\(\int (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\) [498]

Optimal result

Integrand size = 23, antiderivative size = 143 \[ \int (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\frac {(A-i B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (i a+b) d (1+n)}+\frac {(i A-B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (a+i b) d (1+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3620, 3618, 70} \[ \int (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\frac {(A-i B) (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a-i b}\right )}{2 d (n+1) (b+i a)}+\frac {(-B+i A) (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a+i b}\right )}{2 d (n+1) (a+i b)} \]

[In]

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

[Out]

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

Rule 70

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] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 3618

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

Rule 3620

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(c - I*d)/2, Int[(a + b*Tan[e + f*x]
)^m*(1 + I*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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} (A-i B) \int (1+i \tan (c+d x)) (a+b \tan (c+d x))^n \, dx+\frac {1}{2} (A+i B) \int (1-i \tan (c+d x)) (a+b \tan (c+d x))^n \, dx \\ & = -\frac {(i A-B) \text {Subst}\left (\int \frac {(a+i b x)^n}{-1+x} \, dx,x,-i \tan (c+d x)\right )}{2 d}+\frac {(i A+B) \text {Subst}\left (\int \frac {(a-i b x)^n}{-1+x} \, dx,x,i \tan (c+d x)\right )}{2 d} \\ & = -\frac {(i A+B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (a-i b) d (1+n)}-\frac {(A+i B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (i a-b) d (1+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.84 \[ \int (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\frac {i \left (-\frac {(A-i B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-i b}\right )}{a-i b}+\frac {(A+i B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+i b}\right )}{a+i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 d (1+n)} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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