\(\int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{3/2}} \, dx\) [411]

Optimal result

Integrand size = 30, antiderivative size = 280 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{3/2}} \, dx=-\frac {i \sqrt {2} a^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{d e^{3/2}}+\frac {i \sqrt {2} a^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{d e^{3/2}}-\frac {i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)} \left (\sqrt {a}+\cos (c+d x) \left (\sqrt {a}+i \sqrt {a} \tan (c+d x)\right )\right )}\right )}{d e^{3/2}}-\frac {4 i a (a+i a \tan (c+d x))^{3/2}}{3 d (e \sec (c+d x))^{3/2}} \] Output:

-I*2^(1/2)*a^(5/2)*arctan(1-2^(1/2)*e^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/a^(1/ 
2)/(e*sec(d*x+c))^(1/2))/d/e^(3/2)+I*2^(1/2)*a^(5/2)*arctan(1+2^(1/2)*e^(1 
/2)*(a+I*a*tan(d*x+c))^(1/2)/a^(1/2)/(e*sec(d*x+c))^(1/2))/d/e^(3/2)-I*2^( 
1/2)*a^(5/2)*arctanh(2^(1/2)*e^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/(e*sec(d*x+c 
))^(1/2)/(a^(1/2)+cos(d*x+c)*(a^(1/2)+I*a^(1/2)*tan(d*x+c))))/d/e^(3/2)-4/ 
3*I*a*(a+I*a*tan(d*x+c))^(3/2)/d/(e*sec(d*x+c))^(3/2)
 

Mathematica [A] (warning: unable to verify)

Time = 3.55 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.22 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{3/2}} \, dx=\frac {e \left (-\frac {4}{3} i \cos (d x) (\cos (c)-i \sin (c))+\frac {4}{3} (\cos (c)-i \sin (c)) \sin (d x)+\frac {2 \left (\text {arctanh}\left (\frac {\sqrt {1-i \cos (c)+\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1-i \cos (c)-\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \sqrt {-1-i \cos (c)-\sin (c)} \sqrt {1+i \cos (c)-\sin (c)}-\text {arctanh}\left (\frac {\sqrt {1+i \cos (c)-\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1+i \cos (c)+\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \sqrt {1-i \cos (c)+\sin (c)} \sqrt {-1+i \cos (c)+\sin (c)}\right ) (\cos (2 c)-i \sin (2 c)) \sqrt {i+\tan \left (\frac {d x}{2}\right )}}{\sqrt {1+\cos (2 c)+i \sin (2 c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}\right ) (a+i a \tan (c+d x))^{5/2}}{d (e \sec (c+d x))^{5/2} (\cos (d x)+i \sin (d x))^2} \] Input:

Integrate[(a + I*a*Tan[c + d*x])^(5/2)/(e*Sec[c + d*x])^(3/2),x]
 

Output:

(e*(((-4*I)/3)*Cos[d*x]*(Cos[c] - I*Sin[c]) + (4*(Cos[c] - I*Sin[c])*Sin[d 
*x])/3 + (2*(ArcTanh[(Sqrt[1 - I*Cos[c] + Sin[c]]*Sqrt[I - Tan[(d*x)/2]])/ 
(Sqrt[-1 - I*Cos[c] - Sin[c]]*Sqrt[I + Tan[(d*x)/2]])]*Sqrt[-1 - I*Cos[c] 
- Sin[c]]*Sqrt[1 + I*Cos[c] - Sin[c]] - ArcTanh[(Sqrt[1 + I*Cos[c] - Sin[c 
]]*Sqrt[I - Tan[(d*x)/2]])/(Sqrt[-1 + I*Cos[c] + Sin[c]]*Sqrt[I + Tan[(d*x 
)/2]])]*Sqrt[1 - I*Cos[c] + Sin[c]]*Sqrt[-1 + I*Cos[c] + Sin[c]])*(Cos[2*c 
] - I*Sin[2*c])*Sqrt[I + Tan[(d*x)/2]])/(Sqrt[1 + Cos[2*c] + I*Sin[2*c]]*S 
qrt[I - Tan[(d*x)/2]]))*(a + I*a*Tan[c + d*x])^(5/2))/(d*(e*Sec[c + d*x])^ 
(5/2)*(Cos[d*x] + I*Sin[d*x])^2)
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 372, normalized size of antiderivative = 1.33, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3977, 3042, 3976, 826, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{3/2}}dx\)

\(\Big \downarrow \) 3977

\(\displaystyle -\frac {a^2 \int \sqrt {e \sec (c+d x)} \sqrt {i \tan (c+d x) a+a}dx}{e^2}-\frac {4 i a (a+i a \tan (c+d x))^{3/2}}{3 d (e \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a^2 \int \sqrt {e \sec (c+d x)} \sqrt {i \tan (c+d x) a+a}dx}{e^2}-\frac {4 i a (a+i a \tan (c+d x))^{3/2}}{3 d (e \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3976

\(\displaystyle \frac {4 i a^3 \int \frac {\cos (c+d x) (i \tan (c+d x) a+a)}{e \left (a^2+\cos ^2(c+d x) (i \tan (c+d x) a+a)^2\right )}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{d}-\frac {4 i a (a+i a \tan (c+d x))^{3/2}}{3 d (e \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 826

\(\displaystyle \frac {4 i a^3 \left (\frac {\int \frac {a+\cos (c+d x) (i \tan (c+d x) a+a)}{a^2+\cos ^2(c+d x) (i \tan (c+d x) a+a)^2}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 e}-\frac {\int \frac {a-\cos (c+d x) (i \tan (c+d x) a+a)}{a^2+\cos ^2(c+d x) (i \tan (c+d x) a+a)^2}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 e}\right )}{d}-\frac {4 i a (a+i a \tan (c+d x))^{3/2}}{3 d (e \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {4 i a^3 \left (\frac {\frac {\int \frac {1}{\frac {a}{e}-\frac {\sqrt {2} \sqrt {i \tan (c+d x) a+a} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (i \tan (c+d x) a+a)}{e}}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 e}+\frac {\int \frac {1}{\frac {a}{e}+\frac {\sqrt {2} \sqrt {i \tan (c+d x) a+a} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (i \tan (c+d x) a+a)}{e}}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 e}}{2 e}-\frac {\int \frac {a-\cos (c+d x) (i \tan (c+d x) a+a)}{a^2+\cos ^2(c+d x) (i \tan (c+d x) a+a)^2}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 e}\right )}{d}-\frac {4 i a (a+i a \tan (c+d x))^{3/2}}{3 d (e \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {4 i a^3 \left (\frac {\frac {\int \frac {1}{-\frac {\cos (c+d x) (i \tan (c+d x) a+a)}{e}-1}d\left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {i \tan (c+d x) a+a}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\int \frac {1}{-\frac {\cos (c+d x) (i \tan (c+d x) a+a)}{e}-1}d\left (\frac {\sqrt {2} \sqrt {e} \sqrt {i \tan (c+d x) a+a}}{\sqrt {a} \sqrt {e \sec (c+d x)}}+1\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}-\frac {\int \frac {a-\cos (c+d x) (i \tan (c+d x) a+a)}{a^2+\cos ^2(c+d x) (i \tan (c+d x) a+a)^2}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 e}\right )}{d}-\frac {4 i a (a+i a \tan (c+d x))^{3/2}}{3 d (e \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {4 i a^3 \left (\frac {\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}-\frac {\int \frac {a-\cos (c+d x) (i \tan (c+d x) a+a)}{a^2+\cos ^2(c+d x) (i \tan (c+d x) a+a)^2}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 e}\right )}{d}-\frac {4 i a (a+i a \tan (c+d x))^{3/2}}{3 d (e \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {4 i a^3 \left (\frac {\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}-\frac {-\frac {\int -\frac {\sqrt {2} \sqrt {a}-\frac {2 \sqrt {e} \sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{\sqrt {e} \left (\frac {a}{e}-\frac {\sqrt {2} \sqrt {i \tan (c+d x) a+a} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (i \tan (c+d x) a+a)}{e}\right )}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 \sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {a}+\frac {\sqrt {2} \sqrt {e} \sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}\right )}{\sqrt {e} \left (\frac {a}{e}+\frac {\sqrt {2} \sqrt {i \tan (c+d x) a+a} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (i \tan (c+d x) a+a)}{e}\right )}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 \sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}\right )}{d}-\frac {4 i a (a+i a \tan (c+d x))^{3/2}}{3 d (e \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {4 i a^3 \left (\frac {\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}-\frac {\frac {\int \frac {\sqrt {2} \sqrt {a}-\frac {2 \sqrt {e} \sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{\sqrt {e} \left (\frac {a}{e}-\frac {\sqrt {2} \sqrt {i \tan (c+d x) a+a} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (i \tan (c+d x) a+a)}{e}\right )}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 \sqrt {2} \sqrt {a} \sqrt {e}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {a}+\frac {\sqrt {2} \sqrt {e} \sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}\right )}{\sqrt {e} \left (\frac {a}{e}+\frac {\sqrt {2} \sqrt {i \tan (c+d x) a+a} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (i \tan (c+d x) a+a)}{e}\right )}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 \sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}\right )}{d}-\frac {4 i a (a+i a \tan (c+d x))^{3/2}}{3 d (e \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {4 i a^3 \left (\frac {\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}-\frac {\frac {\int \frac {\sqrt {2} \sqrt {a}-\frac {2 \sqrt {e} \sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{\frac {a}{e}-\frac {\sqrt {2} \sqrt {i \tan (c+d x) a+a} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (i \tan (c+d x) a+a)}{e}}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 \sqrt {2} \sqrt {a} e}+\frac {\int \frac {\sqrt {a}+\frac {\sqrt {2} \sqrt {e} \sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{\frac {a}{e}+\frac {\sqrt {2} \sqrt {i \tan (c+d x) a+a} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (i \tan (c+d x) a+a)}{e}}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 \sqrt {a} e}}{2 e}\right )}{d}-\frac {4 i a (a+i a \tan (c+d x))^{3/2}}{3 d (e \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {4 i a^3 \left (\frac {\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}-\frac {\frac {\log \left (\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\log \left (-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}\right )}{d}-\frac {4 i a (a+i a \tan (c+d x))^{3/2}}{3 d (e \sec (c+d x))^{3/2}}\)

Input:

Int[(a + I*a*Tan[c + d*x])^(5/2)/(e*Sec[c + d*x])^(3/2),x]
 

Output:

((4*I)*a^3*((-(ArcTan[1 - (Sqrt[2]*Sqrt[e]*Sqrt[a + I*a*Tan[c + d*x]])/(Sq 
rt[a]*Sqrt[e*Sec[c + d*x]])]/(Sqrt[2]*Sqrt[a]*Sqrt[e])) + ArcTan[1 + (Sqrt 
[2]*Sqrt[e]*Sqrt[a + I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c + d*x]])]/(S 
qrt[2]*Sqrt[a]*Sqrt[e]))/(2*e) - (-1/2*Log[a - (Sqrt[2]*Sqrt[a]*Sqrt[e]*Sq 
rt[a + I*a*Tan[c + d*x]])/Sqrt[e*Sec[c + d*x]] + Cos[c + d*x]*(a + I*a*Tan 
[c + d*x])]/(Sqrt[2]*Sqrt[a]*Sqrt[e]) + Log[a + (Sqrt[2]*Sqrt[a]*Sqrt[e]*S 
qrt[a + I*a*Tan[c + d*x]])/Sqrt[e*Sec[c + d*x]] + Cos[c + d*x]*(a + I*a*Ta 
n[c + d*x])]/(2*Sqrt[2]*Sqrt[a]*Sqrt[e]))/(2*e)))/d - (((4*I)/3)*a*(a + I* 
a*Tan[c + d*x])^(3/2))/(d*(e*Sec[c + d*x])^(3/2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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

rule 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ 
a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] 
 && AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3976

Int[Sqrt[(d_.)*sec[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.) 
*(x_)]], x_Symbol] :> Simp[-4*b*(d^2/f)   Subst[Int[x^2/(a^2 + d^2*x^4), x] 
, x, Sqrt[a + b*Tan[e + f*x]]/Sqrt[d*Sec[e + f*x]]], x] /; FreeQ[{a, b, d, 
e, f}, x] && EqQ[a^2 + b^2, 0]
 

rule 3977

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x 
_)])^(n_), x_Symbol] :> Simp[2*b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^( 
n - 1)/(f*m)), x] - Simp[b^2*((m + 2*n - 2)/(d^2*m))   Int[(d*Sec[e + f*x]) 
^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] 
&& EqQ[a^2 + b^2, 0] && GtQ[n, 1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || 
 (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILtQ[m, 0] & 
& LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) 
&& IntegerQ[2*m]
 

Maple [A] (warning: unable to verify)

Time = 7.65 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.06

method	result	size
default	\(\frac {\left (-\frac {i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )-\sin \left (d x +c \right )+1\right ) \operatorname {arctanh}\left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )}{2}+\frac {i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+\sin \left (d x +c \right )+1\right ) \operatorname {arctanh}\left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )}{2}-\frac {4 i \cos \left (d x +c \right )}{3}-\frac {\sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+\sin \left (d x +c \right )+1\right ) \operatorname {arctanh}\left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )}{2}-\frac {\sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )-\sin \left (d x +c \right )+1\right ) \operatorname {arctanh}\left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )}{2}+\frac {4 \sin \left (d x +c \right )}{3}\right ) a^{2} \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}{d \sqrt {e \sec \left (d x +c \right )}\, e}\)	\(296\)

Input:

int((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/d*(-1/2*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)-sin(d*x+c)+1)*arctanh(1/2 
/(1/(cos(d*x+c)+1))^(1/2)*(-cot(d*x+c)+csc(d*x+c)+1))+1/2*I*(1/(cos(d*x+c) 
+1))^(1/2)*(cos(d*x+c)+sin(d*x+c)+1)*arctanh(1/2*(-cot(d*x+c)+csc(d*x+c)-1 
)/(1/(cos(d*x+c)+1))^(1/2))-4/3*I*cos(d*x+c)-1/2*(1/(cos(d*x+c)+1))^(1/2)* 
(cos(d*x+c)+sin(d*x+c)+1)*arctanh(1/2/(1/(cos(d*x+c)+1))^(1/2)*(-cot(d*x+c 
)+csc(d*x+c)+1))-1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)-sin(d*x+c)+1)*ar 
ctanh(1/2*(-cot(d*x+c)+csc(d*x+c)-1)/(1/(cos(d*x+c)+1))^(1/2))+4/3*sin(d*x 
+c))*a^2*(a*(1+I*tan(d*x+c)))^(1/2)/(e*sec(d*x+c))^(1/2)/e
 

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. 505 vs. \(2 (204) = 408\).

Time = 0.12 (sec) , antiderivative size = 505, normalized size of antiderivative = 1.80 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{3/2}} \, dx=-\frac {3 \, d e^{2} \sqrt {\frac {4 i \, a^{5}}{d^{2} e^{3}}} \log \left (\frac {d e^{2} \sqrt {\frac {4 i \, a^{5}}{d^{2} e^{3}}} + 2 \, {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{a^{2}}\right ) - 3 \, d e^{2} \sqrt {\frac {4 i \, a^{5}}{d^{2} e^{3}}} \log \left (-\frac {d e^{2} \sqrt {\frac {4 i \, a^{5}}{d^{2} e^{3}}} - 2 \, {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{a^{2}}\right ) - 3 \, d e^{2} \sqrt {-\frac {4 i \, a^{5}}{d^{2} e^{3}}} \log \left (\frac {d e^{2} \sqrt {-\frac {4 i \, a^{5}}{d^{2} e^{3}}} + 2 \, {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{a^{2}}\right ) + 3 \, d e^{2} \sqrt {-\frac {4 i \, a^{5}}{d^{2} e^{3}}} \log \left (-\frac {d e^{2} \sqrt {-\frac {4 i \, a^{5}}{d^{2} e^{3}}} - 2 \, {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{a^{2}}\right ) + 8 \, {\left (i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{6 \, d e^{2}} \] Input:

integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(3/2),x, algorithm="fric 
as")
 

Output:

-1/6*(3*d*e^2*sqrt(4*I*a^5/(d^2*e^3))*log((d*e^2*sqrt(4*I*a^5/(d^2*e^3)) + 
 2*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt( 
e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))/a^2) - 3*d*e^2*sqrt( 
4*I*a^5/(d^2*e^3))*log(-(d*e^2*sqrt(4*I*a^5/(d^2*e^3)) - 2*(a^2*e^(2*I*d*x 
 + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2* 
I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))/a^2) - 3*d*e^2*sqrt(-4*I*a^5/(d^2*e^3) 
)*log((d*e^2*sqrt(-4*I*a^5/(d^2*e^3)) + 2*(a^2*e^(2*I*d*x + 2*I*c) + a^2)* 
sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2 
*I*d*x + 1/2*I*c))/a^2) + 3*d*e^2*sqrt(-4*I*a^5/(d^2*e^3))*log(-(d*e^2*sqr 
t(-4*I*a^5/(d^2*e^3)) - 2*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d 
*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c 
))/a^2) + 8*(I*a^2*e^(3*I*d*x + 3*I*c) + I*a^2*e^(I*d*x + I*c))*sqrt(a/(e^ 
(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1 
/2*I*c))/(d*e^2)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((a+I*a*tan(d*x+c))**(5/2)/(e*sec(d*x+c))**(3/2),x)
 

Output:

Timed out
 

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. 1492 vs. \(2 (204) = 408\).

Time = 0.33 (sec) , antiderivative size = 1492, normalized size of antiderivative = 5.33 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(3/2),x, algorithm="maxi 
ma")
 

Output:

-1/12*(-6*I*sqrt(2)*a^2*arctan2(sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), 
cos(2*d*x + 2*c))) + 1, sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d* 
x + 2*c))) + 1) - 6*I*sqrt(2)*a^2*arctan2(sqrt(2)*cos(1/4*arctan2(sin(2*d* 
x + 2*c), cos(2*d*x + 2*c))) + 1, -sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c 
), cos(2*d*x + 2*c))) + 1) - 6*I*sqrt(2)*a^2*arctan2(sqrt(2)*cos(1/4*arcta 
n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 1, sqrt(2)*sin(1/4*arctan2(sin(2 
*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - 6*I*sqrt(2)*a^2*arctan2(sqrt(2)*cos 
(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 1, -sqrt(2)*sin(1/4*ar 
ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - 6*sqrt(2)*a^2*arctan2(sq 
rt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arcta 
n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))), sqrt(2)*cos(1/4*arctan2(sin(2*d*x 
 + 2*c), cos(2*d*x + 2*c))) + cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x 
+ 2*c))) + 1) + 6*sqrt(2)*a^2*arctan2(-sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 
 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 
2*c))), -sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + co 
s(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + 3*I*sqrt(2)*a^2* 
log(2*sqrt(2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/4 
*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2*(sqrt(2)*cos(1/4*arctan2 
(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1)*cos(1/2*arctan2(sin(2*d*x + 2*c 
), cos(2*d*x + 2*c))) + cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2...
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(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
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{3/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:

int((a + a*tan(c + d*x)*1i)^(5/2)/(e/cos(c + d*x))^(3/2),x)
 

Output:

int((a + a*tan(c + d*x)*1i)^(5/2)/(e/cos(c + d*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{3/2}} \, dx=\frac {\sqrt {e}\, \sqrt {a}\, a^{2} \left (-\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{2}}d x \right )+2 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right )^{2}}d x \right ) i +\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\sec \left (d x +c \right )^{2}}d x \right )}{e^{2}} \] Input:

int((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(3/2),x)
 

Output:

(sqrt(e)*sqrt(a)*a**2*( - int((sqrt(sec(c + d*x))*sqrt(tan(c + d*x)*i + 1) 
*tan(c + d*x)**2)/sec(c + d*x)**2,x) + 2*int((sqrt(sec(c + d*x))*sqrt(tan( 
c + d*x)*i + 1)*tan(c + d*x))/sec(c + d*x)**2,x)*i + int((sqrt(sec(c + d*x 
))*sqrt(tan(c + d*x)*i + 1))/sec(c + d*x)**2,x)))/e**2