\(\int \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} \, dx\) [703]

Optimal result

Integrand size = 23, antiderivative size = 201 \[ \int \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\frac {\operatorname {AppellF1}\left (\frac {3}{2},-m,1,\frac {5}{2},\frac {a+b \tan (c+d x)}{a},\frac {a+b \tan (c+d x)}{a-i b}\right ) \tan ^m(c+d x) \left (-\frac {b \tan (c+d x)}{a}\right )^{-m} (a+b \tan (c+d x))^{3/2}}{3 (i a+b) d}-\frac {\operatorname {AppellF1}\left (\frac {3}{2},-m,1,\frac {5}{2},\frac {a+b \tan (c+d x)}{a},\frac {a+b \tan (c+d x)}{a+i b}\right ) \tan ^m(c+d x) \left (-\frac {b \tan (c+d x)}{a}\right )^{-m} (a+b \tan (c+d x))^{3/2}}{3 (i a-b) d} \] Output:

1/3*AppellF1(3/2,-m,1,5/2,(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))^(3/2)/(I*a+b)/d/((-b*tan(d*x+c)/a)^m)-1/3*Appel 
lF1(3/2,-m,1,5/2,(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))^(3/2)/(I*a-b)/d/((-b*tan(d*x+c)/a)^m)
 

Mathematica [F]

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

Integrate[Tan[c + d*x]^m*Sqrt[a + b*Tan[c + d*x]],x]
 

Output:

Integrate[Tan[c + d*x]^m*Sqrt[a + b*Tan[c + d*x]], x]
 

Rubi [A] (warning: unable to verify)

Time = 0.37 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.85, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4058, 615, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4058

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

\(\Big \downarrow \) 615

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

\(\Big \downarrow \) 2009

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

Input:

Int[Tan[c + d*x]^m*Sqrt[a + b*Tan[c + d*x]],x]
 

Output:

((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[a + b*Tan[c + d*x]])/(2*(1 + m)*Sqrt[1 + (b*Ta 
n[c + d*x])/a]) + (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[a + b*Tan[c + d*x]])/(2*(1 + m)* 
Sqrt[1 + (b*Tan[c + d*x])/a]))/d
 

Defintions of rubi rules used

rule 615

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
 x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] 
 /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4058

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), 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/(1 + ff^2*x^2)), x], x, 
 Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a 
*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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