\(\int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^n \, dx\) [716]

Optimal result

Integrand size = 23, antiderivative size = 161 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^n \, dx=\frac {\operatorname {AppellF1}\left (\frac {3}{2},-n,1,\frac {5}{2},-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right ) \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^n \left (\frac {a+b \tan (c+d x)}{a}\right )^{-n}}{3 d}+\frac {\operatorname {AppellF1}\left (\frac {3}{2},-n,1,\frac {5}{2},-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right ) \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^n \left (\frac {a+b \tan (c+d x)}{a}\right )^{-n}}{3 d} \] Output:

1/3*AppellF1(3/2,1,-n,5/2,-I*tan(d*x+c),-b*tan(d*x+c)/a)*tan(d*x+c)^(3/2)* 
(a+b*tan(d*x+c))^n/d/(((a+b*tan(d*x+c))/a)^n)+1/3*AppellF1(3/2,1,-n,5/2,I* 
tan(d*x+c),-b*tan(d*x+c)/a)*tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))^n/d/(((a+b*t 
an(d*x+c))/a)^n)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.98, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4058

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

\(\Big \downarrow \) 615

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {1}{3} \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} \operatorname {AppellF1}\left (\frac {3}{2},1,-n,\frac {5}{2},-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )+\frac {1}{3} \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} \operatorname {AppellF1}\left (\frac {3}{2},1,-n,\frac {5}{2},i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{d}\)

Input:

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

Output:

((AppellF1[3/2, 1, -n, 5/2, (-I)*Tan[c + d*x], -((b*Tan[c + d*x])/a)]*Tan[ 
c + d*x]^(3/2)*(a + b*Tan[c + d*x])^n)/(3*(1 + (b*Tan[c + d*x])/a)^n) + (A 
ppellF1[3/2, 1, -n, 5/2, I*Tan[c + d*x], -((b*Tan[c + d*x])/a)]*Tan[c + d* 
x]^(3/2)*(a + b*Tan[c + d*x])^n)/(3*(1 + (b*Tan[c + d*x])/a)^n))/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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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