\(\int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [359]

Optimal result

Integrand size = 24, antiderivative size = 87 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {i \sec (c+d x)}{2 d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

1/4*I*arctanh(1/2*sec(d*x+c)*a^(1/2)*2^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)+1/2*I*sec(d*x+c)/d/(a
+I*a*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3583, 3570, 212} \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {i \sec (c+d x)}{2 d (a+i a \tan (c+d x))^{3/2}} \]

[In]

Int[Sec[c + d*x]/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3570

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2*(a/(b*f)), Subst[
Int[1/(2 - a*x^2), x], x, Sec[e + f*x]/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 + b^
2, 0]

Rule 3583

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

Rubi steps \begin{align*} \text {integral}& = \frac {i \sec (c+d x)}{2 d (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx}{4 a} \\ & = \frac {i \sec (c+d x)}{2 d (a+i a \tan (c+d x))^{3/2}}+\frac {i \text {Subst}\left (\int \frac {1}{2-a x^2} \, dx,x,\frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}}\right )}{2 a d} \\ & = \frac {i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {i \sec (c+d x)}{2 d (a+i a \tan (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.78 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.09 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\left (2+\frac {2 e^{2 i (c+d x)} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )}{\sqrt {1+e^{2 i (c+d x)}}}\right ) \sec (c+d x)}{4 a d (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \]

[In]

Integrate[Sec[c + d*x]/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

((2 + (2*E^((2*I)*(c + d*x))*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]])/Sqrt[1 + E^((2*I)*(c + d*x))])*Sec[c + d*
x])/(4*a*d*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 415 vs. \(2 (68 ) = 136\).

Time = 8.67 (sec) , antiderivative size = 416, normalized size of antiderivative = 4.78

method	result	size
default	\(-\frac {i \left (2 i \cos \left (d x +c \right ) \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+i \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+2 i \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-2 \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sin \left (d x +c \right )-i \sec \left (d x +c \right ) \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+2 i \sec \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\tan \left (d x +c \right ) \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )\right )}{4 d \left (\tan \left (d x +c \right )-i\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a \left (\cos \left (d x +c \right )+1\right )}\)	\(416\)

[In]

int(sec(d*x+c)/(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/4*I/d/(tan(d*x+c)-I)/(a*(1+I*tan(d*x+c)))^(1/2)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)/a/(cos(d*x+c)+1)*(2*I*co
s(d*x+c)*arctan(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+I*arctan(1/
2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+2*I*(-cos(d*x+c)/(cos(d*x+c)+
1))^(1/2)-2*arctan(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*sin(d*x+
c)-I*sec(d*x+c)*arctan(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+2*I*
sec(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-tan(d*x+c)*arctan(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+c)+1)
/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)))

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. 246 vs. \(2 (64) = 128\).

Time = 0.26 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.83 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {{\left (i \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} + i\right )} e^{\left (-i \, d x - i \, c\right )}}{a d}\right ) - i \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} + i\right )} e^{\left (-i \, d x - i \, c\right )}}{a d}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{2} d} \]

[In]

integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Integral(sec(c + d*x)/(I*a*(tan(c + d*x) - I))**(3/2), x)

Maxima [F]

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

[In]

integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)/(I*a*tan(d*x + c) + a)^(3/2), x)

Giac [F]

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

[In]

integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)/(I*a*tan(d*x + c) + a)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {1}{\cos \left (c+d\,x\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \]

[In]

int(1/(cos(c + d*x)*(a + a*tan(c + d*x)*1i)^(3/2)),x)

[Out]

int(1/(cos(c + d*x)*(a + a*tan(c + d*x)*1i)^(3/2)), x)