\(\int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\) [160]

Optimal result

Integrand size = 24, antiderivative size = 82 \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{a^4 d}+\frac {2 i \sec ^3(c+d x)}{3 a d (a+i a \tan (c+d x))^3}-\frac {2 i \sec (c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3581, 3855} \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{a^4 d}-\frac {2 i \sec (c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {2 i \sec ^3(c+d x)}{3 a d (a+i a \tan (c+d x))^3} \]

[In]

Int[Sec[c + d*x]^5/(a + I*a*Tan[c + d*x])^4,x]

[Out]

ArcTanh[Sin[c + d*x]]/(a^4*d) + (((2*I)/3)*Sec[c + d*x]^3)/(a*d*(a + I*a*Tan[c + d*x])^3) - ((2*I)*Sec[c + d*x
])/(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 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 ^3(c+d x)}{3 a d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx}{a^2} \\ & = \frac {2 i \sec ^3(c+d x)}{3 a d (a+i a \tan (c+d x))^3}-\frac {2 i \sec (c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {\int \sec (c+d x) \, dx}{a^4} \\ & = \frac {\text {arctanh}(\sin (c+d x))}{a^4 d}+\frac {2 i \sec ^3(c+d x)}{3 a d (a+i a \tan (c+d x))^3}-\frac {2 i \sec (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. \(247\) vs. \(2(82)=164\).

Time = 0.61 (sec) , antiderivative size = 247, normalized size of antiderivative = 3.01 \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\sec ^4(c+d x) (\cos (d x)+i \sin (d x))^4 \left (-3 \cos (4 c) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+3 \cos (4 c) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 \cos (3 d x) \sin (c)+6 \cos (d x) \sin (3 c)-3 i \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (4 c)+3 i \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (4 c)+\cos (3 c) (-6 i \cos (d x)-6 \sin (d x))-6 i \sin (3 c) \sin (d x)+2 i \sin (c) \sin (3 d x)+2 \cos (c) (i \cos (3 d x)+\sin (3 d x))\right )}{3 a^4 d (-i+\tan (c+d x))^4} \]

[In]

Integrate[Sec[c + d*x]^5/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(Sec[c + d*x]^4*(Cos[d*x] + I*Sin[d*x])^4*(-3*Cos[4*c]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 3*Cos[4*c]*L
og[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 2*Cos[3*d*x]*Sin[c] + 6*Cos[d*x]*Sin[3*c] - (3*I)*Log[Cos[(c + d*x)/
2] - Sin[(c + d*x)/2]]*Sin[4*c] + (3*I)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[4*c] + Cos[3*c]*((-6*I)*C
os[d*x] - 6*Sin[d*x]) - (6*I)*Sin[3*c]*Sin[d*x] + (2*I)*Sin[c]*Sin[3*d*x] + 2*Cos[c]*(I*Cos[3*d*x] + Sin[3*d*x
])))/(3*a^4*d*(-I + Tan[c + d*x])^4)

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {8 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {16}{3 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4} d}\)	\(71\)
default	\(\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {8 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {16}{3 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4} d}\)	\(71\)
risch	\(-\frac {2 i {\mathrm e}^{-i \left (d x +c \right )}}{a^{4} d}+\frac {2 i {\mathrm e}^{-3 i \left (d x +c \right )}}{3 a^{4} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{4} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{4} d}\)	\(79\)

[In]

int(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.93 \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\left (3 \, e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 3 \, e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{3 \, a^{4} d} \]

[In]

integrate(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\int \frac {\sec ^{5}{\left (c + d x \right )}}{\tan ^{4}{\left (c + d x \right )} - 4 i \tan ^{3}{\left (c + d x \right )} - 6 \tan ^{2}{\left (c + d x \right )} + 4 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

[In]

integrate(sec(d*x+c)**5/(a+I*a*tan(d*x+c))**4,x)

[Out]

Integral(sec(c + d*x)**5/(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 [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.72 \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {-6 i \, \arctan \left (\cos \left (d x + c\right ), \sin \left (d x + c\right ) + 1\right ) - 6 i \, \arctan \left (\cos \left (d x + c\right ), -\sin \left (d x + c\right ) + 1\right ) + 4 i \, \cos \left (3 \, d x + 3 \, c\right ) - 12 i \, \cos \left (d x + c\right ) + 3 \, \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right ) + 4 \, \sin \left (3 \, d x + 3 \, c\right ) - 12 \, \sin \left (d x + c\right )}{6 \, a^{4} d} \]

[In]

integrate(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

1/6*(-6*I*arctan2(cos(d*x + c), sin(d*x + c) + 1) - 6*I*arctan2(cos(d*x + c), -sin(d*x + c) + 1) + 4*I*cos(3*d
*x + 3*c) - 12*I*cos(d*x + c) + 3*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - 3*log(cos(d*x +
c)^2 + sin(d*x + c)^2 - 2*sin(d*x + c) + 1) + 4*sin(3*d*x + 3*c) - 12*sin(d*x + c))/(a^4*d)

Giac [A] (verification not implemented)

none

Time = 0.65 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.87 \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\frac {3 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{4}} - \frac {3 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{4}} + \frac {8 \, {\left (3 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}{a^{4} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{3}}}{3 \, d} \]

[In]

integrate(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.39 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.07 \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {-\frac {8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4}+\frac {8{}\mathrm {i}}{3\,a^4}}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,3{}\mathrm {i}+1\right )} \]

[In]

int(1/(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^4),x)

[Out]

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