3.5.68 \(\int \cos ^2(c+d x) (a+i a \tan (c+d x))^n \, dx\) [468]

Optimal. Leaf size=56 \[ \frac {i a \, _2F_1\left (2,-1+n;n;\frac {1}{2} (1+i \tan (c+d x))\right ) (a+i a \tan (c+d x))^{-1+n}}{4 d (1-n)} \]

[Out]

1/4*I*a*hypergeom([2, -1+n],[n],1/2+1/2*I*tan(d*x+c))*(a+I*a*tan(d*x+c))^(-1+n)/d/(1-n)

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Rubi [A]
time = 0.05, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3568, 70} \begin {gather*} \frac {i a (a+i a \tan (c+d x))^{n-1} \, _2F_1\left (2,n-1;n;\frac {1}{2} (i \tan (c+d x)+1)\right )}{4 d (1-n)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + I*a*Tan[c + d*x])^n,x]

[Out]

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

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+i a \tan (c+d x))^n \, dx &=-\frac {\left (i a^3\right ) \text {Subst}\left (\int \frac {(a+x)^{-2+n}}{(a-x)^2} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=\frac {i a \, _2F_1\left (2,-1+n;n;\frac {1}{2} (1+i \tan (c+d x))\right ) (a+i a \tan (c+d x))^{-1+n}}{4 d (1-n)}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(256\) vs. \(2(56)=112\).
time = 14.46, size = 256, normalized size = 4.57 \begin {gather*} -\frac {i 2^{-3+n} e^{-2 i (c+d n x)} \left (e^{i d x}\right )^n \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^n \left (\left (e^{2 i d (-1+n) x}+e^{2 i (c+d n x)}\right ) n (1+n)+2 e^{2 i (c+d n x)} \left (1+e^{2 i (c+d x)}\right )^n \left (-1+n^2\right ) \, _2F_1\left (n,n;1+n;-e^{2 i (c+d x)}\right )+e^{2 i (2 c+d x+d n x)} \left (1+e^{2 i (c+d x)}\right )^n (-1+n) n \, _2F_1\left (n,1+n;2+n;-e^{2 i (c+d x)}\right )\right ) \sec ^{-n}(c+d x) (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n}{d n \left (-1+n^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a + I*a*Tan[c + d*x])^n,x]

[Out]

((-I)*2^(-3 + n)*(E^(I*d*x))^n*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^n*((E^((2*I)*d*(-1 + n)*x) + E^((2*
I)*(c + d*n*x)))*n*(1 + n) + 2*E^((2*I)*(c + d*n*x))*(1 + E^((2*I)*(c + d*x)))^n*(-1 + n^2)*Hypergeometric2F1[
n, n, 1 + n, -E^((2*I)*(c + d*x))] + E^((2*I)*(2*c + d*x + d*n*x))*(1 + E^((2*I)*(c + d*x)))^n*(-1 + n)*n*Hype
rgeometric2F1[n, 1 + n, 2 + n, -E^((2*I)*(c + d*x))])*(a + I*a*Tan[c + d*x])^n)/(d*E^((2*I)*(c + d*n*x))*n*(-1
+ n^2)*Sec[c + d*x]^n*(Cos[d*x] + I*Sin[d*x])^n)

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Maple [F]
time = 0.97, size = 0, normalized size = 0.00 \[\int \left (\cos ^{2}\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+I*a*tan(d*x+c))^n,x)

[Out]

int(cos(d*x+c)^2*(a+I*a*tan(d*x+c))^n,x)

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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(cos(d*x+c)^2*(a+I*a*tan(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)^n*cos(d*x + c)^2, x)

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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(cos(d*x+c)^2*(a+I*a*tan(d*x+c))^n,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n} \cos ^{2}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(a+I*a*tan(d*x+c))**n,x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**n*cos(c + d*x)**2, x)

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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(cos(d*x+c)^2*(a+I*a*tan(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^n*cos(d*x + c)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int {\cos \left (c+d\,x\right )}^2\,{\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(cos(c + d*x)^2*(a + a*tan(c + d*x)*1i)^n,x)

[Out]

int(cos(c + d*x)^2*(a + a*tan(c + d*x)*1i)^n, x)

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