3.151 \(\int \frac{\sec ^8(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=90 \[ -\frac{i (a-i a \tan (c+d x))^3}{3 a^7 d}-\frac{i (a-i a \tan (c+d x))^2}{a^6 d}-\frac{4 \tan (c+d x)}{a^4 d}+\frac{8 i \log (\cos (c+d x))}{a^4 d}+\frac{8 x}{a^4} \]

[Out]

(8*x)/a^4 + ((8*I)*Log[Cos[c + d*x]])/(a^4*d) - (4*Tan[c + d*x])/(a^4*d) - (I*(a - I*a*Tan[c + d*x])^2)/(a^6*d
) - ((I/3)*(a - I*a*Tan[c + d*x])^3)/(a^7*d)

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Rubi [A]  time = 0.0560963, antiderivative size = 90, 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} \[ -\frac{i (a-i a \tan (c+d x))^3}{3 a^7 d}-\frac{i (a-i a \tan (c+d x))^2}{a^6 d}-\frac{4 \tan (c+d x)}{a^4 d}+\frac{8 i \log (\cos (c+d x))}{a^4 d}+\frac{8 x}{a^4} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^4,x]

[Out]

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

Mathematica [A]  time = 0.669694, size = 168, normalized size = 1.87 \[ \frac{\sec (c) \sec ^3(c+d x) (12 \sin (2 c+d x)-11 \sin (2 c+3 d x)+6 d x \cos (2 c+3 d x)+6 d x \cos (4 c+3 d x)+6 i \cos (2 c+3 d x) \log (\cos (c+d x))+6 \cos (d x) (3 i \log (\cos (c+d x))+3 d x+i)+6 \cos (2 c+d x) (3 i \log (\cos (c+d x))+3 d x+i)+6 i \cos (4 c+3 d x) \log (\cos (c+d x))-21 \sin (d x))}{6 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(Sec[c]*Sec[c + d*x]^3*(6*d*x*Cos[2*c + 3*d*x] + 6*d*x*Cos[4*c + 3*d*x] + 6*Cos[d*x]*(I + 3*d*x + (3*I)*Log[Co
s[c + d*x]]) + 6*Cos[2*c + d*x]*(I + 3*d*x + (3*I)*Log[Cos[c + d*x]]) + (6*I)*Cos[2*c + 3*d*x]*Log[Cos[c + d*x
]] + (6*I)*Cos[4*c + 3*d*x]*Log[Cos[c + d*x]] - 21*Sin[d*x] + 12*Sin[2*c + d*x] - 11*Sin[2*c + 3*d*x]))/(6*a^4
*d)

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Maple [A]  time = 0.085, size = 68, normalized size = 0.8 \begin{align*} -7\,{\frac{\tan \left ( dx+c \right ) }{{a}^{4}d}}+{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\,{a}^{4}d}}+{\frac{2\,i \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{{a}^{4}d}}-{\frac{8\,i\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{{a}^{4}d}} \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))^4,x)

[Out]

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

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Maxima [A]  time = 0.995178, size = 72, normalized size = 0.8 \begin{align*} \frac{\frac{\tan \left (d x + c\right )^{3} + 6 i \, \tan \left (d x + c\right )^{2} - 21 \, \tan \left (d x + c\right )}{a^{4}} - \frac{24 i \, \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a^{4}}}{3 \, 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))^4,x, algorithm="maxima")

[Out]

1/3*((tan(d*x + c)^3 + 6*I*tan(d*x + c)^2 - 21*tan(d*x + c))/a^4 - 24*I*log(I*tan(d*x + c) + 1)/a^4)/d

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Fricas [A]  time = 2.48496, size = 463, normalized size = 5.14 \begin{align*} \frac{48 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} + 48 \, d x +{\left (144 \, d x - 24 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (144 \, d x - 60 i\right )} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (24 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 72 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 72 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 24 i\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 44 i}{3 \,{\left (a^{4} d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} 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))^4,x, algorithm="fricas")

[Out]

1/3*(48*d*x*e^(6*I*d*x + 6*I*c) + 48*d*x + (144*d*x - 24*I)*e^(4*I*d*x + 4*I*c) + (144*d*x - 60*I)*e^(2*I*d*x
+ 2*I*c) + (24*I*e^(6*I*d*x + 6*I*c) + 72*I*e^(4*I*d*x + 4*I*c) + 72*I*e^(2*I*d*x + 2*I*c) + 24*I)*log(e^(2*I*
d*x + 2*I*c) + 1) - 44*I)/(a^4*d*e^(6*I*d*x + 6*I*c) + 3*a^4*d*e^(4*I*d*x + 4*I*c) + 3*a^4*d*e^(2*I*d*x + 2*I*
c) + a^4*d)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \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))**4,x)

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.20122, size = 211, normalized size = 2.34 \begin{align*} -\frac{2 \,{\left (\frac{24 i \, \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - i\right )}{a^{4}} - \frac{12 i \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac{12 i \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac{22 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 21 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 78 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 46 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 78 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 21 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 22 i}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{3} a^{4}}\right )}}{3 \, 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))^4,x, algorithm="giac")

[Out]

-2/3*(24*I*log(tan(1/2*d*x + 1/2*c) - I)/a^4 - 12*I*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 12*I*log(abs(tan(
1/2*d*x + 1/2*c) - 1))/a^4 + (22*I*tan(1/2*d*x + 1/2*c)^6 - 21*tan(1/2*d*x + 1/2*c)^5 - 78*I*tan(1/2*d*x + 1/2
*c)^4 + 46*tan(1/2*d*x + 1/2*c)^3 + 78*I*tan(1/2*d*x + 1/2*c)^2 - 21*tan(1/2*d*x + 1/2*c) - 22*I)/((tan(1/2*d*
x + 1/2*c)^2 - 1)^3*a^4))/d