3.3.20 \(\int \tan ^2(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\) [220]

3.3.20.1 Optimal result

Integrand size = 34, antiderivative size = 164 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=-\frac {2 B (a+i a \tan (c+d x))^n}{d n (2+n)}+\frac {(i A+B) \operatorname {Hypergeometric2F1}\left (1,n,1+n,\frac {1}{2} (1+i \tan (c+d x))\right ) (a+i a \tan (c+d x))^n}{2 d n}+\frac {B \tan ^2(c+d x) (a+i a \tan (c+d x))^n}{d (2+n)}-\frac {(B n+i A (2+n)) (a+i a \tan (c+d x))^{1+n}}{a d (1+n) (2+n)} \]

output

-2*B*(a+I*a*tan(d*x+c))^n/d/n/(2+n)+1/2*(I*A+B)*hypergeom([1, n],[1+n],1/2 
+1/2*I*tan(d*x+c))*(a+I*a*tan(d*x+c))^n/d/n+B*tan(d*x+c)^2*(a+I*a*tan(d*x+ 
c))^n/d/(2+n)-(B*n+I*A*(2+n))*(a+I*a*tan(d*x+c))^(1+n)/a/d/(1+n)/(2+n)
 

3.3.20.2 Mathematica [A] (verified)

Time = 1.48 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.84 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\frac {(a+i a \tan (c+d x))^n \left (-4 B-4 i A n-4 B n-2 i A n^2-2 B n^2+(i A+B) \left (2+3 n+n^2\right ) \operatorname {Hypergeometric2F1}\left (1,n,1+n,\frac {1}{2} (1+i \tan (c+d x))\right )+2 n (-i B n+A (2+n)) \tan (c+d x)+2 B n (1+n) \tan ^2(c+d x)\right )}{2 d n (1+n) (2+n)} \]

input

Integrate[Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^n*(A + B*Tan[c + d*x]),x]
 

output

((a + I*a*Tan[c + d*x])^n*(-4*B - (4*I)*A*n - 4*B*n - (2*I)*A*n^2 - 2*B*n^ 
2 + (I*A + B)*(2 + 3*n + n^2)*Hypergeometric2F1[1, n, 1 + n, (1 + I*Tan[c 
+ d*x])/2] + 2*n*((-I)*B*n + A*(2 + n))*Tan[c + d*x] + 2*B*n*(1 + n)*Tan[c 
 + d*x]^2))/(2*d*n*(1 + n)*(2 + n))
 

3.3.20.3 Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {3042, 4080, 25, 3042, 4075, 3042, 4010, 3042, 3962, 78}

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.

\(\displaystyle \int \tan ^2(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (c+d x)^2 (a+i a \tan (c+d x))^n (A+B \tan (c+d x))dx\)

\(\Big \downarrow \) 4080

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4075

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4010

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3962

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

\(\Big \downarrow \) 78

\(\displaystyle \frac {B \tan ^2(c+d x) (a+i a \tan (c+d x))^n}{d (n+2)}-\frac {-\frac {i a (n+2) (A-i B) (a+i a \tan (c+d x))^n \operatorname {Hypergeometric2F1}\left (1,n,n+1,\frac {i \tan (c+d x) a+a}{2 a}\right )}{2 d n}+\frac {(B n+i A (n+2)) (a+i a \tan (c+d x))^{n+1}}{d (n+1)}+\frac {2 a B (a+i a \tan (c+d x))^n}{d n}}{a (n+2)}\)

input

Int[Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^n*(A + B*Tan[c + d*x]),x]
 

output

(B*Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^n)/(d*(2 + n)) - ((2*a*B*(a + I*a 
*Tan[c + d*x])^n)/(d*n) - ((I/2)*a*(A - I*B)*(2 + n)*Hypergeometric2F1[1, 
n, 1 + n, (a + I*a*Tan[c + d*x])/(2*a)]*(a + I*a*Tan[c + d*x])^n)/(d*n) + 
((B*n + I*A*(2 + n))*(a + I*a*Tan[c + d*x])^(1 + n))/(d*(1 + n)))/(a*(2 + 
n))
 

3.3.20.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 78

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

rule 3042

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

rule 3962

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[-b/d   S 
ubst[Int[(a + x)^(n - 1)/(a - x), x], x, b*Tan[c + d*x]], x] /; FreeQ[{a, b 
, c, d, n}, x] && EqQ[a^2 + b^2, 0]
 

rule 4010

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

rule 4075

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

rule 4080

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[B*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] + Simp[ 
1/(a*(m + n))   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Sim 
p[a*A*c*(m + n) - B*(b*c*m + a*d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*T 
an[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c 
 - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[n, 0]
 

3.3.20.4 Maple [F]

\[\int \left (\tan ^{2}\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{n} \left (A +B \tan \left (d x +c \right )\right )d x\]

input

int(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)
 

output

int(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)
 

3.3.20.5 Fricas [F]

\[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{2} \,d x } \]

input

integrate(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm= 
"fricas")
 

output

integral(-((A - I*B)*e^(6*I*d*x + 6*I*c) - (A - 3*I*B)*e^(4*I*d*x + 4*I*c) 
 - (A + 3*I*B)*e^(2*I*d*x + 2*I*c) + A + I*B)*(2*a*e^(2*I*d*x + 2*I*c)/(e^ 
(2*I*d*x + 2*I*c) + 1))^n/(e^(6*I*d*x + 6*I*c) + 3*e^(4*I*d*x + 4*I*c) + 3 
*e^(2*I*d*x + 2*I*c) + 1), x)
 

3.3.20.6 Sympy [F]

\[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n} \left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{2}{\left (c + d x \right )}\, dx \]

input

integrate(tan(d*x+c)**2*(a+I*a*tan(d*x+c))**n*(A+B*tan(d*x+c)),x)
 

output

Integral((I*a*(tan(c + d*x) - I))**n*(A + B*tan(c + d*x))*tan(c + d*x)**2, 
 x)
 

3.3.20.7 Maxima [F]

\[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{2} \,d x } \]

input

integrate(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm= 
"maxima")
 

output

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^n*tan(d*x + c)^2, x)
 

3.3.20.8 Giac [F]

\[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{2} \,d x } \]

input

integrate(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm= 
"giac")
 

output

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^n*tan(d*x + c)^2, x)
 

3.3.20.9 Mupad [F(-1)]

Timed out. \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^2\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x \]

input

int(tan(c + d*x)^2*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^n,x)
 

output

int(tan(c + d*x)^2*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^n, x)