3.5.51 \(\int (e \sec (c+d x))^m (a+i a \tan (c+d x))^3 \, dx\) [451]

3.5.51.1 Optimal result

Integrand size = 26, antiderivative size = 86 \[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^3 \, dx=\frac {i 2^{3+\frac {m}{2}} a^3 \operatorname {Hypergeometric2F1}\left (-2-\frac {m}{2},\frac {m}{2},\frac {2+m}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^m (1+i \tan (c+d x))^{-m/2}}{d m} \]

output

I*2^(3+1/2*m)*a^3*hypergeom([1/2*m, -2-1/2*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))
 

3.5.51.2 Mathematica [A] (verified)

Time = 1.55 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.71 \[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^3 \, dx=-\frac {i a^3 (e \sec (c+d x))^m \left (-3 i (2+m) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\sec ^2(c+d x)\right ) \tan (c+d x)-i (2+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\sec ^2(c+d x)\right ) \tan (c+d x)+\left (-8-4 m+m \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}\right )}{d m (2+m) \sqrt {-\tan ^2(c+d x)}} \]

input

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

output

((-I)*a^3*(e*Sec[c + d*x])^m*((-3*I)*(2 + m)*Hypergeometric2F1[-1/2, m/2, 
(2 + m)/2, Sec[c + d*x]^2]*Tan[c + d*x] - I*(2 + m)*Hypergeometric2F1[1/2, 
 m/2, (2 + m)/2, Sec[c + d*x]^2]*Tan[c + d*x] + (-8 - 4*m + m*Sec[c + d*x] 
^2)*Sqrt[-Tan[c + d*x]^2]))/(d*m*(2 + m)*Sqrt[-Tan[c + d*x]^2])
 

3.5.51.3 Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.02, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3986

\(\displaystyle (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \int (a-i a \tan (c+d x))^{m/2} (i \tan (c+d x) a+a)^{\frac {m+6}{2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \int (a-i a \tan (c+d x))^{m/2} (i \tan (c+d x) a+a)^{\frac {m+6}{2}}dx\)

\(\Big \downarrow \) 4006

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 79

\(\displaystyle \frac {i a^3 2^{\frac {m}{2}+3} (1+i \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-m-4),\frac {m}{2},\frac {m+2}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{d m}\)

input

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

output

(I*2^(3 + m/2)*a^3*Hypergeometric2F1[(-4 - m)/2, m/2, (2 + m)/2, (1 - I*Ta 
n[c + d*x])/2]*(e*Sec[c + d*x])^m)/(d*m*(1 + I*Tan[c + d*x])^(m/2))
 

3.5.51.3.1 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]
 

3.5.51.4 Maple [F]

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

input

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

output

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

3.5.51.5 Fricas [F]

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

input

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

output

integral(8*a^3*(2*e*e^(I*d*x + I*c)/(e^(2*I*d*x + 2*I*c) + 1))^m*e^(6*I*d* 
x + 6*I*c)/(e^(6*I*d*x + 6*I*c) + 3*e^(4*I*d*x + 4*I*c) + 3*e^(2*I*d*x + 2 
*I*c) + 1), x)
 

3.5.51.6 Sympy [F]

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

input

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

output

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

3.5.51.7 Maxima [F]

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

input

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

output

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

3.5.51.8 Giac [F]

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

input

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

output

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

3.5.51.9 Mupad [F(-1)]

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

input

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

output

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