Integrand size = 26, antiderivative size = 86 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {2} a^{5/2} d}+\frac {i \sec (c+d x)}{a d (a+i a \tan (c+d x))^{3/2}} \] Output:
-1/2*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^(5/2)/d+I*sec(d*x+c)/a/d/(a+I*a*tan(d*x+c))^(3/2)
Time = 1.33 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.73 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {i e^{-\frac {1}{2} i (2 c+d x)} \left (-1-e^{2 i (c+d x)}+e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \sec ^3(c+d x) \left (\cos \left (c+\frac {d x}{2}\right )+i \sin \left (c+\frac {d x}{2}\right )\right )}{2 a^2 d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \] Input:
Integrate[Sec[c + d*x]^3/(a + I*a*Tan[c + d*x])^(5/2),x]
Output:
((I/2)*(-1 - E^((2*I)*(c + d*x)) + E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*( c + d*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]])*Sec[c + d*x]^3*(Cos[c + (d*x)/2] + I*Sin[c + (d*x)/2]))/(a^2*d*E^((I/2)*(2*c + d*x))*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])
Time = 0.51 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.47, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3042, 3982, 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.
\(\displaystyle \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^3}{(a+i a \tan (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 3982 |
\(\displaystyle \frac {2 i \sec (c+d x)}{a d (a+i a \tan (c+d x))^{3/2}}-\frac {2 \int \frac {\sec (c+d x)}{(i \tan (c+d x) a+a)^{3/2}}dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 i \sec (c+d x)}{a d (a+i a \tan (c+d x))^{3/2}}-\frac {2 \int \frac {\sec (c+d x)}{(i \tan (c+d x) a+a)^{3/2}}dx}{a}\) |
\(\Big \downarrow \) 3983 |
\(\displaystyle \frac {2 i \sec (c+d x)}{a d (a+i a \tan (c+d x))^{3/2}}-\frac {2 \left (\frac {\int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx}{4 a}+\frac {i \sec (c+d x)}{2 d (a+i a \tan (c+d x))^{3/2}}\right )}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 i \sec (c+d x)}{a d (a+i a \tan (c+d x))^{3/2}}-\frac {2 \left (\frac {\int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx}{4 a}+\frac {i \sec (c+d x)}{2 d (a+i a \tan (c+d x))^{3/2}}\right )}{a}\) |
\(\Big \downarrow \) 3970 |
\(\displaystyle \frac {2 i \sec (c+d x)}{a d (a+i a \tan (c+d x))^{3/2}}-\frac {2 \left (\frac {i \int \frac {1}{2-\frac {a \sec ^2(c+d x)}{i \tan (c+d x) a+a}}d\frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}}{2 a d}+\frac {i \sec (c+d x)}{2 d (a+i a \tan (c+d x))^{3/2}}\right )}{a}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 i \sec (c+d x)}{a d (a+i a \tan (c+d x))^{3/2}}-\frac {2 \left (\frac {i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {i \sec (c+d x)}{2 d (a+i a \tan (c+d x))^{3/2}}\right )}{a}\) |
Input:
Int[Sec[c + d*x]^3/(a + I*a*Tan[c + d*x])^(5/2),x]
Output:
((2*I)*Sec[c + d*x])/(a*d*(a + I*a*Tan[c + d*x])^(3/2)) - (2*(((I/2)*ArcTa nh[(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))))/a
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[d^2*(d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + f*x])^(n + 1)/(b*f*(m + n - 1))), x] + Simp[d^2*((m - 2)/(a*(m + n - 1))) 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] && LtQ[n, 0] && GtQ[m, 1] && !IL tQ[m + n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
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. 259 vs. \(2 (71 ) = 142\).
Time = 7.18 (sec) , antiderivative size = 260, normalized size of antiderivative = 3.02
method | result | size |
default | \(\frac {-2 i \cos \left (d x +c \right ) \operatorname {arctanh}\left (\frac {\left (i-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {2}}{2 \sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )-i \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+2 \sin \left (d x +c \right ) \operatorname {arctanh}\left (\frac {\left (i-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {2}}{2 \sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )+\sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+1\right )}{2 d \left (-i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a^{2} \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\) | \(260\) |
Input:
int(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
1/2/d/(-I*cos(d*x+c)+sin(d*x+c)-I)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)/a^2/ (a*(1+I*tan(d*x+c)))^(1/2)*(-2*I*cos(d*x+c)*arctanh(1/2/(cot(d*x+c)^2-2*co t(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 ))-I*sin(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+2*sin(d*x+c)* 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^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1 /2)*(cos(d*x+c)+1))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 245 vs. \(2 (67) = 134\).
Time = 0.09 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.85 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\left (i \, \sqrt {2} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {2 \, {\left ({\left (i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2} d}\right ) - i \, \sqrt {2} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {2 \, {\left ({\left (-i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2} d}\right ) - 2 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{3} d} \] Input:
integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")
Output:
1/4*(I*sqrt(2)*a^3*d*sqrt(1/(a^5*d^2))*e^(2*I*d*x + 2*I*c)*log(2*((I*a^2*d *e^(2*I*d*x + 2*I*c) + I*a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/( a^5*d^2)) - I)*e^(-I*d*x - I*c)/(a^2*d)) - I*sqrt(2)*a^3*d*sqrt(1/(a^5*d^2 ))*e^(2*I*d*x + 2*I*c)*log(2*((-I*a^2*d*e^(2*I*d*x + 2*I*c) - I*a^2*d)*sqr t(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^5*d^2)) - I)*e^(-I*d*x - I*c)/(a^ 2*d)) - 2*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^3*d)
\[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(sec(d*x+c)**3/(a+I*a*tan(d*x+c))**(5/2),x)
Output:
Integral(sec(c + d*x)**3/(I*a*(tan(c + d*x) - I))**(5/2), x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 827 vs. \(2 (67) = 134\).
Time = 0.32 (sec) , antiderivative size = 827, normalized size of antiderivative = 9.62 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")
Output:
-1/8*(4*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1) ^(1/4)*((-I*sqrt(2)*cos(2*d*x + 2*c) - sqrt(2)*sin(2*d*x + 2*c))*cos(1/2*a rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + (sqrt(2)*cos(2*d*x + 2*c ) - I*sqrt(2)*sin(2*d*x + 2*c))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d* x + 2*c) + 1)))*sqrt(a) - (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*c os(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*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) - 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*ar ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) + I*sqrt(2)*log(sqrt(c os(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(1/2*a rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 + sqrt(cos(2*d*x + 2*...
Exception generated. \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(5/2),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. \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^3\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \] Input:
int(1/(cos(c + d*x)^3*(a + a*tan(c + d*x)*1i)^(5/2)),x)
Output:
int(1/(cos(c + d*x)^3*(a + a*tan(c + d*x)*1i)^(5/2)), x)
\[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\int \frac {\sec \left (d x +c \right )^{3}}{\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -\sqrt {\tan \left (d x +c \right ) i +1}}d x}{\sqrt {a}\, a^{2}} \] Input:
int(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(5/2),x)
Output:
( - int(sec(c + d*x)**3/(sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 - 2*sqrt (tan(c + d*x)*i + 1)*tan(c + d*x)*i - sqrt(tan(c + d*x)*i + 1)),x))/(sqrt( a)*a**2)