3.11.48 \(\int \frac {(c-i c \tan (e+f x))^n}{(a+i a \tan (e+f x))^2} \, dx\) [1048]

Optimal. Leaf size=52 \[ \frac {i \, _2F_1\left (3,n;1+n;\frac {1}{2} (1-i \tan (e+f x))\right ) (c-i c \tan (e+f x))^n}{8 a^2 f n} \]

[Out]

1/8*I*hypergeom([3, n],[1+n],1/2-1/2*I*tan(f*x+e))*(c-I*c*tan(f*x+e))^n/a^2/f/n

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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 (c-i c \tan (e+f x))^n \, _2F_1\left (3,n;n+1;\frac {1}{2} (1-i \tan (e+f x))\right )}{8 a^2 f n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/8)*Hypergeometric2F1[3, n, 1 + n, (1 - I*Tan[e + f*x])/2]*(c - I*c*Tan[e + f*x])^n)/(a^2*f*n)

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 {(c-i c \tan (e+f x))^n}{(a+i a \tan (e+f x))^2} \, dx &=\frac {\int \cos ^4(e+f x) (c-i c \tan (e+f x))^{2+n} \, dx}{a^2 c^2}\\ &=\frac {\left (i c^3\right ) \text {Subst}\left (\int \frac {(c+x)^{-1+n}}{(c-x)^3} \, dx,x,-i c \tan (e+f x)\right )}{a^2 f}\\ &=\frac {i \, _2F_1\left (3,n;1+n;\frac {1}{2} (1-i \tan (e+f x))\right ) (c-i c \tan (e+f x))^n}{8 a^2 f n}\\ \end {align*}

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Mathematica [F]
time = 180.00, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

$Aborted

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Maple [F]
time = 2.69, size = 0, normalized size = 0.00 \[\int \frac {\left (c -i c \tan \left (f x +e \right )\right )^{n}}{\left (a +i a \tan \left (f x +e \right )\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-I*c*tan(f*x+e))^n/(a+I*a*tan(f*x+e))^2,x)

[Out]

int((c-I*c*tan(f*x+e))^n/(a+I*a*tan(f*x+e))^2,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((c-I*c*tan(f*x+e))^n/(a+I*a*tan(f*x+e))^2,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((c-I*c*tan(f*x+e))^n/(a+I*a*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {\left (- i c \tan {\left (e + f x \right )} + c\right )^{n}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))**n/(a+I*a*tan(f*x+e))**2,x)

[Out]

-Integral((-I*c*tan(e + f*x) + c)**n/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1), x)/a**2

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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((c-I*c*tan(f*x+e))^n/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((-I*c*tan(f*x + e) + c)^n/(I*a*tan(f*x + e) + a)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^n}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c*tan(e + f*x)*1i)^n/(a + a*tan(e + f*x)*1i)^2,x)

[Out]

int((c - c*tan(e + f*x)*1i)^n/(a + a*tan(e + f*x)*1i)^2, x)

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