\(\int \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\) [627]

Optimal result

Integrand size = 35, antiderivative size = 269 \[ \int \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {(a+i b)^2 (i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {i a-b} d}+\frac {2 b^{3/2} B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {(i a+b)^{3/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {2 a A \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{d} \] Output:

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

Mathematica [A] (verified)

Time = 1.64 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.99 \[ \int \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {\cot (c+d x)} \left (\sqrt [4]{-1} (-a+i b)^{3/2} (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\tan (c+d x)}-\sqrt [4]{-1} (a+i b)^{3/2} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\tan (c+d x)}-2 a A \sqrt {a+b \tan (c+d x)}+\frac {2 \sqrt {a} b^{3/2} B \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {\tan (c+d x)} \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \] Input:

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

Output:

(Sqrt[Cot[c + d*x]]*((-1)^(1/4)*(-a + I*b)^(3/2)*(A - I*B)*ArcTan[((-1)^(1 
/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Tan[ 
c + d*x]] - (-1)^(1/4)*(a + I*b)^(3/2)*(A + I*B)*ArcTan[((-1)^(1/4)*Sqrt[a 
 + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Tan[c + d*x]] - 
 2*a*A*Sqrt[a + b*Tan[c + d*x]] + (2*Sqrt[a]*b^(3/2)*B*ArcSinh[(Sqrt[b]*Sq 
rt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[Tan[c + d*x]]*Sqrt[1 + (b*Tan[c + d*x])/a] 
)/Sqrt[a + b*Tan[c + d*x]]))/d
 

Rubi [A] (verified)

Time = 2.04 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.82, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4729, 3042, 4088, 27, 3042, 4138, 2035, 2257, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4729

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4088

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4138

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 2257

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

rule 2257

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol 
] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a 
, c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
 

rule 3042

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

rule 4088

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*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x 
])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Simp[1/(d*(n + 1)*(c^2 + d^2)) 
  Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d* 
(b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1) + a*d*(n 
 + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[ 
e + f*x] - b*(d*(A*b*c + a*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n 
 + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && 
 NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] & 
& LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4138

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] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S 
imp[ff/f   Subst[Int[(a + b*ff*x)^m*(c + d*ff*x)^n*((A + B*ff*x + C*ff^2*x^ 
2)/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f 
, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + 
 d^2, 0]
 

rule 4729

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cot[a 
+ b*x])^m*(c*Tan[a + b*x])^m   Int[ActivateTrig[u]/(c*Tan[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ[u, 
x]
 

Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 2.08 (sec) , antiderivative size = 2397204, normalized size of antiderivative = 8911.54

\[\text {output too large to display}\]

Input:

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

Output:

result too large to display
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 12204 vs. \(2 (215) = 430\).

Time = 4.12 (sec) , antiderivative size = 24440, normalized size of antiderivative = 90.86 \[ \int \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F(-2)]

Exception generated. \[ \int \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{1,[0,9,3]%%%}+%%%{4,[0,7,3]%%%}+%%%{6,[0,5,3]%%%}+%%%{ 
4,[0,3,3]
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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