3.1076 \(\int \frac{(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=129 \[ \frac{i (c-i d)^2}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{x (c-i d)^2}{8 a^3}+\frac{(c+i d) (3 d+i c)}{8 a f (a+i a \tan (e+f x))^2}+\frac{i (c+i d)^2}{6 f (a+i a \tan (e+f x))^3} \]

[Out]

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

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Rubi [A]  time = 0.167825, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3540, 3526, 3479, 8} \[ \frac{i (c-i d)^2}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{x (c-i d)^2}{8 a^3}+\frac{(c+i d) (3 d+i c)}{8 a f (a+i a \tan (e+f x))^2}+\frac{i (c+i d)^2}{6 f (a+i a \tan (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3540

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> -Simp[
(b*(a*c + b*d)^2*(a + b*Tan[e + f*x])^m)/(2*a^3*f*m), x] + Dist[1/(2*a^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Si
mp[a*c^2 - 2*b*c*d + a*d^2 - 2*b*d^2*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0] && LeQ[m, -1] && EqQ[a^2 + b^2, 0]

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

Mathematica [A]  time = 1.24704, size = 256, normalized size = 1.98 \[ \frac{\sec ^3(e+f x) (\cos (f x)+i \sin (f x))^3 \left (12 f x (c-i d)^2 (\cos (3 e)+i \sin (3 e))+6 (3 c+i d) (c-i d) (\cos (e)+i \sin (e)) \sin (2 f x)+3 (c+i d) (d+3 i c) (\cos (e)-i \sin (e)) \cos (4 f x)+6 (3 c+i d) (d+i c) (\cos (e)+i \sin (e)) \cos (2 f x)+2 (c+i d)^2 (\sin (3 e)+i \cos (3 e)) \cos (6 f x)+3 (3 c-i d) (c+i d) (\cos (e)-i \sin (e)) \sin (4 f x)+2 (c+i d)^2 (\cos (3 e)-i \sin (3 e)) \sin (6 f x)\right )}{96 f (a+i a \tan (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]^3*(Cos[f*x] + I*Sin[f*x])^3*(3*(c + I*d)*((3*I)*c + d)*Cos[4*f*x]*(Cos[e] - I*Sin[e]) + 6*(3*c +
I*d)*(I*c + d)*Cos[2*f*x]*(Cos[e] + I*Sin[e]) + 12*(c - I*d)^2*f*x*(Cos[3*e] + I*Sin[3*e]) + 2*(c + I*d)^2*Co
s[6*f*x]*(I*Cos[3*e] + Sin[3*e]) + 6*(c - I*d)*(3*c + I*d)*(Cos[e] + I*Sin[e])*Sin[2*f*x] + 3*(3*c - I*d)*(c +
I*d)*(Cos[e] - I*Sin[e])*Sin[4*f*x] + 2*(c + I*d)^2*(Cos[3*e] - I*Sin[3*e])*Sin[6*f*x]))/(96*f*(a + I*a*Tan[e
+ f*x])^3)

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Maple [B]  time = 0.033, size = 329, normalized size = 2.6 \begin{align*}{\frac{-{\frac{i}{16}}\ln \left ( \tan \left ( fx+e \right ) -i \right ){c}^{2}}{f{a}^{3}}}+{\frac{{\frac{i}{16}}\ln \left ( \tan \left ( fx+e \right ) -i \right ){d}^{2}}{f{a}^{3}}}-{\frac{\ln \left ( \tan \left ( fx+e \right ) -i \right ) cd}{8\,f{a}^{3}}}-{\frac{{\frac{i}{4}}cd}{f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{{c}^{2}}{8\,f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{{d}^{2}}{8\,f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{cd}{4\,f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}-{\frac{{\frac{i}{8}}{c}^{2}}{f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}-{\frac{{\frac{3\,i}{8}}{d}^{2}}{f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}+{\frac{{d}^{2}}{6\,f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{3}}}-{\frac{{\frac{i}{3}}cd}{f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{3}}}-{\frac{{c}^{2}}{6\,f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{3}}}+{\frac{\ln \left ( \tan \left ( fx+e \right ) +i \right ) cd}{8\,f{a}^{3}}}+{\frac{{\frac{i}{16}}\ln \left ( \tan \left ( fx+e \right ) +i \right ){c}^{2}}{f{a}^{3}}}-{\frac{{\frac{i}{16}}\ln \left ( \tan \left ( fx+e \right ) +i \right ){d}^{2}}{f{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*tan(f*x+e))^2/(a+I*a*tan(f*x+e))^3,x)

[Out]

-1/16*I/f/a^3*ln(tan(f*x+e)-I)*c^2+1/16*I/f/a^3*ln(tan(f*x+e)-I)*d^2-1/8/f/a^3*ln(tan(f*x+e)-I)*c*d-1/4*I/f/a^
3/(tan(f*x+e)-I)*c*d+1/8/f/a^3/(tan(f*x+e)-I)*c^2-1/8/f/a^3/(tan(f*x+e)-I)*d^2-1/4/f/a^3/(tan(f*x+e)-I)^2*c*d-
1/8*I/f/a^3/(tan(f*x+e)-I)^2*c^2-3/8*I/f/a^3/(tan(f*x+e)-I)^2*d^2+1/6/f/a^3/(tan(f*x+e)-I)^3*d^2-1/3*I/f/a^3/(
tan(f*x+e)-I)^3*c*d-1/6/f/a^3/(tan(f*x+e)-I)^3*c^2+1/8/f/a^3*ln(tan(f*x+e)+I)*c*d+1/16*I/f/a^3*ln(tan(f*x+e)+I
)*c^2-1/16*I/f/a^3*ln(tan(f*x+e)+I)*d^2

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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((c+d*tan(f*x+e))^2/(a+I*a*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.63784, size = 292, normalized size = 2.26 \begin{align*} \frac{{\left (12 \,{\left (c^{2} - 2 i \, c d - d^{2}\right )} f x e^{\left (6 i \, f x + 6 i \, e\right )} + 2 i \, c^{2} - 4 \, c d - 2 i \, d^{2} +{\left (18 i \, c^{2} + 12 \, c d + 6 i \, d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (9 i \, c^{2} - 6 \, c d + 3 i \, d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{96 \, a^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.25021, size = 403, normalized size = 3.12 \begin{align*} \begin{cases} \frac{\left (\left (512 i a^{6} c^{2} f^{2} e^{6 i e} - 1024 a^{6} c d f^{2} e^{6 i e} - 512 i a^{6} d^{2} f^{2} e^{6 i e}\right ) e^{- 6 i f x} + \left (2304 i a^{6} c^{2} f^{2} e^{8 i e} - 1536 a^{6} c d f^{2} e^{8 i e} + 768 i a^{6} d^{2} f^{2} e^{8 i e}\right ) e^{- 4 i f x} + \left (4608 i a^{6} c^{2} f^{2} e^{10 i e} + 3072 a^{6} c d f^{2} e^{10 i e} + 1536 i a^{6} d^{2} f^{2} e^{10 i e}\right ) e^{- 2 i f x}\right ) e^{- 12 i e}}{24576 a^{9} f^{3}} & \text{for}\: 24576 a^{9} f^{3} e^{12 i e} \neq 0 \\x \left (- \frac{c^{2} - 2 i c d - d^{2}}{8 a^{3}} + \frac{\left (c^{2} e^{6 i e} + 3 c^{2} e^{4 i e} + 3 c^{2} e^{2 i e} + c^{2} - 2 i c d e^{6 i e} - 2 i c d e^{4 i e} + 2 i c d e^{2 i e} + 2 i c d - d^{2} e^{6 i e} + d^{2} e^{4 i e} + d^{2} e^{2 i e} - d^{2}\right ) e^{- 6 i e}}{8 a^{3}}\right ) & \text{otherwise} \end{cases} + \frac{x \left (c^{2} - 2 i c d - d^{2}\right )}{8 a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))**2/(a+I*a*tan(f*x+e))**3,x)

[Out]

Piecewise((((512*I*a**6*c**2*f**2*exp(6*I*e) - 1024*a**6*c*d*f**2*exp(6*I*e) - 512*I*a**6*d**2*f**2*exp(6*I*e)
)*exp(-6*I*f*x) + (2304*I*a**6*c**2*f**2*exp(8*I*e) - 1536*a**6*c*d*f**2*exp(8*I*e) + 768*I*a**6*d**2*f**2*exp
(8*I*e))*exp(-4*I*f*x) + (4608*I*a**6*c**2*f**2*exp(10*I*e) + 3072*a**6*c*d*f**2*exp(10*I*e) + 1536*I*a**6*d**
2*f**2*exp(10*I*e))*exp(-2*I*f*x))*exp(-12*I*e)/(24576*a**9*f**3), Ne(24576*a**9*f**3*exp(12*I*e), 0)), (x*(-(
c**2 - 2*I*c*d - d**2)/(8*a**3) + (c**2*exp(6*I*e) + 3*c**2*exp(4*I*e) + 3*c**2*exp(2*I*e) + c**2 - 2*I*c*d*ex
p(6*I*e) - 2*I*c*d*exp(4*I*e) + 2*I*c*d*exp(2*I*e) + 2*I*c*d - d**2*exp(6*I*e) + d**2*exp(4*I*e) + d**2*exp(2*
I*e) - d**2)*exp(-6*I*e)/(8*a**3)), True)) + x*(c**2 - 2*I*c*d - d**2)/(8*a**3)

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Giac [B]  time = 1.58691, size = 290, normalized size = 2.25 \begin{align*} -\frac{\frac{6 \,{\left (i \, c^{2} + 2 \, c d - i \, d^{2}\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{3}} + \frac{6 \,{\left (-i \, c^{2} - 2 \, c d + i \, d^{2}\right )} \log \left (i \, \tan \left (f x + e\right ) - 1\right )}{a^{3}} + \frac{-11 i \, c^{2} \tan \left (f x + e\right )^{3} - 22 \, c d \tan \left (f x + e\right )^{3} + 11 i \, d^{2} \tan \left (f x + e\right )^{3} - 45 \, c^{2} \tan \left (f x + e\right )^{2} + 90 i \, c d \tan \left (f x + e\right )^{2} + 45 \, d^{2} \tan \left (f x + e\right )^{2} + 69 i \, c^{2} \tan \left (f x + e\right ) + 138 \, c d \tan \left (f x + e\right ) - 21 i \, d^{2} \tan \left (f x + e\right ) + 51 \, c^{2} - 38 i \, c d - 3 \, d^{2}}{a^{3}{\left (\tan \left (f x + e\right ) - i\right )}^{3}}}{96 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^2/(a+I*a*tan(f*x+e))^3,x, algorithm="giac")

[Out]

-1/96*(6*(I*c^2 + 2*c*d - I*d^2)*log(tan(f*x + e) - I)/a^3 + 6*(-I*c^2 - 2*c*d + I*d^2)*log(I*tan(f*x + e) - 1
)/a^3 + (-11*I*c^2*tan(f*x + e)^3 - 22*c*d*tan(f*x + e)^3 + 11*I*d^2*tan(f*x + e)^3 - 45*c^2*tan(f*x + e)^2 +
90*I*c*d*tan(f*x + e)^2 + 45*d^2*tan(f*x + e)^2 + 69*I*c^2*tan(f*x + e) + 138*c*d*tan(f*x + e) - 21*I*d^2*tan(
f*x + e) + 51*c^2 - 38*I*c*d - 3*d^2)/(a^3*(tan(f*x + e) - I)^3))/f