} + 60 \, a^{3}{\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{240 i \, a^{3}}{\cos \left (d x + c\right )^{3}} - \frac{16 i \,{\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} a^{3}}{\cos \left (d x + c\right )^{5}}}{240 \, 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))^3,x, algorithm="maxima")

[Out]

-1/240*(45*a^3*(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)) + 60*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d
*x + c) - 1)) - 240*I*a^3/cos(d*x + c)^3 - 16*I*(5*cos(d*x + c)^2 - 3)*a^3/cos(d*x + c)^5)/d

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Fricas [B]  time = 1.23709, size = 891, normalized size = 7.02 \begin{align*} \frac{-210 i \, a^{3} e^{\left (9 i \, d x + 9 i \, c\right )} + 1580 i \, a^{3} e^{\left (7 i \, d x + 7 i \, c\right )} + 1792 i \, a^{3} e^{\left (5 i \, d x + 5 i \, c\right )} + 980 i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )} + 210 i \, a^{3} e^{\left (i \, d x + i \, c\right )} + 105 \,{\left (a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 105 \,{\left (a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{120 \,{\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, 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))^3,x, algorithm="fricas")

[Out]

1/120*(-210*I*a^3*e^(9*I*d*x + 9*I*c) + 1580*I*a^3*e^(7*I*d*x + 7*I*c) + 1792*I*a^3*e^(5*I*d*x + 5*I*c) + 980*
I*a^3*e^(3*I*d*x + 3*I*c) + 210*I*a^3*e^(I*d*x + I*c) + 105*(a^3*e^(10*I*d*x + 10*I*c) + 5*a^3*e^(8*I*d*x + 8*
I*c) + 10*a^3*e^(6*I*d*x + 6*I*c) + 10*a^3*e^(4*I*d*x + 4*I*c) + 5*a^3*e^(2*I*d*x + 2*I*c) + a^3)*log(e^(I*d*x
+ I*c) + I) - 105*(a^3*e^(10*I*d*x + 10*I*c) + 5*a^3*e^(8*I*d*x + 8*I*c) + 10*a^3*e^(6*I*d*x + 6*I*c) + 10*a^
3*e^(4*I*d*x + 4*I*c) + 5*a^3*e^(2*I*d*x + 2*I*c) + a^3)*log(e^(I*d*x + I*c) - I))/(d*e^(10*I*d*x + 10*I*c) +
5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) + 10*d*e^(4*I*d*x + 4*I*c) + 5*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^{3} \left (\int - 3 \tan ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 i \tan{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int - i \tan ^{3}{\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))**3,x)

[Out]

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

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Giac [A]  time = 1.28311, size = 258, normalized size = 2.03 \begin{align*} \frac{105 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 105 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (15 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 360 i \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} - 390 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 960 i \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 400 i \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 390 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 320 i \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 15 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 136 i \, a^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, 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))^3,x, algorithm="giac")

[Out]

1/120*(105*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 105*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(15*a^3*tan
(1/2*d*x + 1/2*c)^9 - 360*I*a^3*tan(1/2*d*x + 1/2*c)^8 - 390*a^3*tan(1/2*d*x + 1/2*c)^7 + 960*I*a^3*tan(1/2*d*
x + 1/2*c)^6 - 400*I*a^3*tan(1/2*d*x + 1/2*c)^4 + 390*a^3*tan(1/2*d*x + 1/2*c)^3 + 320*I*a^3*tan(1/2*d*x + 1/2
*c)^2 - 15*a^3*tan(1/2*d*x + 1/2*c) - 136*I*a^3)/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d