3.7.89 \(\int (e \cos (c+d x))^m (a+i a \tan (c+d x))^2 \, dx\) [689]

Optimal. Leaf size=86 \[ -\frac {i 2^{2-\frac {m}{2}} a^2 (e \cos (c+d x))^m \, _2F_1\left (\frac {1}{2} (-2+m),-\frac {m}{2};1-\frac {m}{2};\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{m/2}}{d m} \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3596, 3586, 3604, 72, 71} \begin {gather*} -\frac {i a^2 2^{2-\frac {m}{2}} (1+i \tan (c+d x))^{m/2} (e \cos (c+d x))^m \, _2F_1\left (\frac {m-2}{2},-\frac {m}{2};1-\frac {m}{2};\frac {1}{2} (1-i \tan (c+d x))\right )}{d m} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*2^(2 - m/2)*a^2*(e*Cos[c + d*x])^m*Hypergeometric2F1[(-2 + m)/2, -1/2*m, 1 - m/2, (1 - I*Tan[c + d*x])/2
]*(1 + I*Tan[c + d*x])^(m/2))/(d*m)

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 3596

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*Co
s[e + f*x])^m*(d*Sec[e + f*x])^m, Int[(a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e,
f, m, n}, x] &&  !IntegerQ[m]

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

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Mathematica [A]
time = 1.56, size = 144, normalized size = 1.67 \begin {gather*} -\frac {i 2^{2-m} a^2 e^{2 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{-m} \left (e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )\right )^m \cos ^{2-m}(c+d x) (e \cos (c+d x))^m \, _2F_1\left (2-m,2-\frac {m}{2};3-\frac {m}{2};-e^{2 i (c+d x)}\right ) (-i+\tan (c+d x))^2}{d (-4+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.40, size = 0, normalized size = 0.00 \[\int \left (e \cos \left (d x +c \right )\right )^{m} \left (a +i a \tan \left (d x +c \right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2*(Integral(-(e*cos(c + d*x))**m, x) + Integral((e*cos(c + d*x))**m*tan(c + d*x)**2, x) + Integral(-2*I*(e
*cos(c + d*x))**m*tan(c + d*x), 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((e*cos(d*x+c))^m*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (e\,\cos \left (c+d\,x\right )\right )}^m\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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