\(\int \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} \, dx\) [849]

Optimal result

Integrand size = 25, antiderivative size = 185 \[ \int \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} \, dx=\frac {i (i a-b)^{3/2} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {i (i a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {2 a \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{d} \] Output:

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

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.95 \[ \int \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} \, dx=\frac {\sqrt {\cot (c+d x)} \left (\sqrt [4]{-1} (-a+i b)^{3/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\tan (c+d x)}-\sqrt [4]{-1} (a+i b)^{3/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\tan (c+d x)}-2 a \sqrt {a+b \tan (c+d x)}\right )}{d} \] Input:

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

Output:

(Sqrt[Cot[c + d*x]]*((-1)^(1/4)*(-a + I*b)^(3/2)*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)*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*Sqrt[a + b*Tan[ 
c + d*x]]))/d
 

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.99, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 4729, 3042, 4050, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4729

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4050

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4099

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4098

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(((-I)*(a + I*b)^2*ArcTan[(Sqrt[I*a 
- b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a - b]*d) + (I 
*(a - I*b)^2*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + 
 d*x]]])/(Sqrt[I*a + b]*d) - (2*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 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 4050

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

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 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)

Leaf count of result is larger than twice the leaf count of optimal. \(1832\) vs. \(2(151)=302\).

Time = 3.05 (sec) , antiderivative size = 1833, normalized size of antiderivative = 9.91

method	result	size
default	\(\text {Expression too large to display}\)	\(1833\)

Input:

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

Output:

1/2/d*2^(1/2)/a/(-b+(a^2+b^2)^(1/2))^(1/2)*cot(d*x+c)^(3/2)*(a+b*tan(d*x+c 
))^(1/2)*((4*csc(d*x+c)-4*cot(d*x+c))*b*arctan(1/(-b+(a^2+b^2)^(1/2))^(1/2 
)*((b+(a^2+b^2)^(1/2))^(1/2)*(-cot(d*x+c)+csc(d*x+c))+2^(1/2)*((a*cos(d*x+ 
c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2))/(1-cos(d*x+c))*sin(d* 
x+c))*a^2+(-2*csc(d*x+c)+2*cot(d*x+c))*(a^2+b^2)^(1/2)*arctan(1/(-b+(a^2+b 
^2)^(1/2))^(1/2)*(-(b+(a^2+b^2)^(1/2))^(1/2)*(-cot(d*x+c)+csc(d*x+c))+2^(1 
/2)*((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2))/(1-co 
s(d*x+c))*sin(d*x+c))*a^2+(-2*csc(d*x+c)+2*cot(d*x+c))*(a^2+b^2)^(1/2)*arc 
tan(1/(-b+(a^2+b^2)^(1/2))^(1/2)*((b+(a^2+b^2)^(1/2))^(1/2)*(-cot(d*x+c)+c 
sc(d*x+c))+2^(1/2)*((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1)^ 
2)^(1/2))/(1-cos(d*x+c))*sin(d*x+c))*a^2+(4*csc(d*x+c)-4*cot(d*x+c))*b*arc 
tan(1/(-b+(a^2+b^2)^(1/2))^(1/2)*(-(b+(a^2+b^2)^(1/2))^(1/2)*(-cot(d*x+c)+ 
csc(d*x+c))+2^(1/2)*((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1) 
^2)^(1/2))/(1-cos(d*x+c))*sin(d*x+c))*a^2+a^2*ln(1/(1-cos(d*x+c))*(a*(1-co 
s(d*x+c))^2*csc(d*x+c)+2*2^(1/2)*((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/( 
cos(d*x+c)+1)^2)^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)*sin(d*x+c)-2*(a^2+b^2)^(1 
/2)*(1-cos(d*x+c))-2*b*(1-cos(d*x+c))-sin(d*x+c)*a))*(b+(a^2+b^2)^(1/2))^( 
1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*(-cot(d*x+c)+csc(d*x+c))+(-csc(d*x+c)+cot( 
d*x+c))*(-b+(a^2+b^2)^(1/2))^(1/2)*b^2*(b+(a^2+b^2)^(1/2))^(1/2)*ln(1/(1-c 
os(d*x+c))*(a*(1-cos(d*x+c))^2*csc(d*x+c)+2*2^(1/2)*((a*cos(d*x+c)+b*si...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3545 vs. \(2 (145) = 290\).

Time = 0.41 (sec) , antiderivative size = 3545, normalized size of antiderivative = 19.16 \[ \int \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

integrate(cot(d*x+c)^(3/2)*(a+b*tan(d*x+c))^(3/2),x, algorithm="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. \[ \int \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^{3/2}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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