3.473 \(\int \sec (c+d x) (a+i a \tan (c+d x))^n \, dx\)

Optimal. Leaf size=88 \[ \frac {i a 2^{n+\frac {1}{2}} \sec (c+d x) (1+i \tan (c+d x))^{\frac {1}{2}-n} (a+i a \tan (c+d x))^{n-1} \, _2F_1\left (\frac {1}{2},\frac {1}{2}-n;\frac {3}{2};\frac {1}{2} (1-i \tan (c+d x))\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3505, 3523, 70, 69} \[ \frac {i a 2^{n+\frac {1}{2}} \sec (c+d x) (1+i \tan (c+d x))^{\frac {1}{2}-n} (a+i a \tan (c+d x))^{n-1} \text {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-n,\frac {3}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), 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 70

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 3505

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 3523

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

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Mathematica [A]  time = 8.35, size = 134, normalized size = 1.52 \[ -\frac {i 2^{n+1} e^{i (c+d x)} \left (e^{i d x}\right )^n \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^n \, _2F_1\left (\frac {1}{2},1;n+\frac {3}{2};-e^{2 i (c+d x)}\right ) \sec ^{-n}(c+d x) (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n}{d (2 n+1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {2 \, \left (\frac {2 \, a e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{n} e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(2*(2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1))^n*e^(I*d*x + I*c)/(e^(2*I*d*x + 2*I*c) + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.23, size = 0, normalized size = 0.00 \[ \int \sec \left (d x +c \right ) \left (a +i a \tan \left (d x +c \right )\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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