Integrand size = 26, antiderivative size = 86 \[ \int (e \cos (c+d x))^m (a+i a \tan (c+d x))^2 \, dx=-\frac {i 2^{2-\frac {m}{2}} a^2 (e \cos (c+d x))^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-2+m),-\frac {m}{2},1-\frac {m}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{m/2}}{d m} \] Output:
-I*2^(2-1/2*m)*a^2*(e*cos(d*x+c))^m*hypergeom([-1/2*m, -1+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)/d/m
Time = 8.33 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00 \[ \int (e \cos (c+d x))^m (a+i a \tan (c+d x))^2 \, dx=-\frac {i 2^{2-\frac {m}{2}} a^2 (e \cos (c+d x))^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-2+m),-\frac {m}{2},1-\frac {m}{2},-\frac {1}{2} i (i+\tan (c+d x))\right ) (1+i \tan (c+d x))^{m/2}}{d m} \] Input:
Integrate[(e*Cos[c + d*x])^m*(a + I*a*Tan[c + d*x])^2,x]
Output:
((-I)*2^(2 - m/2)*a^2*(e*Cos[c + d*x])^m*Hypergeometric2F1[(-2 + m)/2, -1/ 2*m, 1 - m/2, (-1/2*I)*(I + Tan[c + d*x])]*(1 + I*Tan[c + d*x])^(m/2))/(d* m)
Time = 0.55 (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.
\(\displaystyle \int (a+i a \tan (c+d x))^2 (e \cos (c+d x))^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+i a \tan (c+d x))^2 (e \cos (c+d x))^mdx\) |
\(\Big \downarrow \) 3998 |
\(\displaystyle (e \cos (c+d x))^m (e \sec (c+d x))^m \int (e \sec (c+d x))^{-m} (i \tan (c+d x) a+a)^2dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (e \cos (c+d x))^m (e \sec (c+d x))^m \int (e \sec (c+d x))^{-m} (i \tan (c+d x) a+a)^2dx\) |
\(\Big \downarrow \) 3986 |
\(\displaystyle (a-i a \tan (c+d x))^{m/2} (a+i a \tan (c+d x))^{m/2} (e \cos (c+d x))^m \int (a-i a \tan (c+d x))^{-m/2} (i \tan (c+d x) a+a)^{2-\frac {m}{2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (a-i a \tan (c+d x))^{m/2} (a+i a \tan (c+d x))^{m/2} (e \cos (c+d x))^m \int (a-i a \tan (c+d x))^{-m/2} (i \tan (c+d x) a+a)^{2-\frac {m}{2}}dx\) |
\(\Big \downarrow \) 4006 |
\(\displaystyle \frac {a^2 (a-i a \tan (c+d x))^{m/2} (a+i a \tan (c+d x))^{m/2} (e \cos (c+d x))^m \int (a-i a \tan (c+d x))^{-\frac {m}{2}-1} (i \tan (c+d x) a+a)^{1-\frac {m}{2}}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {a^3 2^{1-\frac {m}{2}} (1+i \tan (c+d x))^{m/2} (a-i a \tan (c+d x))^{m/2} (e \cos (c+d x))^m \int \left (\frac {1}{2} i \tan (c+d x)+\frac {1}{2}\right )^{1-\frac {m}{2}} (a-i a \tan (c+d x))^{-\frac {m}{2}-1}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle -\frac {i a^2 2^{2-\frac {m}{2}} (1+i \tan (c+d x))^{m/2} (e \cos (c+d x))^m \operatorname {Hypergeometric2F1}\left (\frac {m-2}{2},-\frac {m}{2},1-\frac {m}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{d m}\) |
Input:
Int[(e*Cos[c + d*x])^m*(a + I*a*Tan[c + d*x])^2,x]
Output:
((-I)*2^(2 - m/2)*a^2*(e*Cos[c + d*x])^m*Hypergeometric2F1[(-2 + m)/2, -1/ 2*m, 1 - m/2, (1 - I*Tan[c + d*x])/2]*(1 + I*Tan[c + d*x])^(m/2))/(d*m)
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]
\[\int \left (e \cos \left (d x +c \right )\right )^{m} \left (a +i a \tan \left (d x +c \right )\right )^{2}d x\]
Input:
int((e*cos(d*x+c))^m*(a+I*a*tan(d*x+c))^2,x)
Output:
int((e*cos(d*x+c))^m*(a+I*a*tan(d*x+c))^2,x)
\[ \int (e \cos (c+d x))^m (a+i a \tan (c+d x))^2 \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} \left (e \cos \left (d x + c\right )\right )^{m} \,d x } \] Input:
integrate((e*cos(d*x+c))^m*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")
Output:
integral(4*(1/2*(e*e^(2*I*d*x + 2*I*c) + e)*e^(-I*d*x - I*c))^m*a^2*e^(4*I *d*x + 4*I*c)/(e^(4*I*d*x + 4*I*c) + 2*e^(2*I*d*x + 2*I*c) + 1), x)
\[ \int (e \cos (c+d x))^m (a+i a \tan (c+d x))^2 \, dx=- a^{2} \left (\int \left (- \left (e \cos {\left (c + d x \right )}\right )^{m}\right )\, dx + \int \left (e \cos {\left (c + d x \right )}\right )^{m} \tan ^{2}{\left (c + d x \right )}\, dx + \int \left (- 2 i \left (e \cos {\left (c + d x \right )}\right )^{m} \tan {\left (c + d x \right )}\right )\, dx\right ) \] Input:
integrate((e*cos(d*x+c))**m*(a+I*a*tan(d*x+c))**2,x)
Output:
-a**2*(Integral(-(e*cos(c + d*x))**m, x) + Integral((e*cos(c + d*x))**m*ta n(c + d*x)**2, x) + Integral(-2*I*(e*cos(c + d*x))**m*tan(c + d*x), x))
\[ \int (e \cos (c+d x))^m (a+i a \tan (c+d x))^2 \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} \left (e \cos \left (d x + c\right )\right )^{m} \,d x } \] Input:
integrate((e*cos(d*x+c))^m*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")
Output:
integrate((I*a*tan(d*x + c) + a)^2*(e*cos(d*x + c))^m, x)
\[ \int (e \cos (c+d x))^m (a+i a \tan (c+d x))^2 \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} \left (e \cos \left (d x + c\right )\right )^{m} \,d x } \] Input:
integrate((e*cos(d*x+c))^m*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")
Output:
integrate((I*a*tan(d*x + c) + a)^2*(e*cos(d*x + c))^m, x)
Timed out. \[ \int (e \cos (c+d x))^m (a+i a \tan (c+d x))^2 \, dx=\int {\left (e\,\cos \left (c+d\,x\right )\right )}^m\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2 \,d x \] Input:
int((e*cos(c + d*x))^m*(a + a*tan(c + d*x)*1i)^2,x)
Output:
int((e*cos(c + d*x))^m*(a + a*tan(c + d*x)*1i)^2, x)
\[ \int (e \cos (c+d x))^m (a+i a \tan (c+d x))^2 \, dx=e^{m} a^{2} \left (\int \cos \left (d x +c \right )^{m}d x -\left (\int \cos \left (d x +c \right )^{m} \tan \left (d x +c \right )^{2}d x \right )+2 \left (\int \cos \left (d x +c \right )^{m} \tan \left (d x +c \right )d x \right ) i \right ) \] Input:
int((e*cos(d*x+c))^m*(a+I*a*tan(d*x+c))^2,x)
Output:
e**m*a**2*(int(cos(c + d*x)**m,x) - int(cos(c + d*x)**m*tan(c + d*x)**2,x) + 2*int(cos(c + d*x)**m*tan(c + d*x),x)*i)