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

Optimal. Leaf size=91 \[ \frac{a (B+i A) \tan ^2(c+d x)}{2 d}+\frac{a (A-i B) \tan (c+d x)}{d}+\frac{a (B+i A) \log (\cos (c+d x))}{d}-a x (A-i B)+\frac{i a B \tan ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.110829, antiderivative size = 91, 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, 3528, 3525, 3475} \[ \frac{a (B+i A) \tan ^2(c+d x)}{2 d}+\frac{a (A-i B) \tan (c+d x)}{d}+\frac{a (B+i A) \log (\cos (c+d x))}{d}-a x (A-i B)+\frac{i a B \tan ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(A - I*B)*x) + (a*(I*A + B)*Log[Cos[c + d*x]])/d + (a*(A - I*B)*Tan[c + d*x])/d + (a*(I*A + B)*Tan[c + d*x
]^2)/(2*d) + ((I/3)*a*B*Tan[c + d*x]^3)/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 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)) (A+B \tan (c+d x)) \, dx &=\frac{i a B \tan ^3(c+d x)}{3 d}+\int \tan ^2(c+d x) (a (A-i B)+a (i A+B) \tan (c+d x)) \, dx\\ &=\frac{a (i A+B) \tan ^2(c+d x)}{2 d}+\frac{i a B \tan ^3(c+d x)}{3 d}+\int \tan (c+d x) (-a (i A+B)+a (A-i B) \tan (c+d x)) \, dx\\ &=-a (A-i B) x+\frac{a (A-i B) \tan (c+d x)}{d}+\frac{a (i A+B) \tan ^2(c+d x)}{2 d}+\frac{i a B \tan ^3(c+d x)}{3 d}-(a (i A+B)) \int \tan (c+d x) \, dx\\ &=-a (A-i B) x+\frac{a (i A+B) \log (\cos (c+d x))}{d}+\frac{a (A-i B) \tan (c+d x)}{d}+\frac{a (i A+B) \tan ^2(c+d x)}{2 d}+\frac{i a B \tan ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.869767, size = 86, normalized size = 0.95 \[ \frac{a \left (3 (B+i A) \tan ^2(c+d x)-6 (A-i B) \tan ^{-1}(\tan (c+d x))+6 (A-i B) \tan (c+d x)+6 (B+i A) \log (\cos (c+d x))+2 i B \tan ^3(c+d x)\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(-6*(A - I*B)*ArcTan[Tan[c + d*x]] + 6*(I*A + B)*Log[Cos[c + d*x]] + 6*(A - I*B)*Tan[c + d*x] + 3*(I*A + B)
*Tan[c + d*x]^2 + (2*I)*B*Tan[c + d*x]^3))/(6*d)

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Maple [A]  time = 0.012, size = 141, normalized size = 1.6 \begin{align*}{\frac{{\frac{i}{3}}aB \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{{\frac{i}{2}}aA \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{d}}-{\frac{iaB\tan \left ( dx+c \right ) }{d}}+{\frac{aB \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{aA\tan \left ( dx+c \right ) }{d}}-{\frac{{\frac{i}{2}}a\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) A}{d}}-{\frac{a\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) B}{2\,d}}+{\frac{iaB\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d}}-{\frac{aA\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))*(A+B*tan(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.52614, size = 113, normalized size = 1.24 \begin{align*} -\frac{-2 i \, B a \tan \left (d x + c\right )^{3} + 3 \,{\left (-i \, A - B\right )} a \tan \left (d x + c\right )^{2} + 6 \,{\left (d x + c\right )}{\left (A - i \, B\right )} a + 3 \,{\left (i \, A + B\right )} a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) -{\left (6 \, A - 6 i \, B\right )} a \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)^2*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.42001, size = 467, normalized size = 5.13 \begin{align*} \frac{{\left (12 i \, A + 18 \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (18 i \, A + 18 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (6 i \, A + 8 \, B\right )} a +{\left ({\left (3 i \, A + 3 \, B\right )} a e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (9 i \, A + 9 \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (9 i \, A + 9 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (3 i \, A + 3 \, B\right )} a\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\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)^2*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/3*((12*I*A + 18*B)*a*e^(4*I*d*x + 4*I*c) + (18*I*A + 18*B)*a*e^(2*I*d*x + 2*I*c) + (6*I*A + 8*B)*a + ((3*I*A
+ 3*B)*a*e^(6*I*d*x + 6*I*c) + (9*I*A + 9*B)*a*e^(4*I*d*x + 4*I*c) + (9*I*A + 9*B)*a*e^(2*I*d*x + 2*I*c) + (3
*I*A + 3*B)*a)*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)

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Sympy [B]  time = 12.6028, size = 156, normalized size = 1.71 \begin{align*} \frac{a \left (i A + B\right ) \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac{\frac{\left (4 i A a + 6 B a\right ) e^{- 2 i c} e^{4 i d x}}{d} + \frac{\left (6 i A a + 6 B a\right ) e^{- 4 i c} e^{2 i d x}}{d} + \frac{\left (6 i A a + 8 B a\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)**2*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x)

[Out]

a*(I*A + B)*log(exp(2*I*d*x) + exp(-2*I*c))/d + ((4*I*A*a + 6*B*a)*exp(-2*I*c)*exp(4*I*d*x)/d + (6*I*A*a + 6*B
*a)*exp(-4*I*c)*exp(2*I*d*x)/d + (6*I*A*a + 8*B*a)*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))

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Giac [B]  time = 1.64339, size = 383, normalized size = 4.21 \begin{align*} \frac{3 i \, A a e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 3 \, B a e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 9 i \, A a e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 9 \, B a e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 9 i \, A a e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 9 \, B a e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 12 i \, A a e^{\left (4 i \, d x + 4 i \, c\right )} + 18 \, B a e^{\left (4 i \, d x + 4 i \, c\right )} + 18 i \, A a e^{\left (2 i \, d x + 2 i \, c\right )} + 18 \, B a e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, A a \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 3 \, B a \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 6 i \, A a + 8 \, B a}{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)^2*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

1/3*(3*I*A*a*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 3*B*a*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I
*c) + 1) + 9*I*A*a*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 9*B*a*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x
+ 2*I*c) + 1) + 9*I*A*a*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 9*B*a*e^(2*I*d*x + 2*I*c)*log(e^(2
*I*d*x + 2*I*c) + 1) + 12*I*A*a*e^(4*I*d*x + 4*I*c) + 18*B*a*e^(4*I*d*x + 4*I*c) + 18*I*A*a*e^(2*I*d*x + 2*I*c
) + 18*B*a*e^(2*I*d*x + 2*I*c) + 3*I*A*a*log(e^(2*I*d*x + 2*I*c) + 1) + 3*B*a*log(e^(2*I*d*x + 2*I*c) + 1) + 6
*I*A*a + 8*B*a)/(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)