3.55 \(\int \frac{A+B \tan (c+d x)}{(a+i a \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=112 \[ \frac{B+i A}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{x (A-i B)}{8 a^3}+\frac{-B+i A}{6 d (a+i a \tan (c+d x))^3}+\frac{B+i A}{8 a d (a+i a \tan (c+d x))^2} \]

[Out]

((A - I*B)*x)/(8*a^3) + (I*A - B)/(6*d*(a + I*a*Tan[c + d*x])^3) + (I*A + B)/(8*a*d*(a + I*a*Tan[c + d*x])^2)
+ (I*A + B)/(8*d*(a^3 + I*a^3*Tan[c + d*x]))

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Rubi [A]  time = 0.0836763, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3526, 3479, 8} \[ \frac{B+i A}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{x (A-i B)}{8 a^3}+\frac{-B+i A}{6 d (a+i a \tan (c+d x))^3}+\frac{B+i A}{8 a d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[c + d*x])/(a + I*a*Tan[c + d*x])^3,x]

[Out]

((A - I*B)*x)/(8*a^3) + (I*A - B)/(6*d*(a + I*a*Tan[c + d*x])^3) + (I*A + B)/(8*a*d*(a + I*a*Tan[c + d*x])^2)
+ (I*A + B)/(8*d*(a^3 + I*a^3*Tan[c + d*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 3479

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a + b*Tan[c + d*x])^n)/(2*b*d*n), x] +
Dist[1/(2*a), Int[(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0]

Rule 8

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

Rubi steps

\begin{align*} \int \frac{A+B \tan (c+d x)}{(a+i a \tan (c+d x))^3} \, dx &=\frac{i A-B}{6 d (a+i a \tan (c+d x))^3}+\frac{(A-i B) \int \frac{1}{(a+i a \tan (c+d x))^2} \, dx}{2 a}\\ &=\frac{i A-B}{6 d (a+i a \tan (c+d x))^3}+\frac{i A+B}{8 a d (a+i a \tan (c+d x))^2}+\frac{(A-i B) \int \frac{1}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac{i A-B}{6 d (a+i a \tan (c+d x))^3}+\frac{i A+B}{8 a d (a+i a \tan (c+d x))^2}+\frac{i A+B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{(A-i B) \int 1 \, dx}{8 a^3}\\ &=\frac{(A-i B) x}{8 a^3}+\frac{i A-B}{6 d (a+i a \tan (c+d x))^3}+\frac{i A+B}{8 a d (a+i a \tan (c+d x))^2}+\frac{i A+B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.77543, size = 150, normalized size = 1.34 \[ \frac{\sec ^3(c+d x) ((-27 A+3 i B) \cos (c+d x)+2 (6 i A d x-A+6 B d x-i B) \cos (3 (c+d x))-9 i A \sin (c+d x)+2 i A \sin (3 (c+d x))-12 A d x \sin (3 (c+d x))-9 B \sin (c+d x)-2 B \sin (3 (c+d x))+12 i B d x \sin (3 (c+d x)))}{96 a^3 d (\tan (c+d x)-i)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[c + d*x])/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(Sec[c + d*x]^3*((-27*A + (3*I)*B)*Cos[c + d*x] + 2*(-A - I*B + (6*I)*A*d*x + 6*B*d*x)*Cos[3*(c + d*x)] - (9*I
)*A*Sin[c + d*x] - 9*B*Sin[c + d*x] + (2*I)*A*Sin[3*(c + d*x)] - 2*B*Sin[3*(c + d*x)] - 12*A*d*x*Sin[3*(c + d*
x)] + (12*I)*B*d*x*Sin[3*(c + d*x)]))/(96*a^3*d*(-I + Tan[c + d*x])^3)

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Maple [B]  time = 0.03, size = 203, normalized size = 1.8 \begin{align*} -{\frac{A}{6\,{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}-{\frac{{\frac{i}{6}}B}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}-{\frac{{\frac{i}{16}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) A}{{a}^{3}d}}-{\frac{\ln \left ( \tan \left ( dx+c \right ) -i \right ) B}{16\,{a}^{3}d}}-{\frac{{\frac{i}{8}}A}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{B}{8\,{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}+{\frac{A}{8\,{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{{\frac{i}{8}}B}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{B\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{16\,{a}^{3}d}}+{\frac{{\frac{i}{16}}A\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{{a}^{3}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x)

[Out]

-1/6/d/a^3/(tan(d*x+c)-I)^3*A-1/6*I/d/a^3/(tan(d*x+c)-I)^3*B-1/16*I/d/a^3*ln(tan(d*x+c)-I)*A-1/16/d/a^3*ln(tan
(d*x+c)-I)*B-1/8*I/d/a^3/(tan(d*x+c)-I)^2*A-1/8/d/a^3/(tan(d*x+c)-I)^2*B+1/8/d/a^3/(tan(d*x+c)-I)*A-1/8*I/d/a^
3/(tan(d*x+c)-I)*B+1/16/d/a^3*B*ln(tan(d*x+c)+I)+1/16*I/d/a^3*A*ln(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((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.39536, size = 217, normalized size = 1.94 \begin{align*} \frac{{\left (12 \,{\left (A - i \, B\right )} d x e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (18 i \, A + 6 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (9 i \, A - 3 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, A - 2 \, B\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/96*(12*(A - I*B)*d*x*e^(6*I*d*x + 6*I*c) + (18*I*A + 6*B)*e^(4*I*d*x + 4*I*c) + (9*I*A - 3*B)*e^(2*I*d*x + 2
*I*c) + 2*I*A - 2*B)*e^(-6*I*d*x - 6*I*c)/(a^3*d)

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Sympy [A]  time = 4.2097, size = 260, normalized size = 2.32 \begin{align*} \begin{cases} \frac{\left (\left (512 i A a^{6} d^{2} e^{6 i c} - 512 B a^{6} d^{2} e^{6 i c}\right ) e^{- 6 i d x} + \left (2304 i A a^{6} d^{2} e^{8 i c} - 768 B a^{6} d^{2} e^{8 i c}\right ) e^{- 4 i d x} + \left (4608 i A a^{6} d^{2} e^{10 i c} + 1536 B a^{6} d^{2} e^{10 i c}\right ) e^{- 2 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text{for}\: 24576 a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (- \frac{A - i B}{8 a^{3}} + \frac{\left (A e^{6 i c} + 3 A e^{4 i c} + 3 A e^{2 i c} + A - i B e^{6 i c} - i B e^{4 i c} + i B e^{2 i c} + i B\right ) e^{- 6 i c}}{8 a^{3}}\right ) & \text{otherwise} \end{cases} + \frac{x \left (A - i B\right )}{8 a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**3,x)

[Out]

Piecewise((((512*I*A*a**6*d**2*exp(6*I*c) - 512*B*a**6*d**2*exp(6*I*c))*exp(-6*I*d*x) + (2304*I*A*a**6*d**2*ex
p(8*I*c) - 768*B*a**6*d**2*exp(8*I*c))*exp(-4*I*d*x) + (4608*I*A*a**6*d**2*exp(10*I*c) + 1536*B*a**6*d**2*exp(
10*I*c))*exp(-2*I*d*x))*exp(-12*I*c)/(24576*a**9*d**3), Ne(24576*a**9*d**3*exp(12*I*c), 0)), (x*(-(A - I*B)/(8
*a**3) + (A*exp(6*I*c) + 3*A*exp(4*I*c) + 3*A*exp(2*I*c) + A - I*B*exp(6*I*c) - I*B*exp(4*I*c) + I*B*exp(2*I*c
) + I*B)*exp(-6*I*c)/(8*a**3)), True)) + x*(A - I*B)/(8*a**3)

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Giac [A]  time = 1.4069, size = 177, normalized size = 1.58 \begin{align*} -\frac{\frac{6 \,{\left (i \, A + B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} + \frac{6 \,{\left (-i \, A - B\right )} \log \left (i \, \tan \left (d x + c\right ) - 1\right )}{a^{3}} + \frac{-11 i \, A \tan \left (d x + c\right )^{3} - 11 \, B \tan \left (d x + c\right )^{3} - 45 \, A \tan \left (d x + c\right )^{2} + 45 i \, B \tan \left (d x + c\right )^{2} + 69 i \, A \tan \left (d x + c\right ) + 69 \, B \tan \left (d x + c\right ) + 51 \, A - 19 i \, B}{a^{3}{\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/96*(6*(I*A + B)*log(tan(d*x + c) - I)/a^3 + 6*(-I*A - B)*log(I*tan(d*x + c) - 1)/a^3 + (-11*I*A*tan(d*x + c
)^3 - 11*B*tan(d*x + c)^3 - 45*A*tan(d*x + c)^2 + 45*I*B*tan(d*x + c)^2 + 69*I*A*tan(d*x + c) + 69*B*tan(d*x +
c) + 51*A - 19*I*B)/(a^3*(tan(d*x + c) - I)^3))/d