\(\int \frac {(a+b \tan (c+d x))^n (A+B \tan (c+d x))}{\cot ^{\frac {3}{2}}(c+d x)} \, dx\) [664]

Optimal result

Integrand size = 33, antiderivative size = 175 \[ \int \frac {(a+b \tan (c+d x))^n (A+B \tan (c+d x))}{\cot ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(A+i B) \operatorname {AppellF1}\left (\frac {5}{2},-n,1,\frac {7}{2},-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right ) (a+b \tan (c+d x))^n \left (\frac {a+b \tan (c+d x)}{a}\right )^{-n}}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {(A-i B) \operatorname {AppellF1}\left (\frac {5}{2},-n,1,\frac {7}{2},-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right ) (a+b \tan (c+d x))^n \left (\frac {a+b \tan (c+d x)}{a}\right )^{-n}}{5 d \cot ^{\frac {5}{2}}(c+d x)} \] Output:

1/5*(A+I*B)*AppellF1(5/2,1,-n,7/2,-I*tan(d*x+c),-b*tan(d*x+c)/a)*(a+b*tan( 
d*x+c))^n/d/cot(d*x+c)^(5/2)/(((a+b*tan(d*x+c))/a)^n)+1/5*(A-I*B)*AppellF1 
(5/2,1,-n,7/2,I*tan(d*x+c),-b*tan(d*x+c)/a)*(a+b*tan(d*x+c))^n/d/cot(d*x+c 
)^(5/2)/(((a+b*tan(d*x+c))/a)^n)
                                                                                    
                                                                                    
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 1.29 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 4729, 3042, 4086, 3042, 4085, 148, 395, 394}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4729

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4086

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4085

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

\(\Big \downarrow \) 148

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 394

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(A+i B) \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} \operatorname {AppellF1}\left (\frac {5}{2},1,-n,\frac {7}{2},-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{5 d}+\frac {(A-i B) \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} \operatorname {AppellF1}\left (\frac {5}{2},1,-n,\frac {7}{2},i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{5 d}\right )\)

Input:

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

Output:

Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(((A + I*B)*AppellF1[5/2, 1, -n, 7/2 
, (-I)*Tan[c + d*x], -((b*Tan[c + d*x])/a)]*Tan[c + d*x]^(5/2)*(a + b*Tan[ 
c + d*x])^n)/(5*d*(1 + (b*Tan[c + d*x])/a)^n) + ((A - I*B)*AppellF1[5/2, 1 
, -n, 7/2, I*Tan[c + d*x], -((b*Tan[c + d*x])/a)]*Tan[c + d*x]^(5/2)*(a + 
b*Tan[c + d*x])^n)/(5*d*(1 + (b*Tan[c + d*x])/a)^n))
 

Defintions of rubi rules used

rule 148

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))^(p_.), 
x_] :> With[{k = Denominator[m]}, Simp[k/b   Subst[Int[x^(k*(m + 1) - 1)*(c 
 + d*(x^k/b))^n*(e + f*(x^k/b))^p, x], x, (b*x)^(1/k)], x]] /; FreeQ[{b, c, 
 d, e, f, n, p}, x] && FractionQ[m] && IntegerQ[p]
 

rule 394

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 
, -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, 
 d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int 
egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 395

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^ 
FracPart[p])   Int[(e*x)^m*(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ 
[{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 
1] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 3042

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

rule 4085

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[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] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegersQ[2*m, 2*n 
] && EqQ[A^2 + B^2, 0]
 

rule 4086

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] &&  !IntegerQ[m] & 
&  !IntegerQ[n] &&  !IntegersQ[2*m, 2*n] && 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 [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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