3.19 \(\int \sec ^8(c+d x) (a+i a \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=109 \[ \frac{i (a+i a \tan (c+d x))^9}{9 a^7 d}-\frac{3 i (a+i a \tan (c+d x))^8}{4 a^6 d}+\frac{12 i (a+i a \tan (c+d x))^7}{7 a^5 d}-\frac{4 i (a+i a \tan (c+d x))^6}{3 a^4 d} \]

[Out]

(((-4*I)/3)*(a + I*a*Tan[c + d*x])^6)/(a^4*d) + (((12*I)/7)*(a + I*a*Tan[c + d*x])^7)/(a^5*d) - (((3*I)/4)*(a
+ I*a*Tan[c + d*x])^8)/(a^6*d) + ((I/9)*(a + I*a*Tan[c + d*x])^9)/(a^7*d)

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Rubi [A]  time = 0.0662851, antiderivative size = 109, 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} \[ \frac{i (a+i a \tan (c+d x))^9}{9 a^7 d}-\frac{3 i (a+i a \tan (c+d x))^8}{4 a^6 d}+\frac{12 i (a+i a \tan (c+d x))^7}{7 a^5 d}-\frac{4 i (a+i a \tan (c+d x))^6}{3 a^4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(((-4*I)/3)*(a + I*a*Tan[c + d*x])^6)/(a^4*d) + (((12*I)/7)*(a + I*a*Tan[c + d*x])^7)/(a^5*d) - (((3*I)/4)*(a
+ I*a*Tan[c + d*x])^8)/(a^6*d) + ((I/9)*(a + I*a*Tan[c + d*x])^9)/(a^7*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 ^8(c+d x) (a+i a \tan (c+d x))^2 \, dx &=-\frac{i \operatorname{Subst}\left (\int (a-x)^3 (a+x)^5 \, dx,x,i a \tan (c+d x)\right )}{a^7 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (8 a^3 (a+x)^5-12 a^2 (a+x)^6+6 a (a+x)^7-(a+x)^8\right ) \, dx,x,i a \tan (c+d x)\right )}{a^7 d}\\ &=-\frac{4 i (a+i a \tan (c+d x))^6}{3 a^4 d}+\frac{12 i (a+i a \tan (c+d x))^7}{7 a^5 d}-\frac{3 i (a+i a \tan (c+d x))^8}{4 a^6 d}+\frac{i (a+i a \tan (c+d x))^9}{9 a^7 d}\\ \end{align*}

Mathematica [A]  time = 1.24373, size = 99, normalized size = 0.91 \[ \frac{a^2 \sec (c) \sec ^9(c+d x) (-63 \sin (2 c+d x)+84 \sin (2 c+3 d x)+36 \sin (4 c+5 d x)+9 \sin (6 c+7 d x)+\sin (8 c+9 d x)+63 i \cos (2 c+d x)+63 \sin (d x)+63 i \cos (d x))}{504 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(a^2*Sec[c]*Sec[c + d*x]^9*((63*I)*Cos[d*x] + (63*I)*Cos[2*c + d*x] + 63*Sin[d*x] - 63*Sin[2*c + d*x] + 84*Sin
[2*c + 3*d*x] + 36*Sin[4*c + 5*d*x] + 9*Sin[6*c + 7*d*x] + Sin[8*c + 9*d*x]))/(504*d)

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Maple [A]  time = 0.058, size = 141, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ( -{a}^{2} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{9\, \left ( \cos \left ( dx+c \right ) \right ) ^{9}}}+{\frac{2\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{21\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{8\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{105\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{16\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{315\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) +{\frac{{\frac{i}{4}}{a}^{2}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{8}}}-{a}^{2} \left ( -{\frac{16}{35}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{6}}{7}}-{\frac{6\, \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{35}}-{\frac{8\, \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{35}} \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)^8*(a+I*a*tan(d*x+c))^2,x)

[Out]

1/d*(-a^2*(1/9*sin(d*x+c)^3/cos(d*x+c)^9+2/21*sin(d*x+c)^3/cos(d*x+c)^7+8/105*sin(d*x+c)^3/cos(d*x+c)^5+16/315
*sin(d*x+c)^3/cos(d*x+c)^3)+1/4*I*a^2/cos(d*x+c)^8-a^2*(-16/35-1/7*sec(d*x+c)^6-6/35*sec(d*x+c)^4-8/35*sec(d*x
+c)^2)*tan(d*x+c))

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Maxima [A]  time = 1.10516, size = 146, normalized size = 1.34 \begin{align*} -\frac{140 \, a^{2} \tan \left (d x + c\right )^{9} - 315 i \, a^{2} \tan \left (d x + c\right )^{8} + 360 \, a^{2} \tan \left (d x + c\right )^{7} - 1260 i \, a^{2} \tan \left (d x + c\right )^{6} - 1890 i \, a^{2} \tan \left (d x + c\right )^{4} - 840 \, a^{2} \tan \left (d x + c\right )^{3} - 1260 i \, a^{2} \tan \left (d x + c\right )^{2} - 1260 \, a^{2} \tan \left (d x + c\right )}{1260 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/1260*(140*a^2*tan(d*x + c)^9 - 315*I*a^2*tan(d*x + c)^8 + 360*a^2*tan(d*x + c)^7 - 1260*I*a^2*tan(d*x + c)^
6 - 1890*I*a^2*tan(d*x + c)^4 - 840*a^2*tan(d*x + c)^3 - 1260*I*a^2*tan(d*x + c)^2 - 1260*a^2*tan(d*x + c))/d

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Fricas [B]  time = 1.19107, size = 590, normalized size = 5.41 \begin{align*} \frac{8064 i \, a^{2} e^{\left (10 i \, d x + 10 i \, c\right )} + 8064 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + 5376 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 2304 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 576 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 64 i \, a^{2}}{63 \,{\left (d e^{\left (18 i \, d x + 18 i \, c\right )} + 9 \, d e^{\left (16 i \, d x + 16 i \, c\right )} + 36 \, d e^{\left (14 i \, d x + 14 i \, c\right )} + 84 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 126 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 126 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 84 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 36 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 9 \, 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)^8*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/63*(8064*I*a^2*e^(10*I*d*x + 10*I*c) + 8064*I*a^2*e^(8*I*d*x + 8*I*c) + 5376*I*a^2*e^(6*I*d*x + 6*I*c) + 230
4*I*a^2*e^(4*I*d*x + 4*I*c) + 576*I*a^2*e^(2*I*d*x + 2*I*c) + 64*I*a^2)/(d*e^(18*I*d*x + 18*I*c) + 9*d*e^(16*I
*d*x + 16*I*c) + 36*d*e^(14*I*d*x + 14*I*c) + 84*d*e^(12*I*d*x + 12*I*c) + 126*d*e^(10*I*d*x + 10*I*c) + 126*d
*e^(8*I*d*x + 8*I*c) + 84*d*e^(6*I*d*x + 6*I*c) + 36*d*e^(4*I*d*x + 4*I*c) + 9*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 ^{8}{\left (c + d x \right )}\, dx + \int 2 i \tan{\left (c + d x \right )} \sec ^{8}{\left (c + d x \right )}\, dx + \int \sec ^{8}{\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)**8*(a+I*a*tan(d*x+c))**2,x)

[Out]

a**2*(Integral(-tan(c + d*x)**2*sec(c + d*x)**8, x) + Integral(2*I*tan(c + d*x)*sec(c + d*x)**8, x) + Integral
(sec(c + d*x)**8, x))

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Giac [A]  time = 1.20306, size = 146, normalized size = 1.34 \begin{align*} -\frac{28 \, a^{2} \tan \left (d x + c\right )^{9} - 63 i \, a^{2} \tan \left (d x + c\right )^{8} + 72 \, a^{2} \tan \left (d x + c\right )^{7} - 252 i \, a^{2} \tan \left (d x + c\right )^{6} - 378 i \, a^{2} \tan \left (d x + c\right )^{4} - 168 \, a^{2} \tan \left (d x + c\right )^{3} - 252 i \, a^{2} \tan \left (d x + c\right )^{2} - 252 \, a^{2} \tan \left (d x + c\right )}{252 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/252*(28*a^2*tan(d*x + c)^9 - 63*I*a^2*tan(d*x + c)^8 + 72*a^2*tan(d*x + c)^7 - 252*I*a^2*tan(d*x + c)^6 - 3
78*I*a^2*tan(d*x + c)^4 - 168*a^2*tan(d*x + c)^3 - 252*I*a^2*tan(d*x + c)^2 - 252*a^2*tan(d*x + c))/d