3.11.50 \(\int (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^n \, dx\) [1050]

Optimal. Leaf size=66 \[ \frac {i \, _2F_1\left (1,m+n;1+n;\frac {1}{2} (1-i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^n}{2 f n} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 87, normalized size of antiderivative = 1.32, 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 = {3604, 72, 71} \begin {gather*} -\frac {i 2^{n-1} (1-i \tan (e+f x))^{-n} (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^n \, _2F_1\left (m,1-n;m+1;\frac {1}{2} (i \tan (e+f x)+1)\right )}{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])^n,x]

[Out]

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

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 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 (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^n \, dx &=\frac {(a c) \text {Subst}\left (\int (a+i a x)^{-1+m} (c-i c x)^{-1+n} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\left (2^{-1+n} a (c-i c \tan (e+f x))^n \left (\frac {c-i c \tan (e+f x)}{c}\right )^{-n}\right ) \text {Subst}\left (\int \left (\frac {1}{2}-\frac {i x}{2}\right )^{-1+n} (a+i a x)^{-1+m} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {i 2^{-1+n} \, _2F_1\left (m,1-n;1+m;\frac {1}{2} (1+i \tan (e+f x))\right ) (1-i \tan (e+f x))^{-n} (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^n}{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. \(142\) vs. \(2(66)=132\).
time = 13.84, size = 142, normalized size = 2.15 \begin {gather*} -\frac {i 2^{-1+m+n} c \left (e^{i f x}\right )^m \left (\frac {c}{1+e^{2 i (e+f x)}}\right )^{-1+n} \left (\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^m \, _2F_1\left (1,1-n;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}{f m} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]
time = 2.67, size = 0, normalized size = 0.00 \[\int \left (a +i a \tan \left (f x +e \right )\right )^{m} \left (c -i c \tan \left (f x +e \right )\right )^{n}\, dx\]

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))^n,x)

[Out]

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

[Out]

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

[Out]

integral((2*c/(e^(2*I*f*x + 2*I*e) + 1))^n*e^(2*I*f*m*x + 2*I*m*e + m*log(a/c) + m*log(2*c/(e^(2*I*f*x + 2*I*e
) + 1))), 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 (e + f x \right )} - i\right )\right )^{m} \left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{n}\, dx \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))**n,x)

[Out]

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

[Out]

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

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

[Out]

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

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