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

Optimal result

Integrand size = 31, antiderivative size = 60 \[ \int \frac {\sec ^4(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{2 a d}-\frac {i \sec ^3(c+d x)}{3 a d}+\frac {\sec (c+d x) \tan (c+d x)}{2 a d} \]

[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] (verified)

Time = 0.14 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3171, 3169, 3853, 3855, 2686, 30} \[ \int \frac {\sec ^4(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{2 a d}-\frac {i \sec ^3(c+d x)}{3 a d}+\frac {\tan (c+d x) \sec (c+d x)}{2 a d} \]

[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 2686

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 3169

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 3171

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 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 {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 \text {Subst}\left (\int x^2 \, dx,x,\sec (c+d x)\right )}{a d} \\ & = \frac {\text {arctanh}(\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] (verified)

Time = 0.32 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.90 \[ \int \frac {\sec ^4(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=-\frac {i \left (12 i \text {arctanh}\left (\sin (c)+\cos (c) \tan \left (\frac {d x}{2}\right )\right )+\sec ^3(c+d x) (4+3 i \sin (2 (c+d x)))\right )}{12 a d} \]

[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)

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.67

method	result	size
risch	\(-\frac {i \left (3 \,{\mathrm e}^{5 i \left (d x +c \right )}+8 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {\ln \left (i+{\mathrm e}^{i \left (d x +c \right )}\right )}{2 a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a d}\)	\(100\)
norman	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {2 i}{3 a d}+\frac {2 i \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a d}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a d}\)	\(123\)
derivativedivides	\(\frac {\frac {i}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {2 \left (\frac {1}{4}+\frac {i}{4}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {2 \left (\frac {1}{4}+\frac {i}{4}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}-\frac {i}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {2 \left (\frac {1}{4}-\frac {i}{4}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\frac {2 \left (-\frac {1}{4}+\frac {i}{4}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}}{a d}\)	\(138\)
default	\(\frac {\frac {i}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {2 \left (\frac {1}{4}+\frac {i}{4}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {2 \left (\frac {1}{4}+\frac {i}{4}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}-\frac {i}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {2 \left (\frac {1}{4}-\frac {i}{4}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\frac {2 \left (-\frac {1}{4}+\frac {i}{4}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}}{a d}\)	\(138\)

[In]

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

[Out]

-1/3*I/d/a/(exp(2*I*(d*x+c))+1)^3*(3*exp(5*I*(d*x+c))+8*exp(3*I*(d*x+c))-3*exp(I*(d*x+c)))+1/2/a/d*ln(I+exp(I*
(d*x+c)))-1/2/a/d*ln(exp(I*(d*x+c))-I)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 174 vs. \(2 (52) = 104\).

Time = 0.25 (sec) , antiderivative size = 174, normalized size of antiderivative = 2.90 \[ \int \frac {\sec ^4(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {3 \, {\left (e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 3 \, {\left (e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 6 i \, e^{\left (5 i \, d x + 5 i \, c\right )} - 16 i \, e^{\left (3 i \, d x + 3 i \, c\right )} + 6 i \, e^{\left (i \, d x + i \, c\right )}}{6 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \]

[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)

Sympy [F]

\[ \int \frac {\sec ^4(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {\int \frac {\sec ^{4}{\left (c + d x \right )}}{i \sin {\left (c + d x \right )} + \cos {\left (c + d x \right )}}\, dx}{a} \]

[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

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. 186 vs. \(2 (52) = 104\).

Time = 0.20 (sec) , antiderivative size = 186, normalized size of antiderivative = 3.10 \[ \int \frac {\sec ^4(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {\frac {4 \, {\left (\frac {3 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {6 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {3 i \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + 2\right )}}{6 i \, a - \frac {18 i \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {18 i \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {6 i \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a}}{2 \, d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.65 \[ \int \frac {\sec ^4(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {\frac {3 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a} - \frac {3 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a} + \frac {2 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 i\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a}}{6 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 24.86 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.93 \[ \int \frac {\sec ^4(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}+\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{a}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,2{}\mathrm {i}}{a}+\frac {2{}\mathrm {i}}{3\,a}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[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))