3.2.83 \(\int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx\) [183]

3.2.83.1 Optimal result
3.2.83.2 Mathematica [A] (verified)
3.2.83.3 Rubi [A] (verified)
3.2.83.4 Maple [B] (verified)
3.2.83.5 Fricas [B] (verification not implemented)
3.2.83.6 Sympy [F]
3.2.83.7 Maxima [F(-2)]
3.2.83.8 Giac [F(-2)]
3.2.83.9 Mupad [F(-1)]

3.2.83.1 Optimal result

Integrand size = 38, antiderivative size = 191 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (i A+B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {A+i B}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(5 A+3 i B) \sqrt {a+i a \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(7 i A-9 B) \sqrt {a+i a \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}} \]

output
(1/2+1/2*I)*(I*A+B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+ 
c))^(1/2))/d/a^(1/2)+1/3*(7*I*A-9*B)*(a+I*a*tan(d*x+c))^(1/2)/a/d/tan(d*x+ 
c)^(1/2)+(A+I*B)/d/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(3/2)-1/3*(5*A+3*I* 
B)*(a+I*a*tan(d*x+c))^(1/2)/a/d/tan(d*x+c)^(3/2)
 
3.2.83.2 Mathematica [A] (verified)

Time = 3.31 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.79 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {-\frac {3 \sqrt {2} (A-i B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \tan ^2(c+d x)}{\sqrt {i a \tan (c+d x)}}-\frac {2 \left (2 A+(-2 i A+6 B) \tan (c+d x)+(7 A+9 i B) \tan ^2(c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}}}{6 d \tan ^{\frac {3}{2}}(c+d x)} \]

input
Integrate[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d* 
x]]),x]
 
output
((-3*Sqrt[2]*(A - I*B)*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I 
*a*Tan[c + d*x]]]*Tan[c + d*x]^2)/Sqrt[I*a*Tan[c + d*x]] - (2*(2*A + ((-2* 
I)*A + 6*B)*Tan[c + d*x] + (7*A + (9*I)*B)*Tan[c + d*x]^2))/Sqrt[a + I*a*T 
an[c + d*x]])/(6*d*Tan[c + d*x]^(3/2))
 
3.2.83.3 Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 4079, 27, 3042, 4081, 27, 3042, 4081, 27, 3042, 4027, 218}

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.

\(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan (c+d x)^{5/2} \sqrt {a+i a \tan (c+d x)}}dx\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\int \frac {\sqrt {i \tan (c+d x) a+a} (a (5 A+3 i B)-4 a (i A-B) \tan (c+d x))}{2 \tan ^{\frac {5}{2}}(c+d x)}dx}{a^2}+\frac {A+i B}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sqrt {i \tan (c+d x) a+a} (a (5 A+3 i B)-4 a (i A-B) \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)}dx}{2 a^2}+\frac {A+i B}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sqrt {i \tan (c+d x) a+a} (a (5 A+3 i B)-4 a (i A-B) \tan (c+d x))}{\tan (c+d x)^{5/2}}dx}{2 a^2}+\frac {A+i B}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 4081

\(\displaystyle \frac {\frac {2 \int -\frac {\sqrt {i \tan (c+d x) a+a} \left ((7 i A-9 B) a^2+2 (5 A+3 i B) \tan (c+d x) a^2\right )}{2 \tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {2 a (5 A+3 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {A+i B}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((7 i A-9 B) a^2+2 (5 A+3 i B) \tan (c+d x) a^2\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {2 a (5 A+3 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {A+i B}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((7 i A-9 B) a^2+2 (5 A+3 i B) \tan (c+d x) a^2\right )}{\tan (c+d x)^{3/2}}dx}{3 a}-\frac {2 a (5 A+3 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {A+i B}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 4081

\(\displaystyle \frac {-\frac {\frac {2 \int \frac {3 a^3 (A-i B) \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a}-\frac {2 a^2 (-9 B+7 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 a (5 A+3 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {A+i B}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {3 a^2 (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^2 (-9 B+7 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 a (5 A+3 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {A+i B}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {3 a^2 (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^2 (-9 B+7 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 a (5 A+3 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {A+i B}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 4027

\(\displaystyle \frac {-\frac {-\frac {6 i a^4 (A-i B) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {2 a^2 (-9 B+7 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 a (5 A+3 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {A+i B}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {-\frac {\frac {(3-3 i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 (-9 B+7 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 a (5 A+3 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {A+i B}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\)

input
Int[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]),x 
]
 
output
(A + I*B)/(d*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]) + ((-2*a*(5*A 
+ (3*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(3*d*Tan[c + d*x]^(3/2)) - (((3 - 3 
*I)*a^(5/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a 
+ I*a*Tan[c + d*x]]])/d - (2*a^2*((7*I)*A - 9*B)*Sqrt[a + I*a*Tan[c + d*x] 
])/(d*Sqrt[Tan[c + d*x]]))/(3*a))/(2*a^2)
 

3.2.83.3.1 Defintions of rubi rules used

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 218
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4027
Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) 
 + (f_.)*(x_)]], x_Symbol] :> Simp[-2*a*(b/f)   Subst[Int[1/(a*c - b*d - 2* 
a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x] /; 
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && N 
eQ[c^2 + d^2, 0]
 

rule 4079
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(a*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*( 
b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d))   Int[(a + b*Tan[e + f*x])^(m 
 + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m 
- b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free 
Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] 
 && LtQ[m, 0] &&  !GtQ[n, 0]
 

rule 4081
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(A*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*(n + 
1)*(c^2 + d^2))), x] - Simp[1/(a*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + 
f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c* 
m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Fr 
eeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 
0] && LtQ[n, -1]
 
3.2.83.4 Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 739 vs. \(2 (155 ) = 310\).

Time = 0.18 (sec) , antiderivative size = 740, normalized size of antiderivative = 3.87

method result size
derivativedivides \(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (60 i B \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{2}\left (d x +c \right )\right )-36 B \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{3}\left (d x +c \right )\right )+28 i A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{3}\left (d x +c \right )\right )+3 B \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{4}\left (d x +c \right )\right )+36 A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{2}\left (d x +c \right )\right )+3 i A \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{4}\left (d x +c \right )\right )-6 i B \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, a \left (\tan ^{3}\left (d x +c \right )\right )+6 A \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, a \left (\tan ^{3}\left (d x +c \right )\right )-3 i A \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{2}\left (d x +c \right )\right )-3 B \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{2}\left (d x +c \right )\right )+24 B \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+8 A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{12 d a \tan \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (-\tan \left (d x +c \right )+i\right )^{2} \sqrt {-i a}}\) \(740\)
default \(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (60 i B \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{2}\left (d x +c \right )\right )-36 B \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{3}\left (d x +c \right )\right )+28 i A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{3}\left (d x +c \right )\right )+3 B \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{4}\left (d x +c \right )\right )+36 A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{2}\left (d x +c \right )\right )+3 i A \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{4}\left (d x +c \right )\right )-6 i B \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, a \left (\tan ^{3}\left (d x +c \right )\right )+6 A \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, a \left (\tan ^{3}\left (d x +c \right )\right )-3 i A \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{2}\left (d x +c \right )\right )-3 B \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{2}\left (d x +c \right )\right )+24 B \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+8 A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{12 d a \tan \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (-\tan \left (d x +c \right )+i\right )^{2} \sqrt {-i a}}\) \(740\)
parts \(\frac {A \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (3 i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{4}\left (d x +c \right )\right ) a +36 \sqrt {-i a}\, \left (\tan ^{2}\left (d x +c \right )\right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-3 i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{2}\left (d x +c \right )\right ) a +28 i \left (\tan ^{3}\left (d x +c \right )\right ) \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+6 \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{3}\left (d x +c \right )\right ) a +8 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{12 d a \tan \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (-\tan \left (d x +c \right )+i\right )^{2}}-\frac {B \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (2 i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{2}\left (d x +c \right )\right ) a -\sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{3}\left (d x +c \right )\right ) a -20 i \tan \left (d x +c \right ) \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+\sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \tan \left (d x +c \right ) a +12 \sqrt {-i a}\, \left (\tan ^{2}\left (d x +c \right )\right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-8 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{4 d a \sqrt {\tan \left (d x +c \right )}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (-\tan \left (d x +c \right )+i\right )^{2} \sqrt {-i a}}\) \(792\)

input
int((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2),x,method=_R 
ETURNVERBOSE)
 
output
1/12/d*(a*(1+I*tan(d*x+c)))^(1/2)/a/tan(d*x+c)^(3/2)*(60*I*B*(-I*a)^(1/2)* 
(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2-36*B*(-I*a)^(1/2)*(a*ta 
n(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3+28*I*A*(-I*a)^(1/2)*(a*tan(d 
*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3+3*B*2^(1/2)*ln((2*2^(1/2)*(-I*a 
)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x 
+c)+I))*a*tan(d*x+c)^4+36*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^( 
1/2)*tan(d*x+c)^2+3*I*A*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*( 
1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)^4- 
6*I*B*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^( 
1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)^3+6*A*ln((2*2^(1/2)* 
(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(ta 
n(d*x+c)+I))*2^(1/2)*a*tan(d*x+c)^3-3*I*A*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/ 
2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I 
))*a*tan(d*x+c)^2-3*B*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+ 
I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)^2+24 
*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)+8*A*(-I*a 
)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+ 
c)))^(1/2)/(-tan(d*x+c)+I)^2/(-I*a)^(1/2)
 
3.2.83.5 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. 534 vs. \(2 (145) = 290\).

Time = 0.28 (sec) , antiderivative size = 534, normalized size of antiderivative = 2.80 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {3 \, \sqrt {2} {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} - 2 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a d^{2}}} \log \left (\frac {i \, \sqrt {2} a d \sqrt {-\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) - 3 \, \sqrt {2} {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} - 2 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a d^{2}}} \log \left (\frac {-i \, \sqrt {2} a d \sqrt {-\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) + 2 \, \sqrt {2} {\left ({\left (7 \, A + 15 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - {\left (11 \, A + 3 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 15 \, {\left (A + i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, A + 3 i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{12 \, {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} - 2 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )}} \]

input
integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2),x, al 
gorithm="fricas")
 
output
-1/12*(3*sqrt(2)*(a*d*e^(5*I*d*x + 5*I*c) - 2*a*d*e^(3*I*d*x + 3*I*c) + a* 
d*e^(I*d*x + I*c))*sqrt(-(I*A^2 + 2*A*B - I*B^2)/(a*d^2))*log((I*sqrt(2)*a 
*d*sqrt(-(I*A^2 + 2*A*B - I*B^2)/(a*d^2))*e^(I*d*x + I*c) + sqrt(2)*((I*A 
+ B)*e^(2*I*d*x + 2*I*c) + I*A + B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt 
((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))/(4*I*A + 4*B)) - 
 3*sqrt(2)*(a*d*e^(5*I*d*x + 5*I*c) - 2*a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I 
*d*x + I*c))*sqrt(-(I*A^2 + 2*A*B - I*B^2)/(a*d^2))*log((-I*sqrt(2)*a*d*sq 
rt(-(I*A^2 + 2*A*B - I*B^2)/(a*d^2))*e^(I*d*x + I*c) + sqrt(2)*((I*A + B)* 
e^(2*I*d*x + 2*I*c) + I*A + B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I* 
e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))/(4*I*A + 4*B)) + 2*sq 
rt(2)*((7*A + 15*I*B)*e^(6*I*d*x + 6*I*c) - (11*A + 3*I*B)*e^(4*I*d*x + 4* 
I*c) - 15*(A + I*B)*e^(2*I*d*x + 2*I*c) + 3*A + 3*I*B)*sqrt(a/(e^(2*I*d*x 
+ 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1) 
))/(a*d*e^(5*I*d*x + 5*I*c) - 2*a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I 
*c))
 
3.2.83.6 Sympy [F]

\[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {A + B \tan {\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]

input
integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**(1/2)/tan(d*x+c)**(5/2),x)
 
output
Integral((A + B*tan(c + d*x))/(sqrt(I*a*(tan(c + d*x) - I))*tan(c + d*x)** 
(5/2)), x)
 
3.2.83.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: RuntimeError} \]

input
integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2),x, al 
gorithm="maxima")
 
output
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 
3.2.83.8 Giac [F(-2)]

Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: NotImplementedError} \]

input
integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2),x, al 
gorithm="giac")
 
output
Exception raised: NotImplementedError >> unable to parse Giac output: 2*(( 
9*i*sageVARa*sageVARB*sageVARd+6*sageVARa*sageVARd*sageVARA)/9/sageVARa^3/ 
sageVARd^2*sqrt(i*sageVARa*tan(sageVARc+sageVARd*sageVARx)+sageVARa)*sqrt( 
i*sageVARa*tan(sage
 
3.2.83.9 Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \]

input
int((A + B*tan(c + d*x))/(tan(c + d*x)^(5/2)*(a + a*tan(c + d*x)*1i)^(1/2) 
),x)
 
output
int((A + B*tan(c + d*x))/(tan(c + d*x)^(5/2)*(a + a*tan(c + d*x)*1i)^(1/2) 
), x)