3.4.28 \(\int \cot (c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\) [328]

3.4.28.1 Optimal result

Integrand size = 31, antiderivative size = 152 \[ \int \cot (c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {2 a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {(a-i b)^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{3/2} (A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 b B \sqrt {a+b \tan (c+d x)}}{d} \]

output

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

3.4.28.2 Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.95 \[ \int \cot (c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {-2 a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )+(a-i b)^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )+(a+i b)^{3/2} (A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+2 b B \sqrt {a+b \tan (c+d x)}}{d} \]

input

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

output

(-2*a^(3/2)*A*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]] + (a - I*b)^(3/2)* 
(A - I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] + (a + I*b)^(3/2 
)*(A + I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] + 2*b*B*Sqrt[a 
 + b*Tan[c + d*x]])/d
 

3.4.28.3 Rubi [A] (warning: unable to verify)

Time = 1.30 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.91, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.484, Rules used = {3042, 4090, 27, 3042, 4136, 3042, 4022, 3042, 4020, 25, 73, 221, 4117, 73, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4090

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4136

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

\(\displaystyle a^2 A \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx+\frac {(a-i b)^{3/2} (B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{3/2} (-B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}+\frac {2 b B \sqrt {a+b \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {a^2 A \int \frac {\cot (c+d x)}{\sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{d}+\frac {(a-i b)^{3/2} (B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{3/2} (-B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}+\frac {2 b B \sqrt {a+b \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 a^2 A \int \frac {1}{\frac {a+b \tan (c+d x)}{b}-\frac {a}{b}}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {(a-i b)^{3/2} (B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{3/2} (-B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}+\frac {2 b B \sqrt {a+b \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 221

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

input

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

output

((a - I*b)^(3/2)*(I*A + B)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/d - ((a + I 
*b)^(3/2)*(I*A - B)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/d - (2*a^(3/2)*A*A 
rcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]])/d + (2*b*B*Sqrt[a + b*Tan[c + d* 
x]])/d
 

3.4.28.3.1 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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

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

rule 3042

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

rule 4020

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

rule 4022

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2   Int[(a + b*Tan[e + f*x])^m*( 
1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2   Int[(a + b*Tan[e + f*x])^m 
*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c 
 - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]
 

rule 4090

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*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + 
 n))), x] + Simp[1/(d*(m + n))   Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Ta 
n[e + f*x])^n*Simp[a^2*A*d*(m + n) - b*B*(b*c*(m - 1) + a*d*(n + 1)) + d*(m 
 + n)*(2*a*A*b + B*(a^2 - b^2))*Tan[e + f*x] - (b*B*(b*c - a*d)*(m - 1) - b 
*(A*b + a*B)*d*(m + n))*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] && GtQ[m, 1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 1] 
&& ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

rule 4117

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

rule 4136

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) 
+ (f_.)*(x_)]), x_Symbol] :> Simp[1/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^ 
n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Simp[ 
(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^n*((1 + Tan[ 
e + f*x]^2)/(a + b*Tan[e + f*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] & 
&  !GtQ[n, 0] &&  !LeQ[n, -1]
 

3.4.28.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1653\) vs. \(2(126)=252\).

Time = 0.24 (sec) , antiderivative size = 1654, normalized size of antiderivative = 10.88

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1654\)
default	\(\text {Expression too large to display}\)	\(1654\)

input

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

output

2*b*B*(a+b*tan(d*x+c))^(1/2)/d-1/d*b^2/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arcta 
n(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1 
/2)-2*a)^(1/2))*A+1/d*b^2/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan 
(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2 
))*A+1/4/d*b*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan 
(d*x+c)-a-(a^2+b^2)^(1/2))*B*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-1/4/d*b*ln(b*ta 
n(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^ 
(1/2))*B*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+1/4/d/b*ln((a+b*tan(d*x+c))^(1/2)*( 
2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*B*(2*(a^2+b^2 
)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*a-1/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arc 
tan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^ 
(1/2)-2*a)^(1/2))*A*(a^2+b^2)^(1/2)*a+1/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*ar 
ctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2) 
^(1/2)-2*a)^(1/2))*A*(a^2+b^2)^(1/2)*a-1/4/d/b*ln(b*tan(d*x+c)+a+(a+b*tan( 
d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*B*(2*(a^2+b^2 
)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*a-1/4/d*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+ 
c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*A*(2*(a^2+b^2)^(1 
/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)+1/2/d*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1 
/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*A*(2*(a^2+b^2)^(1/2)+2* 
a)^(1/2)*a+1/4/d/b*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2...
 

3.4.28.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3075 vs. \(2 (120) = 240\).

Time = 2.67 (sec) , antiderivative size = 6165, normalized size of antiderivative = 40.56 \[ \int \cot (c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

input

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

output

Too large to include
 

3.4.28.6 Sympy [F]

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

input

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

output

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

3.4.28.7 Maxima [F]

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

input

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

output

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

3.4.28.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.4.28.9 Mupad [B] (verification not implemented)

Time = 11.80 (sec) , antiderivative size = 20255, normalized size of antiderivative = 133.26 \[ \int \cot (c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

input

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

output

(2*B*b*(a + b*tan(c + d*x))^(1/2))/d - atan(((((((32*(4*B*a*b^11*d^4 + 12* 
A*a^2*b^10*d^4 + 12*A*a^4*b^8*d^4 + 4*B*a^3*b^9*d^4))/d^5 - (32*(16*b^10*d 
^4 + 24*a^2*b^8*d^4)*(a + b*tan(c + d*x))^(1/2)*(-(((8*A^2*a^3*d^2 - 8*B^2 
*a^3*d^2 + 16*A*B*b^3*d^2 - 24*A^2*a*b^2*d^2 + 24*B^2*a*b^2*d^2 - 48*A*B*a 
^2*b*d^2)^2/64 - d^4*(A^4*a^6 + A^4*b^6 + B^4*a^6 + B^4*b^6 + 2*A^2*B^2*a^ 
6 + 2*A^2*B^2*b^6 + 3*A^4*a^2*b^4 + 3*A^4*a^4*b^2 + 3*B^4*a^2*b^4 + 3*B^4* 
a^4*b^2 + 6*A^2*B^2*a^2*b^4 + 6*A^2*B^2*a^4*b^2))^(1/2) - A^2*a^3*d^2 + B^ 
2*a^3*d^2 - 2*A*B*b^3*d^2 + 3*A^2*a*b^2*d^2 - 3*B^2*a*b^2*d^2 + 6*A*B*a^2* 
b*d^2)/(4*d^4))^(1/2))/d^4)*(-(((8*A^2*a^3*d^2 - 8*B^2*a^3*d^2 + 16*A*B*b^ 
3*d^2 - 24*A^2*a*b^2*d^2 + 24*B^2*a*b^2*d^2 - 48*A*B*a^2*b*d^2)^2/64 - d^4 
*(A^4*a^6 + A^4*b^6 + B^4*a^6 + B^4*b^6 + 2*A^2*B^2*a^6 + 2*A^2*B^2*b^6 + 
3*A^4*a^2*b^4 + 3*A^4*a^4*b^2 + 3*B^4*a^2*b^4 + 3*B^4*a^4*b^2 + 6*A^2*B^2* 
a^2*b^4 + 6*A^2*B^2*a^4*b^2))^(1/2) - A^2*a^3*d^2 + B^2*a^3*d^2 - 2*A*B*b^ 
3*d^2 + 3*A^2*a*b^2*d^2 - 3*B^2*a*b^2*d^2 + 6*A*B*a^2*b*d^2)/(4*d^4))^(1/2 
) - (32*(a + b*tan(c + d*x))^(1/2)*(28*A^2*a^3*b^10*d^2 - 18*A^2*a^5*b^8*d 
^2 - 28*B^2*a^3*b^10*d^2 + 10*B^2*a^5*b^8*d^2 - 16*A*B*b^13*d^2 + 22*A^2*a 
*b^12*d^2 - 22*B^2*a*b^12*d^2 + 16*A*B*a^2*b^11*d^2 + 64*A*B*a^4*b^9*d^2)) 
/d^4)*(-(((8*A^2*a^3*d^2 - 8*B^2*a^3*d^2 + 16*A*B*b^3*d^2 - 24*A^2*a*b^2*d 
^2 + 24*B^2*a*b^2*d^2 - 48*A*B*a^2*b*d^2)^2/64 - d^4*(A^4*a^6 + A^4*b^6 + 
B^4*a^6 + B^4*b^6 + 2*A^2*B^2*a^6 + 2*A^2*B^2*b^6 + 3*A^4*a^2*b^4 + 3*A...