3.7.63 \(\int \frac {(a+b \tan (c+d x))^n (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx\) [663]

3.7.63.1 Optimal result

Integrand size = 33, antiderivative size = 167 \[ \int \frac {(a+b \tan (c+d x))^n (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\frac {(A+i B) \operatorname {AppellF1}\left (\frac {1}{2},1,-n,\frac {3}{2},-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^n \left (1+\frac {b \tan (c+d x)}{a}\right )^{-n}}{d}+\frac {(A-i B) \operatorname {AppellF1}\left (\frac {1}{2},1,-n,\frac {3}{2},i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^n \left (1+\frac {b \tan (c+d x)}{a}\right )^{-n}}{d} \]

output

(A+I*B)*AppellF1(1/2,1,-n,3/2,-I*tan(d*x+c),-b*tan(d*x+c)/a)*tan(d*x+c)^(1 
/2)*(a+b*tan(d*x+c))^n/d/((1+b*tan(d*x+c)/a)^n)+(A-I*B)*AppellF1(1/2,1,-n, 
3/2,I*tan(d*x+c),-b*tan(d*x+c)/a)*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^n/d/(( 
1+b*tan(d*x+c)/a)^n)
 

3.7.63.2 Mathematica [F]

\[ \int \frac {(a+b \tan (c+d x))^n (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\int \frac {(a+b \tan (c+d x))^n (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx \]

input

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

output

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

3.7.63.3 Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3042, 4086, 3042, 4085, 148, 334, 333}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4086

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4085

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

\(\Big \downarrow \) 148

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

\(\Big \downarrow \) 334

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

\(\Big \downarrow \) 333

\(\displaystyle \frac {(A+i B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},1,-n,\frac {3}{2},-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{d}+\frac {(A-i B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},1,-n,\frac {3}{2},i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{d}\)

input

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

output

((A + I*B)*AppellF1[1/2, 1, -n, 3/2, (-I)*Tan[c + d*x], -((b*Tan[c + d*x]) 
/a)]*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^n)/(d*(1 + (b*Tan[c + d*x])/a 
)^n) + ((A - I*B)*AppellF1[1/2, 1, -n, 3/2, I*Tan[c + d*x], -((b*Tan[c + d 
*x])/a)]*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^n)/(d*(1 + (b*Tan[c + d*x 
])/a)^n)
 

3.7.63.3.1 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 333

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 334

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

3.7.63.4 Maple [F]

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

input

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

output

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

3.7.63.5 Fricas [F]

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

input

integrate((a+b*tan(d*x+c))^n*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x, algorith 
m="fricas")
 

output

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

3.7.63.6 Sympy [F]

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

input

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

output

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

3.7.63.7 Maxima [F(-1)]

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

input

integrate((a+b*tan(d*x+c))^n*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x, algorith 
m="maxima")
 

output

Timed out
 

3.7.63.8 Giac [F(-1)]

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

input

integrate((a+b*tan(d*x+c))^n*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x, algorith 
m="giac")
 

output

Timed out
 

3.7.63.9 Mupad [F(-1)]

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

input

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

output

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