3.310 \(\int (d \tan (e+f x))^n (a+i a \tan (e+f x))^4 \, dx\)

Optimal. Leaf size=189 \[ \frac{8 a^4 (d \tan (e+f x))^{n+1} \, _2F_1(1,n+1;n+2;i \tan (e+f x))}{d f (n+1)}-\frac{2 a^4 \left (2 n^2+11 n+16\right ) (d \tan (e+f x))^{n+1}}{d f (n+1) (n+2) (n+3)}-\frac{\left (a^2+i a^2 \tan (e+f x)\right )^2 (d \tan (e+f x))^{n+1}}{d f (n+3)}-\frac{2 (n+4) \left (a^4+i a^4 \tan (e+f x)\right ) (d \tan (e+f x))^{n+1}}{d f (n+2) (n+3)} \]

[Out]

(-2*a^4*(16 + 11*n + 2*n^2)*(d*Tan[e + f*x])^(1 + n))/(d*f*(1 + n)*(2 + n)*(3 + n)) + (8*a^4*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^2 + I
*a^2*Tan[e + f*x])^2)/(d*f*(3 + n)) - (2*(4 + n)*(d*Tan[e + f*x])^(1 + n)*(a^4 + I*a^4*Tan[e + f*x]))/(d*f*(2
+ n)*(3 + n))

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Rubi [A]  time = 0.530527, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {3556, 3594, 3592, 3537, 12, 64} \[ \frac{8 a^4 (d \tan (e+f x))^{n+1} \, _2F_1(1,n+1;n+2;i \tan (e+f x))}{d f (n+1)}-\frac{2 a^4 \left (2 n^2+11 n+16\right ) (d \tan (e+f x))^{n+1}}{d f (n+1) (n+2) (n+3)}-\frac{\left (a^2+i a^2 \tan (e+f x)\right )^2 (d \tan (e+f x))^{n+1}}{d f (n+3)}-\frac{2 (n+4) \left (a^4+i a^4 \tan (e+f x)\right ) (d \tan (e+f x))^{n+1}}{d f (n+2) (n+3)} \]

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^n*(a + I*a*Tan[e + f*x])^4,x]

[Out]

(-2*a^4*(16 + 11*n + 2*n^2)*(d*Tan[e + f*x])^(1 + n))/(d*f*(1 + n)*(2 + n)*(3 + n)) + (8*a^4*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^2 + I
*a^2*Tan[e + f*x])^2)/(d*f*(3 + n)) - (2*(4 + n)*(d*Tan[e + f*x])^(1 + n)*(a^4 + I*a^4*Tan[e + f*x]))/(d*f*(2
+ n)*(3 + 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 3594

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*B*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(d*f
*(m + n)), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*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] && GtQ[m, 1] &&  !LtQ[n, -1]

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))^4 \, dx &=-\frac{(d \tan (e+f x))^{1+n} \left (a^2+i a^2 \tan (e+f x)\right )^2}{d f (3+n)}+\frac{a \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^2 (2 a d (2+n)+2 i a d (4+n) \tan (e+f x)) \, dx}{d (3+n)}\\ &=-\frac{(d \tan (e+f x))^{1+n} \left (a^2+i a^2 \tan (e+f x)\right )^2}{d f (3+n)}-\frac{2 (4+n) (d \tan (e+f x))^{1+n} \left (a^4+i a^4 \tan (e+f x)\right )}{d f (2+n) (3+n)}+\frac{a \int (d \tan (e+f x))^n (a+i a \tan (e+f x)) \left (2 a^2 d^2 \left (8+9 n+2 n^2\right )+2 i a^2 d^2 \left (16+11 n+2 n^2\right ) \tan (e+f x)\right ) \, dx}{d^2 (2+n) (3+n)}\\ &=-\frac{2 a^4 \left (16+11 n+2 n^2\right ) (d \tan (e+f x))^{1+n}}{d f (1+n) (2+n) (3+n)}-\frac{(d \tan (e+f x))^{1+n} \left (a^2+i a^2 \tan (e+f x)\right )^2}{d f (3+n)}-\frac{2 (4+n) (d \tan (e+f x))^{1+n} \left (a^4+i a^4 \tan (e+f x)\right )}{d f (2+n) (3+n)}+\frac{a \int (d \tan (e+f x))^n \left (8 a^3 d^2 (2+n) (3+n)+8 i a^3 d^2 (2+n) (3+n) \tan (e+f x)\right ) \, dx}{d^2 (2+n) (3+n)}\\ &=-\frac{2 a^4 \left (16+11 n+2 n^2\right ) (d \tan (e+f x))^{1+n}}{d f (1+n) (2+n) (3+n)}-\frac{(d \tan (e+f x))^{1+n} \left (a^2+i a^2 \tan (e+f x)\right )^2}{d f (3+n)}-\frac{2 (4+n) (d \tan (e+f x))^{1+n} \left (a^4+i a^4 \tan (e+f x)\right )}{d f (2+n) (3+n)}+\frac{\left (64 i a^7 d^2 (2+n) (3+n)\right ) \operatorname{Subst}\left (\int \frac{8^{-n} \left (-\frac{i x}{a^3 d (2+n) (3+n)}\right )^n}{-64 a^6 d^4 (2+n)^2 (3+n)^2+8 a^3 d^2 (2+n) (3+n) x} \, dx,x,8 i a^3 d^2 (2+n) (3+n) \tan (e+f x)\right )}{f}\\ &=-\frac{2 a^4 \left (16+11 n+2 n^2\right ) (d \tan (e+f x))^{1+n}}{d f (1+n) (2+n) (3+n)}-\frac{(d \tan (e+f x))^{1+n} \left (a^2+i a^2 \tan (e+f x)\right )^2}{d f (3+n)}-\frac{2 (4+n) (d \tan (e+f x))^{1+n} \left (a^4+i a^4 \tan (e+f x)\right )}{d f (2+n) (3+n)}+\frac{\left (i 8^{2-n} a^7 d^2 (2+n) (3+n)\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{i x}{a^3 d (2+n) (3+n)}\right )^n}{-64 a^6 d^4 (2+n)^2 (3+n)^2+8 a^3 d^2 (2+n) (3+n) x} \, dx,x,8 i a^3 d^2 (2+n) (3+n) \tan (e+f x)\right )}{f}\\ &=-\frac{2 a^4 \left (16+11 n+2 n^2\right ) (d \tan (e+f x))^{1+n}}{d f (1+n) (2+n) (3+n)}+\frac{8 a^4 \, _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^2+i a^2 \tan (e+f x)\right )^2}{d f (3+n)}-\frac{2 (4+n) (d \tan (e+f x))^{1+n} \left (a^4+i a^4 \tan (e+f x)\right )}{d f (2+n) (3+n)}\\ \end{align*}

Mathematica [F]  time = 44.8603, size = 0, normalized size = 0. \[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^4 \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(d*Tan[e + f*x])^n*(a + I*a*Tan[e + f*x])^4,x]

[Out]

Integrate[(d*Tan[e + f*x])^n*(a + I*a*Tan[e + f*x])^4, x]

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Maple [F]  time = 0.253, 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 ) ^{4}\, 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))^4,x)

[Out]

int((d*tan(f*x+e))^n*(a+I*a*tan(f*x+e))^4,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 )}^{4} \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))^4,x, algorithm="maxima")

[Out]

integrate((I*a*tan(f*x + e) + a)^4*(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{16 \, a^{4} \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 (8 i \, f x + 8 i \, e\right )}}{e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, 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))^4,x, algorithm="fricas")

[Out]

integral(16*a^4*((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))^n*e^(8*I*f*x + 8*I*e)/(e^(8*I*f*x
+ 8*I*e) + 4*e^(6*I*f*x + 6*I*e) + 6*e^(4*I*f*x + 4*I*e) + 4*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^{4} \left (\int \left (d \tan{\left (e + f x \right )}\right )^{n}\, dx + \int - 6 \left (d \tan{\left (e + f x \right )}\right )^{n} \tan ^{2}{\left (e + f x \right )}\, dx + \int \left (d \tan{\left (e + f x \right )}\right )^{n} \tan ^{4}{\left (e + f x \right )}\, dx + \int 4 i \left (d \tan{\left (e + f x \right )}\right )^{n} \tan{\left (e + f x \right )}\, dx + \int - 4 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))**4,x)

[Out]

a**4*(Integral((d*tan(e + f*x))**n, x) + Integral(-6*(d*tan(e + f*x))**n*tan(e + f*x)**2, x) + Integral((d*tan
(e + f*x))**n*tan(e + f*x)**4, x) + Integral(4*I*(d*tan(e + f*x))**n*tan(e + f*x), x) + Integral(-4*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 )}^{4} \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))^4,x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^4*(d*tan(f*x + e))^n, x)