3.108 \(\int \frac {\sec ^5(c+d x)}{a+i a \tan (c+d x)} \, dx\)

Optimal. Leaf size=60 \[ -\frac {i \sec ^3(c+d x)}{3 a d}+\frac {\tanh ^{-1}(\sin (c+d x))}{2 a d}+\frac {\tan (c+d x) \sec (c+d x)}{2 a d} \]

[Out]

1/2*arctanh(sin(d*x+c))/a/d-1/3*I*sec(d*x+c)^3/a/d+1/2*sec(d*x+c)*tan(d*x+c)/a/d

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Rubi [A]  time = 0.05, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3501, 3768, 3770} \[ -\frac {i \sec ^3(c+d x)}{3 a d}+\frac {\tanh ^{-1}(\sin (c+d x))}{2 a d}+\frac {\tan (c+d x) \sec (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3501

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

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

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Mathematica [A]  time = 0.25, size = 50, normalized size = 0.83 \[ \frac {12 \tanh ^{-1}\left (\cos (c) \tan \left (\frac {d x}{2}\right )+\sin (c)\right )+(3 \sin (2 (c+d x))-4 i) \sec ^3(c+d x)}{12 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*ArcTanh[Sin[c] + Cos[c]*Tan[(d*x)/2]] + Sec[c + d*x]^3*(-4*I + 3*Sin[2*(c + d*x)]))/(12*a*d)

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fricas [B]  time = 0.45, size = 174, normalized size = 2.90 \[ \frac {3 \, {\left (e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 3 \, {\left (e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 6 i \, e^{\left (5 i \, d x + 5 i \, c\right )} - 16 i \, e^{\left (3 i \, d x + 3 i \, c\right )} + 6 i \, e^{\left (i \, d x + i \, c\right )}}{6 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(e^(6*I*d*x + 6*I*c) + 3*e^(4*I*d*x + 4*I*c) + 3*e^(2*I*d*x + 2*I*c) + 1)*log(e^(I*d*x + I*c) + I) - 3*
(e^(6*I*d*x + 6*I*c) + 3*e^(4*I*d*x + 4*I*c) + 3*e^(2*I*d*x + 2*I*c) + 1)*log(e^(I*d*x + I*c) - I) - 6*I*e^(5*
I*d*x + 5*I*c) - 16*I*e^(3*I*d*x + 3*I*c) + 6*I*e^(I*d*x + I*c))/(a*d*e^(6*I*d*x + 6*I*c) + 3*a*d*e^(4*I*d*x +
4*I*c) + 3*a*d*e^(2*I*d*x + 2*I*c) + a*d)

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giac [A]  time = 0.73, size = 99, normalized size = 1.65 \[ \frac {\frac {3 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a} - \frac {3 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a} + \frac {2 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 i\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*log(tan(1/2*d*x + 1/2*c) + 1)/a - 3*log(tan(1/2*d*x + 1/2*c) - 1)/a + 2*(3*tan(1/2*d*x + 1/2*c)^5 + 6*I
*tan(1/2*d*x + 1/2*c)^4 - 3*tan(1/2*d*x + 1/2*c) + 2*I)/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a))/d

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maple [B]  time = 0.36, size = 258, normalized size = 4.30 \[ \frac {i}{3 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {1}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {i}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {1}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {i}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a d}-\frac {i}{3 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {i}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {i}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.37, size = 186, normalized size = 3.10 \[ \frac {\frac {4 \, {\left (\frac {3 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {6 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {3 i \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + 2\right )}}{6 i \, a - \frac {18 i \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {18 i \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {6 i \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*(3*I*sin(d*x + c)/(cos(d*x + c) + 1) + 6*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 3*I*sin(d*x + c)^5/(cos(
d*x + c) + 1)^5 + 2)/(6*I*a - 18*I*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 18*I*a*sin(d*x + c)^4/(cos(d*x + c)
+ 1)^4 - 6*I*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a - log(sin(d*
x + c)/(cos(d*x + c) + 1) - 1)/a)/d

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mupad [B]  time = 5.38, size = 116, normalized size = 1.93 \[ \frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}+\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{a}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,2{}\mathrm {i}}{a}+\frac {2{}\mathrm {i}}{3\,a}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atanh(tan(c/2 + (d*x)/2))/(a*d) + ((tan(c/2 + (d*x)/2)^4*2i)/a + tan(c/2 + (d*x)/2)^5/a + 2i/(3*a) - tan(c/2 +
(d*x)/2)/a)/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {i \int \frac {\sec ^{5}{\left (c + d x \right )}}{\tan {\left (c + d x \right )} - i}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*Integral(sec(c + d*x)**5/(tan(c + d*x) - I), x)/a

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