3.319 \(\int \frac{(d \tan (e+f x))^n}{a-i a \tan (e+f x)} \, dx\)

Optimal. Leaf size=158 \[ -\frac{i n (d \tan (e+f x))^{n+2} \, _2F_1\left (1,\frac{n+2}{2};\frac{n+4}{2};-\tan ^2(e+f x)\right )}{2 a d^2 f (n+2)}+\frac{(1-n) (d \tan (e+f x))^{n+1} \, _2F_1\left (1,\frac{n+1}{2};\frac{n+3}{2};-\tan ^2(e+f x)\right )}{2 a d f (n+1)}+\frac{(d \tan (e+f x))^{n+1}}{2 d f (a-i a \tan (e+f x))} \]

[Out]

((1 - n)*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(1 + n))/(2*a*d*f*(1 + n
)) - ((I/2)*n*Hypergeometric2F1[1, (2 + n)/2, (4 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(2 + n))/(a*d^2*f*(
2 + n)) + (d*Tan[e + f*x])^(1 + n)/(2*d*f*(a - I*a*Tan[e + f*x]))

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Rubi [A]  time = 0.173778, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3552, 3538, 3476, 364} \[ -\frac{i n (d \tan (e+f x))^{n+2} \, _2F_1\left (1,\frac{n+2}{2};\frac{n+4}{2};-\tan ^2(e+f x)\right )}{2 a d^2 f (n+2)}+\frac{(1-n) (d \tan (e+f x))^{n+1} \, _2F_1\left (1,\frac{n+1}{2};\frac{n+3}{2};-\tan ^2(e+f x)\right )}{2 a d f (n+1)}+\frac{(d \tan (e+f x))^{n+1}}{2 d f (a-i a \tan (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - n)*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(1 + n))/(2*a*d*f*(1 + n
)) - ((I/2)*n*Hypergeometric2F1[1, (2 + n)/2, (4 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(2 + n))/(a*d^2*f*(
2 + n)) + (d*Tan[e + f*x])^(1 + n)/(2*d*f*(a - I*a*Tan[e + f*x]))

Rule 3552

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(a
*(c + d*Tan[e + f*x])^(n + 1))/(2*f*(b*c - a*d)*(a + b*Tan[e + f*x])), x] + Dist[1/(2*a*(b*c - a*d)), Int[(c +
d*Tan[e + f*x])^n*Simp[b*c + a*d*(n - 1) - b*d*n*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x]
&& NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0]

Rule 3538

Int[((b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*T
an[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Tan[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x] && NeQ
[c^2 + d^2, 0] &&  !IntegerQ[2*m]

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

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

Mathematica [A]  time = 0.684499, size = 123, normalized size = 0.78 \[ \frac{\tan (e+f x) (d \tan (e+f x))^n \left (-\frac{(n-1) \, _2F_1\left (1,\frac{n+1}{2};\frac{n+3}{2};-\tan ^2(e+f x)\right )}{a (n+1)}-\frac{i n \tan (e+f x) \, _2F_1\left (1,\frac{n+2}{2};\frac{n+4}{2};-\tan ^2(e+f x)\right )}{a (n+2)}+\frac{1}{a-i a \tan (e+f x)}\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Tan[e + f*x]*(d*Tan[e + f*x])^n*(-(((-1 + n)*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, -Tan[e + f*x]^2])/(a*
(1 + n))) - (I*n*Hypergeometric2F1[1, (2 + n)/2, (4 + n)/2, -Tan[e + f*x]^2]*Tan[e + f*x])/(a*(2 + n)) + (a -
I*a*Tan[e + f*x])^(-1)))/(2*f)

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Maple [F]  time = 0.853, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( d\tan \left ( fx+e \right ) \right ) ^{n}}{a-ia\tan \left ( fx+e \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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