3.918 \(\int (a+i a \tan (e+f x)) (c-i c \tan (e+f x))^4 \, dx\)

Optimal. Leaf size=25 \[ \frac {i a (c-i c \tan (e+f x))^4}{4 f} \]

[Out]

1/4*I*a*(c-I*c*tan(f*x+e))^4/f

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Rubi [A]  time = 0.07, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3522, 3487, 32} \[ \frac {i a (c-i c \tan (e+f x))^4}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/4)*a*(c - I*c*Tan[e + f*x])^4)/f

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 3487

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 3522

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rubi steps

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

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Mathematica [B]  time = 1.56, size = 85, normalized size = 3.40 \[ \frac {a c^4 \sec (e) \sec ^4(e+f x) (2 \sin (e+2 f x)-2 \sin (3 e+2 f x)+\sin (3 e+4 f x)-2 i \cos (e+2 f x)-2 i \cos (3 e+2 f x)-3 \sin (e)-3 i \cos (e))}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*c^4*Sec[e]*Sec[e + f*x]^4*((-3*I)*Cos[e] - (2*I)*Cos[e + 2*f*x] - (2*I)*Cos[3*e + 2*f*x] - 3*Sin[e] + 2*Sin
[e + 2*f*x] - 2*Sin[3*e + 2*f*x] + Sin[3*e + 4*f*x]))/(4*f)

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fricas [B]  time = 0.42, size = 57, normalized size = 2.28 \[ \frac {4 i \, a c^{4}}{f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))*(c-I*c*tan(f*x+e))^4,x, algorithm="fricas")

[Out]

4*I*a*c^4/(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*f*e^(4*I*f*x + 4*I*e) + 4*f*e^(2*I*f*x + 2*I*e)
+ f)

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giac [B]  time = 1.16, size = 61, normalized size = 2.44 \[ \frac {4 i \, a c^{4}}{f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))*(c-I*c*tan(f*x+e))^4,x, algorithm="giac")

[Out]

4*I*a*c^4/(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*f*e^(4*I*f*x + 4*I*e) + 4*f*e^(2*I*f*x + 2*I*e)
+ f)

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maple [B]  time = 0.02, size = 48, normalized size = 1.92 \[ \frac {a \,c^{4} \left (\tan \left (f x +e \right )+\frac {i \left (\tan ^{4}\left (f x +e \right )\right )}{4}-\left (\tan ^{3}\left (f x +e \right )\right )-\frac {3 i \left (\tan ^{2}\left (f x +e \right )\right )}{2}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))*(c-I*c*tan(f*x+e))^4,x)

[Out]

1/f*a*c^4*(tan(f*x+e)+1/4*I*tan(f*x+e)^4-tan(f*x+e)^3-3/2*I*tan(f*x+e)^2)

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maxima [B]  time = 0.70, size = 60, normalized size = 2.40 \[ \frac {i \, a c^{4} \tan \left (f x + e\right )^{4} - 4 \, a c^{4} \tan \left (f x + e\right )^{3} - 6 i \, a c^{4} \tan \left (f x + e\right )^{2} + 4 \, a c^{4} \tan \left (f x + e\right )}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))*(c-I*c*tan(f*x+e))^4,x, algorithm="maxima")

[Out]

1/4*(I*a*c^4*tan(f*x + e)^4 - 4*a*c^4*tan(f*x + e)^3 - 6*I*a*c^4*tan(f*x + e)^2 + 4*a*c^4*tan(f*x + e))/f

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mupad [B]  time = 4.59, size = 78, normalized size = 3.12 \[ -\frac {a\,c^4\,\sin \left (e+f\,x\right )\,\left (-4\,{\cos \left (e+f\,x\right )}^3+{\cos \left (e+f\,x\right )}^2\,\sin \left (e+f\,x\right )\,6{}\mathrm {i}+4\,\cos \left (e+f\,x\right )\,{\sin \left (e+f\,x\right )}^2-{\sin \left (e+f\,x\right )}^3\,1{}\mathrm {i}\right )}{4\,f\,{\cos \left (e+f\,x\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x)*1i)*(c - c*tan(e + f*x)*1i)^4,x)

[Out]

-(a*c^4*sin(e + f*x)*(4*cos(e + f*x)*sin(e + f*x)^2 + cos(e + f*x)^2*sin(e + f*x)*6i - 4*cos(e + f*x)^3 - sin(
e + f*x)^3*1i))/(4*f*cos(e + f*x)^4)

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sympy [B]  time = 0.39, size = 90, normalized size = 3.60 \[ \frac {4 a c^{4}}{- i f e^{8 i e} e^{8 i f x} - 4 i f e^{6 i e} e^{6 i f x} - 6 i f e^{4 i e} e^{4 i f x} - 4 i f e^{2 i e} e^{2 i f x} - i f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))*(c-I*c*tan(f*x+e))**4,x)

[Out]

4*a*c**4/(-I*f*exp(8*I*e)*exp(8*I*f*x) - 4*I*f*exp(6*I*e)*exp(6*I*f*x) - 6*I*f*exp(4*I*e)*exp(4*I*f*x) - 4*I*f
*exp(2*I*e)*exp(2*I*f*x) - I*f)

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