\(\int (a+b \cot (c+d x))^{3/2} (A+B \cot (c+d x)) \, dx\) [96]

Optimal result

Integrand size = 25, antiderivative size = 150 \[ \int (a+b \cot (c+d x))^{3/2} (A+B \cot (c+d x)) \, dx=\frac {(a-i b)^{3/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \cot (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{3/2} (i A-B) \text {arctanh}\left (\frac {\sqrt {a+b \cot (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {2 (A b+a B) \sqrt {a+b \cot (c+d x)}}{d}-\frac {2 B (a+b \cot (c+d x))^{3/2}}{3 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.96 \[ \int (a+b \cot (c+d x))^{3/2} (A+B \cot (c+d x)) \, dx=-\frac {\frac {3 \sqrt {a-\sqrt {-b^2}} \left (-2 a b \left (A \sqrt {-b^2}+b B\right )+a^2 \left (A b-\sqrt {-b^2} B\right )+b^2 \left (-A b+\sqrt {-b^2} B\right )\right ) \text {arctanh}\left (\frac {\sqrt {a+b \cot (c+d x)}}{\sqrt {a-\sqrt {-b^2}}}\right )}{b^2+a \sqrt {-b^2}}+\frac {3 \left (2 a b \left (-A \sqrt {-b^2}+b B\right )-a^2 \left (A b+\sqrt {-b^2} B\right )+b^2 \left (A b+\sqrt {-b^2} B\right )\right ) \text {arctanh}\left (\frac {\sqrt {a+b \cot (c+d x)}}{\sqrt {a+\sqrt {-b^2}}}\right )}{\sqrt {-b^2} \sqrt {a+\sqrt {-b^2}}}+6 (A b+a B) \sqrt {a+b \cot (c+d x)}+2 B (a+b \cot (c+d x))^{3/2}}{3 d} \] Input:

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

Output:

-1/3*((3*Sqrt[a - Sqrt[-b^2]]*(-2*a*b*(A*Sqrt[-b^2] + b*B) + a^2*(A*b - Sq 
rt[-b^2]*B) + b^2*(-(A*b) + Sqrt[-b^2]*B))*ArcTanh[Sqrt[a + b*Cot[c + d*x] 
]/Sqrt[a - Sqrt[-b^2]]])/(b^2 + a*Sqrt[-b^2]) + (3*(2*a*b*(-(A*Sqrt[-b^2]) 
 + b*B) - a^2*(A*b + Sqrt[-b^2]*B) + b^2*(A*b + Sqrt[-b^2]*B))*ArcTanh[Sqr 
t[a + b*Cot[c + d*x]]/Sqrt[a + Sqrt[-b^2]]])/(Sqrt[-b^2]*Sqrt[a + Sqrt[-b^ 
2]]) + 6*(A*b + a*B)*Sqrt[a + b*Cot[c + d*x]] + 2*B*(a + b*Cot[c + d*x])^( 
3/2))/d
 

Rubi [A] (warning: unable to verify)

Time = 0.83 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.89, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4011, 3042, 4011, 3042, 4022, 3042, 4020, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 4011

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

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1652\) vs. \(2(126)=252\).

Time = 0.46 (sec) , antiderivative size = 1653, normalized size of antiderivative = 11.02

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1653\)
default	\(\text {Expression too large to display}\)	\(1653\)
parts	\(\text {Expression too large to display}\)	\(1657\)

Input:

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

Output:

-2/3*B*(a+b*cot(d*x+c))^(3/2)/d-2/d*B*a*(a+b*cot(d*x+c))^(1/2)-2/d*A*b*(a+ 
b*cot(d*x+c))^(1/2)-1/4/d/b*ln(b*cot(d*x+c)+a+(a+b*cot(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^2-1/4/d*ln(b*cot(d*x+c)+a+(a+b*cot(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a) 
^(1/2)+(a^2+b^2)^(1/2))*B*(a^2+b^2)^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+1/ 
2/d*ln(b*cot(d*x+c)+a+(a+b*cot(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+1/d*b/(2*(a^2+b^2)^(1/ 
2)-2*a)^(1/2)*arctan((2*(a+b*cot(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)-2/d*b/(2*(a^2+b^2)^(1 
/2)-2*a)^(1/2)*arctan((2*(a+b*cot(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-1/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)* 
arctan((2*(a+b*cot(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))*B*a^2-1/4/d/b*ln(b*cot(d*x+c)+a-(a+b*cot(d*x+c))^(1/2 
)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*A*(a^2+b^2)^(1/2)*(2*(a^2 
+b^2)^(1/2)+2*a)^(1/2)*a+1/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b* 
cot(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))*B*(a^2+b^2)^(1/2)*a+1/4/d/b*ln(b*cot(d*x+c)+a+(a+b*cot(d*x+c))^(1/2) 
*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*A*(a^2+b^2)^(1/2)*(2*(a^2+ 
b^2)^(1/2)+2*a)^(1/2)*a+1/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*c 
ot(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)...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3252 vs. \(2 (120) = 240\).

Time = 0.42 (sec) , antiderivative size = 3252, normalized size of antiderivative = 21.68 \[ \int (a+b \cot (c+d x))^{3/2} (A+B \cot (c+d x)) \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 24.60 (sec) , antiderivative size = 2823, normalized size of antiderivative = 18.82 \[ \int (a+b \cot (c+d x))^{3/2} (A+B \cot (c+d x)) \, dx=\text {Too large to display} \] Input:

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

Output:

log((16*A^3*a*b^3*(a^2 + b^2)^2)/d^3 - (((16*b^2*(((-A^4*b^2*d^4*(3*a^2 - 
b^2)^2)^(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^2*d^2)/d^4)^(1/2)*(A*b^3 + A*a^2*b 
 + a*d*(((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^2* 
d^2)/d^4)^(1/2)*(a + b*cot(c + d*x))^(1/2)))/d + (16*A^2*b^2*(a + b*cot(c 
+ d*x))^(1/2)*(a^4 + b^4 - 6*a^2*b^2))/d^2)*(((-A^4*b^2*d^4*(3*a^2 - b^2)^ 
2)^(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^2*d^2)/d^4)^(1/2))/2)*((6*A^4*a^2*b^4*d 
^4 - A^4*b^6*d^4 - 9*A^4*a^4*b^2*d^4)^(1/2)/(4*d^4) - (A^2*a^3)/(4*d^2) + 
(3*A^2*a*b^2)/(4*d^2))^(1/2) - log((16*A^3*a*b^3*(a^2 + b^2)^2)/d^3 - (((1 
6*b^2*(((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^2*d 
^2)/d^4)^(1/2)*(A*b^3 + A*a^2*b - a*d*(((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/ 
2) - A^2*a^3*d^2 + 3*A^2*a*b^2*d^2)/d^4)^(1/2)*(a + b*cot(c + d*x))^(1/2)) 
)/d - (16*A^2*b^2*(a + b*cot(c + d*x))^(1/2)*(a^4 + b^4 - 6*a^2*b^2))/d^2) 
*(((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^2*d^2)/d 
^4)^(1/2))/2)*(((6*A^4*a^2*b^4*d^4 - A^4*b^6*d^4 - 9*A^4*a^4*b^2*d^4)^(1/2 
) - A^2*a^3*d^2 + 3*A^2*a*b^2*d^2)/(4*d^4))^(1/2) - log((16*A^3*a*b^3*(a^2 
 + b^2)^2)/d^3 - (((16*b^2*(-((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + A^2*a 
^3*d^2 - 3*A^2*a*b^2*d^2)/d^4)^(1/2)*(A*b^3 + A*a^2*b - a*d*(-((-A^4*b^2*d 
^4*(3*a^2 - b^2)^2)^(1/2) + A^2*a^3*d^2 - 3*A^2*a*b^2*d^2)/d^4)^(1/2)*(a + 
 b*cot(c + d*x))^(1/2)))/d - (16*A^2*b^2*(a + b*cot(c + d*x))^(1/2)*(a^4 + 
 b^4 - 6*a^2*b^2))/d^2)*(-((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + A^2*a...
 

Reduce [F]

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

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

Output:

int(sqrt(cot(c + d*x)*b + a),x)*a**2 + 2*int(sqrt(cot(c + d*x)*b + a)*cot( 
c + d*x),x)*a*b + int(sqrt(cot(c + d*x)*b + a)*cot(c + d*x)**2,x)*b**2