3.45 \(\int \frac{\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=103 \[ \frac{-3 B+i A}{4 a^2 d (1+i \tan (c+d x))}-\frac{x (A+3 i B)}{4 a^2}+\frac{B \log (\cos (c+d x))}{a^2 d}+\frac{(-B+i A) \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

-((A + (3*I)*B)*x)/(4*a^2) + (B*Log[Cos[c + d*x]])/(a^2*d) + (I*A - 3*B)/(4*a^2*d*(1 + I*Tan[c + d*x])) + ((I*
A - B)*Tan[c + d*x]^2)/(4*d*(a + I*a*Tan[c + d*x])^2)

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Rubi [A]  time = 0.210948, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {3595, 3589, 3475, 12, 3526, 8} \[ \frac{-3 B+i A}{4 a^2 d (1+i \tan (c+d x))}-\frac{x (A+3 i B)}{4 a^2}+\frac{B \log (\cos (c+d x))}{a^2 d}+\frac{(-B+i A) \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((A + (3*I)*B)*x)/(4*a^2) + (B*Log[Cos[c + d*x]])/(a^2*d) + (I*A - 3*B)/(4*a^2*d*(1 + I*Tan[c + d*x])) + ((I*
A - B)*Tan[c + d*x]^2)/(4*d*(a + I*a*Tan[c + d*x])^2)

Rule 3595

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((A*b - a*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(2*a*f*
m), x] + Dist[1/(2*a^2*m), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[A*(a*c*m + b*d*n
) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A,
B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]

Rule 3589

Int[(((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]))/((a_.) + (b_.)*tan[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Dist[(B*d)/b, Int[Tan[e + f*x], x], x] + Dist[1/b, Int[Simp[A*b*c + (A*b*d + B*(
b*c - a*d))*Tan[e + f*x], x]/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && 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]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3526

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^m)/(2*a*f*m), x] + Dist[(b*c + a*d)/(2*a*b), Int[(a + b*Tan[e + f*x])^(m + 1),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.872514, size = 185, normalized size = 1.8 \[ \frac{\sec ^2(c+d x) \left (\cos (2 (c+d x)) \left (4 A d x+i A-8 B \log \left (\cos ^2(c+d x)\right )-4 i B d x-B\right )+4 i A d x \sin (2 (c+d x))+A \sin (2 (c+d x))-4 i A+i B \sin (2 (c+d x))+4 B d x \sin (2 (c+d x))-8 i B \sin (2 (c+d x)) \log \left (\cos ^2(c+d x)\right )+16 i B \tan ^{-1}(\tan (d x)) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+8 B\right )}{16 a^2 d (\tan (c+d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^2*((-4*I)*A + 8*B + Cos[2*(c + d*x)]*(I*A - B + 4*A*d*x - (4*I)*B*d*x - 8*B*Log[Cos[c + d*x]^2])
+ (16*I)*B*ArcTan[Tan[d*x]]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]) + A*Sin[2*(c + d*x)] + I*B*Sin[2*(c + d*x
)] + (4*I)*A*d*x*Sin[2*(c + d*x)] + 4*B*d*x*Sin[2*(c + d*x)] - (8*I)*B*Log[Cos[c + d*x]^2]*Sin[2*(c + d*x)]))/
(16*a^2*d*(-I + Tan[c + d*x])^2)

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Maple [A]  time = 0.032, size = 162, normalized size = 1.6 \begin{align*}{\frac{{\frac{i}{4}}A}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{B}{4\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}+{\frac{{\frac{5\,i}{4}}B}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{3\,A}{4\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{{\frac{i}{8}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) A}{{a}^{2}d}}-{\frac{7\,\ln \left ( \tan \left ( dx+c \right ) -i \right ) B}{8\,{a}^{2}d}}-{\frac{B\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{8\,{a}^{2}d}}-{\frac{{\frac{i}{8}}A\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{{a}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*I/d/a^2/(tan(d*x+c)-I)^2*A-1/4/d/a^2/(tan(d*x+c)-I)^2*B+5/4*I/d/a^2/(tan(d*x+c)-I)*B+3/4/d/a^2/(tan(d*x+c)
-I)*A+1/8*I/d/a^2*ln(tan(d*x+c)-I)*A-7/8/d/a^2*ln(tan(d*x+c)-I)*B-1/8/d/a^2*B*ln(tan(d*x+c)+I)-1/8*I/d/a^2*A*l
n(tan(d*x+c)+I)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.44303, size = 240, normalized size = 2.33 \begin{align*} -\frac{{\left (4 \,{\left (A + 7 i \, B\right )} d x e^{\left (4 i \, d x + 4 i \, c\right )} - 16 \, B e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) -{\left (4 i \, A - 8 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A - B\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(4*(A + 7*I*B)*d*x*e^(4*I*d*x + 4*I*c) - 16*B*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - (4*I*A
- 8*B)*e^(2*I*d*x + 2*I*c) + I*A - B)*e^(-4*I*d*x - 4*I*c)/(a^2*d)

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Sympy [A]  time = 4.92274, size = 223, normalized size = 2.17 \begin{align*} \frac{B \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{2} d} + \begin{cases} \frac{\left (\left (- 4 i A a^{2} d e^{2 i c} + 4 B a^{2} d e^{2 i c}\right ) e^{- 4 i d x} + \left (16 i A a^{2} d e^{4 i c} - 32 B a^{2} d e^{4 i c}\right ) e^{- 2 i d x}\right ) e^{- 6 i c}}{64 a^{4} d^{2}} & \text{for}\: 64 a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac{A + 7 i B}{4 a^{2}} - \frac{\left (A e^{4 i c} - 2 A e^{2 i c} + A + 7 i B e^{4 i c} - 4 i B e^{2 i c} + i B\right ) e^{- 4 i c}}{4 a^{2}}\right ) & \text{otherwise} \end{cases} + \frac{x \left (- A - 7 i B\right )}{4 a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*log(exp(2*I*d*x) + exp(-2*I*c))/(a**2*d) + Piecewise((((-4*I*A*a**2*d*exp(2*I*c) + 4*B*a**2*d*exp(2*I*c))*ex
p(-4*I*d*x) + (16*I*A*a**2*d*exp(4*I*c) - 32*B*a**2*d*exp(4*I*c))*exp(-2*I*d*x))*exp(-6*I*c)/(64*a**4*d**2), N
e(64*a**4*d**2*exp(6*I*c), 0)), (x*((A + 7*I*B)/(4*a**2) - (A*exp(4*I*c) - 2*A*exp(2*I*c) + A + 7*I*B*exp(4*I*
c) - 4*I*B*exp(2*I*c) + I*B)*exp(-4*I*c)/(4*a**2)), True)) + x*(-A - 7*I*B)/(4*a**2)

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Giac [A]  time = 1.54296, size = 144, normalized size = 1.4 \begin{align*} -\frac{\frac{2 \,{\left (i \, A + B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac{2 \,{\left (-i \, A + 7 \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} + \frac{3 i \, A \tan \left (d x + c\right )^{2} - 21 \, B \tan \left (d x + c\right )^{2} - 6 \, A \tan \left (d x + c\right ) + 22 i \, B \tan \left (d x + c\right ) + 5 i \, A + 5 \, B}{a^{2}{\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(2*(I*A + B)*log(tan(d*x + c) + I)/a^2 + 2*(-I*A + 7*B)*log(tan(d*x + c) - I)/a^2 + (3*I*A*tan(d*x + c)^
2 - 21*B*tan(d*x + c)^2 - 6*A*tan(d*x + c) + 22*I*B*tan(d*x + c) + 5*I*A + 5*B)/(a^2*(tan(d*x + c) - I)^2))/d