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

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

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3092, 3090, 3768, 3770, 2606, 30} \[ -\frac {i \sec ^3(c+d x)}{3 a d}+\frac {\tanh ^{-1}(\sin (c+d x))}{2 a d}+\frac {\tan (c+d x) \sec (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a*Cos[c + d*x] + I*a*Sin[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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 3090

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[cos[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
IntegerQ[m] && IGtQ[n, 0]

Rule 3092

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> Dist[a^n*b^n, Int[Cos[c + d*x]^m/(b*Cos[c + d*x] + a*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, m},
x] && EqQ[a^2 + b^2, 0] && ILtQ[n, 0]

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 ^4(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx &=-\frac {i \int \sec ^4(c+d x) (i a \cos (c+d x)+a \sin (c+d x)) \, dx}{a^2}\\ &=-\frac {i \int \left (i a \sec ^3(c+d x)+a \sec ^3(c+d x) \tan (c+d x)\right ) \, dx}{a^2}\\ &=-\frac {i \int \sec ^3(c+d x) \tan (c+d x) \, dx}{a}+\frac {\int \sec ^3(c+d x) \, dx}{a}\\ &=\frac {\sec (c+d x) \tan (c+d x)}{2 a d}+\frac {\int \sec (c+d x) \, dx}{2 a}-\frac {i \operatorname {Subst}\left (\int x^2 \, dx,x,\sec (c+d x)\right )}{a d}\\ &=\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*}

________________________________________________________________________________________

Mathematica [A]  time = 0.25, size = 54, normalized size = 0.90 \[ -\frac {i \left ((4+3 i \sin (2 (c+d x))) \sec ^3(c+d x)+12 i \tanh ^{-1}\left (\cos (c) \tan \left (\frac {d x}{2}\right )+\sin (c)\right )\right )}{12 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/12*I)*((12*I)*ArcTanh[Sin[c] + Cos[c]*Tan[(d*x)/2]] + Sec[c + d*x]^3*(4 + (3*I)*Sin[2*(c + d*x)])))/(a*d)

________________________________________________________________________________________

fricas [B]  time = 0.52, 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)^4/(a*cos(d*x+c)+I*a*sin(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)

________________________________________________________________________________________

giac [A]  time = 2.72, 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)^4/(a*cos(d*x+c)+I*a*sin(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

________________________________________________________________________________________

maple [B]  time = 0.23, 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)^4/(a*cos(d*x+c)+I*a*sin(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)

________________________________________________________________________________________

maxima [B]  time = 0.33, 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)^4/(a*cos(d*x+c)+I*a*sin(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

________________________________________________________________________________________

mupad [B]  time = 2.55, 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)^4*(a*cos(c + d*x) + a*sin(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))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec ^{4}{\left (c + d x \right )}}{i \sin {\left (c + d x \right )} + \cos {\left (c + d x \right )}}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4/(a*cos(d*x+c)+I*a*sin(d*x+c)),x)

[Out]

Integral(sec(c + d*x)**4/(I*sin(c + d*x) + cos(c + d*x)), x)/a

________________________________________________________________________________________