3.696 \(\int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) \, dx\)

Optimal. Leaf size=110 \[ -\frac {2 a^3 (A-i B) \tan (e+f x)}{f}-\frac {4 a^3 (B+i A) \log (\cos (e+f x))}{f}+4 a^3 x (A-i B)+\frac {a (B+i A) (a+i a \tan (e+f x))^2}{2 f}+\frac {B (a+i a \tan (e+f x))^3}{3 f} \]

[Out]

4*a^3*(A-I*B)*x-4*a^3*(I*A+B)*ln(cos(f*x+e))/f-2*a^3*(A-I*B)*tan(f*x+e)/f+1/2*a*(I*A+B)*(a+I*a*tan(f*x+e))^2/f
+1/3*B*(a+I*a*tan(f*x+e))^3/f

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Rubi [A]  time = 0.09, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3527, 3478, 3477, 3475} \[ -\frac {2 a^3 (A-i B) \tan (e+f x)}{f}-\frac {4 a^3 (B+i A) \log (\cos (e+f x))}{f}+4 a^3 x (A-i B)+\frac {a (B+i A) (a+i a \tan (e+f x))^2}{2 f}+\frac {B (a+i a \tan (e+f x))^3}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

4*a^3*(A - I*B)*x - (4*a^3*(I*A + B)*Log[Cos[e + f*x]])/f - (2*a^3*(A - I*B)*Tan[e + f*x])/f + (a*(I*A + B)*(a
+ I*a*Tan[e + f*x])^2)/(2*f) + (B*(a + I*a*Tan[e + f*x])^3)/(3*f)

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3477

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(a^2 - b^2)*x, x] + (Dist[2*a*b, Int[Tan[c + d
*x], x], x] + Simp[(b^2*Tan[c + d*x])/d, x]) /; FreeQ[{a, b, c, d}, x]

Rule 3478

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Dist[2*a, Int[(a + b*Tan[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && G
tQ[n, 1]

Rule 3527

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d*
(a + b*Tan[e + f*x])^m)/(f*m), x] + Dist[(b*c + a*d)/b, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] &&  !LtQ[m, 0]

Rubi steps

\begin {align*} \int (a+i a \tan (e+f x))^3 (A+B \tan (e+f x)) \, dx &=\frac {B (a+i a \tan (e+f x))^3}{3 f}-(-A+i B) \int (a+i a \tan (e+f x))^3 \, dx\\ &=\frac {a (i A+B) (a+i a \tan (e+f x))^2}{2 f}+\frac {B (a+i a \tan (e+f x))^3}{3 f}+(2 a (A-i B)) \int (a+i a \tan (e+f x))^2 \, dx\\ &=4 a^3 (A-i B) x-\frac {2 a^3 (A-i B) \tan (e+f x)}{f}+\frac {a (i A+B) (a+i a \tan (e+f x))^2}{2 f}+\frac {B (a+i a \tan (e+f x))^3}{3 f}+\left (4 a^3 (i A+B)\right ) \int \tan (e+f x) \, dx\\ &=4 a^3 (A-i B) x-\frac {4 a^3 (i A+B) \log (\cos (e+f x))}{f}-\frac {2 a^3 (A-i B) \tan (e+f x)}{f}+\frac {a (i A+B) (a+i a \tan (e+f x))^2}{2 f}+\frac {B (a+i a \tan (e+f x))^3}{3 f}\\ \end {align*}

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Mathematica [B]  time = 4.09, size = 331, normalized size = 3.01 \[ \frac {a^3 \sec (e) \sec ^3(e+f x) \left (3 \cos (f x) \left ((-3 B-3 i A) \log \left (\cos ^2(e+f x)\right )+6 A f x-i A-6 i B f x-3 B\right )+3 \cos (2 e+f x) \left ((-3 B-3 i A) \log \left (\cos ^2(e+f x)\right )+6 A f x-i A-6 i B f x-3 B\right )+9 A \sin (2 e+f x)-9 A \sin (2 e+3 f x)+6 A f x \cos (2 e+3 f x)+6 A f x \cos (4 e+3 f x)-3 i A \cos (2 e+3 f x) \log \left (\cos ^2(e+f x)\right )-3 i A \cos (4 e+3 f x) \log \left (\cos ^2(e+f x)\right )-18 A \sin (f x)-15 i B \sin (2 e+f x)+13 i B \sin (2 e+3 f x)-6 i B f x \cos (2 e+3 f x)-6 i B f x \cos (4 e+3 f x)-3 B \cos (2 e+3 f x) \log \left (\cos ^2(e+f x)\right )-3 B \cos (4 e+3 f x) \log \left (\cos ^2(e+f x)\right )+24 i B \sin (f x)\right )}{12 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Sec[e]*Sec[e + f*x]^3*(6*A*f*x*Cos[2*e + 3*f*x] - (6*I)*B*f*x*Cos[2*e + 3*f*x] + 6*A*f*x*Cos[4*e + 3*f*x]
- (6*I)*B*f*x*Cos[4*e + 3*f*x] - (3*I)*A*Cos[2*e + 3*f*x]*Log[Cos[e + f*x]^2] - 3*B*Cos[2*e + 3*f*x]*Log[Cos[
e + f*x]^2] - (3*I)*A*Cos[4*e + 3*f*x]*Log[Cos[e + f*x]^2] - 3*B*Cos[4*e + 3*f*x]*Log[Cos[e + f*x]^2] + 3*Cos[
f*x]*((-I)*A - 3*B + 6*A*f*x - (6*I)*B*f*x + ((-3*I)*A - 3*B)*Log[Cos[e + f*x]^2]) + 3*Cos[2*e + f*x]*((-I)*A
- 3*B + 6*A*f*x - (6*I)*B*f*x + ((-3*I)*A - 3*B)*Log[Cos[e + f*x]^2]) - 18*A*Sin[f*x] + (24*I)*B*Sin[f*x] + 9*
A*Sin[2*e + f*x] - (15*I)*B*Sin[2*e + f*x] - 9*A*Sin[2*e + 3*f*x] + (13*I)*B*Sin[2*e + 3*f*x]))/(12*f)

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fricas [A]  time = 0.77, size = 178, normalized size = 1.62 \[ \frac {{\left (-24 i \, A - 48 \, B\right )} a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-42 i \, A - 66 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-18 i \, A - 26 \, B\right )} a^{3} + {\left ({\left (-12 i \, A - 12 \, B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (-36 i \, A - 36 \, B\right )} a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-36 i \, A - 36 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-12 i \, A - 12 \, B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{3 \, {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e)),x, algorithm="fricas")

[Out]

1/3*((-24*I*A - 48*B)*a^3*e^(4*I*f*x + 4*I*e) + (-42*I*A - 66*B)*a^3*e^(2*I*f*x + 2*I*e) + (-18*I*A - 26*B)*a^
3 + ((-12*I*A - 12*B)*a^3*e^(6*I*f*x + 6*I*e) + (-36*I*A - 36*B)*a^3*e^(4*I*f*x + 4*I*e) + (-36*I*A - 36*B)*a^
3*e^(2*I*f*x + 2*I*e) + (-12*I*A - 12*B)*a^3)*log(e^(2*I*f*x + 2*I*e) + 1))/(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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giac [B]  time = 1.41, size = 333, normalized size = 3.03 \[ \frac {-12 i \, A a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 12 \, B a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 36 i \, A a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 36 \, B a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 36 i \, A a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 36 \, B a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 24 i \, A a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 48 \, B a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 42 i \, A a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 66 \, B a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 12 i \, A a^{3} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 12 \, B a^{3} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 18 i \, A a^{3} - 26 \, B a^{3}}{3 \, {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e)),x, algorithm="giac")

[Out]

1/3*(-12*I*A*a^3*e^(6*I*f*x + 6*I*e)*log(e^(2*I*f*x + 2*I*e) + 1) - 12*B*a^3*e^(6*I*f*x + 6*I*e)*log(e^(2*I*f*
x + 2*I*e) + 1) - 36*I*A*a^3*e^(4*I*f*x + 4*I*e)*log(e^(2*I*f*x + 2*I*e) + 1) - 36*B*a^3*e^(4*I*f*x + 4*I*e)*l
og(e^(2*I*f*x + 2*I*e) + 1) - 36*I*A*a^3*e^(2*I*f*x + 2*I*e)*log(e^(2*I*f*x + 2*I*e) + 1) - 36*B*a^3*e^(2*I*f*
x + 2*I*e)*log(e^(2*I*f*x + 2*I*e) + 1) - 24*I*A*a^3*e^(4*I*f*x + 4*I*e) - 48*B*a^3*e^(4*I*f*x + 4*I*e) - 42*I
*A*a^3*e^(2*I*f*x + 2*I*e) - 66*B*a^3*e^(2*I*f*x + 2*I*e) - 12*I*A*a^3*log(e^(2*I*f*x + 2*I*e) + 1) - 12*B*a^3
*log(e^(2*I*f*x + 2*I*e) + 1) - 18*I*A*a^3 - 26*B*a^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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maple [A]  time = 0.02, size = 160, normalized size = 1.45 \[ -\frac {i a^{3} B \left (\tan ^{3}\left (f x +e \right )\right )}{3 f}-\frac {i a^{3} A \left (\tan ^{2}\left (f x +e \right )\right )}{2 f}+\frac {4 i a^{3} B \tan \left (f x +e \right )}{f}-\frac {3 a^{3} B \left (\tan ^{2}\left (f x +e \right )\right )}{2 f}-\frac {3 a^{3} A \tan \left (f x +e \right )}{f}+\frac {2 i a^{3} A \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{f}+\frac {2 a^{3} B \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{f}-\frac {4 i a^{3} B \arctan \left (\tan \left (f x +e \right )\right )}{f}+\frac {4 a^{3} A \arctan \left (\tan \left (f x +e \right )\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e)),x)

[Out]

-1/3*I/f*a^3*B*tan(f*x+e)^3-1/2*I/f*a^3*A*tan(f*x+e)^2+4*I/f*a^3*B*tan(f*x+e)-3/2/f*a^3*B*tan(f*x+e)^2-3/f*a^3
*A*tan(f*x+e)+2*I/f*a^3*A*ln(1+tan(f*x+e)^2)+2/f*a^3*B*ln(1+tan(f*x+e)^2)-4*I/f*a^3*B*arctan(tan(f*x+e))+4/f*a
^3*A*arctan(tan(f*x+e))

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maxima [A]  time = 0.66, size = 96, normalized size = 0.87 \[ -\frac {2 i \, B a^{3} \tan \left (f x + e\right )^{3} - {\left (-3 i \, A - 9 \, B\right )} a^{3} \tan \left (f x + e\right )^{2} - 24 \, {\left (f x + e\right )} {\left (A - i \, B\right )} a^{3} - 6 \, {\left (2 i \, A + 2 \, B\right )} a^{3} \log \left (\tan \left (f x + e\right )^{2} + 1\right ) + 6 \, {\left (3 \, A - 4 i \, B\right )} a^{3} \tan \left (f x + e\right )}{6 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e)),x, algorithm="maxima")

[Out]

-1/6*(2*I*B*a^3*tan(f*x + e)^3 - (-3*I*A - 9*B)*a^3*tan(f*x + e)^2 - 24*(f*x + e)*(A - I*B)*a^3 - 6*(2*I*A + 2
*B)*a^3*log(tan(f*x + e)^2 + 1) + 6*(3*A - 4*I*B)*a^3*tan(f*x + e))/f

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mupad [B]  time = 8.91, size = 125, normalized size = 1.14 \[ -\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {B\,a^3}{2}+\frac {a^3\,\left (2\,B+A\,1{}\mathrm {i}\right )}{2}\right )}{f}+\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\left (4\,B\,a^3+A\,a^3\,4{}\mathrm {i}\right )}{f}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (B\,a^3\,1{}\mathrm {i}-a^3\,\left (2\,A-B\,1{}\mathrm {i}\right )+a^3\,\left (2\,B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}\right )}{f}-\frac {B\,a^3\,{\mathrm {tan}\left (e+f\,x\right )}^3\,1{}\mathrm {i}}{3\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^3,x)

[Out]

(log(tan(e + f*x) + 1i)*(A*a^3*4i + 4*B*a^3))/f - (tan(e + f*x)^2*((B*a^3)/2 + (a^3*(A*1i + 2*B))/2))/f + (tan
(e + f*x)*(B*a^3*1i - a^3*(2*A - B*1i) + a^3*(A*1i + 2*B)*1i))/f - (B*a^3*tan(e + f*x)^3*1i)/(3*f)

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sympy [B]  time = 0.73, size = 187, normalized size = 1.70 \[ - \frac {4 i a^{3} \left (A - i B\right ) \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{f} + \frac {18 A a^{3} - 26 i B a^{3} + \left (42 A a^{3} e^{2 i e} - 66 i B a^{3} e^{2 i e}\right ) e^{2 i f x} + \left (24 A a^{3} e^{4 i e} - 48 i B a^{3} e^{4 i e}\right ) e^{4 i f x}}{3 i f e^{6 i e} e^{6 i f x} + 9 i f e^{4 i e} e^{4 i f x} + 9 i f e^{2 i e} e^{2 i f x} + 3 i f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**3*(A+B*tan(f*x+e)),x)

[Out]

-4*I*a**3*(A - I*B)*log(exp(2*I*f*x) + exp(-2*I*e))/f + (18*A*a**3 - 26*I*B*a**3 + (42*A*a**3*exp(2*I*e) - 66*
I*B*a**3*exp(2*I*e))*exp(2*I*f*x) + (24*A*a**3*exp(4*I*e) - 48*I*B*a**3*exp(4*I*e))*exp(4*I*f*x))/(3*I*f*exp(6
*I*e)*exp(6*I*f*x) + 9*I*f*exp(4*I*e)*exp(4*I*f*x) + 9*I*f*exp(2*I*e)*exp(2*I*f*x) + 3*I*f)

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