3.5.90 \(\int (e \sec (c+d x))^{3-n} (a+i a \tan (c+d x))^n \, dx\) [490]

Optimal. Leaf size=121 \[ \frac {i 2^{\frac {3+n}{2}} a \, _2F_1\left (\frac {1}{2} (-1-n),\frac {3-n}{2};\frac {5-n}{2};\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{3-n} (1+i \tan (c+d x))^{\frac {1}{2} (-1-n)} (a+i a \tan (c+d x))^{-1+n}}{d (3-n)} \]

[Out]

I*2^(3/2+1/2*n)*a*hypergeom([3/2-1/2*n, -1/2-1/2*n],[5/2-1/2*n],1/2-1/2*I*tan(d*x+c))*(e*sec(d*x+c))^(3-n)*(1+
I*tan(d*x+c))^(-1/2-1/2*n)*(a+I*a*tan(d*x+c))^(-1+n)/d/(3-n)

________________________________________________________________________________________

Rubi [A]
time = 0.16, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3586, 3604, 72, 71} \begin {gather*} \frac {i a 2^{\frac {n+3}{2}} (1+i \tan (c+d x))^{\frac {1}{2} (-n-1)} (a+i a \tan (c+d x))^{n-1} (e \sec (c+d x))^{3-n} \, _2F_1\left (\frac {1}{2} (-n-1),\frac {3-n}{2};\frac {5-n}{2};\frac {1}{2} (1-i \tan (c+d x))\right )}{d (3-n)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(I*2^((3 + n)/2)*a*Hypergeometric2F1[(-1 - n)/2, (3 - n)/2, (5 - n)/2, (1 - I*Tan[c + d*x])/2]*(e*Sec[c + d*x]
)^(3 - n)*(1 + I*Tan[c + d*x])^((-1 - n)/2)*(a + I*a*Tan[c + d*x])^(-1 + n))/(d*(3 - n))

Rule 71

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] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 72

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(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] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 3586

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*S
ec[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 3604

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist
[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]

Rubi steps

\begin {align*} \int (e \sec (c+d x))^{3-n} (a+i a \tan (c+d x))^n \, dx &=\left ((e \sec (c+d x))^{3-n} (a-i a \tan (c+d x))^{\frac {1}{2} (-3+n)} (a+i a \tan (c+d x))^{\frac {1}{2} (-3+n)}\right ) \int (a-i a \tan (c+d x))^{\frac {3-n}{2}} (a+i a \tan (c+d x))^{\frac {3-n}{2}+n} \, dx\\ &=\frac {\left (a^2 (e \sec (c+d x))^{3-n} (a-i a \tan (c+d x))^{\frac {1}{2} (-3+n)} (a+i a \tan (c+d x))^{\frac {1}{2} (-3+n)}\right ) \text {Subst}\left (\int (a-i a x)^{-1+\frac {3-n}{2}} (a+i a x)^{-1+\frac {3-n}{2}+n} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\left (2^{\frac {1}{2}+\frac {n}{2}} a^2 (e \sec (c+d x))^{3-n} (a-i a \tan (c+d x))^{\frac {1}{2} (-3+n)} (a+i a \tan (c+d x))^{\frac {1}{2}+\frac {1}{2} (-3+n)+\frac {n}{2}} \left (\frac {a+i a \tan (c+d x)}{a}\right )^{-\frac {1}{2}-\frac {n}{2}}\right ) \text {Subst}\left (\int \left (\frac {1}{2}+\frac {i x}{2}\right )^{-1+\frac {3-n}{2}+n} (a-i a x)^{-1+\frac {3-n}{2}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {i 2^{\frac {3+n}{2}} a \, _2F_1\left (\frac {1}{2} (-1-n),\frac {3-n}{2};\frac {5-n}{2};\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{3-n} (1+i \tan (c+d x))^{\frac {1}{2} (-1-n)} (a+i a \tan (c+d x))^{-1+n}}{d (3-n)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 7.72, size = 116, normalized size = 0.96 \begin {gather*} \frac {8 e^3 \, _2F_1\left (3,\frac {3-n}{2};\frac {5-n}{2};-\cos (2 (c+d x))+i \sin (2 (c+d x))\right ) \sec (d x) (e \sec (c+d x))^{-n} (i+\tan (d x)) (a+i a \tan (c+d x))^n}{d (-3+n) (\cos (c)+i \sin (c))^3 (-i+\tan (d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*e^3*Hypergeometric2F1[3, (3 - n)/2, (5 - n)/2, -Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]]*Sec[d*x]*(I + Tan[d*
x])*(a + I*a*Tan[c + d*x])^n)/(d*(-3 + n)*(e*Sec[c + d*x])^n*(Cos[c] + I*Sin[c])^3*(-I + Tan[d*x])^2)

________________________________________________________________________________________

Maple [F]
time = 0.76, size = 0, normalized size = 0.00 \[\int \left (e \sec \left (d x +c \right )\right )^{3-n} \left (a +i a \tan \left (d x +c \right )\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*(6*a^n*cos(c*n + (d*n + d)*x + c)*e^3 + 6*I*a^n*e^3*sin(c*n + (d*n + d)*x + c) - (a^n*n*e^3 - 5*a^n*e^3)*cos
(c*n + (d*n + 3*d)*x + 3*c) - 6*((I*a^n*d*n^3 - 7*I*a^n*d*n^2 + 7*I*a^n*d*n + 15*I*a^n*d)*cos(c*n)*e^n - (a^n*
d*n^3 - 7*a^n*d*n^2 + 7*a^n*d*n + 15*a^n*d)*e^n*sin(c*n) + ((I*a^n*d*n^3 - 7*I*a^n*d*n^2 + 7*I*a^n*d*n + 15*I*
a^n*d)*cos(c*n)*e^n - (a^n*d*n^3 - 7*a^n*d*n^2 + 7*a^n*d*n + 15*a^n*d)*e^n*sin(c*n))*cos(6*d*x + 6*c) + 3*((I*
a^n*d*n^3 - 7*I*a^n*d*n^2 + 7*I*a^n*d*n + 15*I*a^n*d)*cos(c*n)*e^n - (a^n*d*n^3 - 7*a^n*d*n^2 + 7*a^n*d*n + 15
*a^n*d)*e^n*sin(c*n))*cos(4*d*x + 4*c) + 3*((I*a^n*d*n^3 - 7*I*a^n*d*n^2 + 7*I*a^n*d*n + 15*I*a^n*d)*cos(c*n)*
e^n - (a^n*d*n^3 - 7*a^n*d*n^2 + 7*a^n*d*n + 15*a^n*d)*e^n*sin(c*n))*cos(2*d*x + 2*c) - ((a^n*d*n^3 - 7*a^n*d*
n^2 + 7*a^n*d*n + 15*a^n*d)*cos(c*n)*e^n - (-I*a^n*d*n^3 + 7*I*a^n*d*n^2 - 7*I*a^n*d*n - 15*I*a^n*d)*e^n*sin(c
*n))*sin(6*d*x + 6*c) - 3*((a^n*d*n^3 - 7*a^n*d*n^2 + 7*a^n*d*n + 15*a^n*d)*cos(c*n)*e^n - (-I*a^n*d*n^3 + 7*I
*a^n*d*n^2 - 7*I*a^n*d*n - 15*I*a^n*d)*e^n*sin(c*n))*sin(4*d*x + 4*c) - 3*((a^n*d*n^3 - 7*a^n*d*n^2 + 7*a^n*d*
n + 15*a^n*d)*cos(c*n)*e^n - (-I*a^n*d*n^3 + 7*I*a^n*d*n^2 - 7*I*a^n*d*n - 15*I*a^n*d)*e^n*sin(c*n))*sin(2*d*x
+ 2*c))*integrate(((cos(8*d*x + 8*c)*e^3 + 4*cos(6*d*x + 6*c)*e^3 + 6*cos(4*d*x + 4*c)*e^3 + 4*cos(2*d*x + 2*
c)*e^3 + e^3)*cos((d*n + d)*x + c) + (e^3*sin(8*d*x + 8*c) + 4*e^3*sin(6*d*x + 6*c) + 6*e^3*sin(4*d*x + 4*c) +
4*e^3*sin(2*d*x + 2*c))*sin((d*n + d)*x + c))/((n^2 - 8*n + 15)*cos(8*d*x + 8*c)^2*e^n + 16*(n^2 - 8*n + 15)*
cos(6*d*x + 6*c)^2*e^n + 36*(n^2 - 8*n + 15)*cos(4*d*x + 4*c)^2*e^n + 16*(n^2 - 8*n + 15)*cos(2*d*x + 2*c)^2*e
^n + (n^2 - 8*n + 15)*e^n*sin(8*d*x + 8*c)^2 + 16*(n^2 - 8*n + 15)*e^n*sin(6*d*x + 6*c)^2 + 36*(n^2 - 8*n + 15
)*e^n*sin(4*d*x + 4*c)^2 + 48*(n^2 - 8*n + 15)*e^n*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 16*(n^2 - 8*n + 15)*e^n
*sin(2*d*x + 2*c)^2 + 8*(n^2 - 8*n + 15)*cos(2*d*x + 2*c)*e^n + 2*(4*(n^2 - 8*n + 15)*cos(6*d*x + 6*c)*e^n + 6
*(n^2 - 8*n + 15)*cos(4*d*x + 4*c)*e^n + 4*(n^2 - 8*n + 15)*cos(2*d*x + 2*c)*e^n + (n^2 - 8*n + 15)*e^n)*cos(8
*d*x + 8*c) + 8*(6*(n^2 - 8*n + 15)*cos(4*d*x + 4*c)*e^n + 4*(n^2 - 8*n + 15)*cos(2*d*x + 2*c)*e^n + (n^2 - 8*
n + 15)*e^n)*cos(6*d*x + 6*c) + 12*(4*(n^2 - 8*n + 15)*cos(2*d*x + 2*c)*e^n + (n^2 - 8*n + 15)*e^n)*cos(4*d*x
+ 4*c) + (n^2 - 8*n + 15)*e^n + 4*(2*(n^2 - 8*n + 15)*e^n*sin(6*d*x + 6*c) + 3*(n^2 - 8*n + 15)*e^n*sin(4*d*x
+ 4*c) + 2*(n^2 - 8*n + 15)*e^n*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + 16*(3*(n^2 - 8*n + 15)*e^n*sin(4*d*x + 4*
c) + 2*(n^2 - 8*n + 15)*e^n*sin(2*d*x + 2*c))*sin(6*d*x + 6*c)), x) + 6*((a^n*d*n^3 - 7*a^n*d*n^2 + 7*a^n*d*n
+ 15*a^n*d)*cos(c*n)*e^n - (-I*a^n*d*n^3 + 7*I*a^n*d*n^2 - 7*I*a^n*d*n - 15*I*a^n*d)*e^n*sin(c*n) + ((a^n*d*n^
3 - 7*a^n*d*n^2 + 7*a^n*d*n + 15*a^n*d)*cos(c*n)*e^n - (-I*a^n*d*n^3 + 7*I*a^n*d*n^2 - 7*I*a^n*d*n - 15*I*a^n*
d)*e^n*sin(c*n))*cos(6*d*x + 6*c) + 3*((a^n*d*n^3 - 7*a^n*d*n^2 + 7*a^n*d*n + 15*a^n*d)*cos(c*n)*e^n - (-I*a^n
*d*n^3 + 7*I*a^n*d*n^2 - 7*I*a^n*d*n - 15*I*a^n*d)*e^n*sin(c*n))*cos(4*d*x + 4*c) + 3*((a^n*d*n^3 - 7*a^n*d*n^
2 + 7*a^n*d*n + 15*a^n*d)*cos(c*n)*e^n - (-I*a^n*d*n^3 + 7*I*a^n*d*n^2 - 7*I*a^n*d*n - 15*I*a^n*d)*e^n*sin(c*n
))*cos(2*d*x + 2*c) - ((-I*a^n*d*n^3 + 7*I*a^n*d*n^2 - 7*I*a^n*d*n - 15*I*a^n*d)*cos(c*n)*e^n + (a^n*d*n^3 - 7
*a^n*d*n^2 + 7*a^n*d*n + 15*a^n*d)*e^n*sin(c*n))*sin(6*d*x + 6*c) - 3*((-I*a^n*d*n^3 + 7*I*a^n*d*n^2 - 7*I*a^n
*d*n - 15*I*a^n*d)*cos(c*n)*e^n + (a^n*d*n^3 - 7*a^n*d*n^2 + 7*a^n*d*n + 15*a^n*d)*e^n*sin(c*n))*sin(4*d*x + 4
*c) - 3*((-I*a^n*d*n^3 + 7*I*a^n*d*n^2 - 7*I*a^n*d*n - 15*I*a^n*d)*cos(c*n)*e^n + (a^n*d*n^3 - 7*a^n*d*n^2 + 7
*a^n*d*n + 15*a^n*d)*e^n*sin(c*n))*sin(2*d*x + 2*c))*integrate(-((e^3*sin(8*d*x + 8*c) + 4*e^3*sin(6*d*x + 6*c
) + 6*e^3*sin(4*d*x + 4*c) + 4*e^3*sin(2*d*x + 2*c))*cos((d*n + d)*x + c) - (cos(8*d*x + 8*c)*e^3 + 4*cos(6*d*
x + 6*c)*e^3 + 6*cos(4*d*x + 4*c)*e^3 + 4*cos(2*d*x + 2*c)*e^3 + e^3)*sin((d*n + d)*x + c))/((n^2 - 8*n + 15)*
cos(8*d*x + 8*c)^2*e^n + 16*(n^2 - 8*n + 15)*cos(6*d*x + 6*c)^2*e^n + 36*(n^2 - 8*n + 15)*cos(4*d*x + 4*c)^2*e
^n + 16*(n^2 - 8*n + 15)*cos(2*d*x + 2*c)^2*e^n + (n^2 - 8*n + 15)*e^n*sin(8*d*x + 8*c)^2 + 16*(n^2 - 8*n + 15
)*e^n*sin(6*d*x + 6*c)^2 + 36*(n^2 - 8*n + 15)*e^n*sin(4*d*x + 4*c)^2 + 48*(n^2 - 8*n + 15)*e^n*sin(4*d*x + 4*
c)*sin(2*d*x + 2*c) + 16*(n^2 - 8*n + 15)*e^n*sin(2*d*x + 2*c)^2 + 8*(n^2 - 8*n + 15)*cos(2*d*x + 2*c)*e^n + 2
*(4*(n^2 - 8*n + 15)*cos(6*d*x + 6*c)*e^n + 6*(n^2 - 8*n + 15)*cos(4*d*x + 4*c)*e^n + 4*(n^2 - 8*n + 15)*cos(2
*d*x + 2*c)*e^n + (n^2 - 8*n + 15)*e^n)*cos(8*d*x + 8*c) + 8*(6*(n^2 - 8*n + 15)*cos(4*d*x + 4*c)*e^n + 4*(n^2
- 8*n + 15)*cos(2*d*x + 2*c)*e^n + (n^2 - 8*n + 15)*e^n)*cos(6*d*x + 6*c) + 12*(4*(n^2 - 8*n + 15)*cos(2*d*x
+ 2*c)*e^n + (n^2 - 8*n + 15)*e^n)*cos(4*d*x + 4*c) + (n^2 - 8*n + 15)*e^n + 4*(2*(n^2 - 8*n + 15)*e^n*sin(6*d
*x + 6*c) + 3*(n^2 - 8*n + 15)*e^n*sin(4*d*x + 4*c) + 2*(n^2 - 8*n + 15)*e^n*sin(2*d*x + 2*c))*sin(8*d*x + 8*c
) + 16*(3*(n^2 - 8*n + 15)*e^n*sin(4*d*x + 4*c)...

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(((-I*n - I)*e^(4*I*d*x + 4*I*c) - 2*I*n*e^(2*I*d*x + 2*I*c) - I*n + I)*(2*e^(I*d*x + I*c + 1)/(e^(2*I*d*x
+ 2*I*c) + 1))^(-n + 3)*e^(I*d*n*x + I*c*n + n*log(a*e^(-1)) + n*log(2*e^(I*d*x + I*c + 1)/(e^(2*I*d*x + 2*I*
c) + 1))) + 8*d*e^(2*I*d*x + 2*I*c)*integral(-1/8*(n^2 + (n^2 - 1)*e^(4*I*d*x + 4*I*c) + 2*(n^2 - 1)*e^(2*I*d*
x + 2*I*c) - 1)*(2*e^(I*d*x + I*c + 1)/(e^(2*I*d*x + 2*I*c) + 1))^(-n + 3)*e^(I*d*n*x + I*c*n - 2*I*d*x + n*lo
g(a*e^(-1)) + n*log(2*e^(I*d*x + I*c + 1)/(e^(2*I*d*x + 2*I*c) + 1)) - 2*I*c), x))*e^(-2*I*d*x - 2*I*c)/d

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (e \sec {\left (c + d x \right )}\right )^{3 - n} \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{3-n}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________