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

Optimal. Leaf size=80 \[ -\frac{a^2 (A-i B) \tan (c+d x)}{d}-\frac{2 a^2 (B+i A) \log (\cos (c+d x))}{d}+2 a^2 x (A-i B)+\frac{B (a+i a \tan (c+d x))^2}{2 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0691261, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3527, 3477, 3475} \[ -\frac{a^2 (A-i B) \tan (c+d x)}{d}-\frac{2 a^2 (B+i A) \log (\cos (c+d x))}{d}+2 a^2 x (A-i B)+\frac{B (a+i a \tan (c+d x))^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

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]

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 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 (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx &=\frac{B (a+i a \tan (c+d x))^2}{2 d}-(-A+i B) \int (a+i a \tan (c+d x))^2 \, dx\\ &=2 a^2 (A-i B) x-\frac{a^2 (A-i B) \tan (c+d x)}{d}+\frac{B (a+i a \tan (c+d x))^2}{2 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{a^2 (A-i B) \tan (c+d x)}{d}+\frac{B (a+i a \tan (c+d x))^2}{2 d}\\ \end{align*}

Mathematica [B]  time = 2.22004, size = 263, normalized size = 3.29 \[ \frac{a^2 \sec (c) \sec ^2(c+d x) (\cos (2 d x)+i \sin (2 d x)) \left (-8 (A-i B) \cos (c) \cos ^2(c+d x) \tan ^{-1}(\tan (3 c+d x))-i \left ((B+i A) \cos (c+2 d x) \left (4 d x-i \log \left (\cos ^2(c+d x)\right )\right )+2 \cos (c) \left ((A-i B) \log \left (\cos ^2(c+d x)\right )+4 i A d x+4 B d x-i B\right )-2 i A \sin (c+2 d x)+4 i A d x \cos (3 c+2 d x)+A \cos (3 c+2 d x) \log \left (\cos ^2(c+d x)\right )+2 i A \sin (c)-4 B \sin (c+2 d x)+4 B d x \cos (3 c+2 d x)-i B \cos (3 c+2 d x) \log \left (\cos ^2(c+d x)\right )+4 B \sin (c)\right )\right )}{4 d (\cos (d x)+i \sin (d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.006, size = 123, normalized size = 1.5 \begin{align*} -{\frac{{a}^{2}B \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{{a}^{2}A\tan \left ( dx+c \right ) }{d}}+{\frac{2\,i{a}^{2}B\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((a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x)

[Out]

-1/2/d*a^2*B*tan(d*x+c)^2-1/d*a^2*A*tan(d*x+c)+2*I/d*a^2*B*tan(d*x+c)+I/d*a^2*A*ln(1+tan(d*x+c)^2)+1/d*a^2*B*l
n(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 = 1.54547, size = 100, normalized size = 1.25 \begin{align*} -\frac{B a^{2} \tan \left (d x + c\right )^{2} - 2 \,{\left (d x + c\right )}{\left (2 \, A - 2 i \, B\right )} a^{2} - 2 \,{\left (i \, A + B\right )} a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) +{\left (2 \, A - 4 i \, B\right )} a^{2} \tan \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.4749, size = 339, normalized size = 4.24 \begin{align*} \frac{{\left (-2 i \, A - 6 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (-2 i \, A - 4 \, B\right )} a^{2} +{\left ({\left (-2 i \, A - 2 \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-4 i \, A - 4 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (-2 i \, A - 2 \, B\right )} a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((-2*I*A - 6*B)*a^2*e^(2*I*d*x + 2*I*c) + (-2*I*A - 4*B)*a^2 + ((-2*I*A - 2*B)*a^2*e^(4*I*d*x + 4*I*c) + (-4*I
*A - 4*B)*a^2*e^(2*I*d*x + 2*I*c) + (-2*I*A - 2*B)*a^2)*log(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(4*I*d*x + 4*I*c) +
2*d*e^(2*I*d*x + 2*I*c) + d)

________________________________________________________________________________________

Sympy [A]  time = 7.08905, size = 121, normalized size = 1.51 \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 (2 i A a^{2} + 4 B a^{2}\right ) e^{- 4 i c}}{d} - \frac{\left (2 i A a^{2} + 6 B a^{2}\right ) e^{- 2 i c} e^{2 i d x}}{d}}{e^{4 i d x} + 2 e^{- 2 i c} e^{2 i d x} + e^{- 4 i c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 + (-(2*I*A*a**2 + 4*B*a**2)*exp(-4*I*c)/d - (2*I*A*a**2 +
6*B*a**2)*exp(-2*I*c)*exp(2*I*d*x)/d)/(exp(4*I*d*x) + 2*exp(-2*I*c)*exp(2*I*d*x) + exp(-4*I*c))

________________________________________________________________________________________

Giac [B]  time = 1.3872, size = 290, normalized size = 3.62 \begin{align*} \frac{-2 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 ) - 2 \, 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 ) - 4 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 ) - 4 \, 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 ) - 2 i \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 6 \, B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i \, A a^{2} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 2 \, B a^{2} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 2 i \, A a^{2} - 4 \, B a^{2}}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*I*A*a^2*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 2*B*a^2*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*
I*c) + 1) - 4*I*A*a^2*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 4*B*a^2*e^(2*I*d*x + 2*I*c)*log(e^(2*
I*d*x + 2*I*c) + 1) - 2*I*A*a^2*e^(2*I*d*x + 2*I*c) - 6*B*a^2*e^(2*I*d*x + 2*I*c) - 2*I*A*a^2*log(e^(2*I*d*x +
2*I*c) + 1) - 2*B*a^2*log(e^(2*I*d*x + 2*I*c) + 1) - 2*I*A*a^2 - 4*B*a^2)/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I
*d*x + 2*I*c) + d)