3.8.33 \(\int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3} \, dx\) [733]

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

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 112, normalized size of antiderivative = 1.00, 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 = {3607, 3560, 8} \begin {gather*} \frac {B+i A}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {x (A-i B)}{8 a^3}+\frac {-B+i A}{6 f (a+i a \tan (e+f x))^3}+\frac {B+i A}{8 a f (a+i a \tan (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 3560

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 3607

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]

Rubi steps

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

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Mathematica [A]
time = 0.47, size = 150, normalized size = 1.34 \begin {gather*} \frac {\sec ^3(e+f x) ((-27 A+3 i B) \cos (e+f x)+2 (-A-i B+6 i A f x+6 B f x) \cos (3 (e+f x))-9 i A \sin (e+f x)-9 B \sin (e+f x)+2 i A \sin (3 (e+f x))-2 B \sin (3 (e+f x))-12 A f x \sin (3 (e+f x))+12 i B f x \sin (3 (e+f x)))}{96 a^3 f (-i+\tan (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.22, size = 110, normalized size = 0.98

method	result	size
derivativedivides	\(\frac {\left (-\frac {i A}{16}-\frac {B}{16}\right ) \ln \left (-i+\tan \left (f x +e \right )\right )-\frac {-\frac {A}{8}+\frac {i B}{8}}{-i+\tan \left (f x +e \right )}-\frac {\frac {i A}{4}+\frac {B}{4}}{2 \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {\frac {A}{2}+\frac {i B}{2}}{3 \left (-i+\tan \left (f x +e \right )\right )^{3}}+\frac {i \left (-i B +A \right ) \ln \left (i+\tan \left (f x +e \right )\right )}{16}}{f \,a^{3}}\)	\(110\)
default	\(\frac {\left (-\frac {i A}{16}-\frac {B}{16}\right ) \ln \left (-i+\tan \left (f x +e \right )\right )-\frac {-\frac {A}{8}+\frac {i B}{8}}{-i+\tan \left (f x +e \right )}-\frac {\frac {i A}{4}+\frac {B}{4}}{2 \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {\frac {A}{2}+\frac {i B}{2}}{3 \left (-i+\tan \left (f x +e \right )\right )^{3}}+\frac {i \left (-i B +A \right ) \ln \left (i+\tan \left (f x +e \right )\right )}{16}}{f \,a^{3}}\)	\(110\)
risch	\(-\frac {i x B}{8 a^{3}}+\frac {x A}{8 a^{3}}+\frac {{\mathrm e}^{-2 i \left (f x +e \right )} B}{16 a^{3} f}+\frac {3 i {\mathrm e}^{-2 i \left (f x +e \right )} A}{16 a^{3} f}-\frac {{\mathrm e}^{-4 i \left (f x +e \right )} B}{32 a^{3} f}+\frac {3 i {\mathrm e}^{-4 i \left (f x +e \right )} A}{32 a^{3} f}-\frac {{\mathrm e}^{-6 i \left (f x +e \right )} B}{48 a^{3} f}+\frac {i {\mathrm e}^{-6 i \left (f x +e \right )} A}{48 a^{3} f}\)	\(128\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

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

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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((A+B*tan(f*x+e))/(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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Fricas [A]
time = 6.15, size = 80, normalized size = 0.71 \begin {gather*} \frac {{\left (12 \, {\left (A - i \, B\right )} f x e^{\left (6 i \, f x + 6 i \, e\right )} - 6 \, {\left (-3 i \, A - B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - 3 \, {\left (-3 i \, A + B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, A - 2 \, B\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{96 \, a^{3} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.27, size = 258, normalized size = 2.30 \begin {gather*} \begin {cases} \frac {\left (\left (512 i A a^{6} f^{2} e^{6 i e} - 512 B a^{6} f^{2} e^{6 i e}\right ) e^{- 6 i f x} + \left (2304 i A a^{6} f^{2} e^{8 i e} - 768 B a^{6} f^{2} e^{8 i e}\right ) e^{- 4 i f x} + \left (4608 i A a^{6} f^{2} e^{10 i e} + 1536 B a^{6} f^{2} e^{10 i e}\right ) e^{- 2 i f x}\right ) e^{- 12 i e}}{24576 a^{9} f^{3}} & \text {for}\: a^{9} f^{3} e^{12 i e} \neq 0 \\x \left (- \frac {A - i B}{8 a^{3}} + \frac {\left (A e^{6 i e} + 3 A e^{4 i e} + 3 A e^{2 i e} + A - i B e^{6 i e} - i B e^{4 i e} + i B e^{2 i e} + i B\right ) e^{- 6 i e}}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (A - i B\right )}{8 a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))**3,x)

[Out]

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

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Giac [A]
time = 0.70, size = 140, normalized size = 1.25 \begin {gather*} -\frac {\frac {6 \, {\left (i \, A + B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{3}} + \frac {6 \, {\left (-i \, A - B\right )} \log \left (i \, \tan \left (f x + e\right ) - 1\right )}{a^{3}} + \frac {-11 i \, A \tan \left (f x + e\right )^{3} - 11 \, B \tan \left (f x + e\right )^{3} - 45 \, A \tan \left (f x + e\right )^{2} + 45 i \, B \tan \left (f x + e\right )^{2} + 69 i \, A \tan \left (f x + e\right ) + 69 \, B \tan \left (f x + e\right ) + 51 \, A - 19 i \, B}{a^{3} {\left (\tan \left (f x + e\right ) - i\right )}^{3}}}{96 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 9.13, size = 111, normalized size = 0.99 \begin {gather*} -\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {B}{8\,a^3}+\frac {A\,1{}\mathrm {i}}{8\,a^3}\right )-\frac {A\,5{}\mathrm {i}}{12\,a^3}-\frac {B}{12\,a^3}+\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {3\,A}{8\,a^3}-\frac {B\,3{}\mathrm {i}}{8\,a^3}\right )}{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 (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,a^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (tan(e + f*x)^2*((A*1i)/(8*a^3) + B/(8*a^3)) - (A*5i)/(12*a^3) - B/(12*a^3) + tan(e + f*x)*((3*A)/(8*a^3) -
(B*3i)/(8*a^3)))/(f*(tan(e + f*x)*3i - 3*tan(e + f*x)^2 - tan(e + f*x)^3*1i + 1)) - (x*(A*1i + B)*1i)/(8*a^3)

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