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

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

[Out]

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

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Rubi [A]  time = 0.53, antiderivative size = 189, normalized size of antiderivative = 1.00, 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 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])))

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 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 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]

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*}

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Mathematica [B]  time = 9.94, size = 1065, normalized size = 5.63 \[ \frac {i 2^{3-n} \left (-\frac {i \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}\right )^n \cos ^4(e+f x) \left (2^n \, _2F_1\left (1,n;n+1;-\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}\right )-\left (1+e^{2 i (e+f x)}\right )^n \, _2F_1\left (n,n;n+1;\frac {1}{2} \left (1-e^{2 i (e+f x)}\right )\right )\right ) (d \tan (e+f x))^n (i \tan (e+f x) a+a)^4 \tan ^{-n}(e+f x)}{\left (e^{2 i e}+e^{4 i e}\right ) f n (\cos (f x)+i \sin (f x))^4}-\frac {8 i e^{-4 i e} \left (-1+e^{2 i (e+f x)}\right )^n \left (-\frac {i \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}\right )^n \left (\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{-n} \cos ^4(e+f x) \left (-\frac {\left (1+e^{2 i e}\right ) \left (-1+e^{2 i (e+f x)}\right ) \, _2F_1\left (1,n+1;n+2;\frac {1-e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}\right ) \left (1+e^{2 i (e+f x)}\right )^{-n-1}}{n+1}-\frac {\, _2F_1\left (1,n;n+1;\frac {1-e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}\right ) \left (1+e^{2 i (e+f x)}\right )^{-n}}{n}+\frac {2^{-n} \, _2F_1\left (n,n;n+1;\frac {1}{2} \left (1-e^{2 i (e+f x)}\right )\right )}{n}\right ) (d \tan (e+f x))^n (i \tan (e+f x) a+a)^4 \tan ^{-n}(e+f x)}{\left (1+e^{2 i e}\right ) f (\cos (f x)+i \sin (f x))^4}+\frac {\cos ^4(e+f x) \left (\frac {(\cos (2 e)+2 i \sin (2 e)-1) (2 i \cos (4 e)+2 \sin (4 e)) \sec ^2(e)}{n+1}+\frac {\sec (e+f x) (2 i \cos (4 e)+2 \sin (4 e)) (-\cos (e-f x)+\cos (e+f x)-2 i \sin (e-f x)+2 i \sin (e+f x)) \sec ^2(e)}{n+1}\right ) (d \tan (e+f x))^n (i \tan (e+f x) a+a)^4}{f (\cos (f x)+i \sin (f x))^4}+\frac {\cos ^4(e+f x) \left (\frac {\sec (e) (\cos (4 e)-i \sin (4 e)) \sin (f x) \sec ^3(e+f x)}{n+3}+\frac {(\cos (4 e)-i \sin (4 e)) \tan (e) \sec ^2(e+f x)}{n+3}+\frac {\sec (e) (2 \cos (4 e)-2 i \sin (4 e)) \sin (f x) \sec (e+f x)}{(n+1) (n+3)}+\frac {(2 \cos (4 e)-2 i \sin (4 e)) \tan (e)}{(n+1) (n+3)}\right ) (d \tan (e+f x))^n (i \tan (e+f x) a+a)^4}{f (\cos (f x)+i \sin (f x))^4}+\frac {\cos ^4(e+f x) \left (\frac {(-2 n+\cos (2 e)-3) (-2 i \cos (4 e)-2 \sin (4 e)) \sec ^2(e)}{(n+1) (n+2)}+\frac {(\cos (e+f x)-\cos (e-f x)) \sec (e+f x) (-2 i \cos (4 e)-2 \sin (4 e)) \sec ^2(e)}{n+1}+\frac {\sec ^2(e+f x) (-4 i \cos (4 e)-4 \sin (4 e))}{n+2}\right ) (d \tan (e+f x))^n (i \tan (e+f x) a+a)^4}{f (\cos (f x)+i \sin (f x))^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {16 \, a^{4} \left (\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{n} e^{\left (8 i \, f x + 8 i \, e\right )}}{e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, e^{\left (2 i \, f x + 2 i \, e\right )} + 1}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{4} \left (d \tan \left (f x + e\right )\right )^{n}\,{d x} \]

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)

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maple [F]  time = 1.13, size = 0, normalized size = 0.00 \[ \int \left (d \tan \left (f x +e \right )\right )^{n} \left (a +i a \tan \left (f x +e \right )\right )^{4}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{4} \left (d \tan \left (f x + e\right )\right )^{n}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(e + f*x))^n*(a + a*tan(e + f*x)*1i)^4,x)

[Out]

int((d*tan(e + f*x))^n*(a + a*tan(e + f*x)*1i)^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{4} \left (\int \left (d \tan {\left (e + f x \right )}\right )^{n}\, dx + \int \left (- 6 \left (d \tan {\left (e + f x \right )}\right )^{n} \tan ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (d \tan {\left (e + f x \right )}\right )^{n} \tan ^{4}{\left (e + f x \right )}\, dx + \int 4 i \left (d \tan {\left (e + f x \right )}\right )^{n} \tan {\left (e + f x \right )}\, dx + \int \left (- 4 i \left (d \tan {\left (e + f x \right )}\right )^{n} \tan ^{3}{\left (e + f x \right )}\right )\, dx\right ) \]

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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