\(\int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx\) [858]

Optimal result

Integrand size = 25, antiderivative size = 248 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\frac {i (i a-b)^{5/2} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {5 a b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {i (i a+b)^{5/2} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {b^2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\cot (c+d x)}} \] Output:

I*(I*a-b)^(5/2)*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+5*a*b^(3/2)*arctanh(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)^(5/2)*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+b^2*(a+b*tan(d*x+c))^(1/2)/d/cot(d*x 
+c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.97 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left ((-1)^{3/4} (-a+i b)^{5/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-(-1)^{3/4} (a+i b)^{5/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+b^2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+\frac {5 a^{3/2} b^{3/2} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \] Input:

Integrate[Sqrt[Cot[c + d*x]]*(a + b*Tan[c + d*x])^(5/2),x]
 

Output:

(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((-1)^(3/4)*(-a + I*b)^(5/2)*ArcTan 
[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] 
- (-1)^(3/4)*(a + I*b)^(5/2)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + 
 d*x]])/Sqrt[a + b*Tan[c + d*x]]] + b^2*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[ 
c + d*x]] + (5*a^(3/2)*b^(3/2)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a 
]]*Sqrt[1 + (b*Tan[c + d*x])/a])/Sqrt[a + b*Tan[c + d*x]]))/d
 

Rubi [A] (verified)

Time = 1.13 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.83, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4729, 3042, 4049, 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.

\(\displaystyle \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2}dx\)

\(\Big \downarrow \) 4729

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {(a+b \tan (c+d x))^{5/2}}{\sqrt {\tan (c+d x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {(a+b \tan (c+d x))^{5/2}}{\sqrt {\tan (c+d x)}}dx\)

\(\Big \downarrow \) 4049

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\int \frac {5 a b^2 \tan ^2(c+d x)+2 b \left (3 a^2-b^2\right ) \tan (c+d x)+a \left (2 a^2-b^2\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {b^2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{2} \int \frac {5 a b^2 \tan ^2(c+d x)+2 b \left (3 a^2-b^2\right ) \tan (c+d x)+a \left (2 a^2-b^2\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {b^2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{2} \int \frac {5 a b^2 \tan (c+d x)^2+2 b \left (3 a^2-b^2\right ) \tan (c+d x)+a \left (2 a^2-b^2\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {b^2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )\)

\(\Big \downarrow \) 4138

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \frac {5 a b^2 \tan ^2(c+d x)+2 b \left (3 a^2-b^2\right ) \tan (c+d x)+a \left (2 a^2-b^2\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{2 d}+\frac {b^2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )\)

\(\Big \downarrow \) 2035

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \frac {5 a b^2 \tan ^2(c+d x)+2 b \left (3 a^2-b^2\right ) \tan (c+d x)+a \left (2 a^2-b^2\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{d}+\frac {b^2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )\)

\(\Big \downarrow \) 2257

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \left (\frac {5 a b^2}{\sqrt {a+b \tan (c+d x)}}+\frac {2 \left (a^3-3 b^2 a+b \left (3 a^2-b^2\right ) \tan (c+d x)\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}\right )d\sqrt {\tan (c+d x)}}{d}+\frac {b^2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {b^2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {i (-b+i a)^{5/2} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+5 a b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+i (b+i a)^{5/2} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\right )\)

Input:

Int[Sqrt[Cot[c + d*x]]*(a + b*Tan[c + d*x])^(5/2),x]
 

Output:

Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((I*(I*a - b)^(5/2)*ArcTan[(Sqrt[I*a 
 - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + 5*a*b^(3/2)*ArcTanh[ 
(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + I*(I*a + b)^(5/2) 
*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + 
 (b^2*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/d)
 

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 4049

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c 
+ d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[1/(d*(m + n - 1)) 
 Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n 
- 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ 
e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2, x], x], 
x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2 
, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || I 
ntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])) 
)
 

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)

Leaf count of result is larger than twice the leaf count of optimal. \(2580\) vs. \(2(202)=404\).

Time = 3.77 (sec) , antiderivative size = 2581, normalized size of antiderivative = 10.41

method	result	size
default	\(\text {Expression too large to display}\)	\(2581\)

Input:

int(cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/4/d*2^(1/2)/a/(-b+(a^2+b^2)^(1/2))^(1/2)*(a+b*tan(d*x+c))^(1/2)*cot(d*x+ 
c)^(1/2)/(cos(d*x+c)+1)/((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c 
)+1)^2)^(1/2)*(-10*arctanh(1/b^(1/2)*((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+ 
c)/(cos(d*x+c)+1)^2)^(1/2)*sin(d*x+c)/(-1+cos(d*x+c)))*2^(1/2)*b^(3/2)*(-b 
+(a^2+b^2)^(1/2))^(1/2)*a^2*sin(d*x+c)+(a^2+b^2)^(1/2)*ln(-(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)-a*cot(d*x+c)*cos(d*x+c)+2*a*cot(d*x+c)-2*(a^2+b^2)^ 
(1/2)*cos(d*x+c)-2*cos(d*x+c)*b+sin(d*x+c)*a-a*csc(d*x+c)+2*(a^2+b^2)^(1/2 
)+2*b)/(-1+cos(d*x+c)))*(b+(a^2+b^2)^(1/2))^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/ 
2)*a^2*sin(d*x+c)-(a^2+b^2)^(1/2)*ln(-(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 
)-a*cot(d*x+c)*cos(d*x+c)+2*a*cot(d*x+c)-2*(a^2+b^2)^(1/2)*cos(d*x+c)-2*co 
s(d*x+c)*b+sin(d*x+c)*a-a*csc(d*x+c)+2*(a^2+b^2)^(1/2)+2*b)/(-1+cos(d*x+c) 
))*(b+(a^2+b^2)^(1/2))^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*b^2*sin(d*x+c)-(a^ 
2+b^2)^(1/2)*ln((a*cot(d*x+c)*cos(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*cot(d*x+c)+2*(a^2+b^2)^(1/2)*cos(d*x+c)+2*cos(d*x+c)*b-sin(d*x+c)* 
a+a*csc(d*x+c)-2*(a^2+b^2)^(1/2)-2*b)/(-1+cos(d*x+c)))*(b+(a^2+b^2)^(1/2)) 
^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*a^2*sin(d*x+c)+(a^2+b^2)^(1/2)*ln((a*cot 
(d*x+c)*cos(d*x+c)+2*2^(1/2)*((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(c...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4793 vs. \(2 (196) = 392\).

Time = 1.14 (sec) , antiderivative size = 9618, normalized size of antiderivative = 38.78 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \] Input:

integrate(cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:

integrate(cot(d*x+c)**(1/2)*(a+b*tan(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sqrt {\cot \left (d x + c\right )} \,d x } \] Input:

integrate(cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

integrate((b*tan(d*x + c) + a)^(5/2)*sqrt(cot(d*x + c)), x)
 

Giac [F(-2)]

Exception generated. \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(5/2),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,[0,14,5]%%%}+%%%{6,[0,12,5]%%%}+%%%{15,[0,10,5]%%%}+ 
%%%{20,[0
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\int \sqrt {\mathrm {cot}\left (c+d\,x\right )}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2} \,d x \] Input:

int(cot(c + d*x)^(1/2)*(a + b*tan(c + d*x))^(5/2),x)
 

Output:

int(cot(c + d*x)^(1/2)*(a + b*tan(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\int \sqrt {\cot \left (d x +c \right )}\, \left (a +\tan \left (d x +c \right ) b \right )^{\frac {5}{2}}d x \] Input:

int(cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(5/2),x)
 

Output:

int(cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(5/2),x)