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

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 90, normalized size of antiderivative = 1.00, 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 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])

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]

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 = 0.78, 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)

________________________________________________________________________________________

fricas [A]  time = 0.50, size = 153, normalized size = 1.70 \[ \frac {48 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} + 48 \, d x + {\left (144 \, d x - 24 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (144 \, d x - 60 i\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (24 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 72 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 72 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 24 i\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 44 i}{3 \, {\left (a^{4} d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )}} \]

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)

________________________________________________________________________________________

giac [A]  time = 1.99, size = 154, normalized size = 1.71 \[ -\frac {2 \, {\left (-\frac {12 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{4}} + \frac {24 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}{a^{4}} - \frac {12 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{4}} + \frac {22 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 78 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 46 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 78 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 22 i}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{4}}\right )}}{3 \, d} \]

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*(-12*I*log(tan(1/2*d*x + 1/2*c) + 1)/a^4 + 24*I*log(tan(1/2*d*x + 1/2*c) - I)/a^4 - 12*I*log(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 + 4
6*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

________________________________________________________________________________________

maple [A]  time = 0.40, size = 68, normalized size = 0.76 \[ -\frac {7 \tan \left (d x +c \right )}{a^{4} d}+\frac {\tan ^{3}\left (d x +c \right )}{3 a^{4} d}+\frac {2 i \left (\tan ^{2}\left (d x +c \right )\right )}{a^{4} d}-\frac {8 i \ln \left (\tan \left (d x +c \right )-i\right )}{a^{4} d} \]

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

________________________________________________________________________________________

maxima [A]  time = 0.77, size = 53, normalized size = 0.59 \[ \frac {\frac {\tan \left (d x + c\right )^{3} + 6 i \, \tan \left (d x + c\right )^{2} - 21 \, \tan \left (d x + c\right )}{a^{4}} - \frac {24 i \, \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a^{4}}}{3 \, d} \]

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

________________________________________________________________________________________

mupad [B]  time = 3.38, size = 60, normalized size = 0.67 \[ -\frac {\frac {7\,\mathrm {tan}\left (c+d\,x\right )}{a^4}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,a^4}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,8{}\mathrm {i}}{a^4}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,2{}\mathrm {i}}{a^4}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^8*(a + a*tan(c + d*x)*1i)^4),x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec ^{8}{\left (c + d x \right )}}{\tan ^{4}{\left (c + d x \right )} - 4 i \tan ^{3}{\left (c + d x \right )} - 6 \tan ^{2}{\left (c + d x \right )} + 4 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

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]

Integral(sec(c + d*x)**8/(tan(c + d*x)**4 - 4*I*tan(c + d*x)**3 - 6*tan(c + d*x)**2 + 4*I*tan(c + d*x) + 1), x
)/a**4

________________________________________________________________________________________