Integrand size = 26, antiderivative size = 86 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {2 i \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}-\frac {2 i \sec (c+d x)}{a d \sqrt {a+i a \tan (c+d x)}} \] Output:
2*I*2^(1/2)*arctanh(1/2*a^(1/2)*sec(d*x+c)*2^(1/2)/(a+I*a*tan(d*x+c))^(1/2 ))/a^(3/2)/d-2*I*sec(d*x+c)/a/d/(a+I*a*tan(d*x+c))^(1/2)
Time = 0.90 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.17 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {8 e^{3 i (c+d x)} \left (-1+\sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )}{a d \left (1+e^{2 i (c+d x)}\right )^2 (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \] Input:
Integrate[Sec[c + d*x]^3/(a + I*a*Tan[c + d*x])^(3/2),x]
Output:
(8*E^((3*I)*(c + d*x))*(-1 + Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]]))/(a*d*(1 + E^((2*I)*(c + d*x)))^2*(-I + Tan[c + d *x])*Sqrt[a + I*a*Tan[c + d*x]])
Time = 0.39 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3042, 3972, 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.
\(\displaystyle \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^3}{(a+i a \tan (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 3972 |
\(\displaystyle \frac {2 \int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx}{a}-\frac {2 i \sec (c+d x)}{a d \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx}{a}-\frac {2 i \sec (c+d x)}{a d \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 3970 |
\(\displaystyle \frac {4 i \int \frac {1}{2-\frac {a \sec ^2(c+d x)}{i \tan (c+d x) a+a}}d\frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}}{a d}-\frac {2 i \sec (c+d x)}{a d \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 i \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}-\frac {2 i \sec (c+d x)}{a d \sqrt {a+i a \tan (c+d x)}}\) |
Input:
Int[Sec[c + d*x]^3/(a + I*a*Tan[c + d*x])^(3/2),x]
Output:
((2*I)*Sqrt[2]*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(a^(3/2)*d) - ((2*I)*Sec[c + d*x])/(a*d*Sqrt[a + I*a*Tan[c + d* x]])
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[2*d^2*(d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + f*x])^(n + 1)/(b*f*(m - 2))), x] + Simp[2*(d^2/a) 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] && EqQ[m/2 + n, 0] && LtQ[n, -1]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (71 ) = 142\).
Time = 6.52 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.97
method | result | size |
default | \(-\frac {2 \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )+i\right )^{3} \left (i-\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (i-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{2 \sqrt {\csc \left (d x +c \right )^{2} \left (1-\cos \left (d x +c \right )\right )^{2}-1}}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d a \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (\csc \left (d x +c \right )^{2} \left (1-\cos \left (d x +c \right )\right )^{2}-1\right )^{2} \left (-i+\tan \left (d x +c \right )\right )}\) | \(169\) |
Input:
int(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
-2/d/a/(a*(1+I*tan(d*x+c)))^(1/2)/(csc(d*x+c)^2*(1-cos(d*x+c))^2-1)^2*(cot (d*x+c)-csc(d*x+c)+I)^3*(I-2^(1/2)*arctanh(1/2/(csc(d*x+c)^2*(1-cos(d*x+c) )^2-1)^(1/2)*2^(1/2)*(I-cot(d*x+c)+csc(d*x+c)))*(-2*cos(d*x+c)/(cos(d*x+c) +1))^(1/2)+csc(d*x+c)-cot(d*x+c))/(-I+tan(d*x+c))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (67) = 134\).
Time = 0.09 (sec) , antiderivative size = 196, normalized size of antiderivative = 2.28 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {-i \, \sqrt {2} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} \log \left (-\frac {8 \, {\left ({\left (i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{a d}\right ) + i \, \sqrt {2} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} \log \left (-\frac {8 \, {\left ({\left (-i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{a d}\right ) - 2 i \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{a^{2} d} \] Input:
integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
Output:
(-I*sqrt(2)*a^2*d*sqrt(1/(a^3*d^2))*log(-8*((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(2)*a^2*d*sqrt(1/(a^3*d^2))*log(-8*((-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)) - 2*I*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)) )/(a^2*d)
\[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(sec(d*x+c)**3/(a+I*a*tan(d*x+c))**(3/2),x)
Output:
Integral(sec(c + d*x)**3/(I*a*(tan(c + d*x) - I))**(3/2), x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 813 vs. \(2 (67) = 134\).
Time = 0.26 (sec) , antiderivative size = 813, normalized size of antiderivative = 9.45 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:
integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
Output:
-1/2*((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^( 1/4)*(2*sqrt(2)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2 *d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^ (1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) - 2*sq rt(2)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c ) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (co s(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos( 1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 1) - I*sqrt(2)*log( sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos (1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(1/2*arctan2(sin (2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 + 2*(cos(2*d*x + 2*c)^2 + sin(2*d* x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c ), cos(2*d*x + 2*c) + 1)) + 1) + I*sqrt(2)*log(sqrt(cos(2*d*x + 2*c)^2 + s in(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(1/2*arctan2(sin(2*d*x + 2* c), cos(2*d*x + 2*c) + 1))^2 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^ 2 + 2*cos(2*d*x + 2*c) + 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 - 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)...
Exception generated. \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(sec(d*x+c)^3/(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. \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^3\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \] Input:
int(1/(cos(c + d*x)^3*(a + a*tan(c + d*x)*1i)^(3/2)),x)
Output:
int(1/(cos(c + d*x)^3*(a + a*tan(c + d*x)*1i)^(3/2)), x)
\[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{3} i -3 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{3} \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) \tan \left (d x +c \right )^{2} d -3 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{3} \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) d +\left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{3} \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) \tan \left (d x +c \right )^{2} d i +\left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{3} \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) d i \right )}{a^{2} d \left (\tan \left (d x +c \right )^{2}+1\right )} \] Input:
int(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(3/2),x)
Output:
(sqrt(a)*( - 2*sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**3*i - 3*int((sqrt(ta n(c + d*x)*i + 1)*sec(c + d*x)**3*tan(c + d*x)**2)/(tan(c + d*x)**3*i + ta n(c + d*x)**2 + tan(c + d*x)*i + 1),x)*tan(c + d*x)**2*d - 3*int((sqrt(tan (c + d*x)*i + 1)*sec(c + d*x)**3*tan(c + d*x)**2)/(tan(c + d*x)**3*i + tan (c + d*x)**2 + tan(c + d*x)*i + 1),x)*d + int((sqrt(tan(c + d*x)*i + 1)*se c(c + d*x)**3*tan(c + d*x))/(tan(c + d*x)**3*i + tan(c + d*x)**2 + tan(c + d*x)*i + 1),x)*tan(c + d*x)**2*d*i + int((sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**3*tan(c + d*x))/(tan(c + d*x)**3*i + tan(c + d*x)**2 + tan(c + d*x )*i + 1),x)*d*i))/(a**2*d*(tan(c + d*x)**2 + 1))