3.28 \(\int \sec ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=118 \[ \frac{7 i a^2 \sec ^5(c+d x)}{30 d}+\frac{7 a^2 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{i \sec ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{6 d}+\frac{7 a^2 \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac{7 a^2 \tan (c+d x) \sec (c+d x)}{16 d} \]

[Out]

(7*a^2*ArcTanh[Sin[c + d*x]])/(16*d) + (((7*I)/30)*a^2*Sec[c + d*x]^5)/d + (7*a^2*Sec[c + d*x]*Tan[c + d*x])/(
16*d) + (7*a^2*Sec[c + d*x]^3*Tan[c + d*x])/(24*d) + ((I/6)*Sec[c + d*x]^5*(a^2 + I*a^2*Tan[c + d*x]))/d

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Rubi [A]  time = 0.0871915, antiderivative size = 118, 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} \[ \frac{7 i a^2 \sec ^5(c+d x)}{30 d}+\frac{7 a^2 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{i \sec ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{6 d}+\frac{7 a^2 \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac{7 a^2 \tan (c+d x) \sec (c+d x)}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*a^2*ArcTanh[Sin[c + d*x]])/(16*d) + (((7*I)/30)*a^2*Sec[c + d*x]^5)/d + (7*a^2*Sec[c + d*x]*Tan[c + d*x])/(
16*d) + (7*a^2*Sec[c + d*x]^3*Tan[c + d*x])/(24*d) + ((I/6)*Sec[c + d*x]^5*(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 ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx &=\frac{i \sec ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{6 d}+\frac{1}{6} (7 a) \int \sec ^5(c+d x) (a+i a \tan (c+d x)) \, dx\\ &=\frac{7 i a^2 \sec ^5(c+d x)}{30 d}+\frac{i \sec ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{6 d}+\frac{1}{6} \left (7 a^2\right ) \int \sec ^5(c+d x) \, dx\\ &=\frac{7 i a^2 \sec ^5(c+d x)}{30 d}+\frac{7 a^2 \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac{i \sec ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{6 d}+\frac{1}{8} \left (7 a^2\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac{7 i a^2 \sec ^5(c+d x)}{30 d}+\frac{7 a^2 \sec (c+d x) \tan (c+d x)}{16 d}+\frac{7 a^2 \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac{i \sec ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{6 d}+\frac{1}{16} \left (7 a^2\right ) \int \sec (c+d x) \, dx\\ &=\frac{7 a^2 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{7 i a^2 \sec ^5(c+d x)}{30 d}+\frac{7 a^2 \sec (c+d x) \tan (c+d x)}{16 d}+\frac{7 a^2 \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac{i \sec ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{6 d}\\ \end{align*}

Mathematica [A]  time = 1.01147, size = 159, normalized size = 1.35 \[ \frac{a^2 (\cos (2 c)-i \sin (2 c)) (\tan (c+d x)-i)^2 \sec ^4(c+d x) \left (150 \sin (c+d x)-35 (17 \sin (3 (c+d x))+3 \sin (5 (c+d x)))-1536 i \cos (c+d x)+1680 \cos ^6(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{3840 d (\cos (d x)+i \sin (d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sec[c + d*x]^4*(Cos[2*c] - I*Sin[2*c])*((-1536*I)*Cos[c + d*x] + 1680*Cos[c + d*x]^6*(Log[Cos[(c + d*x)/2
] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 150*Sin[c + d*x] - 35*(17*Sin[3*(c + d*x)]
+ 3*Sin[5*(c + d*x)]))*(-I + Tan[c + d*x])^2)/(3840*d*(Cos[d*x] + I*Sin[d*x])^2)

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Maple [A]  time = 0.053, size = 169, normalized size = 1.4 \begin{align*} -{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{6\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}-{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{16\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{a}^{2}\sin \left ( dx+c \right ) }{16\,d}}+{\frac{7\,{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}}+{\frac{{\frac{2\,i}{5}}{a}^{2}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{{a}^{2} \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(a+I*a*tan(d*x+c))^2,x)

[Out]

-1/6/d*a^2*sin(d*x+c)^3/cos(d*x+c)^6-1/8/d*a^2*sin(d*x+c)^3/cos(d*x+c)^4-1/16/d*a^2*sin(d*x+c)^3/cos(d*x+c)^2-
1/16*a^2*sin(d*x+c)/d+7/16/d*a^2*ln(sec(d*x+c)+tan(d*x+c))+2/5*I/d*a^2/cos(d*x+c)^5+1/4*a^2*sec(d*x+c)^3*tan(d
*x+c)/d+3/8*a^2*sec(d*x+c)*tan(d*x+c)/d

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Maxima [A]  time = 1.08428, size = 244, normalized size = 2.07 \begin{align*} -\frac{5 \, a^{2}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{5} - 8 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 30 \, a^{2}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - \frac{192 i \, a^{2}}{\cos \left (d x + c\right )^{5}}}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/480*(5*a^2*(2*(3*sin(d*x + c)^5 - 8*sin(d*x + c)^3 - 3*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3
*sin(d*x + c)^2 - 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) + 30*a^2*(2*(3*sin(d*x + c)^3 - 5*si
n(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 192
*I*a^2/cos(d*x + c)^5)/d

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Fricas [B]  time = 1.25052, size = 1062, normalized size = 9. \begin{align*} \frac{-210 i \, a^{2} e^{\left (11 i \, d x + 11 i \, c\right )} - 1190 i \, a^{2} e^{\left (9 i \, d x + 9 i \, c\right )} + 3372 i \, a^{2} e^{\left (7 i \, d x + 7 i \, c\right )} + 2772 i \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + 1190 i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 210 i \, a^{2} e^{\left (i \, d x + i \, c\right )} + 105 \,{\left (a^{2} e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a^{2} e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, 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 ) - 105 \,{\left (a^{2} e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a^{2} e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, 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 )}{240 \,{\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, 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)^5*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/240*(-210*I*a^2*e^(11*I*d*x + 11*I*c) - 1190*I*a^2*e^(9*I*d*x + 9*I*c) + 3372*I*a^2*e^(7*I*d*x + 7*I*c) + 27
72*I*a^2*e^(5*I*d*x + 5*I*c) + 1190*I*a^2*e^(3*I*d*x + 3*I*c) + 210*I*a^2*e^(I*d*x + I*c) + 105*(a^2*e^(12*I*d
*x + 12*I*c) + 6*a^2*e^(10*I*d*x + 10*I*c) + 15*a^2*e^(8*I*d*x + 8*I*c) + 20*a^2*e^(6*I*d*x + 6*I*c) + 15*a^2*
e^(4*I*d*x + 4*I*c) + 6*a^2*e^(2*I*d*x + 2*I*c) + a^2)*log(e^(I*d*x + I*c) + I) - 105*(a^2*e^(12*I*d*x + 12*I*
c) + 6*a^2*e^(10*I*d*x + 10*I*c) + 15*a^2*e^(8*I*d*x + 8*I*c) + 20*a^2*e^(6*I*d*x + 6*I*c) + 15*a^2*e^(4*I*d*x
+ 4*I*c) + 6*a^2*e^(2*I*d*x + 2*I*c) + a^2)*log(e^(I*d*x + I*c) - I))/(d*e^(12*I*d*x + 12*I*c) + 6*d*e^(10*I*
d*x + 10*I*c) + 15*d*e^(8*I*d*x + 8*I*c) + 20*d*e^(6*I*d*x + 6*I*c) + 15*d*e^(4*I*d*x + 4*I*c) + 6*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 ^{5}{\left (c + d x \right )}\, dx + \int 2 i \tan{\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx + \int \sec ^{5}{\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)**5*(a+I*a*tan(d*x+c))**2,x)

[Out]

a**2*(Integral(-tan(c + d*x)**2*sec(c + d*x)**5, x) + Integral(2*I*tan(c + d*x)*sec(c + d*x)**5, x) + Integral
(sec(c + d*x)**5, x))

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Giac [B]  time = 1.23203, size = 323, normalized size = 2.74 \begin{align*} \frac{105 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 105 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (135 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} - 480 i \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{10} - 445 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 480 i \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} - 330 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 960 i \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 330 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 960 i \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 445 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 96 i \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 135 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 96 i \, a^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{6}}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(105*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 105*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(135*a^2*ta
n(1/2*d*x + 1/2*c)^11 - 480*I*a^2*tan(1/2*d*x + 1/2*c)^10 - 445*a^2*tan(1/2*d*x + 1/2*c)^9 + 480*I*a^2*tan(1/2
*d*x + 1/2*c)^8 - 330*a^2*tan(1/2*d*x + 1/2*c)^7 - 960*I*a^2*tan(1/2*d*x + 1/2*c)^6 - 330*a^2*tan(1/2*d*x + 1/
2*c)^5 + 960*I*a^2*tan(1/2*d*x + 1/2*c)^4 - 445*a^2*tan(1/2*d*x + 1/2*c)^3 - 96*I*a^2*tan(1/2*d*x + 1/2*c)^2 +
135*a^2*tan(1/2*d*x + 1/2*c) + 96*I*a^2)/(tan(1/2*d*x + 1/2*c)^2 - 1)^6)/d