3.7.45 \(\int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx\) [645]

3.7.45.1 Optimal result

Integrand size = 35, antiderivative size = 210 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx=-\frac {(A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a-b)^{3/2} d}+\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a+b)^{3/2} d}-\frac {2 (A b-a B)}{\left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}} \]

output

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

3.7.45.2 Mathematica [A] (verified)

Time = 2.35 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.23 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (-\frac {\sqrt [4]{-1} a (a+i b) (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {\sqrt [4]{-1} a (a-i b) (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}+\frac {2 b (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{\sqrt {a+b \tan (c+d x)}}+2 (-A b+a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}\right )}{a \left (a^2+b^2\right ) d} \]

input

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

output

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

3.7.45.3 Rubi [A] (verified)

Time = 1.19 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.06, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 4729, 3042, 4091, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4729

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4091

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4099

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4098

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

input

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

output

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

3.7.45.3.1 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 4091

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*b - a*B)*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^n/(f*(m + 
 1)*(a^2 + b^2))), x] + Simp[1/(b*(m + 1)*(a^2 + b^2))   Int[(a + b*Tan[e + 
 f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[b*B*(b*c*(m + 1) + a*d*n) 
+ A*b*(a*c*(m + 1) - b*d*n) - b*(A*(b*c - a*d) - B*(a*c + b*d))*(m + 1)*Tan 
[e + f*x] - b*d*(A*b - a*B)*(m + 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] && LtQ[m, -1] && LtQ[0, n, 1] && (IntegerQ[m] || Integers 
Q[2*m, 2*n])
 

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]
 

3.7.45.4 Maple [B] (warning: unable to verify)

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

Time = 1.32 (sec) , antiderivative size = 1559507, normalized size of antiderivative = 7426.22

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

input

int((A+B*tan(d*x+c))/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(3/2),x)
 

output

result too large to display
 

3.7.45.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 18672 vs. \(2 (170) = 340\).

Time = 7.01 (sec) , antiderivative size = 18672, normalized size of antiderivative = 88.91 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

input

integrate((A+B*tan(d*x+c))/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(3/2),x, algo 
rithm="fricas")
 

output

Too large to include
 

3.7.45.6 Sympy [F]

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

input

integrate((A+B*tan(d*x+c))/cot(d*x+c)**(1/2)/(a+b*tan(d*x+c))**(3/2),x)
 

output

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

3.7.45.7 Maxima [F]

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

input

integrate((A+B*tan(d*x+c))/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(3/2),x, algo 
rithm="maxima")
 

output

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

3.7.45.8 Giac [F(-1)]

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

input

integrate((A+B*tan(d*x+c))/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(3/2),x, algo 
rithm="giac")
 

output

Timed out
 

3.7.45.9 Mupad [F(-1)]

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

input

int((A + B*tan(c + d*x))/(cot(c + d*x)^(1/2)*(a + b*tan(c + d*x))^(3/2)),x 
)
 

output

int((A + B*tan(c + d*x))/(cot(c + d*x)^(1/2)*(a + b*tan(c + d*x))^(3/2)), 
x)