\(\int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\) [570]

Optimal result

Integrand size = 36, antiderivative size = 158 \[ \int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\frac {2 (A-i B) \operatorname {AppellF1}\left (\frac {1}{2},1-n,1,\frac {3}{2},-i \tan (c+d x),i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n}{d \sqrt {\cot (c+d x)}}+\frac {2 i B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},-i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n}{d \sqrt {\cot (c+d x)}} \]

[Out]

2*(A-I*B)*AppellF1(1/2,1-n,1,3/2,-I*tan(d*x+c),I*tan(d*x+c))*(a+I*a*tan(d*x+c))^n/d/cot(d*x+c)^(1/2)/((1+I*tan
(d*x+c))^n)+2*I*B*hypergeom([1/2, 1-n],[3/2],-I*tan(d*x+c))*(a+I*a*tan(d*x+c))^n/d/cot(d*x+c)^(1/2)/((1+I*tan(
d*x+c))^n)

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4326, 3682, 3645, 129, 441, 440, 3680, 68, 66} \[ \int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\frac {2 (A-i B) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \operatorname {AppellF1}\left (\frac {1}{2},1-n,1,\frac {3}{2},-i \tan (c+d x),i \tan (c+d x)\right )}{d \sqrt {\cot (c+d x)}}+\frac {2 i B (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},-i \tan (c+d x)\right )}{d \sqrt {\cot (c+d x)}} \]

[In]

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

[Out]

(2*(A - I*B)*AppellF1[1/2, 1 - n, 1, 3/2, (-I)*Tan[c + d*x], I*Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^n)/(d*Sqrt
[Cot[c + d*x]]*(1 + I*Tan[c + d*x])^n) + ((2*I)*B*Hypergeometric2F1[1/2, 1 - n, 3/2, (-I)*Tan[c + d*x]]*(a + I
*a*Tan[c + d*x])^n)/(d*Sqrt[Cot[c + d*x]]*(1 + I*Tan[c + d*x])^n)

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-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 68

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

Rule 129

Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]
}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + b*(x^k/e))^m*(c + d*(x^k/e))^n, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 441

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^F
racPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 3645

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

Rule 3680

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b*(B/f), Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x
]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && EqQ[A*b + a*B,
0]

Rule 3682

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

Rule 4326

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx \\ & = -\left (\left ((-A+i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a+i a \tan (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx\right )+\frac {\left (i B \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a-i a \tan (c+d x)) (a+i a \tan (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx}{a} \\ & = -\frac {\left (i a^2 (-A+i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {(a+x)^{-1+n}}{\sqrt {-\frac {i x}{a}} \left (-a^2+a x\right )} \, dx,x,i a \tan (c+d x)\right )}{d}+\frac {\left (i a B \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {(a+i a x)^{-1+n}}{\sqrt {x}} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\left (2 a^3 (-A+i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {\left (a+i a x^2\right )^{-1+n}}{-a^2+i a^2 x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}+\frac {\left (i B \sqrt {\cot (c+d x)} (1+i \tan (c+d x))^{-n} \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n\right ) \text {Subst}\left (\int \frac {(1+i x)^{-1+n}}{\sqrt {x}} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {2 i B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},-i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n}{d \sqrt {\cot (c+d x)}}+\frac {\left (2 a^2 (-A+i B) \sqrt {\cot (c+d x)} (1+i \tan (c+d x))^{-n} \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n\right ) \text {Subst}\left (\int \frac {\left (1+i x^2\right )^{-1+n}}{-a^2+i a^2 x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = \frac {2 (A-i B) \operatorname {AppellF1}\left (\frac {1}{2},1-n,1,\frac {3}{2},-i \tan (c+d x),i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n}{d \sqrt {\cot (c+d x)}}+\frac {2 i B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},-i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n}{d \sqrt {\cot (c+d x)}} \\ \end{align*}

Mathematica [F]

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

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(((A - 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*sqrt(
(I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))/(e^(2*I*d*x + 2*I*c) + 1), x)

Sympy [F]

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

[In]

integrate(cot(d*x+c)**(1/2)*(a+I*a*tan(d*x+c))**n*(A+B*tan(d*x+c)),x)

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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