3.315 \(\int \frac{(d \tan (e+f x))^n}{(a+i a \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=209 \[ \frac{i (2-n) n (d \tan (e+f x))^{n+2} \, _2F_1\left (1,\frac{n+2}{2};\frac{n+4}{2};-\tan ^2(e+f x)\right )}{4 a^2 d^2 f (n+2)}+\frac{(1-n)^2 (d \tan (e+f x))^{n+1} \, _2F_1\left (1,\frac{n+1}{2};\frac{n+3}{2};-\tan ^2(e+f x)\right )}{4 a^2 d f (n+1)}+\frac{(2-n) (d \tan (e+f x))^{n+1}}{4 a^2 d f (1+i \tan (e+f x))}+\frac{(d \tan (e+f x))^{n+1}}{4 d f (a+i a \tan (e+f x))^2} \]

[Out]

((1 - n)^2*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(1 + n))/(4*a^2*d*f*(1
+ n)) + ((2 - n)*(d*Tan[e + f*x])^(1 + n))/(4*a^2*d*f*(1 + I*Tan[e + f*x])) + ((I/4)*(2 - n)*n*Hypergeometric
2F1[1, (2 + n)/2, (4 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(2 + n))/(a^2*d^2*f*(2 + n)) + (d*Tan[e + f*x])
^(1 + n)/(4*d*f*(a + I*a*Tan[e + f*x])^2)

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Rubi [A]  time = 0.361672, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3559, 3596, 3538, 3476, 364} \[ \frac{i (2-n) n (d \tan (e+f x))^{n+2} \, _2F_1\left (1,\frac{n+2}{2};\frac{n+4}{2};-\tan ^2(e+f x)\right )}{4 a^2 d^2 f (n+2)}+\frac{(1-n)^2 (d \tan (e+f x))^{n+1} \, _2F_1\left (1,\frac{n+1}{2};\frac{n+3}{2};-\tan ^2(e+f x)\right )}{4 a^2 d f (n+1)}+\frac{(2-n) (d \tan (e+f x))^{n+1}}{4 a^2 d f (1+i \tan (e+f x))}+\frac{(d \tan (e+f x))^{n+1}}{4 d f (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - n)^2*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(1 + n))/(4*a^2*d*f*(1
+ n)) + ((2 - n)*(d*Tan[e + f*x])^(1 + n))/(4*a^2*d*f*(1 + I*Tan[e + f*x])) + ((I/4)*(2 - n)*n*Hypergeometric
2F1[1, (2 + n)/2, (4 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(2 + n))/(a^2*d^2*f*(2 + n)) + (d*Tan[e + f*x])
^(1 + n)/(4*d*f*(a + I*a*Tan[e + f*x])^2)

Rule 3559

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d))
, Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m + n + 1) + b*d*(m + n + 1)*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] && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3596

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 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))^2} \, dx &=\frac{(d \tan (e+f x))^{1+n}}{4 d f (a+i a \tan (e+f x))^2}+\frac{\int \frac{(d \tan (e+f x))^n (a d (3-n)-i a d (1-n) \tan (e+f x))}{a+i a \tan (e+f x)} \, dx}{4 a^2 d}\\ &=\frac{(2-n) (d \tan (e+f x))^{1+n}}{4 a^2 d f (1+i \tan (e+f x))}+\frac{(d \tan (e+f x))^{1+n}}{4 d f (a+i a \tan (e+f x))^2}+\frac{\int (d \tan (e+f x))^n \left (2 a^2 d^2 (1-n)^2+2 i a^2 d^2 (2-n) n \tan (e+f x)\right ) \, dx}{8 a^4 d^2}\\ &=\frac{(2-n) (d \tan (e+f x))^{1+n}}{4 a^2 d f (1+i \tan (e+f x))}+\frac{(d \tan (e+f x))^{1+n}}{4 d f (a+i a \tan (e+f x))^2}+\frac{(1-n)^2 \int (d \tan (e+f x))^n \, dx}{4 a^2}+\frac{(i (2-n) n) \int (d \tan (e+f x))^{1+n} \, dx}{4 a^2 d}\\ &=\frac{(2-n) (d \tan (e+f x))^{1+n}}{4 a^2 d f (1+i \tan (e+f x))}+\frac{(d \tan (e+f x))^{1+n}}{4 d f (a+i a \tan (e+f x))^2}+\frac{\left (d (1-n)^2\right ) \operatorname{Subst}\left (\int \frac{x^n}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{4 a^2 f}+\frac{(i (2-n) n) \operatorname{Subst}\left (\int \frac{x^{1+n}}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{4 a^2 f}\\ &=\frac{(1-n)^2 \, _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}}{4 a^2 d f (1+n)}+\frac{(2-n) (d \tan (e+f x))^{1+n}}{4 a^2 d f (1+i \tan (e+f x))}+\frac{i (2-n) 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}}{4 a^2 d^2 f (2+n)}+\frac{(d \tan (e+f x))^{1+n}}{4 d f (a+i a \tan (e+f x))^2}\\ \end{align*}

Mathematica [F]  time = 5.33841, size = 0, normalized size = 0. \[ \int \frac{(d \tan (e+f x))^n}{(a+i a \tan (e+f x))^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.631, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( d\tan \left ( fx+e \right ) \right ) ^{n}}{ \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{2}}}\, 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))^2,x)

[Out]

int((d*tan(f*x+e))^n/(a+I*a*tan(f*x+e))^2,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))^2,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 (4 i \, f x + 4 i \, e\right )} + 2 \, e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{4 \, a^{2}}, 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))^2,x, algorithm="fricas")

[Out]

integral(1/4*((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(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(-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))**2,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}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}\,{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))^2,x, algorithm="giac")

[Out]

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