Integrand size = 24, antiderivative size = 87 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {i \sec (c+d x)}{2 d (a+i a \tan (c+d x))^{3/2}} \] Output:
1/4*I*arctanh(1/2*a^(1/2)*sec(d*x+c)*2^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*2^( 1/2)/a^(3/2)/d+1/2*I*sec(d*x+c)/d/(a+I*a*tan(d*x+c))^(3/2)
Time = 0.67 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.09 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\left (2+\frac {2 e^{2 i (c+d x)} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )}{\sqrt {1+e^{2 i (c+d x)}}}\right ) \sec (c+d x)}{4 a d (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \] Input:
Integrate[Sec[c + d*x]/(a + I*a*Tan[c + d*x])^(3/2),x]
Output:
((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]])
Time = 0.36 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3042, 3983, 3042, 3970, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 3983 |
\(\displaystyle \frac {\int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx}{4 a}+\frac {i \sec (c+d x)}{2 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx}{4 a}+\frac {i \sec (c+d x)}{2 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3970 |
\(\displaystyle \frac {i \int \frac {1}{2-\frac {a \sec ^2(c+d x)}{i \tan (c+d x) a+a}}d\frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}}{2 a d}+\frac {i \sec (c+d x)}{2 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {i \sec (c+d x)}{2 d (a+i a \tan (c+d x))^{3/2}}\) |
Input:
Int[Sec[c + d*x]/(a + I*a*Tan[c + d*x])^(3/2),x]
Output:
((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))
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_S ymbol] :> Simp[-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 ]
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] + Simp[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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 342 vs. \(2 (68 ) = 136\).
Time = 7.80 (sec) , antiderivative size = 343, normalized size of antiderivative = 3.94
method | result | size |
default | \(-\frac {\sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (2 \cos \left (d x +c \right )-\sec \left (d x +c \right )\right )+\left (-4 \cos \left (d x +c \right )-2\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\operatorname {arctanh}\left (\frac {\left (i-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {2}}{2 \sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right ) \left (2 \sin \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 i \sin \left (d x +c \right ) \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {2}-2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+i \operatorname {arctanh}\left (\frac {\left (i-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {2}}{2 \sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right ) \left (-2 \cos \left (d x +c \right )-1+\sec \left (d x +c \right )\right )}{4 d \left (-i+\tan \left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a \left (\cos \left (d x +c \right )+1\right )}\) | \(343\) |
Input:
int(sec(d*x+c)/(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/4/d/(-I+tan(d*x+c))/(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^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)* (2*cos(d*x+c)-sec(d*x+c))+(-4*cos(d*x+c)-2)*(-cos(d*x+c)/(cos(d*x+c)+1))^( 1/2)+arctanh(1/2/(cot(d*x+c)^2-2*cot(d*x+c)*csc(d*x+c)+csc(d*x+c)^2-1)^(1/ 2)*(I-cot(d*x+c)+csc(d*x+c))*2^(1/2))*(2*sin(d*x+c)+tan(d*x+c))+2*I*sin(d* x+c)*((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*2^(1/2)-2*(-cos(d*x+c)/(cos(d*x +c)+1))^(1/2))+I*arctanh(1/2/(cot(d*x+c)^2-2*cot(d*x+c)*csc(d*x+c)+csc(d*x +c)^2-1)^(1/2)*(I-cot(d*x+c)+csc(d*x+c))*2^(1/2))*(-2*cos(d*x+c)-1+sec(d*x +c)))
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 = 0.09 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.83 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {{\left (i \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} + i\right )} e^{\left (-i \, d x - i \, c\right )}}{a d}\right ) - i \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} + i\right )} e^{\left (-i \, d x - i \, c\right )}}{a d}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{2} d} \] Input:
integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
Output:
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*s qrt(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)
\[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {\sec {\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))**(3/2),x)
Output:
Integral(sec(c + d*x)/(I*a*(tan(c + d*x) - I))**(3/2), x)
\[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int { \frac {\sec \left (d x + c\right )}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
Output:
integrate(sec(d*x + c)/(I*a*tan(d*x + c) + a)^(3/2), x)
Exception generated. \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad Argument TypeDone
Timed out. \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {1}{\cos \left (c+d\,x\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \] Input:
int(1/(cos(c + d*x)*(a + a*tan(c + d*x)*1i)^(3/2)),x)
Output:
int(1/(cos(c + d*x)*(a + a*tan(c + d*x)*1i)^(3/2)), x)
\[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\int \frac {\sec \left (d x +c \right )}{\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i +\sqrt {\tan \left (d x +c \right ) i +1}}d x}{\sqrt {a}\, a} \] Input:
int(sec(d*x+c)/(a+I*a*tan(d*x+c))^(3/2),x)
Output:
int(sec(c + d*x)/(sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*i + sqrt(tan(c + d *x)*i + 1)),x)/(sqrt(a)*a)