3.11.80 \(\int \frac {(c+d \tan (e+f x))^3}{a+i a \tan (e+f x)} \, dx\) [1080]

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

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3631, 3606, 3556} \begin {gather*} \frac {x \left (c^3-3 i c^2 d+3 c d^2+3 i d^3\right )}{2 a}-\frac {d^2 (c+3 i d) \tan (e+f x)}{2 a f}+\frac {d^2 (-d+3 i c) \log (\cos (e+f x))}{a f}+\frac {(-d+i c) (c+d \tan (e+f x))^2}{2 f (a+i a \tan (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3606

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 3631

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*
c - a*d)*((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*(a + b*Tan[e + f*x]))), x] + Dist[1/(2*a^2), Int[(c + d*Tan[e +
f*x])^(n - 2)*Simp[a*c^2 + a*d^2*(n - 1) - b*c*d*n - d*(a*c*(n - 2) + b*d*n)*Tan[e + f*x], x], 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] && GtQ[n, 1]

Rubi steps

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

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Mathematica [A]
time = 2.04, size = 236, normalized size = 1.83 \begin {gather*} \frac {\sec (e+f x) (\cos (f x)+i \sin (f x)) \left (-4 (3 c+i d) d^2 f x (\cos (e)+i \sin (e))+2 \left (c^3-3 i c^2 d+3 c d^2+3 i d^3\right ) f x (\cos (e)+i \sin (e))+4 (3 c+i d) d^2 \text {ArcTan}(\tan (f x)) (\cos (e)+i \sin (e))+2 i (3 c+i d) d^2 \log \left (\cos ^2(e+f x)\right ) (\cos (e)+i \sin (e))+(c+i d)^3 \cos (2 f x) (i \cos (e)+\sin (e))+(c+i d)^3 (\cos (e)-i \sin (e)) \sin (2 f x)+4 d^3 \sec (e+f x) \sin (f x) (-i+\tan (e))\right )}{4 f (a+i a \tan (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]*(Cos[f*x] + I*Sin[f*x])*(-4*(3*c + I*d)*d^2*f*x*(Cos[e] + I*Sin[e]) + 2*(c^3 - (3*I)*c^2*d + 3*c
*d^2 + (3*I)*d^3)*f*x*(Cos[e] + I*Sin[e]) + 4*(3*c + I*d)*d^2*ArcTan[Tan[f*x]]*(Cos[e] + I*Sin[e]) + (2*I)*(3*
c + I*d)*d^2*Log[Cos[e + f*x]^2]*(Cos[e] + I*Sin[e]) + (c + I*d)^3*Cos[2*f*x]*(I*Cos[e] + Sin[e]) + (c + I*d)^
3*(Cos[e] - I*Sin[e])*Sin[2*f*x] + 4*d^3*Sec[e + f*x]*Sin[f*x]*(-I + Tan[e])))/(4*f*(a + I*a*Tan[e + f*x]))

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Maple [A]
time = 0.24, size = 133, normalized size = 1.03

method	result	size
derivativedivides	\(\frac {-i d^{3} \tan \left (f x +e \right )+\left (-\frac {3}{4} c^{2} d +\frac {5}{4} d^{3}-\frac {1}{4} i c^{3}-\frac {9}{4} i c \,d^{2}\right ) \ln \left (\tan \left (f x +e \right )-i\right )-\frac {-\frac {3}{2} i c^{2} d +\frac {1}{2} i d^{3}-\frac {1}{2} c^{3}+\frac {3}{2} c \,d^{2}}{\tan \left (f x +e \right )-i}-\frac {i \left (3 i c^{2} d -i d^{3}-c^{3}+3 c \,d^{2}\right ) \ln \left (\tan \left (f x +e \right )+i\right )}{4}}{f a}\)	\(133\)
default	\(\frac {-i d^{3} \tan \left (f x +e \right )+\left (-\frac {3}{4} c^{2} d +\frac {5}{4} d^{3}-\frac {1}{4} i c^{3}-\frac {9}{4} i c \,d^{2}\right ) \ln \left (\tan \left (f x +e \right )-i\right )-\frac {-\frac {3}{2} i c^{2} d +\frac {1}{2} i d^{3}-\frac {1}{2} c^{3}+\frac {3}{2} c \,d^{2}}{\tan \left (f x +e \right )-i}-\frac {i \left (3 i c^{2} d -i d^{3}-c^{3}+3 c \,d^{2}\right ) \ln \left (\tan \left (f x +e \right )+i\right )}{4}}{f a}\)	\(133\)
risch	\(-\frac {3 i x \,c^{2} d}{2 a}+\frac {5 i x \,d^{3}}{2 a}+\frac {c^{3} x}{2 a}+\frac {9 x c \,d^{2}}{2 a}-\frac {3 \,{\mathrm e}^{-2 i \left (f x +e \right )} c^{2} d}{4 a f}+\frac {{\mathrm e}^{-2 i \left (f x +e \right )} d^{3}}{4 a f}+\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} c^{3}}{4 a f}-\frac {3 i {\mathrm e}^{-2 i \left (f x +e \right )} c \,d^{2}}{4 a f}+\frac {6 d^{2} c e}{a f}+\frac {2 i d^{3} e}{a f}+\frac {2 d^{3}}{f a \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {3 i d^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) c}{a f}-\frac {d^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{a f}\)	\(224\)

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)),x,method=_RETURNVERBOSE)

[Out]

1/f/a*(-I*d^3*tan(f*x+e)+(-3/4*c^2*d+5/4*d^3-1/4*I*c^3-9/4*I*c*d^2)*ln(tan(f*x+e)-I)-(-3/2*I*c^2*d+1/2*I*d^3-1
/2*c^3+3/2*c*d^2)/(tan(f*x+e)-I)-1/4*I*(3*I*c^2*d-I*d^3-c^3+3*c*d^2)*ln(tan(f*x+e)+I))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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)),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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Fricas [A]
time = 0.77, size = 204, normalized size = 1.58 \begin {gather*} \frac {2 \, {\left (c^{3} - 3 i \, c^{2} d + 9 \, c d^{2} + 5 i \, d^{3}\right )} f x e^{\left (4 i \, f x + 4 i \, e\right )} + i \, c^{3} - 3 \, c^{2} d - 3 i \, c d^{2} + d^{3} + {\left (i \, c^{3} - 3 \, c^{2} d - 3 i \, c d^{2} + 9 \, d^{3} + 2 \, {\left (c^{3} - 3 i \, c^{2} d + 9 \, c d^{2} + 5 i \, d^{3}\right )} f x\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 4 \, {\left ({\left (-3 i \, c d^{2} + d^{3}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-3 i \, c d^{2} + d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{4 \, {\left (a f e^{\left (4 i \, f x + 4 i \, e\right )} + a f e^{\left (2 i \, f x + 2 i \, e\right )}\right )}} \end {gather*}

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)),x, algorithm="fricas")

[Out]

1/4*(2*(c^3 - 3*I*c^2*d + 9*c*d^2 + 5*I*d^3)*f*x*e^(4*I*f*x + 4*I*e) + I*c^3 - 3*c^2*d - 3*I*c*d^2 + d^3 + (I*
c^3 - 3*c^2*d - 3*I*c*d^2 + 9*d^3 + 2*(c^3 - 3*I*c^2*d + 9*c*d^2 + 5*I*d^3)*f*x)*e^(2*I*f*x + 2*I*e) - 4*((-3*
I*c*d^2 + d^3)*e^(4*I*f*x + 4*I*e) + (-3*I*c*d^2 + d^3)*e^(2*I*f*x + 2*I*e))*log(e^(2*I*f*x + 2*I*e) + 1))/(a*
f*e^(4*I*f*x + 4*I*e) + a*f*e^(2*I*f*x + 2*I*e))

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Sympy [A]
time = 0.44, size = 260, normalized size = 2.02 \begin {gather*} \frac {2 d^{3}}{a f e^{2 i e} e^{2 i f x} + a f} + \begin {cases} \frac {\left (i c^{3} - 3 c^{2} d - 3 i c d^{2} + d^{3}\right ) e^{- 2 i e} e^{- 2 i f x}}{4 a f} & \text {for}\: a f e^{2 i e} \neq 0 \\x \left (- \frac {c^{3} - 3 i c^{2} d + 9 c d^{2} + 5 i d^{3}}{2 a} + \frac {\left (c^{3} e^{2 i e} + c^{3} - 3 i c^{2} d e^{2 i e} + 3 i c^{2} d + 9 c d^{2} e^{2 i e} - 3 c d^{2} + 5 i d^{3} e^{2 i e} - i d^{3}\right ) e^{- 2 i e}}{2 a}\right ) & \text {otherwise} \end {cases} + \frac {i d^{2} \cdot \left (3 c + i d\right ) \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a f} + \frac {x \left (c^{3} - 3 i c^{2} d + 9 c d^{2} + 5 i d^{3}\right )}{2 a} \end {gather*}

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)),x)

[Out]

2*d**3/(a*f*exp(2*I*e)*exp(2*I*f*x) + a*f) + Piecewise(((I*c**3 - 3*c**2*d - 3*I*c*d**2 + d**3)*exp(-2*I*e)*ex
p(-2*I*f*x)/(4*a*f), Ne(a*f*exp(2*I*e), 0)), (x*(-(c**3 - 3*I*c**2*d + 9*c*d**2 + 5*I*d**3)/(2*a) + (c**3*exp(
2*I*e) + c**3 - 3*I*c**2*d*exp(2*I*e) + 3*I*c**2*d + 9*c*d**2*exp(2*I*e) - 3*c*d**2 + 5*I*d**3*exp(2*I*e) - I*
d**3)*exp(-2*I*e)/(2*a)), True)) + I*d**2*(3*c + I*d)*log(exp(2*I*f*x) + exp(-2*I*e))/(a*f) + x*(c**3 - 3*I*c*
*2*d + 9*c*d**2 + 5*I*d**3)/(2*a)

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Giac [A]
time = 0.73, size = 186, normalized size = 1.44 \begin {gather*} -\frac {\frac {4 i \, d^{3} \tan \left (f x + e\right )}{a} - \frac {{\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} + \frac {{\left (i \, c^{3} + 3 \, c^{2} d + 9 i \, c d^{2} - 5 \, d^{3}\right )} \log \left (-i \, \tan \left (f x + e\right ) - 1\right )}{a} + \frac {-i \, c^{3} \tan \left (f x + e\right ) - 3 \, c^{2} d \tan \left (f x + e\right ) - 9 i \, c d^{2} \tan \left (f x + e\right ) + 5 \, d^{3} \tan \left (f x + e\right ) - 3 \, c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} - 3 i \, d^{3}}{a {\left (\tan \left (f x + e\right ) - i\right )}}}{4 \, f} \end {gather*}

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)),x, algorithm="giac")

[Out]

-1/4*(4*I*d^3*tan(f*x + e)/a - (I*c^3 + 3*c^2*d - 3*I*c*d^2 - d^3)*log(tan(f*x + e) + I)/a + (I*c^3 + 3*c^2*d
+ 9*I*c*d^2 - 5*d^3)*log(-I*tan(f*x + e) - 1)/a + (-I*c^3*tan(f*x + e) - 3*c^2*d*tan(f*x + e) - 9*I*c*d^2*tan(
f*x + e) + 5*d^3*tan(f*x + e) - 3*c^3 - 3*I*c^2*d - 3*c*d^2 - 3*I*d^3)/(a*(tan(f*x + e) - I)))/f

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Mupad [B]
time = 5.66, size = 175, normalized size = 1.36 \begin {gather*} -\frac {\frac {3\,c^2\,d-d^3}{2\,a}+\frac {\left (d^3\,1{}\mathrm {i}+3\,c\,d^2\right )\,1{}\mathrm {i}}{2\,a}-\frac {-d^3+c^3\,1{}\mathrm {i}}{2\,a}}{f\,\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}-\frac {d^3\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}{a\,f}-\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\left (-c^3\,1{}\mathrm {i}-3\,c^2\,d+c\,d^2\,3{}\mathrm {i}+d^3\right )}{4\,a\,f}-\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,\left (c^3\,1{}\mathrm {i}+3\,c^2\,d+c\,d^2\,9{}\mathrm {i}-5\,d^3\right )}{4\,a\,f} \end {gather*}

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),x)

[Out]

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

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