3.20 \(\int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=82 \[ -\frac{i (a+i a \tan (c+d x))^7}{7 a^5 d}+\frac{2 i (a+i a \tan (c+d x))^6}{3 a^4 d}-\frac{4 i (a+i a \tan (c+d x))^5}{5 a^3 d} \]

[Out]

(((-4*I)/5)*(a + I*a*Tan[c + d*x])^5)/(a^3*d) + (((2*I)/3)*(a + I*a*Tan[c + d*x])^6)/(a^4*d) - ((I/7)*(a + I*a
*Tan[c + d*x])^7)/(a^5*d)

________________________________________________________________________________________

Rubi [A]  time = 0.056461, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {3487, 43} \[ -\frac{i (a+i a \tan (c+d x))^7}{7 a^5 d}+\frac{2 i (a+i a \tan (c+d x))^6}{3 a^4 d}-\frac{4 i (a+i a \tan (c+d x))^5}{5 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(((-4*I)/5)*(a + I*a*Tan[c + d*x])^5)/(a^3*d) + (((2*I)/3)*(a + I*a*Tan[c + d*x])^6)/(a^4*d) - ((I/7)*(a + I*a
*Tan[c + d*x])^7)/(a^5*d)

Rule 3487

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx &=-\frac{i \operatorname{Subst}\left (\int (a-x)^2 (a+x)^4 \, dx,x,i a \tan (c+d x)\right )}{a^5 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (4 a^2 (a+x)^4-4 a (a+x)^5+(a+x)^6\right ) \, dx,x,i a \tan (c+d x)\right )}{a^5 d}\\ &=-\frac{4 i (a+i a \tan (c+d x))^5}{5 a^3 d}+\frac{2 i (a+i a \tan (c+d x))^6}{3 a^4 d}-\frac{i (a+i a \tan (c+d x))^7}{7 a^5 d}\\ \end{align*}

Mathematica [A]  time = 0.880934, size = 90, normalized size = 1.1 \[ \frac{a^2 \sec (c) \sec ^7(c+d x) (-35 \sin (2 c+d x)+42 \sin (2 c+3 d x)+14 \sin (4 c+5 d x)+2 \sin (6 c+7 d x)+35 i \cos (2 c+d x)+35 \sin (d x)+35 i \cos (d x))}{210 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(a^2*Sec[c]*Sec[c + d*x]^7*((35*I)*Cos[d*x] + (35*I)*Cos[2*c + d*x] + 35*Sin[d*x] - 35*Sin[2*c + d*x] + 42*Sin
[2*c + 3*d*x] + 14*Sin[4*c + 5*d*x] + 2*Sin[6*c + 7*d*x]))/(210*d)

________________________________________________________________________________________

Maple [A]  time = 0.056, size = 113, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ( -{a}^{2} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{7\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{35\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{8\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{105\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) +{\frac{{\frac{i}{3}}{a}^{2}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}-{a}^{2} \left ( -{\frac{8}{15}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15}} \right ) \tan \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^2,x)

[Out]

1/d*(-a^2*(1/7*sin(d*x+c)^3/cos(d*x+c)^7+4/35*sin(d*x+c)^3/cos(d*x+c)^5+8/105*sin(d*x+c)^3/cos(d*x+c)^3)+1/3*I
*a^2/cos(d*x+c)^6-a^2*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c))

________________________________________________________________________________________

Maxima [A]  time = 1.09504, size = 128, normalized size = 1.56 \begin{align*} -\frac{15 \, a^{2} \tan \left (d x + c\right )^{7} - 35 i \, a^{2} \tan \left (d x + c\right )^{6} + 21 \, a^{2} \tan \left (d x + c\right )^{5} - 105 i \, a^{2} \tan \left (d x + c\right )^{4} - 35 \, a^{2} \tan \left (d x + c\right )^{3} - 105 i \, a^{2} \tan \left (d x + c\right )^{2} - 105 \, a^{2} \tan \left (d x + c\right )}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/105*(15*a^2*tan(d*x + c)^7 - 35*I*a^2*tan(d*x + c)^6 + 21*a^2*tan(d*x + c)^5 - 105*I*a^2*tan(d*x + c)^4 - 3
5*a^2*tan(d*x + c)^3 - 105*I*a^2*tan(d*x + c)^2 - 105*a^2*tan(d*x + c))/d

________________________________________________________________________________________

Fricas [B]  time = 1.06656, size = 464, normalized size = 5.66 \begin{align*} \frac{4480 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + 4480 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 2688 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 896 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 128 i \, a^{2}}{105 \,{\left (d e^{\left (14 i \, d x + 14 i \, c\right )} + 7 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 21 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 35 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 35 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 21 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 7 \, 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)^6*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/105*(4480*I*a^2*e^(8*I*d*x + 8*I*c) + 4480*I*a^2*e^(6*I*d*x + 6*I*c) + 2688*I*a^2*e^(4*I*d*x + 4*I*c) + 896*
I*a^2*e^(2*I*d*x + 2*I*c) + 128*I*a^2)/(d*e^(14*I*d*x + 14*I*c) + 7*d*e^(12*I*d*x + 12*I*c) + 21*d*e^(10*I*d*x
+ 10*I*c) + 35*d*e^(8*I*d*x + 8*I*c) + 35*d*e^(6*I*d*x + 6*I*c) + 21*d*e^(4*I*d*x + 4*I*c) + 7*d*e^(2*I*d*x +
2*I*c) + d)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int - \tan ^{2}{\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\, dx + \int 2 i \tan{\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\, dx + \int \sec ^{6}{\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)**6*(a+I*a*tan(d*x+c))**2,x)

[Out]

a**2*(Integral(-tan(c + d*x)**2*sec(c + d*x)**6, x) + Integral(2*I*tan(c + d*x)*sec(c + d*x)**6, x) + Integral
(sec(c + d*x)**6, x))

________________________________________________________________________________________

Giac [A]  time = 1.18313, size = 128, normalized size = 1.56 \begin{align*} -\frac{15 \, a^{2} \tan \left (d x + c\right )^{7} - 35 i \, a^{2} \tan \left (d x + c\right )^{6} + 21 \, a^{2} \tan \left (d x + c\right )^{5} - 105 i \, a^{2} \tan \left (d x + c\right )^{4} - 35 \, a^{2} \tan \left (d x + c\right )^{3} - 105 i \, a^{2} \tan \left (d x + c\right )^{2} - 105 \, a^{2} \tan \left (d x + c\right )}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/105*(15*a^2*tan(d*x + c)^7 - 35*I*a^2*tan(d*x + c)^6 + 21*a^2*tan(d*x + c)^5 - 105*I*a^2*tan(d*x + c)^4 - 3
5*a^2*tan(d*x + c)^3 - 105*I*a^2*tan(d*x + c)^2 - 105*a^2*tan(d*x + c))/d