3.11.55 \(\int \frac {(a+i a \tan (e+f x))^m}{c-i c \tan (e+f x)} \, dx\) [1055]

Optimal. Leaf size=52 \[ -\frac {i \, _2F_1\left (2,m;1+m;\frac {1}{2} (1+i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m}{4 c f m} \]

[Out]

-1/4*I*hypergeom([2, m],[1+m],1/2+1/2*I*tan(f*x+e))*(a+I*a*tan(f*x+e))^m/c/f/m

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Rubi [A]
time = 0.09, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3603, 3568, 70} \begin {gather*} -\frac {i (a+i a \tan (e+f x))^m \, _2F_1\left (2,m;m+1;\frac {1}{2} (i \tan (e+f x)+1)\right )}{4 c f m} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^m/(c - I*c*Tan[e + f*x]),x]

[Out]

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

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]

Rule 3603

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (e+f x))^m}{c-i c \tan (e+f x)} \, dx &=\frac {\int \cos ^2(e+f x) (a+i a \tan (e+f x))^{1+m} \, dx}{a c}\\ &=-\frac {\left (i a^2\right ) \text {Subst}\left (\int \frac {(a+x)^{-1+m}}{(a-x)^2} \, dx,x,i a \tan (e+f x)\right )}{c f}\\ &=-\frac {i \, _2F_1\left (2,m;1+m;\frac {1}{2} (1+i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m}{4 c f m}\\ \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. \(133\) vs. \(2(52)=104\).
time = 16.05, size = 133, normalized size = 2.56 \begin {gather*} -\frac {i 2^{-2+m} \left (e^{i f x}\right )^m \left (\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^m \left (1+e^{2 i (e+f x)}\right )^2 \, _2F_1\left (1,2;1+m;-e^{2 i (e+f x)}\right ) \sec ^{-m}(e+f x) (\cos (f x)+i \sin (f x))^{-m} (a+i a \tan (e+f x))^m}{c f m} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^m/(c - I*c*Tan[e + f*x]),x]

[Out]

((-I)*2^(-2 + m)*(E^(I*f*x))^m*(E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x))))^m*(1 + E^((2*I)*(e + f*x)))^2*Hyper
geometric2F1[1, 2, 1 + m, -E^((2*I)*(e + f*x))]*(a + I*a*Tan[e + f*x])^m)/(c*f*m*Sec[e + f*x]^m*(Cos[f*x] + I*
Sin[f*x])^m)

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Maple [F]
time = 0.99, size = 0, normalized size = 0.00 \[\int \frac {\left (a +i a \tan \left (f x +e \right )\right )^{m}}{c -i c \tan \left (f x +e \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^m/(c-I*c*tan(f*x+e)),x)

[Out]

int((a+I*a*tan(f*x+e))^m/(c-I*c*tan(f*x+e)),x)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^m/(c-I*c*tan(f*x+e)),x, algorithm="maxima")

[Out]

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

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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((a+I*a*tan(f*x+e))^m/(c-I*c*tan(f*x+e)),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**m/(c-I*c*tan(f*x+e)),x)

[Out]

I*Integral((I*a*tan(e + f*x) + a)**m/(tan(e + f*x) + I), x)/c

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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((a+I*a*tan(f*x+e))^m/(c-I*c*tan(f*x+e)),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^m/(-I*c*tan(f*x + e) + c), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^m}{c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x)*1i)^m/(c - c*tan(e + f*x)*1i),x)

[Out]

int((a + a*tan(e + f*x)*1i)^m/(c - c*tan(e + f*x)*1i), x)

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