Integrand size = 26, antiderivative size = 79 \[ \int \frac {(a+i a \tan (c+d x))^m}{\sqrt {\tan (c+d x)}} \, dx=\frac {2 \operatorname {AppellF1}\left (\frac {1}{2},1-m,1,\frac {3}{2},-i \tan (c+d x),i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-m} \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^m}{d} \]
2*AppellF1(1/2,1-m,1,3/2,-I*tan(d*x+c),I*tan(d*x+c))*tan(d*x+c)^(1/2)*(a+I *a*tan(d*x+c))^m/d/((1+I*tan(d*x+c))^m)
\[ \int \frac {(a+i a \tan (c+d x))^m}{\sqrt {\tan (c+d x)}} \, dx=\int \frac {(a+i a \tan (c+d x))^m}{\sqrt {\tan (c+d x)}} \, dx \]
Time = 0.31 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.23, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3042, 4047, 25, 27, 148, 27, 334, 333}
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.
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^m}{\sqrt {\tan (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^m}{\sqrt {\tan (c+d x)}}dx\) |
\(\Big \downarrow \) 4047 |
\(\displaystyle \frac {i a^2 \int -\frac {(i \tan (c+d x) a+a)^{m-1}}{a \sqrt {\tan (c+d x)} (a-i a \tan (c+d x))}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {i a^2 \int \frac {(i \tan (c+d x) a+a)^{m-1}}{a \sqrt {\tan (c+d x)} (a-i a \tan (c+d x))}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {i a \int \frac {(i \tan (c+d x) a+a)^{m-1}}{\sqrt {\tan (c+d x)} (a-i a \tan (c+d x))}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 148 |
\(\displaystyle \frac {2 a^2 \int \frac {\left (a-i a^3 \tan ^2(c+d x)\right )^{m-1}}{a \left (i a^2 \tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 a \int \frac {\left (a-i a^3 \tan ^2(c+d x)\right )^{m-1}}{i a^2 \tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 334 |
\(\displaystyle \frac {2 \left (1-i a^2 \tan ^2(c+d x)\right )^{-m} \left (a-i a^3 \tan ^2(c+d x)\right )^m \int \frac {\left (1-i a^2 \tan ^2(c+d x)\right )^{m-1}}{i a^2 \tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 333 |
\(\displaystyle \frac {2 i a \tan (c+d x) \left (1-i a^2 \tan ^2(c+d x)\right )^{-m} \left (a-i a^3 \tan ^2(c+d x)\right )^m \operatorname {AppellF1}\left (\frac {1}{2},1,1-m,\frac {3}{2},-i a^2 \tan ^2(c+d x),i a^2 \tan ^2(c+d x)\right )}{d}\) |
((2*I)*a*AppellF1[1/2, 1, 1 - m, 3/2, (-I)*a^2*Tan[c + d*x]^2, I*a^2*Tan[c + d*x]^2]*Tan[c + d*x]*(a - I*a^3*Tan[c + d*x]^2)^m)/(d*(1 - I*a^2*Tan[c + d*x]^2)^m)
3.4.34.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))^(p_.), x_] :> With[{k = Denominator[m]}, Simp[k/b Subst[Int[x^(k*(m + 1) - 1)*(c + d*(x^k/b))^n*(e + f*(x^k/b))^p, x], x, (b*x)^(1/k)], x]] /; FreeQ[{b, c, d, e, f, n, p}, x] && FractionQ[m] && IntegerQ[p]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[ (1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && !(IntegerQ[p] || GtQ[a, 0])
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[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]
\[\int \frac {\left (a +i a \tan \left (d x +c \right )\right )^{m}}{\sqrt {\tan \left (d x +c \right )}}d x\]
\[ \int \frac {(a+i a \tan (c+d x))^m}{\sqrt {\tan (c+d x)}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{m}}{\sqrt {\tan \left (d x + c\right )}} \,d x } \]
integral((2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1))^m*sqrt((-I*e^ (2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1), x)
\[ \int \frac {(a+i a \tan (c+d x))^m}{\sqrt {\tan (c+d x)}} \, dx=\int \frac {\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{m}}{\sqrt {\tan {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {(a+i a \tan (c+d x))^m}{\sqrt {\tan (c+d x)}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{m}}{\sqrt {\tan \left (d x + c\right )}} \,d x } \]
\[ \int \frac {(a+i a \tan (c+d x))^m}{\sqrt {\tan (c+d x)}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{m}}{\sqrt {\tan \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^m}{\sqrt {\tan (c+d x)}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^m}{\sqrt {\mathrm {tan}\left (c+d\,x\right )}} \,d x \]