\(\int \frac {(a+b \tan (c+d x))^n}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\) [718]

Optimal result

Integrand size = 23, antiderivative size = 157 \[ \int \frac {(a+b \tan (c+d x))^n}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {\operatorname {AppellF1}\left (-\frac {1}{2},-n,1,\frac {1}{2},-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right ) (a+b \tan (c+d x))^n \left (\frac {a+b \tan (c+d x)}{a}\right )^{-n}}{d \sqrt {\tan (c+d x)}}-\frac {\operatorname {AppellF1}\left (-\frac {1}{2},-n,1,\frac {1}{2},-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right ) (a+b \tan (c+d x))^n \left (\frac {a+b \tan (c+d x)}{a}\right )^{-n}}{d \sqrt {\tan (c+d x)}} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.97, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4058

\(\displaystyle \frac {\int \frac {(a+b \tan (c+d x))^n}{\tan ^{\frac {3}{2}}(c+d x) \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 615

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

integrate((a+b*tan(d*x+c))^n/tan(d*x+c)^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate((a+b*tan(d*x+c))^n/tan(d*x+c)^(3/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((a+b*tan(d*x+c))^n/tan(d*x+c)^(3/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a+b \tan (c+d x))^n}{\tan ^{\frac {3}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*sqrt(tan(c + d*x))*(tan(c + d*x)*b + a)**n*tan(c + d*x)*a + 4*sqrt(tan( 
c + d*x))*(tan(c + d*x)*b + a)**n*b*n + 2*sqrt(tan(c + d*x))*(tan(c + d*x) 
*b + a)**n*b - 8*int((sqrt(tan(c + d*x))*(tan(c + d*x)*b + a)**n*tan(c + d 
*x)**2)/(4*tan(c + d*x)*b*n**2 - tan(c + d*x)*b + 4*a*n**2 - a),x)*tan(c + 
 d*x)*a*b*d*n**3 - 4*int((sqrt(tan(c + d*x))*(tan(c + d*x)*b + a)**n*tan(c 
 + d*x)**2)/(4*tan(c + d*x)*b*n**2 - tan(c + d*x)*b + 4*a*n**2 - a),x)*tan 
(c + d*x)*a*b*d*n**2 + 2*int((sqrt(tan(c + d*x))*(tan(c + d*x)*b + a)**n*t 
an(c + d*x)**2)/(4*tan(c + d*x)*b*n**2 - tan(c + d*x)*b + 4*a*n**2 - a),x) 
*tan(c + d*x)*a*b*d*n + int((sqrt(tan(c + d*x))*(tan(c + d*x)*b + a)**n*ta 
n(c + d*x)**2)/(4*tan(c + d*x)*b*n**2 - tan(c + d*x)*b + 4*a*n**2 - a),x)* 
tan(c + d*x)*a*b*d - 4*int((sqrt(tan(c + d*x))*(tan(c + d*x)*b + a)**n*tan 
(c + d*x))/(4*tan(c + d*x)*b*n**2 - tan(c + d*x)*b + 4*a*n**2 - a),x)*tan( 
c + d*x)*a**2*d*n**2 + int((sqrt(tan(c + d*x))*(tan(c + d*x)*b + a)**n*tan 
(c + d*x))/(4*tan(c + d*x)*b*n**2 - tan(c + d*x)*b + 4*a*n**2 - a),x)*tan( 
c + d*x)*a**2*d - 8*int((sqrt(tan(c + d*x))*(tan(c + d*x)*b + a)**n*tan(c 
+ d*x))/(2*tan(c + d*x)*b*n - tan(c + d*x)*b + 2*a*n - a),x)*tan(c + d*x)* 
b**2*d*n**3 + 4*int((sqrt(tan(c + d*x))*(tan(c + d*x)*b + a)**n*tan(c + d* 
x))/(2*tan(c + d*x)*b*n - tan(c + d*x)*b + 2*a*n - a),x)*tan(c + d*x)*b**2 
*d*n**2 + 2*int((sqrt(tan(c + d*x))*(tan(c + d*x)*b + a)**n*tan(c + d*x))/ 
(2*tan(c + d*x)*b*n - tan(c + d*x)*b + 2*a*n - a),x)*tan(c + d*x)*b**2*...