\(\int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx\) [490]

Optimal result

Integrand size = 33, antiderivative size = 209 \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {(A-i B) \operatorname {AppellF1}\left (\frac {1}{2},-m,1,\frac {3}{2},\frac {a+b \tan (c+d x)}{a},\frac {a+b \tan (c+d x)}{a-i b}\right ) \tan ^m(c+d x) \left (-\frac {b \tan (c+d x)}{a}\right )^{-m} \sqrt {a+b \tan (c+d x)}}{(i a+b) d}+\frac {(i A-B) \operatorname {AppellF1}\left (\frac {1}{2},-m,1,\frac {3}{2},\frac {a+b \tan (c+d x)}{a},\frac {a+b \tan (c+d x)}{a+i b}\right ) \tan ^m(c+d x) \left (-\frac {b \tan (c+d x)}{a}\right )^{-m} \sqrt {a+b \tan (c+d x)}}{(a+i b) d} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.62 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3042, 4086, 3042, 4085, 152, 150}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4086

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4085

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

\(\Big \downarrow \) 152

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

\(\Big \downarrow \) 150

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 152

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) 
Int[(b*x)^m*(1 + d*(x/c))^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d, e, f, m, 
 n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(tan(c + d*x)**m*sqrt(tan(c + d*x)*b + a),x)