Integrand size = 28, antiderivative size = 105 \[ \int (e \cos (c+d x))^m \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {i 2^{\frac {1}{2}-\frac {m}{2}} a (e \cos (c+d x))^m \operatorname {Hypergeometric2F1}\left (-\frac {m}{2},\frac {1+m}{2},1-\frac {m}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{\frac {1+m}{2}}}{d m \sqrt {a+i a \tan (c+d x)}} \]
-I*2^(-1/2*m+1/2)*a*(e*cos(d*x+c))^m*hypergeom([-1/2*m, 1/2+1/2*m],[1-1/2* m],1/2-1/2*I*tan(d*x+c))*(1+I*tan(d*x+c))^(1/2+1/2*m)/d/m/(a+I*a*tan(d*x+c ))^(1/2)
Time = 2.50 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.16 \[ \int (e \cos (c+d x))^m \sqrt {a+i a \tan (c+d x)} \, dx=\frac {i 2^{-m} \left (1+e^{2 i (c+d x)}\right )^{\frac {1}{2}-m} \left (e e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )\right )^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-m,\frac {1-m}{2},\frac {3-m}{2},-e^{2 i (c+d x)}\right ) \sqrt {a+i a \tan (c+d x)}}{d (-1+m)} \]
(I*(1 + E^((2*I)*(c + d*x)))^(1/2 - m)*((e*(1 + E^((2*I)*(c + d*x))))/E^(I *(c + d*x)))^m*Hypergeometric2F1[1/2 - m, (1 - m)/2, (3 - m)/2, -E^((2*I)* (c + d*x))]*Sqrt[a + I*a*Tan[c + d*x]])/(2^m*d*(-1 + m))
Time = 0.61 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3998, 3042, 3986, 3042, 4006, 80, 79}
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.
\(\displaystyle \int \sqrt {a+i a \tan (c+d x)} (e \cos (c+d x))^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {a+i a \tan (c+d x)} (e \cos (c+d x))^mdx\) |
\(\Big \downarrow \) 3998 |
\(\displaystyle (e \cos (c+d x))^m (e \sec (c+d x))^m \int (e \sec (c+d x))^{-m} \sqrt {i \tan (c+d x) a+a}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (e \cos (c+d x))^m (e \sec (c+d x))^m \int (e \sec (c+d x))^{-m} \sqrt {i \tan (c+d x) a+a}dx\) |
\(\Big \downarrow \) 3986 |
\(\displaystyle (a-i a \tan (c+d x))^{m/2} (a+i a \tan (c+d x))^{m/2} (e \cos (c+d x))^m \int (a-i a \tan (c+d x))^{-m/2} (i \tan (c+d x) a+a)^{\frac {1-m}{2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (a-i a \tan (c+d x))^{m/2} (a+i a \tan (c+d x))^{m/2} (e \cos (c+d x))^m \int (a-i a \tan (c+d x))^{-m/2} (i \tan (c+d x) a+a)^{\frac {1-m}{2}}dx\) |
\(\Big \downarrow \) 4006 |
\(\displaystyle \frac {a^2 (a-i a \tan (c+d x))^{m/2} (a+i a \tan (c+d x))^{m/2} (e \cos (c+d x))^m \int (a-i a \tan (c+d x))^{-\frac {m}{2}-1} (i \tan (c+d x) a+a)^{\frac {1}{2} (-m-1)}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {a^2 2^{-\frac {m}{2}-\frac {1}{2}} (1+i \tan (c+d x))^{\frac {m+1}{2}} (a-i a \tan (c+d x))^{m/2} (a+i a \tan (c+d x))^{\frac {1}{2} (-m-1)+\frac {m}{2}} (e \cos (c+d x))^m \int \left (\frac {1}{2} i \tan (c+d x)+\frac {1}{2}\right )^{\frac {1}{2} (-m-1)} (a-i a \tan (c+d x))^{-\frac {m}{2}-1}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle -\frac {i a 2^{\frac {1}{2}-\frac {m}{2}} (1+i \tan (c+d x))^{\frac {m+1}{2}} (a+i a \tan (c+d x))^{\frac {1}{2} (-m-1)+\frac {m}{2}} (e \cos (c+d x))^m \operatorname {Hypergeometric2F1}\left (-\frac {m}{2},\frac {m+1}{2},1-\frac {m}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{d m}\) |
((-I)*2^(1/2 - m/2)*a*(e*Cos[c + d*x])^m*Hypergeometric2F1[-1/2*m, (1 + m) /2, 1 - m/2, (1 - I*Tan[c + d*x])/2]*(1 + I*Tan[c + d*x])^((1 + m)/2)*(a + I*a*Tan[c + d*x])^((-1 - m)/2 + m/2))/(d*m)
3.7.93.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) ^FracPart[n]) Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) ), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !Integ erQ[n] && (RationalQ[m] || !SimplerQ[n + 1, m + 1])
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_.), x_Symbol] :> Simp[(d*Sec[e + f*x])^m/((a + b*Tan[e + f*x])^(m/ 2)*(a - b*Tan[e + f*x])^(m/2)) Int[(a + b*Tan[e + f*x])^(m/2 + n)*(a - b* Tan[e + f*x])^(m/2), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x _)])^(n_.), x_Symbol] :> Simp[(d*Cos[e + f*x])^m*(d*Sec[e + f*x])^m Int[( a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m , n}, x] && !IntegerQ[m]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(c/f) Subst[Int[(a + b*x)^(m - 1)*( c + d*x)^(n - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]
\[\int \left (e \cos \left (d x +c \right )\right )^{m} \sqrt {a +i a \tan \left (d x +c \right )}d x\]
\[ \int (e \cos (c+d x))^m \sqrt {a+i a \tan (c+d x)} \, dx=\int { \sqrt {i \, a \tan \left (d x + c\right ) + a} \left (e \cos \left (d x + c\right )\right )^{m} \,d x } \]
integral(sqrt(2)*(1/2*(e*e^(2*I*d*x + 2*I*c) + e)*e^(-I*d*x - I*c))^m*sqrt (a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c), x)
\[ \int (e \cos (c+d x))^m \sqrt {a+i a \tan (c+d x)} \, dx=\int \left (e \cos {\left (c + d x \right )}\right )^{m} \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}\, dx \]
\[ \int (e \cos (c+d x))^m \sqrt {a+i a \tan (c+d x)} \, dx=\int { \sqrt {i \, a \tan \left (d x + c\right ) + a} \left (e \cos \left (d x + c\right )\right )^{m} \,d x } \]
\[ \int (e \cos (c+d x))^m \sqrt {a+i a \tan (c+d x)} \, dx=\int { \sqrt {i \, a \tan \left (d x + c\right ) + a} \left (e \cos \left (d x + c\right )\right )^{m} \,d x } \]
Timed out. \[ \int (e \cos (c+d x))^m \sqrt {a+i a \tan (c+d x)} \, dx=\int {\left (e\,\cos \left (c+d\,x\right )\right )}^m\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]