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

Optimal. Leaf size=107 \[ \frac{a^2 (B+i A) \tan (c+d x)}{d}-\frac{2 a^2 (A-i B) \log (\cos (c+d x))}{d}-2 a^2 x (B+i A)+\frac{A (a+i a \tan (c+d x))^2}{2 d}-\frac{i B (a+i a \tan (c+d x))^3}{3 a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.115465, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3592, 3527, 3477, 3475} \[ \frac{a^2 (B+i A) \tan (c+d x)}{d}-\frac{2 a^2 (A-i B) \log (\cos (c+d x))}{d}-2 a^2 x (B+i A)+\frac{A (a+i a \tan (c+d x))^2}{2 d}-\frac{i B (a+i a \tan (c+d x))^3}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [B]  time = 3.91559, size = 273, normalized size = 2.55 \[ \frac{(a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \left ((A-i B) \cos ^3(c+d x) (-4 d x \sin (2 c)-4 i d x \cos (2 c))-(A-i B) (\cos (2 c)-i \sin (2 c)) \cos ^3(c+d x) \log \left (\cos ^2(c+d x)\right )+2 (B+i A) (\cos (2 c)-i \sin (2 c)) \cos ^3(c+d x) \tan ^{-1}(\tan (3 c+d x))+\frac{1}{3} (6 A-7 i B) \sec (c) (\sin (2 c)+i \cos (2 c)) \sin (d x) \cos ^2(c+d x)-\frac{1}{6} (\cos (2 c)-i \sin (2 c)) (3 A+2 B \tan (c)-6 i B) \cos (c+d x)+\frac{1}{3} B \cos (c) (\tan (c)+i)^2 \sin (d x)\right )}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*(I*A + B)*ArcTan[Tan[3*c + d*x]]*Cos[c + d*x]^3*(Cos[2*c] - I*Sin[2*c]) - (A - I*B)*Cos[c + d*x]^3*Log[Cos
[c + d*x]^2]*(Cos[2*c] - I*Sin[2*c]) + (A - I*B)*Cos[c + d*x]^3*((-4*I)*d*x*Cos[2*c] - 4*d*x*Sin[2*c]) + ((6*A
- (7*I)*B)*Cos[c + d*x]^2*Sec[c]*(I*Cos[2*c] + Sin[2*c])*Sin[d*x])/3 + (B*Cos[c]*Sin[d*x]*(I + Tan[c])^2)/3 -
(Cos[c + d*x]*(Cos[2*c] - I*Sin[2*c])*(3*A - (6*I)*B + 2*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 = 158, normalized size = 1.5 \begin{align*}{\frac{i{a}^{2}B \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{d}}-{\frac{{a}^{2}B \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{2\,i{a}^{2}A\tan \left ( dx+c \right ) }{d}}-{\frac{{a}^{2}A \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+2\,{\frac{{a}^{2}B\tan \left ( dx+c \right ) }{d}}-{\frac{i{a}^{2}B\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{d}}+{\frac{{a}^{2}A\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{d}}-{\frac{2\,i{a}^{2}A\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d}}-2\,{\frac{{a}^{2}B\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)*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.70474, size = 126, normalized size = 1.18 \begin{align*} -\frac{2 \, B a^{2} \tan \left (d x + c\right )^{3} +{\left (3 \, A - 6 i \, B\right )} a^{2} \tan \left (d x + c\right )^{2} + 12 \,{\left (d x + c\right )}{\left (i \, A + B\right )} a^{2} - 6 \,{\left (A - i \, B\right )} a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 12 \,{\left (-i \, A - B\right )} a^{2} \tan \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*B*a^2*tan(d*x + c)^3 + (3*A - 6*I*B)*a^2*tan(d*x + c)^2 + 12*(d*x + c)*(I*A + B)*a^2 - 6*(A - I*B)*a^2
*log(tan(d*x + c)^2 + 1) + 12*(-I*A - B)*a^2*tan(d*x + c))/d

________________________________________________________________________________________

Fricas [A]  time = 1.32, size = 474, normalized size = 4.43 \begin{align*} -\frac{2 \,{\left (3 \,{\left (3 \, A - 5 i \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \,{\left (5 \, A - 6 i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (6 \, A - 7 i \, B\right )} a^{2} + 3 \,{\left ({\left (A - i \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \,{\left (A - i \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \,{\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (A - i \, B\right )} a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{3 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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)*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-2/3*(3*(3*A - 5*I*B)*a^2*e^(4*I*d*x + 4*I*c) + 3*(5*A - 6*I*B)*a^2*e^(2*I*d*x + 2*I*c) + (6*A - 7*I*B)*a^2 +
3*((A - I*B)*a^2*e^(6*I*d*x + 6*I*c) + 3*(A - I*B)*a^2*e^(4*I*d*x + 4*I*c) + 3*(A - I*B)*a^2*e^(2*I*d*x + 2*I*
c) + (A - I*B)*a^2)*log(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*
I*d*x + 2*I*c) + d)

________________________________________________________________________________________

Sympy [A]  time = 15.407, size = 172, normalized size = 1.61 \begin{align*} \frac{2 a^{2} \left (- A + i B\right ) \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac{- \frac{\left (6 A a^{2} - 10 i B a^{2}\right ) e^{- 2 i c} e^{4 i d x}}{d} - \frac{\left (10 A a^{2} - 12 i B a^{2}\right ) e^{- 4 i c} e^{2 i d x}}{d} - \frac{\left (12 A a^{2} - 14 i B a^{2}\right ) e^{- 6 i c}}{3 d}}{e^{6 i d x} + 3 e^{- 2 i c} e^{4 i d x} + 3 e^{- 4 i c} e^{2 i d x} + e^{- 6 i c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a**2*(-A + I*B)*log(exp(2*I*d*x) + exp(-2*I*c))/d + (-(6*A*a**2 - 10*I*B*a**2)*exp(-2*I*c)*exp(4*I*d*x)/d -
(10*A*a**2 - 12*I*B*a**2)*exp(-4*I*c)*exp(2*I*d*x)/d - (12*A*a**2 - 14*I*B*a**2)*exp(-6*I*c)/(3*d))/(exp(6*I*d
*x) + 3*exp(-2*I*c)*exp(4*I*d*x) + 3*exp(-4*I*c)*exp(2*I*d*x) + exp(-6*I*c))

________________________________________________________________________________________

Giac [B]  time = 1.47892, size = 421, normalized size = 3.93 \begin{align*} -\frac{6 \, 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 ) - 6 i \, 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 ) + 18 \, 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 ) - 18 i \, 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 ) + 18 \, 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 ) - 18 i \, 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 ) + 18 \, A a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 30 i \, B a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 30 \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 36 i \, B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 6 \, A a^{2} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 6 i \, B a^{2} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 12 \, A a^{2} - 14 i \, B a^{2}}{3 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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)*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/3*(6*A*a^2*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 6*I*B*a^2*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x
+ 2*I*c) + 1) + 18*A*a^2*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 18*I*B*a^2*e^(4*I*d*x + 4*I*c)*log
(e^(2*I*d*x + 2*I*c) + 1) + 18*A*a^2*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 18*I*B*a^2*e^(2*I*d*x
+ 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 18*A*a^2*e^(4*I*d*x + 4*I*c) - 30*I*B*a^2*e^(4*I*d*x + 4*I*c) + 30*A*a
^2*e^(2*I*d*x + 2*I*c) - 36*I*B*a^2*e^(2*I*d*x + 2*I*c) + 6*A*a^2*log(e^(2*I*d*x + 2*I*c) + 1) - 6*I*B*a^2*log
(e^(2*I*d*x + 2*I*c) + 1) + 12*A*a^2 - 14*I*B*a^2)/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2
*I*d*x + 2*I*c) + d)