3.5.61 \(\int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx\) [461]

3.5.61.1 Optimal result

Integrand size = 35, antiderivative size = 216 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\frac {(A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{3/2} d}-\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{3/2} d}-\frac {2 A}{a d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {2 b \left (a^2 A+2 A b^2-a b B\right ) \sqrt {\tan (c+d x)}}{a^2 \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}} \]

output

(A+I*B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/(I*a 
-b)^(3/2)/d-(A-I*B)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c) 
)^(1/2))/(I*a+b)^(3/2)/d-2*A/a/d/tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2)-2 
*b*(A*a^2+2*A*b^2-B*a*b)*tan(d*x+c)^(1/2)/a^2/(a^2+b^2)/d/(a+b*tan(d*x+c)) 
^(1/2)
 

3.5.61.2 Mathematica [A] (verified)

Time = 2.33 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.15 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\frac {\frac {\sqrt [4]{-1} \left (\frac {(a+i b) (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}-\frac {(a-i b) (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}\right )}{a^2+b^2}-\frac {2 A}{a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {2 b \left (a^2 A+2 A b^2-a b B\right ) \sqrt {\tan (c+d x)}}{a^2 \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{d} \]

input

Integrate[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^(3 
/2)),x]
 

output

(((-1)^(1/4)*(((a + I*b)*(A - I*B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[ 
Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[-a + I*b] - ((a - I*b)*(A + 
 I*B)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[ 
c + d*x]]])/Sqrt[a + I*b]))/(a^2 + b^2) - (2*A)/(a*Sqrt[Tan[c + d*x]]*Sqrt 
[a + b*Tan[c + d*x]]) - (2*b*(a^2*A + 2*A*b^2 - a*b*B)*Sqrt[Tan[c + d*x]]) 
/(a^2*(a^2 + b^2)*Sqrt[a + b*Tan[c + d*x]]))/d
 

3.5.61.3 Rubi [A] (verified)

Time = 1.32 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.19, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 4092, 27, 3042, 4132, 27, 3042, 4099, 3042, 4098, 104, 216, 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.

\(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan (c+d x)^{3/2} (a+b \tan (c+d x))^{3/2}}dx\)

\(\Big \downarrow \) 4092

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4132

\(\displaystyle -\frac {\frac {2 \int \frac {(A b-a B) a^2+(a A+b B) \tan (c+d x) a^2}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}+\frac {2 b \left (a^2 A-a b B+2 A b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{a}-\frac {2 A}{a d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\int \frac {(A b-a B) a^2+(a A+b B) \tan (c+d x) a^2}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}+\frac {2 b \left (a^2 A-a b B+2 A b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{a}-\frac {2 A}{a d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\int \frac {(A b-a B) a^2+(a A+b B) \tan (c+d x) a^2}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}+\frac {2 b \left (a^2 A-a b B+2 A b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{a}-\frac {2 A}{a d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}\)

\(\Big \downarrow \) 4099

\(\displaystyle -\frac {2 A}{a d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {\frac {2 b \left (a^2 A-a b B+2 A b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {1}{2} a^2 (b+i a) (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} a^2 (-b+i a) (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 A}{a d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {\frac {2 b \left (a^2 A-a b B+2 A b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {1}{2} a^2 (b+i a) (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} a^2 (-b+i a) (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}}{a}\)

\(\Big \downarrow \) 4098

\(\displaystyle -\frac {2 A}{a d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {\frac {2 b \left (a^2 A-a b B+2 A b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {a^2 (b+i a) (A+i B) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}-\frac {a^2 (-b+i a) (A-i B) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}}{a \left (a^2+b^2\right )}}{a}\)

\(\Big \downarrow \) 104

\(\displaystyle -\frac {2 A}{a d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {\frac {2 b \left (a^2 A-a b B+2 A b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {a^2 (b+i a) (A+i B) \int \frac {1}{\frac {(i a-b) \tan (c+d x)}{a+b \tan (c+d x)}+1}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}-\frac {a^2 (-b+i a) (A-i B) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}}{a \left (a^2+b^2\right )}}{a}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {2 A}{a d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {\frac {2 b \left (a^2 A-a b B+2 A b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {a^2 (b+i a) (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}-\frac {a^2 (-b+i a) (A-i B) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}}{a \left (a^2+b^2\right )}}{a}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {2 A}{a d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {\frac {2 b \left (a^2 A-a b B+2 A b^2\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {a^2 (b+i a) (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}-\frac {a^2 (-b+i a) (A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}}{a \left (a^2+b^2\right )}}{a}\)

input

Int[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^(3/2)),x 
]
 

output

(-2*A)/(a*d*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]]) - (((a^2*(I*a + b 
)*(A + I*B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d 
*x]]])/(Sqrt[I*a - b]*d) - (a^2*(I*a - b)*(A - I*B)*ArcTanh[(Sqrt[I*a + b] 
*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a + b]*d))/(a*(a^2 
 + b^2)) + (2*b*(a^2*A + 2*A*b^2 - a*b*B)*Sqrt[Tan[c + d*x]])/(a*(a^2 + b^ 
2)*d*Sqrt[a + b*Tan[c + d*x]]))/a
 

3.5.61.3.1 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 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 216

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

rule 219

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])
 

rule 3042

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

rule 4092

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[b*(A*b - a*B)*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1) 
/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^ 
2 + b^2))   Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b* 
B*(b*c*(m + 1) + a*d*(n + 1)) + A*(a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2 
)) - (A*b - a*B)*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(A*b - a*B)*(m + n 
+ 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && 
 NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] 
&& (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] 
 || (EqQ[c, 0] && NeQ[a, 0])))
 

rule 4098

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[A^2/f   Subst[Int[(a + b*x)^m*((c + d*x)^n/(A - B*x)), x], x, Tan[e + f* 
x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && 
 NeQ[a^2 + b^2, 0] && EqQ[A^2 + B^2, 0]
 

rule 4099

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(A + I*B)/2   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 - I*T 
an[e + f*x]), x], x] + Simp[(A - I*B)/2   Int[(a + b*Tan[e + f*x])^m*(c + d 
*Tan[e + f*x])^n*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A 
, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2 + B^2, 
0]
 

rule 4132

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*(b*B - a*C))*(a + b*Tan[e + 
 f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + 
b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 + b^2))   Int[(a + b*Tan[e + 
f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1) - b^2*d* 
(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d 
)*(A*b - a*B - b*C)*Tan[e + f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Ta 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ 
[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] && 
!(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

3.5.61.4 Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(5197\) vs. \(2(186)=372\).

Time = 3.93 (sec) , antiderivative size = 5198, normalized size of antiderivative = 24.06

method	result	size
default	\(\text {Expression too large to display}\)	\(5198\)
parts	\(\text {Expression too large to display}\)	\(1560457\)

input

int((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(3/2),x,method=_RET 
URNVERBOSE)
 

output

result too large to display
 

3.5.61.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 18689 vs. \(2 (180) = 360\).

Time = 7.04 (sec) , antiderivative size = 18689, normalized size of antiderivative = 86.52 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

input

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(3/2),x, algo 
rithm="fricas")
 

output

Too large to include
 

3.5.61.6 Sympy [F]

\[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {A + B \tan {\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]

input

integrate((A+B*tan(d*x+c))/tan(d*x+c)**(3/2)/(a+b*tan(d*x+c))**(3/2),x)
 

output

Integral((A + B*tan(c + d*x))/((a + b*tan(c + d*x))**(3/2)*tan(c + d*x)**( 
3/2)), x)
 

3.5.61.7 Maxima [F(-1)]

Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]

input

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(3/2),x, algo 
rithm="maxima")
 

output

Timed out
 

3.5.61.8 Giac [F(-1)]

Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]

input

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(3/2),x, algo 
rithm="giac")
 

output

Timed out
 

3.5.61.9 Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\text {Hanged} \]

input

int((A + B*tan(c + d*x))/(tan(c + d*x)^(3/2)*(a + b*tan(c + d*x))^(3/2)),x 
)
 

output

\text{Hanged}