3.1082 \(\int \frac {(c+d \tan (e+f x))^3}{(a+i a \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=140 \[ \frac {(c+i d) (c-3 i d) (d+i c)}{8 a^3 f (1+i \tan (e+f x))}+\frac {x (c-i d)^3}{8 a^3}+\frac {i (c+d \tan (e+f x))^3}{6 f (a+i a \tan (e+f x))^3}+\frac {(c+i d)^2 (d+i c)}{8 a f (a+i a \tan (e+f x))^2} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 140, normalized size of antiderivative = 1.00, 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 = {3546, 3540, 3526, 8} \[ \frac {(c+i d) (c-3 i d) (d+i c)}{8 a^3 f (1+i \tan (e+f x))}+\frac {x (c-i d)^3}{8 a^3}+\frac {i (c+d \tan (e+f x))^3}{6 f (a+i a \tan (e+f x))^3}+\frac {(c+i d)^2 (d+i c)}{8 a f (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c - I*d)^3*x)/(8*a^3) + ((c + I*d)*(c - (3*I)*d)*(I*c + d))/(8*a^3*f*(1 + I*Tan[e + f*x])) + ((c + I*d)^2*(I
*c + d))/(8*a*f*(a + I*a*Tan[e + f*x])^2) + ((I/6)*(c + d*Tan[e + f*x])^3)/(f*(a + I*a*Tan[e + f*x])^3)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 3546

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

Rubi steps

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

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Mathematica [A]  time = 2.02, size = 260, normalized size = 1.86 \[ \frac {\sec ^3(e+f x) (\cos (f x)+i \sin (f x))^3 \left (12 f x (c-i d)^3 (\cos (3 e)+i \sin (3 e))+18 i (c+i d) (c-i d)^2 (\cos (e)+i \sin (e)) \cos (2 f x)+18 (c+i d) (c-i d)^2 (\cos (e)+i \sin (e)) \sin (2 f x)+9 (c+i d)^2 (c-i d) (\cos (e)-i \sin (e)) \sin (4 f x)+9 (c+i d)^2 (d+i c) (\cos (e)-i \sin (e)) \cos (4 f x)+2 (c+i d)^3 (\sin (3 e)+i \cos (3 e)) \cos (6 f x)+2 (c+i d)^3 (\cos (3 e)-i \sin (3 e)) \sin (6 f x)\right )}{96 f (a+i a \tan (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]^3*(Cos[f*x] + I*Sin[f*x])^3*(9*(c + I*d)^2*(I*c + d)*Cos[4*f*x]*(Cos[e] - I*Sin[e]) + (18*I)*(c
- I*d)^2*(c + I*d)*Cos[2*f*x]*(Cos[e] + I*Sin[e]) + 12*(c - I*d)^3*f*x*(Cos[3*e] + I*Sin[3*e]) + 2*(c + I*d)^3
*Cos[6*f*x]*(I*Cos[3*e] + Sin[3*e]) + 18*(c - I*d)^2*(c + I*d)*(Cos[e] + I*Sin[e])*Sin[2*f*x] + 9*(c - I*d)*(c
+ I*d)^2*(Cos[e] - I*Sin[e])*Sin[4*f*x] + 2*(c + I*d)^3*(Cos[3*e] - I*Sin[3*e])*Sin[6*f*x]))/(96*f*(a + I*a*T
an[e + f*x])^3)

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fricas [A]  time = 0.42, size = 141, normalized size = 1.01 \[ \frac {{\left ({\left (12 \, c^{3} - 36 i \, c^{2} d - 36 \, c d^{2} + 12 i \, d^{3}\right )} f x e^{\left (6 i \, f x + 6 i \, e\right )} + 2 i \, c^{3} - 6 \, c^{2} d - 6 i \, c d^{2} + 2 \, d^{3} + {\left (18 i \, c^{3} + 18 \, c^{2} d + 18 i \, c d^{2} + 18 \, d^{3}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (9 i \, c^{3} - 9 \, c^{2} d + 9 i \, c d^{2} - 9 \, d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{96 \, a^{3} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*((12*c^3 - 36*I*c^2*d - 36*c*d^2 + 12*I*d^3)*f*x*e^(6*I*f*x + 6*I*e) + 2*I*c^3 - 6*c^2*d - 6*I*c*d^2 + 2*
d^3 + (18*I*c^3 + 18*c^2*d + 18*I*c*d^2 + 18*d^3)*e^(4*I*f*x + 4*I*e) + (9*I*c^3 - 9*c^2*d + 9*I*c*d^2 - 9*d^3
)*e^(2*I*f*x + 2*I*e))*e^(-6*I*f*x - 6*I*e)/(a^3*f)

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giac [B]  time = 1.48, size = 286, normalized size = 2.04 \[ -\frac {\frac {6 \, {\left (i \, c^{3} + 3 \, c^{2} d - 3 i \, c d^{2} - d^{3}\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{3}} + \frac {6 \, {\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} \log \left (i \, \tan \left (f x + e\right ) - 1\right )}{a^{3}} + \frac {-11 i \, c^{3} \tan \left (f x + e\right )^{3} - 33 \, c^{2} d \tan \left (f x + e\right )^{3} + 33 i \, c d^{2} \tan \left (f x + e\right )^{3} + 11 \, d^{3} \tan \left (f x + e\right )^{3} - 45 \, c^{3} \tan \left (f x + e\right )^{2} + 135 i \, c^{2} d \tan \left (f x + e\right )^{2} + 135 \, c d^{2} \tan \left (f x + e\right )^{2} + 51 i \, d^{3} \tan \left (f x + e\right )^{2} + 69 i \, c^{3} \tan \left (f x + e\right ) + 207 \, c^{2} d \tan \left (f x + e\right ) - 63 i \, c d^{2} \tan \left (f x + e\right ) + 75 \, d^{3} \tan \left (f x + e\right ) + 51 \, c^{3} - 57 i \, c^{2} d - 9 \, c d^{2} - 29 i \, d^{3}}{a^{3} {\left (\tan \left (f x + e\right ) - i\right )}^{3}}}{96 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(6*(I*c^3 + 3*c^2*d - 3*I*c*d^2 - d^3)*log(tan(f*x + e) - I)/a^3 + 6*(-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3
)*log(I*tan(f*x + e) - 1)/a^3 + (-11*I*c^3*tan(f*x + e)^3 - 33*c^2*d*tan(f*x + e)^3 + 33*I*c*d^2*tan(f*x + e)^
3 + 11*d^3*tan(f*x + e)^3 - 45*c^3*tan(f*x + e)^2 + 135*I*c^2*d*tan(f*x + e)^2 + 135*c*d^2*tan(f*x + e)^2 + 51
*I*d^3*tan(f*x + e)^2 + 69*I*c^3*tan(f*x + e) + 207*c^2*d*tan(f*x + e) - 63*I*c*d^2*tan(f*x + e) + 75*d^3*tan(
f*x + e) + 51*c^3 - 57*I*c^2*d - 9*c*d^2 - 29*I*d^3)/(a^3*(tan(f*x + e) - I)^3))/f

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maple [B]  time = 0.25, size = 454, normalized size = 3.24 \[ \frac {3 \ln \left (\tan \left (f x +e \right )+i\right ) c^{2} d}{16 f \,a^{3}}-\frac {\ln \left (\tan \left (f x +e \right )+i\right ) d^{3}}{16 f \,a^{3}}+\frac {i \ln \left (\tan \left (f x +e \right )+i\right ) c^{3}}{16 f \,a^{3}}-\frac {i c^{2} d}{2 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}+\frac {c^{3}}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}-\frac {3 c \,d^{2}}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}-\frac {3 i c^{2} d}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}+\frac {i d^{3}}{6 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}+\frac {3 i \ln \left (\tan \left (f x +e \right )-i\right ) c \,d^{2}}{16 f \,a^{3}}+\frac {c \,d^{2}}{2 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}-\frac {9 i c \,d^{2}}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {c^{3}}{6 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{3}}-\frac {7 i d^{3}}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )}+\frac {\ln \left (\tan \left (f x +e \right )-i\right ) d^{3}}{16 f \,a^{3}}-\frac {3 i \ln \left (\tan \left (f x +e \right )+i\right ) c \,d^{2}}{16 f \,a^{3}}-\frac {3 \ln \left (\tan \left (f x +e \right )-i\right ) c^{2} d}{16 f \,a^{3}}-\frac {3 c^{2} d}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}+\frac {5 d^{3}}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {i \ln \left (\tan \left (f x +e \right )-i\right ) c^{3}}{16 f \,a^{3}}-\frac {i c^{3}}{8 f \,a^{3} \left (\tan \left (f x +e \right )-i\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/16/f/a^3*ln(tan(f*x+e)+I)*c^2*d-1/16/f/a^3*ln(tan(f*x+e)+I)*d^3+1/16*I/f/a^3*ln(tan(f*x+e)+I)*c^3-1/2*I/f/a^
3/(tan(f*x+e)-I)^3*c^2*d+1/8/f/a^3/(tan(f*x+e)-I)*c^3-3/8/f/a^3/(tan(f*x+e)-I)*c*d^2-3/8*I/f/a^3/(tan(f*x+e)-I
)*c^2*d+1/6*I/f/a^3/(tan(f*x+e)-I)^3*d^3+3/16*I/f/a^3*ln(tan(f*x+e)-I)*c*d^2+1/2/f/a^3/(tan(f*x+e)-I)^3*c*d^2-
9/8*I/f/a^3/(tan(f*x+e)-I)^2*c*d^2-1/6/f/a^3/(tan(f*x+e)-I)^3*c^3-7/8*I/f/a^3/(tan(f*x+e)-I)*d^3+1/16/f/a^3*ln
(tan(f*x+e)-I)*d^3-3/16*I/f/a^3*ln(tan(f*x+e)+I)*c*d^2-3/16/f/a^3*ln(tan(f*x+e)-I)*c^2*d-3/8/f/a^3/(tan(f*x+e)
-I)^2*c^2*d+5/8/f/a^3/(tan(f*x+e)-I)^2*d^3-1/16*I/f/a^3*ln(tan(f*x+e)-I)*c^3-1/8*I/f/a^3/(tan(f*x+e)-I)^2*c^3

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 5.61, size = 183, normalized size = 1.31 \[ \frac {\frac {5\,d^3}{12\,a^3}-\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {3\,c^3}{8\,a^3}-\frac {d^3\,9{}\mathrm {i}}{8\,a^3}+\frac {3\,c\,d^2}{8\,a^3}-\frac {c^2\,d\,9{}\mathrm {i}}{8\,a^3}\right )-{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {7\,d^3}{8\,a^3}+\frac {3\,c^2\,d}{8\,a^3}+\frac {c^3\,1{}\mathrm {i}}{8\,a^3}-\frac {c\,d^2\,3{}\mathrm {i}}{8\,a^3}\right )+\frac {c^2\,d}{4\,a^3}+\frac {c^3\,5{}\mathrm {i}}{12\,a^3}+\frac {c\,d^2\,1{}\mathrm {i}}{4\,a^3}}{f\,\left (-{\mathrm {tan}\left (e+f\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (e+f\,x\right )}^2+\mathrm {tan}\left (e+f\,x\right )\,3{}\mathrm {i}+1\right )}+\frac {x\,{\left (d+c\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{8\,a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*tan(e + f*x))^3/(a + a*tan(e + f*x)*1i)^3,x)

[Out]

((c^3*5i)/(12*a^3) - tan(e + f*x)*((3*c^3)/(8*a^3) - (d^3*9i)/(8*a^3) + (3*c*d^2)/(8*a^3) - (c^2*d*9i)/(8*a^3)
) + (5*d^3)/(12*a^3) - tan(e + f*x)^2*((c^3*1i)/(8*a^3) + (7*d^3)/(8*a^3) - (c*d^2*3i)/(8*a^3) + (3*c^2*d)/(8*
a^3)) + (c*d^2*1i)/(4*a^3) + (c^2*d)/(4*a^3))/(f*(tan(e + f*x)*3i - 3*tan(e + f*x)^2 - tan(e + f*x)^3*1i + 1))
+ (x*(c*1i + d)^3*1i)/(8*a^3)

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sympy [A]  time = 0.96, size = 558, normalized size = 3.99 \[ \begin {cases} - \frac {\left (\left (- 512 i a^{6} c^{3} f^{2} e^{6 i e} + 1536 a^{6} c^{2} d f^{2} e^{6 i e} + 1536 i a^{6} c d^{2} f^{2} e^{6 i e} - 512 a^{6} d^{3} f^{2} e^{6 i e}\right ) e^{- 6 i f x} + \left (- 2304 i a^{6} c^{3} f^{2} e^{8 i e} + 2304 a^{6} c^{2} d f^{2} e^{8 i e} - 2304 i a^{6} c d^{2} f^{2} e^{8 i e} + 2304 a^{6} d^{3} f^{2} e^{8 i e}\right ) e^{- 4 i f x} + \left (- 4608 i a^{6} c^{3} f^{2} e^{10 i e} - 4608 a^{6} c^{2} d f^{2} e^{10 i e} - 4608 i a^{6} c d^{2} f^{2} e^{10 i e} - 4608 a^{6} d^{3} f^{2} e^{10 i e}\right ) e^{- 2 i f x}\right ) e^{- 12 i e}}{24576 a^{9} f^{3}} & \text {for}\: 24576 a^{9} f^{3} e^{12 i e} \neq 0 \\x \left (- \frac {c^{3} - 3 i c^{2} d - 3 c d^{2} + i d^{3}}{8 a^{3}} + \frac {\left (c^{3} e^{6 i e} + 3 c^{3} e^{4 i e} + 3 c^{3} e^{2 i e} + c^{3} - 3 i c^{2} d e^{6 i e} - 3 i c^{2} d e^{4 i e} + 3 i c^{2} d e^{2 i e} + 3 i c^{2} d - 3 c d^{2} e^{6 i e} + 3 c d^{2} e^{4 i e} + 3 c d^{2} e^{2 i e} - 3 c d^{2} + i d^{3} e^{6 i e} - 3 i d^{3} e^{4 i e} + 3 i d^{3} e^{2 i e} - i d^{3}\right ) e^{- 6 i e}}{8 a^{3}}\right ) & \text {otherwise} \end {cases} - \frac {x \left (- c^{3} + 3 i c^{2} d + 3 c d^{2} - i d^{3}\right )}{8 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-((-512*I*a**6*c**3*f**2*exp(6*I*e) + 1536*a**6*c**2*d*f**2*exp(6*I*e) + 1536*I*a**6*c*d**2*f**2*ex
p(6*I*e) - 512*a**6*d**3*f**2*exp(6*I*e))*exp(-6*I*f*x) + (-2304*I*a**6*c**3*f**2*exp(8*I*e) + 2304*a**6*c**2*
d*f**2*exp(8*I*e) - 2304*I*a**6*c*d**2*f**2*exp(8*I*e) + 2304*a**6*d**3*f**2*exp(8*I*e))*exp(-4*I*f*x) + (-460
8*I*a**6*c**3*f**2*exp(10*I*e) - 4608*a**6*c**2*d*f**2*exp(10*I*e) - 4608*I*a**6*c*d**2*f**2*exp(10*I*e) - 460
8*a**6*d**3*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**3 - 3*I*c**2*d - 3*c*d**2 + I*d**3)/(8*a**3) + (c**3*exp(6*I*e) + 3*c**3*exp(4*I*e) + 3*c**3*exp(
2*I*e) + c**3 - 3*I*c**2*d*exp(6*I*e) - 3*I*c**2*d*exp(4*I*e) + 3*I*c**2*d*exp(2*I*e) + 3*I*c**2*d - 3*c*d**2*
exp(6*I*e) + 3*c*d**2*exp(4*I*e) + 3*c*d**2*exp(2*I*e) - 3*c*d**2 + I*d**3*exp(6*I*e) - 3*I*d**3*exp(4*I*e) +
3*I*d**3*exp(2*I*e) - I*d**3)*exp(-6*I*e)/(8*a**3)), True)) - x*(-c**3 + 3*I*c**2*d + 3*c*d**2 - I*d**3)/(8*a*
*3)

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