3.44 \(\int \frac{\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=142 \[ \frac{(A+2 i B) \tan ^2(c+d x)}{2 a^2 d (1+i \tan (c+d x))}+\frac{3 (-3 B+i A) \tan (c+d x)}{4 a^2 d}+\frac{(A+2 i B) \log (\cos (c+d x))}{a^2 d}-\frac{3 x (-3 B+i A)}{4 a^2}+\frac{(-B+i A) \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

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

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Rubi [A]  time = 0.280072, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.088, Rules used = {3595, 3525, 3475} \[ \frac{(A+2 i B) \tan ^2(c+d x)}{2 a^2 d (1+i \tan (c+d x))}+\frac{3 (-3 B+i A) \tan (c+d x)}{4 a^2 d}+\frac{(A+2 i B) \log (\cos (c+d x))}{a^2 d}-\frac{3 x (-3 B+i A)}{4 a^2}+\frac{(-B+i A) \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(-3*(I*A - 3*B)*x)/(4*a^2) + ((A + (2*I)*B)*Log[Cos[c + d*x]])/(a^2*d) + (3*(I*A - 3*B)*Tan[c + d*x])/(4*a^2*d
) + ((A + (2*I)*B)*Tan[c + d*x]^2)/(2*a^2*d*(1 + I*Tan[c + d*x])) + ((I*A - B)*Tan[c + d*x]^3)/(4*d*(a + I*a*T
an[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 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 \frac{\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx &=\frac{(i A-B) \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{\int \frac{\tan ^2(c+d x) (3 a (i A-B)+a (A+5 i B) \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac{(A+2 i B) \tan ^2(c+d x)}{2 a^2 d (1+i \tan (c+d x))}+\frac{(i A-B) \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \tan (c+d x) \left (-8 a^2 (A+2 i B)+6 a^2 (i A-3 B) \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=-\frac{3 (i A-3 B) x}{4 a^2}+\frac{3 (i A-3 B) \tan (c+d x)}{4 a^2 d}+\frac{(A+2 i B) \tan ^2(c+d x)}{2 a^2 d (1+i \tan (c+d x))}+\frac{(i A-B) \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{(A+2 i B) \int \tan (c+d x) \, dx}{a^2}\\ &=-\frac{3 (i A-3 B) x}{4 a^2}+\frac{(A+2 i B) \log (\cos (c+d x))}{a^2 d}+\frac{3 (i A-3 B) \tan (c+d x)}{4 a^2 d}+\frac{(A+2 i B) \tan ^2(c+d x)}{2 a^2 d (1+i \tan (c+d x))}+\frac{(i A-B) \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}\\ \end{align*}

Mathematica [B]  time = 6.91042, size = 956, normalized size = 6.73 \[ \frac{i \sec (c) \sec ^2(c+d x) (-B \cos (2 c-d x)+B \cos (2 c+d x)-i B \sin (2 c-d x)+i B \sin (2 c+d x)) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{2 d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}+\frac{x \sec (c+d x) (-\tan (c) A+i A-2 B-2 i B \tan (c)+(A+2 i B) (-\cos (2 c)-i \sin (2 c)) \tan (c)) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{(A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}-\frac{(2 A+3 i B) \cos (2 d x) \sec (c+d x) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{4 d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}+\frac{\sec (c+d x) (A \cos (c)+2 i B \cos (c)+i A \sin (c)-2 B \sin (c)) \left (\tan ^{-1}(\tan (d x)) \sin (c)-i \tan ^{-1}(\tan (d x)) \cos (c)\right ) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}+\frac{\sec (c+d x) (A \cos (c)+2 i B \cos (c)+i A \sin (c)-2 B \sin (c)) \left (\frac{1}{2} \cos (c) \log \left (\cos ^2(c+d x)\right )+\frac{1}{2} i \sin (c) \log \left (\cos ^2(c+d x)\right )\right ) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}+\frac{(A+i B) \cos (4 d x) \sec (c+d x) \left (\frac{1}{16} \cos (2 c)-\frac{1}{16} i \sin (2 c)\right ) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}+\frac{(3 B-i A) \sec (c+d x) \left (\frac{3}{4} d x \cos (2 c)+\frac{3}{4} i d x \sin (2 c)\right ) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}+\frac{i (2 A+3 i B) \sec (c+d x) \sin (2 d x) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{4 d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2}+\frac{(B-i A) \sec (c+d x) \left (\frac{1}{16} \cos (2 c)-\frac{1}{16} i \sin (2 c)\right ) \sin (4 d x) (A+B \tan (c+d x)) (\cos (d x)+i \sin (d x))^2}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Tan[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^2,x]

[Out]

-((2*A + (3*I)*B)*Cos[2*d*x]*Sec[c + d*x]*(Cos[d*x] + I*Sin[d*x])^2*(A + B*Tan[c + d*x]))/(4*d*(A*Cos[c + d*x]
+ B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^2) + (Sec[c + d*x]*(A*Cos[c] + (2*I)*B*Cos[c] + I*A*Sin[c] - 2*B*Sin
[c])*((-I)*ArcTan[Tan[d*x]]*Cos[c] + ArcTan[Tan[d*x]]*Sin[c])*(Cos[d*x] + I*Sin[d*x])^2*(A + B*Tan[c + d*x]))/
(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^2) + (Sec[c + d*x]*(A*Cos[c] + (2*I)*B*Cos[c] + I*
A*Sin[c] - 2*B*Sin[c])*((Cos[c]*Log[Cos[c + d*x]^2])/2 + (I/2)*Log[Cos[c + d*x]^2]*Sin[c])*(Cos[d*x] + I*Sin[d
*x])^2*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^2) + ((A + I*B)*Cos[4
*d*x]*Sec[c + d*x]*(Cos[2*c]/16 - (I/16)*Sin[2*c])*(Cos[d*x] + I*Sin[d*x])^2*(A + B*Tan[c + d*x]))/(d*(A*Cos[c
+ d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^2) + (((-I)*A + 3*B)*Sec[c + d*x]*((3*d*x*Cos[2*c])/4 + ((3*I
)/4)*d*x*Sin[2*c])*(Cos[d*x] + I*Sin[d*x])^2*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I
*a*Tan[c + d*x])^2) + ((I/4)*(2*A + (3*I)*B)*Sec[c + d*x]*(Cos[d*x] + I*Sin[d*x])^2*Sin[2*d*x]*(A + B*Tan[c +
d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^2) + (((-I)*A + B)*Sec[c + d*x]*(Cos[2*c]/1
6 - (I/16)*Sin[2*c])*(Cos[d*x] + I*Sin[d*x])^2*Sin[4*d*x]*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c +
d*x])*(a + I*a*Tan[c + d*x])^2) + ((I/2)*Sec[c]*Sec[c + d*x]^2*(Cos[d*x] + I*Sin[d*x])^2*(-(B*Cos[2*c - d*x])
+ B*Cos[2*c + d*x] - I*B*Sin[2*c - d*x] + I*B*Sin[2*c + d*x])*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Si
n[c + d*x])*(a + I*a*Tan[c + d*x])^2) + (x*Sec[c + d*x]*(Cos[d*x] + I*Sin[d*x])^2*(I*A - 2*B - A*Tan[c] - (2*I
)*B*Tan[c] + (A + (2*I)*B)*(-Cos[2*c] - I*Sin[2*c])*Tan[c])*(A + B*Tan[c + d*x]))/((A*Cos[c + d*x] + B*Sin[c +
d*x])*(a + I*a*Tan[c + d*x])^2)

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Maple [A]  time = 0.033, size = 177, normalized size = 1.3 \begin{align*} -{\frac{B\tan \left ( dx+c \right ) }{{a}^{2}d}}+{\frac{{\frac{5\,i}{4}}A}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{7\,B}{4\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{A}{4\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{{\frac{i}{4}}B}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{7\,\ln \left ( \tan \left ( dx+c \right ) -i \right ) A}{8\,{a}^{2}d}}-{\frac{{\frac{17\,i}{8}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) B}{{a}^{2}d}}-{\frac{A\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{8\,{a}^{2}d}}+{\frac{{\frac{i}{8}}B\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)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^2,x)

[Out]

-1/d/a^2*B*tan(d*x+c)+5/4*I/d/a^2/(tan(d*x+c)-I)*A-7/4/d/a^2/(tan(d*x+c)-I)*B-1/4/d/a^2/(tan(d*x+c)-I)^2*A-1/4
*I/d/a^2/(tan(d*x+c)-I)^2*B-7/8/d/a^2*ln(tan(d*x+c)-I)*A-17/8*I/d/a^2*ln(tan(d*x+c)-I)*B-1/8/d/a^2*A*ln(tan(d*
x+c)+I)+1/8*I/d/a^2*B*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(tan(d*x+c)^3*(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 = 1.46559, size = 424, normalized size = 2.99 \begin{align*} \frac{{\left (-28 i \, A + 68 \, B\right )} d x e^{\left (6 i \, d x + 6 i \, c\right )} +{\left ({\left (-28 i \, A + 68 \, B\right )} d x - 8 \, A - 44 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} -{\left (7 \, A + 11 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 16 \,{\left ({\left (A + 2 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (A + 2 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + A + i \, B}{16 \,{\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/16*((-28*I*A + 68*B)*d*x*e^(6*I*d*x + 6*I*c) + ((-28*I*A + 68*B)*d*x - 8*A - 44*I*B)*e^(4*I*d*x + 4*I*c) - (
7*A + 11*I*B)*e^(2*I*d*x + 2*I*c) + 16*((A + 2*I*B)*e^(6*I*d*x + 6*I*c) + (A + 2*I*B)*e^(4*I*d*x + 4*I*c))*log
(e^(2*I*d*x + 2*I*c) + 1) + A + I*B)/(a^2*d*e^(6*I*d*x + 6*I*c) + a^2*d*e^(4*I*d*x + 4*I*c))

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Sympy [A]  time = 14.4321, size = 223, normalized size = 1.57 \begin{align*} - \frac{2 i B e^{- 2 i c}}{a^{2} d \left (e^{2 i d x} + e^{- 2 i c}\right )} - \frac{\left (\begin{cases} 7 i A x e^{4 i c} + \frac{2 A e^{2 i c} e^{- 2 i d x}}{d} - \frac{A e^{- 4 i d x}}{4 d} - 17 B x e^{4 i c} + \frac{3 i B e^{2 i c} e^{- 2 i d x}}{d} - \frac{i B e^{- 4 i d x}}{4 d} & \text{for}\: d \neq 0 \\x \left (7 i A e^{4 i c} - 4 i A e^{2 i c} + i A - 17 B e^{4 i c} + 6 B e^{2 i c} - B\right ) & \text{otherwise} \end{cases}\right ) e^{- 4 i c}}{4 a^{2}} + \frac{\left (A + 2 i B\right ) \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**2,x)

[Out]

-2*I*B*exp(-2*I*c)/(a**2*d*(exp(2*I*d*x) + exp(-2*I*c))) - Piecewise((7*I*A*x*exp(4*I*c) + 2*A*exp(2*I*c)*exp(
-2*I*d*x)/d - A*exp(-4*I*d*x)/(4*d) - 17*B*x*exp(4*I*c) + 3*I*B*exp(2*I*c)*exp(-2*I*d*x)/d - I*B*exp(-4*I*d*x)
/(4*d), Ne(d, 0)), (x*(7*I*A*exp(4*I*c) - 4*I*A*exp(2*I*c) + I*A - 17*B*exp(4*I*c) + 6*B*exp(2*I*c) - B), True
))*exp(-4*I*c)/(4*a**2) + (A + 2*I*B)*log(exp(2*I*d*x) + exp(-2*I*c))/(a**2*d)

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Giac [A]  time = 1.85702, size = 162, normalized size = 1.14 \begin{align*} -\frac{\frac{2 \,{\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac{2 \,{\left (7 \, A + 17 i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} + \frac{16 \, B \tan \left (d x + c\right )}{a^{2}} - \frac{21 \, A \tan \left (d x + c\right )^{2} + 51 i \, B \tan \left (d x + c\right )^{2} - 22 i \, A \tan \left (d x + c\right ) + 74 \, B \tan \left (d x + c\right ) - 5 \, A - 27 i \, 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)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/16*(2*(A - I*B)*log(tan(d*x + c) + I)/a^2 + 2*(7*A + 17*I*B)*log(tan(d*x + c) - I)/a^2 + 16*B*tan(d*x + c)/
a^2 - (21*A*tan(d*x + c)^2 + 51*I*B*tan(d*x + c)^2 - 22*I*A*tan(d*x + c) + 74*B*tan(d*x + c) - 5*A - 27*I*B)/(
a^2*(tan(d*x + c) - I)^2))/d