3.690 \(\int (e \cos (c+d x))^m (a+i a \tan (c+d x)) \, dx\)

Optimal. Leaf size=82 \[ -\frac{i a 2^{1-\frac{m}{2}} (1+i \tan (c+d x))^{m/2} (e \cos (c+d x))^m \, _2F_1\left (-\frac{m}{2},\frac{m}{2};1-\frac{m}{2};\frac{1}{2} (1-i \tan (c+d x))\right )}{d m} \]

[Out]

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

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Rubi [A]  time = 0.174969, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {3515, 3505, 3523, 70, 69} \[ -\frac{i a 2^{1-\frac{m}{2}} (1+i \tan (c+d x))^{m/2} (e \cos (c+d x))^m \, _2F_1\left (-\frac{m}{2},\frac{m}{2};1-\frac{m}{2};\frac{1}{2} (1-i \tan (c+d x))\right )}{d m} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3515

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 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]

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 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]))

Rubi steps

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

Mathematica [A]  time = 7.57816, size = 131, normalized size = 1.6 \[ -\frac{a 2^{1-m} e^{i (c+2 d x)} \left (e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )\right )^m (\tan (c+d x)-i) (\cos (d x)-i \sin (d x)) \, _2F_1\left (1,\frac{m+2}{2};2-\frac{m}{2};-e^{2 i (c+d x)}\right ) \cos ^{1-m}(c+d x) (e \cos (c+d x))^m}{d (m-2)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.506, size = 0, normalized size = 0. \begin{align*} \int \left ( e\cos \left ( dx+c \right ) \right ) ^{m} \left ( a+ia\tan \left ( dx+c \right ) \right ) \, dx \end{align*}

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)),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )} \left (e \cos \left (d x + c\right )\right )^{m}\,{d x} \end{align*}

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)),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{2 \, \left (\frac{1}{2} \,{\left (e e^{\left (2 i \, d x + 2 i \, c\right )} + e\right )} e^{\left (-i \, d x - i \, c\right )}\right )^{m} a e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \end{align*}

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)),x, algorithm="fricas")

[Out]

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

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

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)),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )} \left (e \cos \left (d x + c\right )\right )^{m}\,{d x} \end{align*}

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)),x, algorithm="giac")

[Out]

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