\(\int (e \sec (c+d x))^m (a+i a \tan (c+d x))^2 \, dx\) [452]

Optimal result

Integrand size = 26, antiderivative size = 86 \[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^2 \, dx=\frac {i 2^{2+\frac {m}{2}} a^2 \operatorname {Hypergeometric2F1}\left (-1-\frac {m}{2},\frac {m}{2},\frac {2+m}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^m (1+i \tan (c+d x))^{-m/2}}{d m} \] Output:

I*2^(2+1/2*m)*a^2*hypergeom([1/2*m, -1-1/2*m],[1+1/2*m],1/2-1/2*I*tan(d*x+ 
c))*(e*sec(d*x+c))^m/d/m/((1+I*tan(d*x+c))^(1/2*m))
 

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.33 \[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^2 \, dx=\frac {a^2 (e \sec (c+d x))^m \left (2 i+\cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}+\cot (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}\right )}{d m} \] Input:

Integrate[(e*Sec[c + d*x])^m*(a + I*a*Tan[c + d*x])^2,x]
 

Output:

(a^2*(e*Sec[c + d*x])^m*(2*I + Cot[c + d*x]*Hypergeometric2F1[-1/2, m/2, ( 
2 + m)/2, Sec[c + d*x]^2]*Sqrt[-Tan[c + d*x]^2] + Cot[c + d*x]*Hypergeomet 
ric2F1[1/2, m/2, (2 + m)/2, Sec[c + d*x]^2]*Sqrt[-Tan[c + d*x]^2]))/(d*m)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.33, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {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.

\(\displaystyle \int (a+i a \tan (c+d x))^2 (e \sec (c+d x))^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+i a \tan (c+d x))^2 (e \sec (c+d x))^mdx\)

\(\Big \downarrow \) 3986

\(\displaystyle (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \int (a-i a \tan (c+d x))^{m/2} (i \tan (c+d x) a+a)^{\frac {m+4}{2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \int (a-i a \tan (c+d x))^{m/2} (i \tan (c+d x) a+a)^{\frac {m+4}{2}}dx\)

\(\Big \downarrow \) 4006

\(\displaystyle \frac {a^2 (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \int (a-i a \tan (c+d x))^{\frac {m-2}{2}} (i \tan (c+d x) a+a)^{\frac {m+2}{2}}d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {a 2^{\frac {m}{2}-1} (1-i \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \int \left (\frac {1}{2}-\frac {1}{2} i \tan (c+d x)\right )^{\frac {m-2}{2}} (i \tan (c+d x) a+a)^{\frac {m+2}{2}}d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 79

\(\displaystyle -\frac {i 2^{m/2} (1-i \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{\frac {m+4}{2}-\frac {m}{2}} (e \sec (c+d x))^m \operatorname {Hypergeometric2F1}\left (\frac {2-m}{2},\frac {m+4}{2},\frac {m+6}{2},\frac {1}{2} (i \tan (c+d x)+1)\right )}{d (m+4)}\)

Input:

Int[(e*Sec[c + d*x])^m*(a + I*a*Tan[c + d*x])^2,x]
 

Output:

((-I)*2^(m/2)*Hypergeometric2F1[(2 - m)/2, (4 + m)/2, (6 + m)/2, (1 + I*Ta 
n[c + d*x])/2]*(e*Sec[c + d*x])^m*(a + I*a*Tan[c + d*x])^(-1/2*m + (4 + m) 
/2))/(d*(4 + m)*(1 - I*Tan[c + d*x])^(m/2))
 

Defintions of rubi rules used

rule 79

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]))
 

rule 80

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])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3986

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]
 

rule 4006

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]
 

Maple [F]

\[\int \left (e \sec \left (d x +c \right )\right )^{m} \left (a +i a \tan \left (d x +c \right )\right )^{2}d x\]

Input:

int((e*sec(d*x+c))^m*(a+I*a*tan(d*x+c))^2,x)
 

Output:

int((e*sec(d*x+c))^m*(a+I*a*tan(d*x+c))^2,x)
 

Fricas [F]

\[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^2 \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} \left (e \sec \left (d x + c\right )\right )^{m} \,d x } \] Input:

integrate((e*sec(d*x+c))^m*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")
 

Output:

integral(4*a^2*(2*e*e^(I*d*x + I*c)/(e^(2*I*d*x + 2*I*c) + 1))^m*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)
 

Sympy [F]

\[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^2 \, dx=- a^{2} \left (\int \left (- \left (e \sec {\left (c + d x \right )}\right )^{m}\right )\, dx + \int \left (e \sec {\left (c + d x \right )}\right )^{m} \tan ^{2}{\left (c + d x \right )}\, dx + \int \left (- 2 i \left (e \sec {\left (c + d x \right )}\right )^{m} \tan {\left (c + d x \right )}\right )\, dx\right ) \] Input:

integrate((e*sec(d*x+c))**m*(a+I*a*tan(d*x+c))**2,x)
 

Output:

-a**2*(Integral(-(e*sec(c + d*x))**m, x) + Integral((e*sec(c + d*x))**m*ta 
n(c + d*x)**2, x) + Integral(-2*I*(e*sec(c + d*x))**m*tan(c + d*x), x))
 

Maxima [F]

\[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^2 \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} \left (e \sec \left (d x + c\right )\right )^{m} \,d x } \] Input:

integrate((e*sec(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*sec(d*x + c))^m, x)
 

Giac [F]

\[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^2 \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} \left (e \sec \left (d x + c\right )\right )^{m} \,d x } \] Input:

integrate((e*sec(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*sec(d*x + c))^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^2 \, dx=\int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^m\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2 \,d x \] 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)
 

Reduce [F]

\[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^2 \, dx=\frac {e^{m} a^{2} \left (2 \sec \left (d x +c \right )^{m} i +\left (\int \sec \left (d x +c \right )^{m}d x \right ) d m -\left (\int \sec \left (d x +c \right )^{m} \tan \left (d x +c \right )^{2}d x \right ) d m \right )}{d m} \] Input:

int((e*sec(d*x+c))^m*(a+I*a*tan(d*x+c))^2,x)
 

Output:

(e**m*a**2*(2*sec(c + d*x)**m*i + int(sec(c + d*x)**m,x)*d*m - int(sec(c + 
 d*x)**m*tan(c + d*x)**2,x)*d*m))/(d*m)