3.1178 \(\int \frac{(c+d \tan (e+f x))^n}{(a+i a \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=273 \[ \frac{\left (c^2+2 i c d (1-n)-d^2 \left (2 n^2-4 n+1\right )\right ) (c+d \tan (e+f x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{c+d \tan (e+f x)}{c+i d}\right )}{8 a^2 f (n+1) (-d+i c)^3}+\frac{(c+d \tan (e+f x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{c+d \tan (e+f x)}{c-i d}\right )}{8 a^2 f (n+1) (d+i c)}+\frac{(-d (2-n)+i c) (c+d \tan (e+f x))^{n+1}}{4 a^2 f (c+i d)^2 (1+i \tan (e+f x))}-\frac{(c+d \tan (e+f x))^{n+1}}{4 f (-d+i c) (a+i a \tan (e+f x))^2} \]

[Out]

(Hypergeometric2F1[1, 1 + n, 2 + n, (c + d*Tan[e + f*x])/(c - I*d)]*(c + d*Tan[e + f*x])^(1 + n))/(8*a^2*(I*c
+ d)*f*(1 + n)) + ((c^2 + (2*I)*c*d*(1 - n) - d^2*(1 - 4*n + 2*n^2))*Hypergeometric2F1[1, 1 + n, 2 + n, (c + d
*Tan[e + f*x])/(c + I*d)]*(c + d*Tan[e + f*x])^(1 + n))/(8*a^2*(I*c - d)^3*f*(1 + n)) + ((I*c - d*(2 - n))*(c
+ d*Tan[e + f*x])^(1 + n))/(4*a^2*(c + I*d)^2*f*(1 + I*Tan[e + f*x])) - (c + d*Tan[e + f*x])^(1 + n)/(4*(I*c -
d)*f*(a + I*a*Tan[e + f*x])^2)

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Rubi [A]  time = 0.556523, antiderivative size = 273, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3559, 3596, 3539, 3537, 68} \[ \frac{\left (c^2+2 i c d (1-n)-d^2 \left (2 n^2-4 n+1\right )\right ) (c+d \tan (e+f x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{c+d \tan (e+f x)}{c+i d}\right )}{8 a^2 f (n+1) (-d+i c)^3}+\frac{(c+d \tan (e+f x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{c+d \tan (e+f x)}{c-i d}\right )}{8 a^2 f (n+1) (d+i c)}+\frac{(-d (2-n)+i c) (c+d \tan (e+f x))^{n+1}}{4 a^2 f (c+i d)^2 (1+i \tan (e+f x))}-\frac{(c+d \tan (e+f x))^{n+1}}{4 f (-d+i c) (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Hypergeometric2F1[1, 1 + n, 2 + n, (c + d*Tan[e + f*x])/(c - I*d)]*(c + d*Tan[e + f*x])^(1 + n))/(8*a^2*(I*c
+ d)*f*(1 + n)) + ((c^2 + (2*I)*c*d*(1 - n) - d^2*(1 - 4*n + 2*n^2))*Hypergeometric2F1[1, 1 + n, 2 + n, (c + d
*Tan[e + f*x])/(c + I*d)]*(c + d*Tan[e + f*x])^(1 + n))/(8*a^2*(I*c - d)^3*f*(1 + n)) + ((I*c - d*(2 - n))*(c
+ d*Tan[e + f*x])^(1 + n))/(4*a^2*(c + I*d)^2*f*(1 + I*Tan[e + f*x])) - (c + d*Tan[e + f*x])^(1 + n)/(4*(I*c -
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 3539

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

Rule 3537

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

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

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

Mathematica [F]  time = 7.39494, size = 0, normalized size = 0. \[ \int \frac{(c+d \tan (e+f x))^n}{(a+i a \tan (e+f x))^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.244, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c+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((c+d*tan(f*x+e))^n/(a+I*a*tan(f*x+e))^2,x)

[Out]

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

[Out]

integral(1/4*(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + 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((c+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 ) + c\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((c+d*tan(f*x+e))^n/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")

[Out]

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