3.311 \(\int (d \tan (e+f x))^n (a+i a \tan (e+f x))^3 \, dx\)

Optimal. Leaf size=127 \[ \frac{4 a^3 (d \tan (e+f x))^{n+1} \, _2F_1(1,n+1;n+2;i \tan (e+f x))}{d f (n+1)}-\frac{a^3 (2 n+5) (d \tan (e+f x))^{n+1}}{d f (n+1) (n+2)}-\frac{\left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{n+1}}{d f (n+2)} \]

[Out]

-((a^3*(5 + 2*n)*(d*Tan[e + f*x])^(1 + n))/(d*f*(1 + n)*(2 + n))) + (4*a^3*Hypergeometric2F1[1, 1 + n, 2 + n,
I*Tan[e + f*x]]*(d*Tan[e + f*x])^(1 + n))/(d*f*(1 + n)) - ((d*Tan[e + f*x])^(1 + n)*(a^3 + I*a^3*Tan[e + f*x])
)/(d*f*(2 + n))

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Rubi [A]  time = 0.259031, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3556, 3592, 3537, 12, 64} \[ \frac{4 a^3 (d \tan (e+f x))^{n+1} \, _2F_1(1,n+1;n+2;i \tan (e+f x))}{d f (n+1)}-\frac{a^3 (2 n+5) (d \tan (e+f x))^{n+1}}{d f (n+1) (n+2)}-\frac{\left (a^3+i a^3 \tan (e+f x)\right ) (d \tan (e+f x))^{n+1}}{d f (n+2)} \]

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^n*(a + I*a*Tan[e + f*x])^3,x]

[Out]

-((a^3*(5 + 2*n)*(d*Tan[e + f*x])^(1 + n))/(d*f*(1 + n)*(2 + n))) + (4*a^3*Hypergeometric2F1[1, 1 + n, 2 + n,
I*Tan[e + f*x]]*(d*Tan[e + f*x])^(1 + n))/(d*f*(1 + n)) - ((d*Tan[e + f*x])^(1 + n)*(a^3 + I*a^3*Tan[e + f*x])
)/(d*f*(2 + n))

Rule 3556

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*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] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3592

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

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rubi steps

\begin{align*} \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^3 \, dx &=-\frac{(d \tan (e+f x))^{1+n} \left (a^3+i a^3 \tan (e+f x)\right )}{d f (2+n)}+\frac{a \int (d \tan (e+f x))^n (a+i a \tan (e+f x)) (a d (3+2 n)+i a d (5+2 n) \tan (e+f x)) \, dx}{d (2+n)}\\ &=-\frac{a^3 (5+2 n) (d \tan (e+f x))^{1+n}}{d f (1+n) (2+n)}-\frac{(d \tan (e+f x))^{1+n} \left (a^3+i a^3 \tan (e+f x)\right )}{d f (2+n)}+\frac{a \int (d \tan (e+f x))^n \left (4 a^2 d (2+n)+4 i a^2 d (2+n) \tan (e+f x)\right ) \, dx}{d (2+n)}\\ &=-\frac{a^3 (5+2 n) (d \tan (e+f x))^{1+n}}{d f (1+n) (2+n)}-\frac{(d \tan (e+f x))^{1+n} \left (a^3+i a^3 \tan (e+f x)\right )}{d f (2+n)}+\frac{\left (16 i a^5 d (2+n)\right ) \operatorname{Subst}\left (\int \frac{4^{-n} \left (-\frac{i x}{a^2 (2+n)}\right )^n}{-16 a^4 d^2 (2+n)^2+4 a^2 d (2+n) x} \, dx,x,4 i a^2 d (2+n) \tan (e+f x)\right )}{f}\\ &=-\frac{a^3 (5+2 n) (d \tan (e+f x))^{1+n}}{d f (1+n) (2+n)}-\frac{(d \tan (e+f x))^{1+n} \left (a^3+i a^3 \tan (e+f x)\right )}{d f (2+n)}+\frac{\left (i 4^{2-n} a^5 d (2+n)\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{i x}{a^2 (2+n)}\right )^n}{-16 a^4 d^2 (2+n)^2+4 a^2 d (2+n) x} \, dx,x,4 i a^2 d (2+n) \tan (e+f x)\right )}{f}\\ &=-\frac{a^3 (5+2 n) (d \tan (e+f x))^{1+n}}{d f (1+n) (2+n)}+\frac{4 a^3 \, _2F_1(1,1+n;2+n;i \tan (e+f x)) (d \tan (e+f x))^{1+n}}{d f (1+n)}-\frac{(d \tan (e+f x))^{1+n} \left (a^3+i a^3 \tan (e+f x)\right )}{d f (2+n)}\\ \end{align*}

Mathematica [A]  time = 3.3599, size = 214, normalized size = 1.69 \[ \frac{e^{-3 i e} 2^{-n} \left (-\frac{i \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}\right )^{n+1} \cos ^3(e+f x) (a+i a \tan (e+f x))^3 \left (2 (n+2) \left (1+e^{2 i (e+f x)}\right )^{n+2} \, _2F_1\left (n+1,n+1;n+2;\frac{1}{2} \left (1-e^{2 i (e+f x)}\right )\right )-2^n \left ((4 n+7) e^{2 i (e+f x)}+2 n+5\right )\right ) \tan ^{-n}(e+f x) (d \tan (e+f x))^n}{f (n+1) (n+2) \left (1+e^{2 i (e+f x)}\right ) (\cos (f x)+i \sin (f x))^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*Tan[e + f*x])^n*(a + I*a*Tan[e + f*x])^3,x]

[Out]

((((-I)*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x))))^(1 + n)*Cos[e + f*x]^3*(-(2^n*(5 + 2*n + E^((2*
I)*(e + f*x))*(7 + 4*n))) + 2*(1 + E^((2*I)*(e + f*x)))^(2 + n)*(2 + n)*Hypergeometric2F1[1 + n, 1 + n, 2 + n,
(1 - E^((2*I)*(e + f*x)))/2])*(d*Tan[e + f*x])^n*(a + I*a*Tan[e + f*x])^3)/(2^n*E^((3*I)*e)*(1 + E^((2*I)*(e
+ f*x)))*f*(1 + n)*(2 + n)*(Cos[f*x] + I*Sin[f*x])^3*Tan[e + f*x]^n)

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Maple [F]  time = 0.24, size = 0, normalized size = 0. \begin{align*} \int \left ( d\tan \left ( fx+e \right ) \right ) ^{n} \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{3}\, 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))^3,x)

[Out]

int((d*tan(f*x+e))^n*(a+I*a*tan(f*x+e))^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} \left (d \tan \left (f x + e\right )\right )^{n}\,{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))^3,x, algorithm="maxima")

[Out]

integrate((I*a*tan(f*x + e) + a)^3*(d*tan(f*x + e))^n, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{8 \, a^{3} \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} e^{\left (6 i \, f x + 6 i \, e\right )}}{e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, e^{\left (2 i \, f x + 2 i \, e\right )} + 1}, 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))^3,x, algorithm="fricas")

[Out]

integral(8*a^3*((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))^n*e^(6*I*f*x + 6*I*e)/(e^(6*I*f*x
+ 6*I*e) + 3*e^(4*I*f*x + 4*I*e) + 3*e^(2*I*f*x + 2*I*e) + 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int \left (d \tan{\left (e + f x \right )}\right )^{n}\, dx + \int - 3 \left (d \tan{\left (e + f x \right )}\right )^{n} \tan ^{2}{\left (e + f x \right )}\, dx + \int 3 i \left (d \tan{\left (e + f x \right )}\right )^{n} \tan{\left (e + f x \right )}\, dx + \int - i \left (d \tan{\left (e + f x \right )}\right )^{n} \tan ^{3}{\left (e + f x \right )}\, dx\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))**3,x)

[Out]

a**3*(Integral((d*tan(e + f*x))**n, x) + Integral(-3*(d*tan(e + f*x))**n*tan(e + f*x)**2, x) + Integral(3*I*(d
*tan(e + f*x))**n*tan(e + f*x), x) + Integral(-I*(d*tan(e + f*x))**n*tan(e + f*x)**3, x))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} \left (d \tan \left (f x + e\right )\right )^{n}\,{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))^3,x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^3*(d*tan(f*x + e))^n, x)