3.692 \(\int \frac {(e \cos (c+d x))^m}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=86 \[ -\frac {i 2^{-\frac {m}{2}-2} (1+i \tan (c+d x))^{m/2} (e \cos (c+d x))^m \, _2F_1\left (-\frac {m}{2},\frac {m+6}{2};1-\frac {m}{2};\frac {1}{2} (1-i \tan (c+d x))\right )}{a^2 d m} \]

[Out]

-I*2^(-2-1/2*m)*(e*cos(d*x+c))^m*hypergeom([-1/2*m, 3+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)/a^2/d/m

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Rubi [A]  time = 0.24, 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 = {3515, 3505, 3523, 70, 69} \[ -\frac {i 2^{-\frac {m}{2}-2} (1+i \tan (c+d x))^{m/2} (e \cos (c+d x))^m \, _2F_1\left (-\frac {m}{2},\frac {m+6}{2};1-\frac {m}{2};\frac {1}{2} (1-i \tan (c+d x))\right )}{a^2 d m} \]

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)*(e*Cos[c + d*x])^m*Hypergeometric2F1[-m/2, (6 + m)/2, 1 - m/2, (1 - I*Tan[c + d*x])/2]*(1 +
I*Tan[c + d*x])^(m/2))/(a^2*d*m)

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 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 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 \frac {(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 \frac {(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 ) \operatorname {Subst}\left (\int (a-i a x)^{-1-\frac {m}{2}} (a+i a x)^{-3-\frac {m}{2}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\left (2^{-3-\frac {m}{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 )^{-3-\frac {m}{2}} (a-i a x)^{-1-\frac {m}{2}} \, dx,x,\tan (c+d x)\right )}{a d}\\ &=-\frac {i 2^{-2-\frac {m}{2}} (e \cos (c+d x))^m \, _2F_1\left (-\frac {m}{2},\frac {6+m}{2};1-\frac {m}{2};\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{m/2}}{a^2 d m}\\ \end {align*}

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Mathematica [A]  time = 71.09, size = 154, normalized size = 1.79 \[ -\frac {i 2^{-m-2} e^{-2 i (c+2 d x)} \left (1+e^{2 i (c+d x)}\right )^3 \left (e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )\right )^m (\cos (d x)+i \sin (d x))^2 \, _2F_1\left (1,\frac {m+2}{2};-\frac {m}{2}-1;-e^{2 i (c+d x)}\right ) \cos ^{-m-2}(c+d x) (e \cos (c+d x))^m}{a^2 d (m+4) (\tan (c+d x)-i)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (\frac {1}{2} \, {\left (e e^{\left (2 i \, d x + 2 i \, c\right )} + e\right )} e^{\left (-i \, d x - i \, c\right )}\right )^{m} {\left (e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{4 \, a^{2}}, x\right ) \]

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

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

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

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maple [F]  time = 4.94, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \cos \left (d x +c \right )\right )^{m}}{\left (a +i a \tan \left (d x +c \right )\right )^{2}}\, dx \]

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(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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

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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {\left (e \cos {\left (c + d x \right )}\right )^{m}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx}{a^{2}} \]

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]

-Integral((e*cos(c + d*x))**m/(tan(c + d*x)**2 - 2*I*tan(c + d*x) - 1), x)/a**2

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