\(\int \frac {(d \tan (e+f x))^n}{a+b \tan (e+f x)} \, dx\) [700]

Optimal result

Integrand size = 23, antiderivative size = 181 \[ \int \frac {(d \tan (e+f x))^n}{a+b \tan (e+f x)} \, dx=\frac {a \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{1+n}}{\left (a^2+b^2\right ) d f (1+n)}+\frac {b^2 \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,-\frac {b \tan (e+f x)}{a}\right ) (d \tan (e+f x))^{1+n}}{a \left (a^2+b^2\right ) d f (1+n)}-\frac {b \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{2+n}}{\left (a^2+b^2\right ) d^2 f (2+n)} \] Output:

a*hypergeom([1, 1/2+1/2*n],[3/2+1/2*n],-tan(f*x+e)^2)*(d*tan(f*x+e))^(1+n) 
/(a^2+b^2)/d/f/(1+n)+b^2*hypergeom([1, 1+n],[2+n],-b*tan(f*x+e)/a)*(d*tan( 
f*x+e))^(1+n)/a/(a^2+b^2)/d/f/(1+n)-b*hypergeom([1, 1+1/2*n],[2+1/2*n],-ta 
n(f*x+e)^2)*(d*tan(f*x+e))^(2+n)/(a^2+b^2)/d^2/f/(2+n)
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.78 \[ \int \frac {(d \tan (e+f x))^n}{a+b \tan (e+f x)} \, dx=\frac {\tan (e+f x) (d \tan (e+f x))^n \left (a^2 (2+n) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right )+b \left (b (2+n) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,-\frac {b \tan (e+f x)}{a}\right )-a (1+n) \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)\right )\right )}{a \left (a^2+b^2\right ) f (1+n) (2+n)} \] Input:

Integrate[(d*Tan[e + f*x])^n/(a + b*Tan[e + f*x]),x]
 

Output:

(Tan[e + f*x]*(d*Tan[e + f*x])^n*(a^2*(2 + n)*Hypergeometric2F1[1, (1 + n) 
/2, (3 + n)/2, -Tan[e + f*x]^2] + b*(b*(2 + n)*Hypergeometric2F1[1, 1 + n, 
 2 + n, -((b*Tan[e + f*x])/a)] - a*(1 + n)*Hypergeometric2F1[1, (2 + n)/2, 
 (4 + n)/2, -Tan[e + f*x]^2]*Tan[e + f*x])))/(a*(a^2 + b^2)*f*(1 + n)*(2 + 
 n))
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.96, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4057, 3042, 4021, 3042, 3957, 278, 4117, 74}

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.

\(\displaystyle \int \frac {(d \tan (e+f x))^n}{a+b \tan (e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(d \tan (e+f x))^n}{a+b \tan (e+f x)}dx\)

\(\Big \downarrow \) 4057

\(\displaystyle \frac {b^2 \int \frac {(d \tan (e+f x))^n \left (\tan ^2(e+f x)+1\right )}{a+b \tan (e+f x)}dx}{a^2+b^2}+\frac {\int (d \tan (e+f x))^n (a-b \tan (e+f x))dx}{a^2+b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \int \frac {(d \tan (e+f x))^n \left (\tan (e+f x)^2+1\right )}{a+b \tan (e+f x)}dx}{a^2+b^2}+\frac {\int (d \tan (e+f x))^n (a-b \tan (e+f x))dx}{a^2+b^2}\)

\(\Big \downarrow \) 4021

\(\displaystyle \frac {b^2 \int \frac {(d \tan (e+f x))^n \left (\tan (e+f x)^2+1\right )}{a+b \tan (e+f x)}dx}{a^2+b^2}+\frac {a \int (d \tan (e+f x))^ndx-\frac {b \int (d \tan (e+f x))^{n+1}dx}{d}}{a^2+b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \int \frac {(d \tan (e+f x))^n \left (\tan (e+f x)^2+1\right )}{a+b \tan (e+f x)}dx}{a^2+b^2}+\frac {a \int (d \tan (e+f x))^ndx-\frac {b \int (d \tan (e+f x))^{n+1}dx}{d}}{a^2+b^2}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {\frac {a d \int \frac {(d \tan (e+f x))^n}{\tan ^2(e+f x) d^2+d^2}d(d \tan (e+f x))}{f}-\frac {b \int \frac {(d \tan (e+f x))^{n+1}}{\tan ^2(e+f x) d^2+d^2}d(d \tan (e+f x))}{f}}{a^2+b^2}+\frac {b^2 \int \frac {(d \tan (e+f x))^n \left (\tan (e+f x)^2+1\right )}{a+b \tan (e+f x)}dx}{a^2+b^2}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {b^2 \int \frac {(d \tan (e+f x))^n \left (\tan (e+f x)^2+1\right )}{a+b \tan (e+f x)}dx}{a^2+b^2}+\frac {\frac {a (d \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2},\frac {n+3}{2},-\tan ^2(e+f x)\right )}{d f (n+1)}-\frac {b (d \tan (e+f x))^{n+2} \operatorname {Hypergeometric2F1}\left (1,\frac {n+2}{2},\frac {n+4}{2},-\tan ^2(e+f x)\right )}{d^2 f (n+2)}}{a^2+b^2}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {b^2 \int \frac {(d \tan (e+f x))^n}{a+b \tan (e+f x)}d\tan (e+f x)}{f \left (a^2+b^2\right )}+\frac {\frac {a (d \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2},\frac {n+3}{2},-\tan ^2(e+f x)\right )}{d f (n+1)}-\frac {b (d \tan (e+f x))^{n+2} \operatorname {Hypergeometric2F1}\left (1,\frac {n+2}{2},\frac {n+4}{2},-\tan ^2(e+f x)\right )}{d^2 f (n+2)}}{a^2+b^2}\)

\(\Big \downarrow \) 74

\(\displaystyle \frac {\frac {a (d \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2},\frac {n+3}{2},-\tan ^2(e+f x)\right )}{d f (n+1)}-\frac {b (d \tan (e+f x))^{n+2} \operatorname {Hypergeometric2F1}\left (1,\frac {n+2}{2},\frac {n+4}{2},-\tan ^2(e+f x)\right )}{d^2 f (n+2)}}{a^2+b^2}+\frac {b^2 (d \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,-\frac {b \tan (e+f x)}{a}\right )}{a d f (n+1) \left (a^2+b^2\right )}\)

Input:

Int[(d*Tan[e + f*x])^n/(a + b*Tan[e + f*x]),x]
 

Output:

(b^2*Hypergeometric2F1[1, 1 + n, 2 + n, -((b*Tan[e + f*x])/a)]*(d*Tan[e + 
f*x])^(1 + n))/(a*(a^2 + b^2)*d*f*(1 + n)) + ((a*Hypergeometric2F1[1, (1 + 
 n)/2, (3 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(1 + n))/(d*f*(1 + n)) 
 - (b*Hypergeometric2F1[1, (2 + n)/2, (4 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e 
 + f*x])^(2 + n))/(d^2*f*(2 + n)))/(a^2 + b^2)
 

Defintions of rubi rules used

rule 74

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x 
)^(m + 1)/(b*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] 
/; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[c, 0] 
 &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))
 

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 3042

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

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 

rule 4021

Int[((b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Tan[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Tan[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x] && NeQ[c^ 
2 + d^2, 0] &&  !IntegerQ[2*m]
 

rule 4057

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

rule 4117

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
 Simp[A/f   Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; 
FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
 

Maple [F]

\[\int \frac {\left (d \tan \left (f x +e \right )\right )^{n}}{a +b \tan \left (f x +e \right )}d x\]

Input:

int((d*tan(f*x+e))^n/(a+b*tan(f*x+e)),x)
 

Output:

int((d*tan(f*x+e))^n/(a+b*tan(f*x+e)),x)
 

Fricas [F]

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

integrate((d*tan(f*x+e))^n/(a+b*tan(f*x+e)),x, algorithm="fricas")
 

Output:

integral((d*tan(f*x + e))^n/(b*tan(f*x + e) + a), x)
 

Sympy [F]

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

integrate((d*tan(f*x+e))**n/(a+b*tan(f*x+e)),x)
 

Output:

Integral((d*tan(e + f*x))**n/(a + b*tan(e + f*x)), x)
 

Maxima [F]

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

integrate((d*tan(f*x+e))^n/(a+b*tan(f*x+e)),x, algorithm="maxima")
 

Output:

integrate((d*tan(f*x + e))^n/(b*tan(f*x + e) + a), x)
 

Giac [F]

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

integrate((d*tan(f*x+e))^n/(a+b*tan(f*x+e)),x, algorithm="giac")
 

Output:

integrate((d*tan(f*x + e))^n/(b*tan(f*x + e) + a), x)
 

Mupad [F(-1)]

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

int((d*tan(e + f*x))^n/(a + b*tan(e + f*x)),x)
 

Output:

int((d*tan(e + f*x))^n/(a + b*tan(e + f*x)), x)
 

Reduce [F]

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

int((d*tan(f*x+e))^n/(a+b*tan(f*x+e)),x)
 

Output:

(d**n*(tan(e + f*x)**n - int(tan(e + f*x)**n/(tan(e + f*x)**2*b + tan(e + 
f*x)*a),x)*a*f*n - int((tan(e + f*x)**n*tan(e + f*x)**2)/(tan(e + f*x)*b + 
 a),x)*b*f*n - int((tan(e + f*x)**n*tan(e + f*x))/(tan(e + f*x)*b + a),x)* 
a*f*n))/(b*f*n)