\(\int (e \sec (c+d x))^m (a+i a \tan (c+d x))^{3/2} \, dx\) [459]

Optimal result

Integrand size = 28, antiderivative size = 107 \[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^{3/2} \, dx=\frac {i 2^{\frac {3+m}{2}} a \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1-m),\frac {m}{2},\frac {2+m}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^m (1+i \tan (c+d x))^{\frac {1}{2} (-1-m)} \sqrt {a+i a \tan (c+d x)}}{d m} \] Output:

I*2^(3/2+1/2*m)*a*hypergeom([1/2*m, -1/2-1/2*m],[1+1/2*m],1/2-1/2*I*tan(d* 
x+c))*(e*sec(d*x+c))^m*(1+I*tan(d*x+c))^(-1/2-1/2*m)*(a+I*a*tan(d*x+c))^(1 
/2)/d/m
 

Mathematica [A] (verified)

Time = 1.62 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.74 \[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {i 2^{\frac {3}{2}+m} e^{i (c+2 d x)} \sqrt {e^{i d x}} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{\frac {1}{2}+m} \left (1+e^{2 i (c+d x)}\right )^{\frac {1}{2}+m} \operatorname {Hypergeometric2F1}\left (\frac {3}{2}+m,\frac {3+m}{2},\frac {5+m}{2},-e^{2 i (c+d x)}\right ) \sec ^{-\frac {3}{2}-m}(c+d x) (e \sec (c+d x))^m (a+i a \tan (c+d x))^{3/2}}{d (3+m) (\cos (d x)+i \sin (d x))^{3/2}} \] Input:

Integrate[(e*Sec[c + d*x])^m*(a + I*a*Tan[c + d*x])^(3/2),x]
 

Output:

((-I)*2^(3/2 + m)*E^(I*(c + 2*d*x))*Sqrt[E^(I*d*x)]*(E^(I*(c + d*x))/(1 + 
E^((2*I)*(c + d*x))))^(1/2 + m)*(1 + E^((2*I)*(c + d*x)))^(1/2 + m)*Hyperg 
eometric2F1[3/2 + m, (3 + m)/2, (5 + m)/2, -E^((2*I)*(c + d*x))]*Sec[c + d 
*x]^(-3/2 - m)*(e*Sec[c + d*x])^m*(a + I*a*Tan[c + d*x])^(3/2))/(d*(3 + m) 
*(Cos[d*x] + I*Sin[d*x])^(3/2))
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, 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.

\(\displaystyle \int (a+i a \tan (c+d x))^{3/2} (e \sec (c+d x))^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+i a \tan (c+d x))^{3/2} (e \sec (c+d x))^mdx\)

\(\Big \downarrow \) 3986

\(\displaystyle (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \int (a-i a \tan (c+d x))^{m/2} (i \tan (c+d x) a+a)^{\frac {m+3}{2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \int (a-i a \tan (c+d x))^{m/2} (i \tan (c+d x) a+a)^{\frac {m+3}{2}}dx\)

\(\Big \downarrow \) 4006

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 79

\(\displaystyle -\frac {i 2^{m/2} (1-i \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{\frac {m+3}{2}-\frac {m}{2}} (e \sec (c+d x))^m \operatorname {Hypergeometric2F1}\left (\frac {2-m}{2},\frac {m+3}{2},\frac {m+5}{2},\frac {1}{2} (i \tan (c+d x)+1)\right )}{d (m+3)}\)

Input:

Int[(e*Sec[c + d*x])^m*(a + I*a*Tan[c + d*x])^(3/2),x]
 

Output:

((-I)*2^(m/2)*Hypergeometric2F1[(2 - m)/2, (3 + m)/2, (5 + 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 + (3 + m) 
/2))/(d*(3 + 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]

\[\int \left (e \sec \left (d x +c \right )\right )^{m} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}d x\]

Input:

int((e*sec(d*x+c))^m*(a+I*a*tan(d*x+c))^(3/2),x)
 

Output:

int((e*sec(d*x+c))^m*(a+I*a*tan(d*x+c))^(3/2),x)
 

Fricas [F]

\[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^{3/2} \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \left (e \sec \left (d x + c\right )\right )^{m} \,d x } \] Input:

integrate((e*sec(d*x+c))^m*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

integral(2*sqrt(2)*a*(2*e*e^(I*d*x + I*c)/(e^(2*I*d*x + 2*I*c) + 1))^m*sqr 
t(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(3*I*d*x + 3*I*c)/(e^(2*I*d*x + 2*I*c) + 
1), x)
 

Sympy [F]

\[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^{3/2} \, dx=\int \left (e \sec {\left (c + d x \right )}\right )^{m} \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}\, dx \] Input:

integrate((e*sec(d*x+c))**m*(a+I*a*tan(d*x+c))**(3/2),x)
 

Output:

Integral((e*sec(c + d*x))**m*(I*a*(tan(c + d*x) - I))**(3/2), x)
 

Maxima [F]

\[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^{3/2} \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \left (e \sec \left (d x + c\right )\right )^{m} \,d x } \] Input:

integrate((e*sec(d*x+c))^m*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

integrate((I*a*tan(d*x + c) + a)^(3/2)*(e*sec(d*x + c))^m, x)
 

Giac [F(-2)]

Exception generated. \[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((e*sec(d*x+c))^m*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [F(-1)]

Timed out. \[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^{3/2} \, dx=\int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^m\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \] Input:

int((e/cos(c + d*x))^m*(a + a*tan(c + d*x)*1i)^(3/2),x)
 

Output:

int((e/cos(c + d*x))^m*(a + a*tan(c + d*x)*1i)^(3/2), x)
 

Reduce [F]

\[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^{3/2} \, dx=\frac {2 e^{m} \sqrt {a}\, a i \left (-\sec \left (d x +c \right )^{m} \sqrt {\tan \left (d x +c \right ) i +1}+\left (\int \sec \left (d x +c \right )^{m} \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )d x \right ) d m +\left (\int \sec \left (d x +c \right )^{m} \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )d x \right ) d \right )}{d} \] Input:

int((e*sec(d*x+c))^m*(a+I*a*tan(d*x+c))^(3/2),x)
 

Output:

(2*e**m*sqrt(a)*a*i*( - sec(c + d*x)**m*sqrt(tan(c + d*x)*i + 1) + int(sec 
(c + d*x)**m*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x),x)*d*m + int(sec(c + d* 
x)**m*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x),x)*d))/d