\(\int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\) [159]

Optimal result

Integrand size = 24, antiderivative size = 107 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {15 \text {arctanh}(\sin (c+d x))}{2 a^4 d}-\frac {15 \sec (c+d x) \tan (c+d x)}{2 a^4 d}+\frac {2 i \sec ^5(c+d x)}{a d (a+i a \tan (c+d x))^3}+\frac {10 i \sec ^3(c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )} \]

[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))

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3581, 3853, 3855} \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {15 \text {arctanh}(\sin (c+d x))}{2 a^4 d}+\frac {10 i \sec ^3(c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {15 \tan (c+d x) \sec (c+d x)}{2 a^4 d}+\frac {2 i \sec ^5(c+d x)}{a d (a+i a \tan (c+d x))^3} \]

[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 3581

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 3853

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 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \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 \text {arctanh}(\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*}

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(988\) vs. \(2(107)=214\).

Time = 6.64 (sec) , antiderivative size = 988, normalized size of antiderivative = 9.23 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\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 (a+i a \tan (c+d x))^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 (a+i a \tan (c+d x))^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 (a+i a \tan (c+d x))^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 (a+i a \tan (c+d x))^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 (a+i a \tan (c+d x))^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 (a+i a \tan (c+d x))^4}+\frac {\sec ^4(c+d x) (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}+\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 (a+i a \tan (c+d x))^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 (a+i a \tan (c+d x))^4}+\frac {4 \sec ^4(c+d x) (\cos (d x)+i \sin (d x))^4 \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 )}{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 ) (a+i a \tan (c+d x))^4}+\frac {4 \sec ^4(c+d x) (\cos (d x)+i \sin (d x))^4 \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 )}{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 ) (a+i a \tan (c+d x))^4} \]

[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)

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {8 i {\mathrm e}^{-i \left (d x +c \right )}}{a^{4} d}+\frac {i \left (7 \,{\mathrm e}^{3 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \,a^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a^{4} d}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a^{4} d}\)	\(107\)
derivativedivides	\(\frac {\frac {16}{-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 \left (\frac {1}{4}-2 i\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {15 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+\frac {2 \left (\frac {1}{4}+2 i\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {15 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}}{a^{4} d}\)	\(118\)
default	\(\frac {\frac {16}{-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 \left (\frac {1}{4}-2 i\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {15 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+\frac {2 \left (\frac {1}{4}+2 i\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {15 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}}{a^{4} d}\)	\(118\)

[In]

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

[Out]

8*I/a^4/d*exp(-I*(d*x+c))+I/d/a^4/(exp(2*I*(d*x+c))+1)^2*(7*exp(3*I*(d*x+c))+9*exp(I*(d*x+c)))-15/2/a^4/d*ln(e
xp(I*(d*x+c))+I)+15/2/a^4/d*ln(exp(I*(d*x+c))-I)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.50 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {15 \, {\left (e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, e^{\left (3 i \, d x + 3 i \, c\right )} + e^{\left (i \, d x + i \, c\right )}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 15 \, {\left (e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, e^{\left (3 i \, d x + 3 i \, c\right )} + e^{\left (i \, d x + i \, c\right )}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 30 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 50 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 16 i}{2 \, {\left (a^{4} d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{4} d e^{\left (i \, d x + i \, c\right )}\right )}} \]

[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))

Sympy [F]

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

[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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 457 vs. \(2 (95) = 190\).

Time = 0.32 (sec) , antiderivative size = 457, normalized size of antiderivative = 4.27 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {30 \, {\left (\cos \left (5 \, d x + 5 \, c\right ) + 2 \, \cos \left (3 \, d x + 3 \, c\right ) + \cos \left (d x + c\right ) + i \, \sin \left (5 \, d x + 5 \, c\right ) + 2 i \, \sin \left (3 \, d x + 3 \, c\right ) + i \, \sin \left (d x + c\right )\right )} \arctan \left (\cos \left (d x + c\right ), \sin \left (d x + c\right ) + 1\right ) + 30 \, {\left (\cos \left (5 \, d x + 5 \, c\right ) + 2 \, \cos \left (3 \, d x + 3 \, c\right ) + \cos \left (d x + c\right ) + i \, \sin \left (5 \, d x + 5 \, c\right ) + 2 i \, \sin \left (3 \, d x + 3 \, c\right ) + i \, \sin \left (d x + c\right )\right )} \arctan \left (\cos \left (d x + c\right ), -\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (i \, \cos \left (5 \, d x + 5 \, c\right ) + 2 i \, \cos \left (3 \, d x + 3 \, c\right ) + i \, \cos \left (d x + c\right ) - \sin \left (5 \, d x + 5 \, c\right ) - 2 \, \sin \left (3 \, d x + 3 \, c\right ) - \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 ) + 15 \, {\left (-i \, \cos \left (5 \, d x + 5 \, c\right ) - 2 i \, \cos \left (3 \, d x + 3 \, c\right ) - i \, \cos \left (d x + c\right ) + \sin \left (5 \, d x + 5 \, c\right ) + 2 \, \sin \left (3 \, d x + 3 \, c\right ) + \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}{-4 \, {\left (i \, a^{4} \cos \left (5 \, d x + 5 \, c\right ) + 2 i \, a^{4} \cos \left (3 \, d x + 3 \, c\right ) + i \, a^{4} \cos \left (d x + c\right ) - a^{4} \sin \left (5 \, d x + 5 \, c\right ) - 2 \, a^{4} \sin \left (3 \, d x + 3 \, c\right ) - a^{4} \sin \left (d x + c\right )\right )} d} \]

[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) + 2*cos(3*d*x + 3*c) + cos(d*x + c) + I*sin(5*d*x + 5*c) + 2*I*sin(3*d*x + 3*c) + I*sin(
d*x + c))*arctan2(cos(d*x + c), sin(d*x + c) + 1) + 30*(cos(5*d*x + 5*c) + 2*cos(3*d*x + 3*c) + cos(d*x + c) +
I*sin(5*d*x + 5*c) + 2*I*sin(3*d*x + 3*c) + I*sin(d*x + c))*arctan2(cos(d*x + c), -sin(d*x + c) + 1) + 15*(I*
cos(5*d*x + 5*c) + 2*I*cos(3*d*x + 3*c) + I*cos(d*x + c) - sin(5*d*x + 5*c) - 2*sin(3*d*x + 3*c) - 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) - 2*I*cos(3*d*x + 3*c)
- I*cos(d*x + c) + sin(5*d*x + 5*c) + 2*sin(3*d*x + 3*c) + 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)

Giac [A] (verification not implemented)

none

Time = 0.77 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.06 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {\frac {15 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{4}} - \frac {15 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{4}} - \frac {2 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 i\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{4}} - \frac {32}{a^{4} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}}}{2 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 6.26 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.51 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {15\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}+\frac {\frac {9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{a^4}-\frac {7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,39{}\mathrm {i}}{a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,17{}\mathrm {i}}{a^4}+\frac {24{}\mathrm {i}}{a^4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,1{}\mathrm {i}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,2{}\mathrm {i}-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}+1\right )} \]

[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)