Integrand size = 26, antiderivative size = 86 \[ \int \frac {(e \cos (c+d x))^m}{a+i a \tan (c+d x)} \, dx=-\frac {i 2^{-1-\frac {m}{2}} (e \cos (c+d x))^m \operatorname {Hypergeometric2F1}\left (-\frac {m}{2},\frac {4+m}{2},1-\frac {m}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{m/2}}{a d m} \]
-I*2^(-1-1/2*m)*(e*cos(d*x+c))^m*hypergeom([-1/2*m, 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*m)/a/d/m
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(433\) vs. \(2(86)=172\).
Time = 7.52 (sec) , antiderivative size = 433, normalized size of antiderivative = 5.03 \[ \int \frac {(e \cos (c+d x))^m}{a+i a \tan (c+d x)} \, dx=-\frac {2^{-m/2} \cos (c+d x) (e \cos (c+d x))^m \left (1-2 \cos ^2(c+d x)+i \sin (2 (c+d x))\right )^{m/2} \left (2^{m/2} (2+m) \operatorname {Hypergeometric2F1}\left (1+m,\frac {2+m}{2},2+m,2 \cos (c+d x) (\cos (c+d x)-i \sin (c+d x))\right ) ((\cos (c)-i \sin (c)) \sin (c) (i+\tan (d x)))^{m/2}-2 (1+m) \operatorname {Hypergeometric2F1}\left (-1-\frac {m}{2},\frac {m}{2},-\frac {m}{2},\frac {1}{2} (1+i \tan (c+d x))\right ) ((\cos (c)+i \sin (c)) \sin (c) (-i+\tan (d x)))^{m/2} (1-i \tan (c+d x))^{m/2}\right )}{a d (1+m) (2+m) \left (-i \sin (c+d x) \left (\left (1-2 \cos ^2(c+d x)+i \sin (2 (c+d x))\right )^{m/2} ((\cos (c)+i \sin (c)) \sin (c) (-i+\tan (d x)))^{m/2}-((\cos (c)-i \sin (c)) \sin (c) (i+\tan (d x)))^{m/2}\right )+\cos (c+d x) \left (\left (1-2 \cos ^2(c+d x)+i \sin (2 (c+d x))\right )^{m/2} ((\cos (c)+i \sin (c)) \sin (c) (-i+\tan (d x)))^{m/2}+((\cos (c)-i \sin (c)) \sin (c) (i+\tan (d x)))^{m/2}\right )\right ) (-i+\tan (c+d x))} \]
-((Cos[c + d*x]*(e*Cos[c + d*x])^m*(1 - 2*Cos[c + d*x]^2 + I*Sin[2*(c + d* x)])^(m/2)*(2^(m/2)*(2 + m)*Hypergeometric2F1[1 + m, (2 + m)/2, 2 + m, 2*C os[c + d*x]*(Cos[c + d*x] - I*Sin[c + d*x])]*((Cos[c] - I*Sin[c])*Sin[c]*( I + Tan[d*x]))^(m/2) - 2*(1 + m)*Hypergeometric2F1[-1 - m/2, m/2, -1/2*m, (1 + I*Tan[c + d*x])/2]*((Cos[c] + I*Sin[c])*Sin[c]*(-I + Tan[d*x]))^(m/2) *(1 - I*Tan[c + d*x])^(m/2)))/(2^(m/2)*a*d*(1 + m)*(2 + m)*((-I)*Sin[c + d *x]*((1 - 2*Cos[c + d*x]^2 + I*Sin[2*(c + d*x)])^(m/2)*((Cos[c] + I*Sin[c] )*Sin[c]*(-I + Tan[d*x]))^(m/2) - ((Cos[c] - I*Sin[c])*Sin[c]*(I + Tan[d*x ]))^(m/2)) + Cos[c + d*x]*((1 - 2*Cos[c + d*x]^2 + I*Sin[2*(c + d*x)])^(m/ 2)*((Cos[c] + I*Sin[c])*Sin[c]*(-I + Tan[d*x]))^(m/2) + ((Cos[c] - I*Sin[c ])*Sin[c]*(I + Tan[d*x]))^(m/2)))*(-I + Tan[c + d*x])))
Time = 0.58 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, 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.
\(\displaystyle \int \frac {(e \cos (c+d x))^m}{a+i a \tan (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(e \cos (c+d x))^m}{a+i a \tan (c+d x)}dx\) |
\(\Big \downarrow \) 3998 |
\(\displaystyle (e \cos (c+d x))^m (e \sec (c+d x))^m \int \frac {(e \sec (c+d x))^{-m}}{i \tan (c+d x) a+a}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (e \cos (c+d x))^m (e \sec (c+d x))^m \int \frac {(e \sec (c+d x))^{-m}}{i \tan (c+d x) a+a}dx\) |
\(\Big \downarrow \) 3986 |
\(\displaystyle (a-i a \tan (c+d x))^{m/2} (a+i a \tan (c+d x))^{m/2} (e \cos (c+d x))^m \int (a-i a \tan (c+d x))^{-m/2} (i \tan (c+d x) a+a)^{-\frac {m}{2}-1}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (a-i a \tan (c+d x))^{m/2} (a+i a \tan (c+d x))^{m/2} (e \cos (c+d x))^m \int (a-i a \tan (c+d x))^{-m/2} (i \tan (c+d x) a+a)^{-\frac {m}{2}-1}dx\) |
\(\Big \downarrow \) 4006 |
\(\displaystyle \frac {a^2 (a-i a \tan (c+d x))^{m/2} (a+i a \tan (c+d x))^{m/2} (e \cos (c+d x))^m \int (a-i a \tan (c+d x))^{-\frac {m}{2}-1} (i \tan (c+d x) a+a)^{-\frac {m}{2}-2}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {2^{-\frac {m}{2}-2} (1+i \tan (c+d x))^{m/2} (a-i a \tan (c+d x))^{m/2} (e \cos (c+d x))^m \int \left (\frac {1}{2} i \tan (c+d x)+\frac {1}{2}\right )^{-\frac {m}{2}-2} (a-i a \tan (c+d x))^{-\frac {m}{2}-1}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle -\frac {i 2^{-\frac {m}{2}-1} (1+i \tan (c+d x))^{m/2} (e \cos (c+d x))^m \operatorname {Hypergeometric2F1}\left (-\frac {m}{2},\frac {m+4}{2},1-\frac {m}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{a d m}\) |
((-I)*2^(-1 - m/2)*(e*Cos[c + d*x])^m*Hypergeometric2F1[-1/2*m, (4 + m)/2, 1 - m/2, (1 - I*Tan[c + d*x])/2]*(1 + I*Tan[c + d*x])^(m/2))/(a*d*m)
3.7.91.3.1 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]
\[\int \frac {\left (e \cos \left (d x +c \right )\right )^{m}}{a +i a \tan \left (d x +c \right )}d x\]
\[ \int \frac {(e \cos (c+d x))^m}{a+i a \tan (c+d x)} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{m}}{i \, a \tan \left (d x + c\right ) + a} \,d x } \]
integral(1/2*(1/2*(e*e^(2*I*d*x + 2*I*c) + e)*e^(-I*d*x - I*c))^m*(e^(2*I* d*x + 2*I*c) + 1)*e^(-2*I*d*x - 2*I*c)/a, x)
\[ \int \frac {(e \cos (c+d x))^m}{a+i a \tan (c+d x)} \, dx=- \frac {i \int \frac {\left (e \cos {\left (c + d x \right )}\right )^{m}}{\tan {\left (c + d x \right )} - i}\, dx}{a} \]
Exception generated. \[ \int \frac {(e \cos (c+d x))^m}{a+i a \tan (c+d x)} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {(e \cos (c+d x))^m}{a+i a \tan (c+d x)} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{m}}{i \, a \tan \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(e \cos (c+d x))^m}{a+i a \tan (c+d x)} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^m}{a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]