3.10.8 \(\int (a+i a \tan (e+f x)) (c-i c \tan (e+f x))^3 \, dx\) [908]

Optimal. Leaf size=25 \[ \frac {i a (c-i c \tan (e+f x))^3}{3 f} \]

[Out]

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

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Rubi [A]
time = 0.05, 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 = {3603, 3568, 32} \begin {gather*} \frac {i a (c-i c \tan (e+f x))^3}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/3)*a*(c - I*c*Tan[e + f*x])^3)/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 3568

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 3603

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

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(55\) vs. \(2(25)=50\).
time = 0.20, size = 55, normalized size = 2.20 \begin {gather*} \frac {a c^3 \left (3 f x-3 \text {ArcTan}(\tan (e+f x))+3 \tan (e+f x)-3 i \tan ^2(e+f x)-\tan ^3(e+f x)\right )}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*c^3*(3*f*x - 3*ArcTan[Tan[e + f*x]] + 3*Tan[e + f*x] - (3*I)*Tan[e + f*x]^2 - Tan[e + f*x]^3))/(3*f)

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Maple [A]
time = 0.07, size = 21, normalized size = 0.84

method	result	size
derivativedivides	\(-\frac {a \,c^{3} \left (\tan \left (f x +e \right )+i\right )^{3}}{3 f}\)	\(21\)
default	\(-\frac {a \,c^{3} \left (\tan \left (f x +e \right )+i\right )^{3}}{3 f}\)	\(21\)
risch	\(\frac {8 i a \,c^{3}}{3 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3}}\)	\(24\)
norman	\(\frac {a \,c^{3} \tan \left (f x +e \right )}{f}-\frac {a \,c^{3} \left (\tan ^{3}\left (f x +e \right )\right )}{3 f}-\frac {i a \,c^{3} \left (\tan ^{2}\left (f x +e \right )\right )}{f}\)	\(51\)

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))^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (20) = 40\).
time = 0.53, size = 48, normalized size = 1.92 \begin {gather*} -\frac {a c^{3} \tan \left (f x + e\right )^{3} + 3 i \, a c^{3} \tan \left (f x + e\right )^{2} - 3 \, a c^{3} \tan \left (f x + e\right )}{3 \, f} \end {gather*}

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))^3,x, algorithm="maxima")

[Out]

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

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (20) = 40\).
time = 1.55, size = 48, normalized size = 1.92 \begin {gather*} \frac {8 i \, a c^{3}}{3 \, {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end {gather*}

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))^3,x, algorithm="fricas")

[Out]

8/3*I*a*c^3/(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e) + f)

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Sympy [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (19) = 38\).
time = 0.15, size = 66, normalized size = 2.64 \begin {gather*} \frac {8 i a c^{3}}{3 f e^{6 i e} e^{6 i f x} + 9 f e^{4 i e} e^{4 i f x} + 9 f e^{2 i e} e^{2 i f x} + 3 f} \end {gather*}

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))**3,x)

[Out]

8*I*a*c**3/(3*f*exp(6*I*e)*exp(6*I*f*x) + 9*f*exp(4*I*e)*exp(4*I*f*x) + 9*f*exp(2*I*e)*exp(2*I*f*x) + 3*f)

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (20) = 40\).
time = 0.57, size = 48, normalized size = 1.92 \begin {gather*} \frac {8 i \, a c^{3}}{3 \, {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end {gather*}

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))^3,x, algorithm="giac")

[Out]

8/3*I*a*c^3/(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e) + f)

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Mupad [B]
time = 4.73, size = 34, normalized size = 1.36 \begin {gather*} -\frac {a\,c^3\,\mathrm {tan}\left (e+f\,x\right )\,\left ({\mathrm {tan}\left (e+f\,x\right )}^2+\mathrm {tan}\left (e+f\,x\right )\,3{}\mathrm {i}-3\right )}{3\,f} \end {gather*}

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)^3,x)

[Out]

-(a*c^3*tan(e + f*x)*(tan(e + f*x)*3i + tan(e + f*x)^2 - 3))/(3*f)

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