3.159 \(\int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=107 \[ -\frac {15 \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac {10 i \sec ^3(c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {15 \tan (c+d x) \sec (c+d x)}{2 a^4 d}+\frac {2 i \sec ^5(c+d x)}{a d (a+i a \tan (c+d x))^3} \]

[Out]

-15/2*arctanh(sin(d*x+c))/a^4/d-15/2*sec(d*x+c)*tan(d*x+c)/a^4/d+2*I*sec(d*x+c)^5/a/d/(a+I*a*tan(d*x+c))^3+10*
I*sec(d*x+c)^3/d/(a^4+I*a^4*tan(d*x+c))

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Rubi [A]  time = 0.10, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3500, 3768, 3770} \[ -\frac {15 \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac {10 i \sec ^3(c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {15 \tan (c+d x) \sec (c+d x)}{2 a^4 d}+\frac {2 i \sec ^5(c+d x)}{a d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^7/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(-15*ArcTanh[Sin[c + d*x]])/(2*a^4*d) - (15*Sec[c + d*x]*Tan[c + d*x])/(2*a^4*d) + ((2*I)*Sec[c + d*x]^5)/(a*d
*(a + I*a*Tan[c + d*x])^3) + ((10*I)*Sec[c + d*x]^3)/(d*(a^4 + I*a^4*Tan[c + d*x]))

Rule 3500

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*d^2
*(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1))/(b*f*(m + 2*n)), x] - Dist[(d^2*(m - 2))/(b^2*(m + 2*n
)), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a
^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (Integers
Q[n, m + 1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]

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 \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^4} \, dx &=\frac {2 i \sec ^5(c+d x)}{a d (a+i a \tan (c+d x))^3}-\frac {5 \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^2} \, dx}{a^2}\\ &=\frac {2 i \sec ^5(c+d x)}{a d (a+i a \tan (c+d x))^3}+\frac {10 i \sec ^3(c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {15 \int \sec ^3(c+d x) \, dx}{a^4}\\ &=-\frac {15 \sec (c+d x) \tan (c+d x)}{2 a^4 d}+\frac {2 i \sec ^5(c+d x)}{a d (a+i a \tan (c+d x))^3}+\frac {10 i \sec ^3(c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {15 \int \sec (c+d x) \, dx}{2 a^4}\\ &=-\frac {15 \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac {15 \sec (c+d x) \tan (c+d x)}{2 a^4 d}+\frac {2 i \sec ^5(c+d x)}{a d (a+i a \tan (c+d x))^3}+\frac {10 i \sec ^3(c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 6.19, size = 988, normalized size = 9.23 \[ \frac {15 \cos (4 c) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^4(c+d x) (\cos (d x)+i \sin (d x))^4}{2 d (i \tan (c+d x) a+a)^4}-\frac {15 \cos (4 c) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^4(c+d x) (\cos (d x)+i \sin (d x))^4}{2 d (i \tan (c+d x) a+a)^4}+\frac {\cos (d x) \sec ^4(c+d x) (8 i \cos (3 c)-8 \sin (3 c)) (\cos (d x)+i \sin (d x))^4}{d (i \tan (c+d x) a+a)^4}+\frac {\sec (c) \sec ^4(c+d x) (4 i \cos (4 c)-4 \sin (4 c)) (\cos (d x)+i \sin (d x))^4}{d (i \tan (c+d x) a+a)^4}+\frac {15 i \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^4(c+d x) \sin (4 c) (\cos (d x)+i \sin (d x))^4}{2 d (i \tan (c+d x) a+a)^4}-\frac {15 i \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^4(c+d x) \sin (4 c) (\cos (d x)+i \sin (d x))^4}{2 d (i \tan (c+d x) a+a)^4}+\frac {\sec ^4(c+d x) (8 \cos (3 c)+8 i \sin (3 c)) \sin (d x) (\cos (d x)+i \sin (d x))^4}{d (i \tan (c+d x) a+a)^4}+\frac {4 \sec ^4(c+d x) \left (\frac {1}{2} \cos \left (4 c-\frac {d x}{2}\right )-\frac {1}{2} \cos \left (4 c+\frac {d x}{2}\right )+\frac {1}{2} i \sin \left (4 c-\frac {d x}{2}\right )-\frac {1}{2} i \sin \left (4 c+\frac {d x}{2}\right )\right ) (\cos (d x)+i \sin (d x))^4}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) (i \tan (c+d x) a+a)^4}+\frac {4 \sec ^4(c+d x) \left (-\frac {1}{2} \cos \left (4 c-\frac {d x}{2}\right )+\frac {1}{2} \cos \left (4 c+\frac {d x}{2}\right )-\frac {1}{2} i \sin \left (4 c-\frac {d x}{2}\right )+\frac {1}{2} i \sin \left (4 c+\frac {d x}{2}\right )\right ) (\cos (d x)+i \sin (d x))^4}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) (i \tan (c+d x) a+a)^4}+\frac {\sec ^4(c+d x) \left (\frac {1}{4} \cos (4 c)+\frac {1}{4} i \sin (4 c)\right ) (\cos (d x)+i \sin (d x))^4}{d \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2 (i \tan (c+d x) a+a)^4}+\frac {\sec ^4(c+d x) \left (-\frac {1}{4} \cos (4 c)-\frac {1}{4} i \sin (4 c)\right ) (\cos (d x)+i \sin (d x))^4}{d \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2 (i \tan (c+d x) a+a)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^7/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(15*Cos[4*c]*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^4*(Cos[d*x] + I*Sin[d*x])^4)/(2*d*(a +
I*a*Tan[c + d*x])^4) - (15*Cos[4*c]*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^4*(Cos[d*x] + I*
Sin[d*x])^4)/(2*d*(a + I*a*Tan[c + d*x])^4) + (Cos[d*x]*Sec[c + d*x]^4*((8*I)*Cos[3*c] - 8*Sin[3*c])*(Cos[d*x]
+ I*Sin[d*x])^4)/(d*(a + I*a*Tan[c + d*x])^4) + (Sec[c]*Sec[c + d*x]^4*((4*I)*Cos[4*c] - 4*Sin[4*c])*(Cos[d*x
] + I*Sin[d*x])^4)/(d*(a + I*a*Tan[c + d*x])^4) + (((15*I)/2)*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec
[c + d*x]^4*Sin[4*c]*(Cos[d*x] + I*Sin[d*x])^4)/(d*(a + I*a*Tan[c + d*x])^4) - (((15*I)/2)*Log[Cos[c/2 + (d*x)
/2] + Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^4*Sin[4*c]*(Cos[d*x] + I*Sin[d*x])^4)/(d*(a + I*a*Tan[c + d*x])^4) + (S
ec[c + d*x]^4*(8*Cos[3*c] + (8*I)*Sin[3*c])*(Cos[d*x] + I*Sin[d*x])^4*Sin[d*x])/(d*(a + I*a*Tan[c + d*x])^4) +
(Sec[c + d*x]^4*(Cos[4*c]/4 + (I/4)*Sin[4*c])*(Cos[d*x] + I*Sin[d*x])^4)/(d*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (
d*x)/2])^2*(a + I*a*Tan[c + d*x])^4) + (Sec[c + d*x]^4*(-1/4*Cos[4*c] - (I/4)*Sin[4*c])*(Cos[d*x] + I*Sin[d*x]
)^4)/(d*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2*(a + I*a*Tan[c + d*x])^4) + (4*Sec[c + d*x]^4*(Cos[d*x] +
I*Sin[d*x])^4*(Cos[4*c - (d*x)/2]/2 - Cos[4*c + (d*x)/2]/2 + (I/2)*Sin[4*c - (d*x)/2] - (I/2)*Sin[4*c + (d*x)/
2]))/(d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])*(a + I*a*Tan[c + d*x])^4) + (4*Sec[c +
d*x]^4*(Cos[d*x] + I*Sin[d*x])^4*(-1/2*Cos[4*c - (d*x)/2] + Cos[4*c + (d*x)/2]/2 - (I/2)*Sin[4*c - (d*x)/2] +
(I/2)*Sin[4*c + (d*x)/2]))/(d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])*(a + I*a*Tan[c
+ d*x])^4)

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fricas [A]  time = 0.46, size = 160, normalized size = 1.50 \[ -\frac {15 \, {\left (e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, e^{\left (3 i \, d x + 3 i \, c\right )} + e^{\left (i \, d x + i \, c\right )}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 15 \, {\left (e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, e^{\left (3 i \, d x + 3 i \, c\right )} + e^{\left (i \, d x + i \, c\right )}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 30 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 50 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 16 i}{2 \, {\left (a^{4} d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{4} d e^{\left (i \, d x + i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

-1/2*(15*(e^(5*I*d*x + 5*I*c) + 2*e^(3*I*d*x + 3*I*c) + e^(I*d*x + I*c))*log(e^(I*d*x + I*c) + I) - 15*(e^(5*I
*d*x + 5*I*c) + 2*e^(3*I*d*x + 3*I*c) + e^(I*d*x + I*c))*log(e^(I*d*x + I*c) - I) - 30*I*e^(4*I*d*x + 4*I*c) -
50*I*e^(2*I*d*x + 2*I*c) - 16*I)/(a^4*d*e^(5*I*d*x + 5*I*c) + 2*a^4*d*e^(3*I*d*x + 3*I*c) + a^4*d*e^(I*d*x +
I*c))

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giac [A]  time = 2.02, size = 113, normalized size = 1.06 \[ -\frac {\frac {15 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{4}} - \frac {15 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{4}} - \frac {2 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 i\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{4}} - \frac {32}{a^{4} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

-1/2*(15*log(tan(1/2*d*x + 1/2*c) + 1)/a^4 - 15*log(tan(1/2*d*x + 1/2*c) - 1)/a^4 - 2*(tan(1/2*d*x + 1/2*c)^3
- 8*I*tan(1/2*d*x + 1/2*c)^2 + tan(1/2*d*x + 1/2*c) + 8*I)/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^4) - 32/(a^4*(tan
(1/2*d*x + 1/2*c) - I)))/d

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maple [A]  time = 0.42, size = 192, normalized size = 1.79 \[ \frac {1}{2 a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {4 i}{a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {1}{2 a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {15 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \,a^{4}}+\frac {1}{2 a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {4 i}{a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{2 a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {15 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{4}}+\frac {16}{a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^4,x)

[Out]

1/2/a^4/d/(tan(1/2*d*x+1/2*c)-1)-4*I/a^4/d/(tan(1/2*d*x+1/2*c)-1)+1/2/a^4/d/(tan(1/2*d*x+1/2*c)-1)^2+15/2/d/a^
4*ln(tan(1/2*d*x+1/2*c)-1)+1/2/a^4/d/(tan(1/2*d*x+1/2*c)+1)+4*I/a^4/d/(tan(1/2*d*x+1/2*c)+1)-1/2/a^4/d/(tan(1/
2*d*x+1/2*c)+1)^2-15/2/d/a^4*ln(tan(1/2*d*x+1/2*c)+1)+16/a^4/d/(tan(1/2*d*x+1/2*c)-I)

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maxima [B]  time = 0.96, size = 467, normalized size = 4.36 \[ \frac {{\left (30 \, \cos \left (5 \, d x + 5 \, c\right ) + 60 \, \cos \left (3 \, d x + 3 \, c\right ) + 30 \, \cos \left (d x + c\right ) + 30 i \, \sin \left (5 \, d x + 5 \, c\right ) + 60 i \, \sin \left (3 \, d x + 3 \, c\right ) + 30 i \, \sin \left (d x + c\right )\right )} \arctan \left (\cos \left (d x + c\right ), \sin \left (d x + c\right ) + 1\right ) + {\left (30 \, \cos \left (5 \, d x + 5 \, c\right ) + 60 \, \cos \left (3 \, d x + 3 \, c\right ) + 30 \, \cos \left (d x + c\right ) + 30 i \, \sin \left (5 \, d x + 5 \, c\right ) + 60 i \, \sin \left (3 \, d x + 3 \, c\right ) + 30 i \, \sin \left (d x + c\right )\right )} \arctan \left (\cos \left (d x + c\right ), -\sin \left (d x + c\right ) + 1\right ) - {\left (-15 i \, \cos \left (5 \, d x + 5 \, c\right ) - 30 i \, \cos \left (3 \, d x + 3 \, c\right ) - 15 i \, \cos \left (d x + c\right ) + 15 \, \sin \left (5 \, d x + 5 \, c\right ) + 30 \, \sin \left (3 \, d x + 3 \, c\right ) + 15 \, \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - {\left (15 i \, \cos \left (5 \, d x + 5 \, c\right ) + 30 i \, \cos \left (3 \, d x + 3 \, c\right ) + 15 i \, \cos \left (d x + c\right ) - 15 \, \sin \left (5 \, d x + 5 \, c\right ) - 30 \, \sin \left (3 \, d x + 3 \, c\right ) - 15 \, \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right ) + 60 \, \cos \left (4 \, d x + 4 \, c\right ) + 100 \, \cos \left (2 \, d x + 2 \, c\right ) + 60 i \, \sin \left (4 \, d x + 4 \, c\right ) + 100 i \, \sin \left (2 \, d x + 2 \, c\right ) + 32}{{\left (-4 i \, a^{4} \cos \left (5 \, d x + 5 \, c\right ) - 8 i \, a^{4} \cos \left (3 \, d x + 3 \, c\right ) - 4 i \, a^{4} \cos \left (d x + c\right ) + 4 \, a^{4} \sin \left (5 \, d x + 5 \, c\right ) + 8 \, a^{4} \sin \left (3 \, d x + 3 \, c\right ) + 4 \, a^{4} \sin \left (d x + c\right )\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

((30*cos(5*d*x + 5*c) + 60*cos(3*d*x + 3*c) + 30*cos(d*x + c) + 30*I*sin(5*d*x + 5*c) + 60*I*sin(3*d*x + 3*c)
+ 30*I*sin(d*x + c))*arctan2(cos(d*x + c), sin(d*x + c) + 1) + (30*cos(5*d*x + 5*c) + 60*cos(3*d*x + 3*c) + 30
*cos(d*x + c) + 30*I*sin(5*d*x + 5*c) + 60*I*sin(3*d*x + 3*c) + 30*I*sin(d*x + c))*arctan2(cos(d*x + c), -sin(
d*x + c) + 1) - (-15*I*cos(5*d*x + 5*c) - 30*I*cos(3*d*x + 3*c) - 15*I*cos(d*x + c) + 15*sin(5*d*x + 5*c) + 30
*sin(3*d*x + 3*c) + 15*sin(d*x + c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - (15*I*cos(5*d
*x + 5*c) + 30*I*cos(3*d*x + 3*c) + 15*I*cos(d*x + c) - 15*sin(5*d*x + 5*c) - 30*sin(3*d*x + 3*c) - 15*sin(d*x
+ c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*sin(d*x + c) + 1) + 60*cos(4*d*x + 4*c) + 100*cos(2*d*x + 2*c)
+ 60*I*sin(4*d*x + 4*c) + 100*I*sin(2*d*x + 2*c) + 32)/((-4*I*a^4*cos(5*d*x + 5*c) - 8*I*a^4*cos(3*d*x + 3*c)
- 4*I*a^4*cos(d*x + c) + 4*a^4*sin(5*d*x + 5*c) + 8*a^4*sin(3*d*x + 3*c) + 4*a^4*sin(d*x + c))*d)

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mupad [B]  time = 5.52, size = 162, normalized size = 1.51 \[ -\frac {15\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}+\frac {\frac {9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{a^4}-\frac {7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,39{}\mathrm {i}}{a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,17{}\mathrm {i}}{a^4}+\frac {24{}\mathrm {i}}{a^4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,1{}\mathrm {i}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,2{}\mathrm {i}-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^7*(a + a*tan(c + d*x)*1i)^4),x)

[Out]

((9*tan(c/2 + (d*x)/2)^3)/a^4 - (tan(c/2 + (d*x)/2)^2*39i)/a^4 + (tan(c/2 + (d*x)/2)^4*17i)/a^4 + 24i/a^4 - (7
*tan(c/2 + (d*x)/2))/a^4)/(d*(tan(c/2 + (d*x)/2)*1i - 2*tan(c/2 + (d*x)/2)^2 - tan(c/2 + (d*x)/2)^3*2i + tan(c
/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^5*1i + 1)) - (15*atanh(tan(c/2 + (d*x)/2)))/(a^4*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec ^{7}{\left (c + d x \right )}}{\tan ^{4}{\left (c + d x \right )} - 4 i \tan ^{3}{\left (c + d x \right )} - 6 \tan ^{2}{\left (c + d x \right )} + 4 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**7/(a+I*a*tan(d*x+c))**4,x)

[Out]

Integral(sec(c + d*x)**7/(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

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