3.9 \(\int \tan ^2(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=141 \[ -\frac{a^2 (4 A-5 i B) \tan ^3(c+d x)}{12 d}+\frac{a^2 (B+i A) \tan ^2(c+d x)}{d}+\frac{2 a^2 (A-i B) \tan (c+d x)}{d}+\frac{2 a^2 (B+i A) \log (\cos (c+d x))}{d}-2 a^2 x (A-i B)+\frac{i B \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d} \]

[Out]

-2*a^2*(A - I*B)*x + (2*a^2*(I*A + B)*Log[Cos[c + d*x]])/d + (2*a^2*(A - I*B)*Tan[c + d*x])/d + (a^2*(I*A + B)
*Tan[c + d*x]^2)/d - (a^2*(4*A - (5*I)*B)*Tan[c + d*x]^3)/(12*d) + ((I/4)*B*Tan[c + d*x]^3*(a^2 + I*a^2*Tan[c
+ d*x]))/d

________________________________________________________________________________________

Rubi [A]  time = 0.252153, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {3594, 3592, 3528, 3525, 3475} \[ -\frac{a^2 (4 A-5 i B) \tan ^3(c+d x)}{12 d}+\frac{a^2 (B+i A) \tan ^2(c+d x)}{d}+\frac{2 a^2 (A-i B) \tan (c+d x)}{d}+\frac{2 a^2 (B+i A) \log (\cos (c+d x))}{d}-2 a^2 x (A-i B)+\frac{i B \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

-2*a^2*(A - I*B)*x + (2*a^2*(I*A + B)*Log[Cos[c + d*x]])/d + (2*a^2*(A - I*B)*Tan[c + d*x])/d + (a^2*(I*A + B)
*Tan[c + d*x]^2)/d - (a^2*(4*A - (5*I)*B)*Tan[c + d*x]^3)/(12*d) + ((I/4)*B*Tan[c + d*x]^3*(a^2 + I*a^2*Tan[c
+ d*x]))/d

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 3528

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] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3525

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

Rule 3475

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

Rubi steps

\begin{align*} \int \tan ^2(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx &=\frac{i B \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\frac{1}{4} \int \tan ^2(c+d x) (a+i a \tan (c+d x)) (a (4 A-3 i B)+a (4 i A+5 B) \tan (c+d x)) \, dx\\ &=-\frac{a^2 (4 A-5 i B) \tan ^3(c+d x)}{12 d}+\frac{i B \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\frac{1}{4} \int \tan ^2(c+d x) \left (8 a^2 (A-i B)+8 a^2 (i A+B) \tan (c+d x)\right ) \, dx\\ &=\frac{a^2 (i A+B) \tan ^2(c+d x)}{d}-\frac{a^2 (4 A-5 i B) \tan ^3(c+d x)}{12 d}+\frac{i B \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\frac{1}{4} \int \tan (c+d x) \left (-8 a^2 (i A+B)+8 a^2 (A-i B) \tan (c+d x)\right ) \, dx\\ &=-2 a^2 (A-i B) x+\frac{2 a^2 (A-i B) \tan (c+d x)}{d}+\frac{a^2 (i A+B) \tan ^2(c+d x)}{d}-\frac{a^2 (4 A-5 i B) \tan ^3(c+d x)}{12 d}+\frac{i B \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}-\left (2 a^2 (i A+B)\right ) \int \tan (c+d x) \, dx\\ &=-2 a^2 (A-i B) x+\frac{2 a^2 (i A+B) \log (\cos (c+d x))}{d}+\frac{2 a^2 (A-i B) \tan (c+d x)}{d}+\frac{a^2 (i A+B) \tan ^2(c+d x)}{d}-\frac{a^2 (4 A-5 i B) \tan ^3(c+d x)}{12 d}+\frac{i B \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}\\ \end{align*}

Mathematica [B]  time = 6.32311, size = 305, normalized size = 2.16 \[ \frac{(a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \left (-4 d x (A-i B) (\cos (2 c)-i \sin (2 c)) \cos ^3(c+d x)+(B+i A) (\cos (2 c)-i \sin (2 c)) \cos ^3(c+d x) \log \left (\cos ^2(c+d x)\right )+2 (A-i B) (\cos (2 c)-i \sin (2 c)) \cos ^3(c+d x) \tan ^{-1}(\tan (3 c+d x))+\frac{1}{3} (7 A-8 i B) \sec (c) (\cos (2 c)-i \sin (2 c)) \sin (d x) \cos ^2(c+d x)-\frac{1}{6} (\cos (2 c)-i \sin (2 c)) (2 (A-2 i B) \tan (c)-6 i A-9 B) \cos (c+d x)+\frac{1}{3} (A-2 i B) \cos (c) (\tan (c)+i)^2 \sin (d x)-\frac{1}{4} B (\cos (2 c)-i \sin (2 c)) \sec (c+d x)\right )}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

((-4*(A - I*B)*d*x*Cos[c + d*x]^3*(Cos[2*c] - I*Sin[2*c]) + 2*(A - I*B)*ArcTan[Tan[3*c + d*x]]*Cos[c + d*x]^3*
(Cos[2*c] - I*Sin[2*c]) + (I*A + B)*Cos[c + d*x]^3*Log[Cos[c + d*x]^2]*(Cos[2*c] - I*Sin[2*c]) - (B*Sec[c + d*
x]*(Cos[2*c] - I*Sin[2*c]))/4 + ((7*A - (8*I)*B)*Cos[c + d*x]^2*Sec[c]*(Cos[2*c] - I*Sin[2*c])*Sin[d*x])/3 + (
(A - (2*I)*B)*Cos[c]*Sin[d*x]*(I + Tan[c])^2)/3 - (Cos[c + d*x]*(Cos[2*c] - I*Sin[2*c])*((-6*I)*A - 9*B + 2*(A
- (2*I)*B)*Tan[c]))/6)*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c +
d*x] + B*Sin[c + d*x]))

________________________________________________________________________________________

Maple [A]  time = 0.006, size = 193, normalized size = 1.4 \begin{align*}{\frac{{\frac{2\,i}{3}}{a}^{2}B \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{d}}-{\frac{{a}^{2}B \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{i{a}^{2}A \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{d}}-{\frac{{a}^{2}A \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-{\frac{2\,i{a}^{2}B\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}B \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{d}}+2\,{\frac{{a}^{2}A\tan \left ( dx+c \right ) }{d}}-{\frac{i{a}^{2}A\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{d}}-{\frac{{a}^{2}B\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{d}}+{\frac{2\,i{a}^{2}B\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d}}-2\,{\frac{{a}^{2}A\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.61527, size = 158, normalized size = 1.12 \begin{align*} -\frac{3 \, B a^{2} \tan \left (d x + c\right )^{4} +{\left (4 \, A - 8 i \, B\right )} a^{2} \tan \left (d x + c\right )^{3} + 12 \,{\left (-i \, A - B\right )} a^{2} \tan \left (d x + c\right )^{2} + 12 \,{\left (d x + c\right )}{\left (2 \, A - 2 i \, B\right )} a^{2} - 12 \,{\left (-i \, A - B\right )} a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) -{\left (24 \, A - 24 i \, B\right )} a^{2} \tan \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/12*(3*B*a^2*tan(d*x + c)^4 + (4*A - 8*I*B)*a^2*tan(d*x + c)^3 + 12*(-I*A - B)*a^2*tan(d*x + c)^2 + 12*(d*x
+ c)*(2*A - 2*I*B)*a^2 - 12*(-I*A - B)*a^2*log(tan(d*x + c)^2 + 1) - (24*A - 24*I*B)*a^2*tan(d*x + c))/d

________________________________________________________________________________________

Fricas [A]  time = 1.38267, size = 643, normalized size = 4.56 \begin{align*} \frac{{\left (30 i \, A + 42 \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (66 i \, A + 72 \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (50 i \, A + 58 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (14 i \, A + 16 \, B\right )} a^{2} +{\left ({\left (6 i \, A + 6 \, B\right )} a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} +{\left (24 i \, A + 24 \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (36 i \, A + 36 \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (24 i \, A + 24 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (6 i \, A + 6 \, B\right )} a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}{3 \,{\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/3*((30*I*A + 42*B)*a^2*e^(6*I*d*x + 6*I*c) + (66*I*A + 72*B)*a^2*e^(4*I*d*x + 4*I*c) + (50*I*A + 58*B)*a^2*e
^(2*I*d*x + 2*I*c) + (14*I*A + 16*B)*a^2 + ((6*I*A + 6*B)*a^2*e^(8*I*d*x + 8*I*c) + (24*I*A + 24*B)*a^2*e^(6*I
*d*x + 6*I*c) + (36*I*A + 36*B)*a^2*e^(4*I*d*x + 4*I*c) + (24*I*A + 24*B)*a^2*e^(2*I*d*x + 2*I*c) + (6*I*A + 6
*B)*a^2)*log(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I
*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)

________________________________________________________________________________________

Sympy [A]  time = 42.0235, size = 221, normalized size = 1.57 \begin{align*} \frac{2 a^{2} \left (i A + B\right ) \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac{\frac{\left (10 i A a^{2} + 14 B a^{2}\right ) e^{- 2 i c} e^{6 i d x}}{d} + \frac{\left (14 i A a^{2} + 16 B a^{2}\right ) e^{- 8 i c}}{3 d} + \frac{\left (22 i A a^{2} + 24 B a^{2}\right ) e^{- 4 i c} e^{4 i d x}}{d} + \frac{\left (50 i A a^{2} + 58 B a^{2}\right ) e^{- 6 i c} e^{2 i d x}}{3 d}}{e^{8 i d x} + 4 e^{- 2 i c} e^{6 i d x} + 6 e^{- 4 i c} e^{4 i d x} + 4 e^{- 6 i c} e^{2 i d x} + e^{- 8 i c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**2*(a+I*a*tan(d*x+c))**2*(A+B*tan(d*x+c)),x)

[Out]

2*a**2*(I*A + B)*log(exp(2*I*d*x) + exp(-2*I*c))/d + ((10*I*A*a**2 + 14*B*a**2)*exp(-2*I*c)*exp(6*I*d*x)/d + (
14*I*A*a**2 + 16*B*a**2)*exp(-8*I*c)/(3*d) + (22*I*A*a**2 + 24*B*a**2)*exp(-4*I*c)*exp(4*I*d*x)/d + (50*I*A*a*
*2 + 58*B*a**2)*exp(-6*I*c)*exp(2*I*d*x)/(3*d))/(exp(8*I*d*x) + 4*exp(-2*I*c)*exp(6*I*d*x) + 6*exp(-4*I*c)*exp
(4*I*d*x) + 4*exp(-6*I*c)*exp(2*I*d*x) + exp(-8*I*c))

________________________________________________________________________________________

Giac [B]  time = 1.70968, size = 551, normalized size = 3.91 \begin{align*} \frac{6 i \, A a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 6 \, B a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 24 i \, A a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 24 \, B a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 36 i \, A a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 36 \, B a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 24 i \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 24 \, B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 30 i \, A a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 42 \, B a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 66 i \, A a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 72 \, B a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 50 i \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 58 \, B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 6 i \, A a^{2} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 6 \, B a^{2} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 14 i \, A a^{2} + 16 \, B a^{2}}{3 \,{\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

1/3*(6*I*A*a^2*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 6*B*a^2*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x +
2*I*c) + 1) + 24*I*A*a^2*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 24*B*a^2*e^(6*I*d*x + 6*I*c)*log(
e^(2*I*d*x + 2*I*c) + 1) + 36*I*A*a^2*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 36*B*a^2*e^(4*I*d*x +
4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 24*I*A*a^2*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 24*B*a^2*
e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 30*I*A*a^2*e^(6*I*d*x + 6*I*c) + 42*B*a^2*e^(6*I*d*x + 6*I*
c) + 66*I*A*a^2*e^(4*I*d*x + 4*I*c) + 72*B*a^2*e^(4*I*d*x + 4*I*c) + 50*I*A*a^2*e^(2*I*d*x + 2*I*c) + 58*B*a^2
*e^(2*I*d*x + 2*I*c) + 6*I*A*a^2*log(e^(2*I*d*x + 2*I*c) + 1) + 6*B*a^2*log(e^(2*I*d*x + 2*I*c) + 1) + 14*I*A*
a^2 + 16*B*a^2)/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x +
2*I*c) + d)