3.155 \(\int \frac {1}{(a+i a \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=116 \[ \frac {i}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {x}{16 a^4}+\frac {i}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {i}{12 a d (a+i a \tan (c+d x))^3}+\frac {i}{8 d (a+i a \tan (c+d x))^4} \]

[Out]

1/16*x/a^4+1/8*I/d/(a+I*a*tan(d*x+c))^4+1/12*I/a/d/(a+I*a*tan(d*x+c))^3+1/16*I/d/(a^2+I*a^2*tan(d*x+c))^2+1/16
*I/d/(a^4+I*a^4*tan(d*x+c))

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Rubi [A]  time = 0.06, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3479, 8} \[ \frac {i}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {i}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {x}{16 a^4}+\frac {i}{12 a d (a+i a \tan (c+d x))^3}+\frac {i}{8 d (a+i a \tan (c+d x))^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(-4),x]

[Out]

x/(16*a^4) + (I/8)/(d*(a + I*a*Tan[c + d*x])^4) + (I/12)/(a*d*(a + I*a*Tan[c + d*x])^3) + (I/16)/(d*(a^2 + I*a
^2*Tan[c + d*x])^2) + (I/16)/(d*(a^4 + I*a^4*Tan[c + d*x]))

Rule 8

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

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]

Rubi steps

\begin {align*} \int \frac {1}{(a+i a \tan (c+d x))^4} \, dx &=\frac {i}{8 d (a+i a \tan (c+d x))^4}+\frac {\int \frac {1}{(a+i a \tan (c+d x))^3} \, dx}{2 a}\\ &=\frac {i}{8 d (a+i a \tan (c+d x))^4}+\frac {i}{12 a d (a+i a \tan (c+d x))^3}+\frac {\int \frac {1}{(a+i a \tan (c+d x))^2} \, dx}{4 a^2}\\ &=\frac {i}{8 d (a+i a \tan (c+d x))^4}+\frac {i}{12 a d (a+i a \tan (c+d x))^3}+\frac {i}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {\int \frac {1}{a+i a \tan (c+d x)} \, dx}{8 a^3}\\ &=\frac {i}{8 d (a+i a \tan (c+d x))^4}+\frac {i}{12 a d (a+i a \tan (c+d x))^3}+\frac {i}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {i}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {\int 1 \, dx}{16 a^4}\\ &=\frac {x}{16 a^4}+\frac {i}{8 d (a+i a \tan (c+d x))^4}+\frac {i}{12 a d (a+i a \tan (c+d x))^3}+\frac {i}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {i}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 98, normalized size = 0.84 \[ \frac {\sec ^4(c+d x) (-32 \sin (2 (c+d x))+24 i d x \sin (4 (c+d x))+3 \sin (4 (c+d x))+64 i \cos (2 (c+d x))+3 (8 d x+i) \cos (4 (c+d x))+36 i)}{384 a^4 d (\tan (c+d x)-i)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^(-4),x]

[Out]

(Sec[c + d*x]^4*(36*I + (64*I)*Cos[2*(c + d*x)] + 3*(I + 8*d*x)*Cos[4*(c + d*x)] - 32*Sin[2*(c + d*x)] + 3*Sin
[4*(c + d*x)] + (24*I)*d*x*Sin[4*(c + d*x)]))/(384*a^4*d*(-I + Tan[c + d*x])^4)

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fricas [A]  time = 0.40, size = 65, normalized size = 0.56 \[ \frac {{\left (24 \, d x e^{\left (8 i \, d x + 8 i \, c\right )} + 48 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 36 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 16 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{384 \, a^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/384*(24*d*x*e^(8*I*d*x + 8*I*c) + 48*I*e^(6*I*d*x + 6*I*c) + 36*I*e^(4*I*d*x + 4*I*c) + 16*I*e^(2*I*d*x + 2*
I*c) + 3*I)*e^(-8*I*d*x - 8*I*c)/(a^4*d)

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giac [A]  time = 0.89, size = 92, normalized size = 0.79 \[ -\frac {-\frac {12 i \, \log \left (-i \, \tan \left (d x + c\right ) + 1\right )}{a^{4}} + \frac {12 i \, \log \left (-i \, \tan \left (d x + c\right ) - 1\right )}{a^{4}} + \frac {-25 i \, \tan \left (d x + c\right )^{4} - 124 \, \tan \left (d x + c\right )^{3} + 246 i \, \tan \left (d x + c\right )^{2} + 252 \, \tan \left (d x + c\right ) - 153 i}{a^{4} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{384 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

-1/384*(-12*I*log(-I*tan(d*x + c) + 1)/a^4 + 12*I*log(-I*tan(d*x + c) - 1)/a^4 + (-25*I*tan(d*x + c)^4 - 124*t
an(d*x + c)^3 + 246*I*tan(d*x + c)^2 + 252*tan(d*x + c) - 153*I)/(a^4*(tan(d*x + c) - I)^4))/d

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maple [A]  time = 0.11, size = 118, normalized size = 1.02 \[ \frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{32 d \,a^{4}}+\frac {i}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {i \ln \left (\tan \left (d x +c \right )-i\right )}{32 a^{4} d}-\frac {i}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {1}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {1}{16 a^{4} d \left (\tan \left (d x +c \right )-i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+I*a*tan(d*x+c))^4,x)

[Out]

1/32*I/d/a^4*ln(tan(d*x+c)+I)+1/8*I/d/a^4/(tan(d*x+c)-I)^4-1/32*I/d/a^4*ln(tan(d*x+c)-I)-1/16*I/d/a^4/(tan(d*x
+c)-I)^2-1/12/d/a^4/(tan(d*x+c)-I)^3+1/16/a^4/d/(tan(d*x+c)-I)

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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(1/(a+I*a*tan(d*x+c))^4,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 = 3.50, size = 60, normalized size = 0.52 \[ \frac {x}{16\,a^4}-\frac {-\frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{16}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{4}+\frac {19\,\mathrm {tan}\left (c+d\,x\right )}{48}-\frac {1}{3}{}\mathrm {i}}{a^4\,d\,{\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + a*tan(c + d*x)*1i)^4,x)

[Out]

x/(16*a^4) - ((19*tan(c + d*x))/48 + (tan(c + d*x)^2*1i)/4 - tan(c + d*x)^3/16 - 1i/3)/(a^4*d*(tan(c + d*x)*1i
+ 1)^4)

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sympy [A]  time = 0.41, size = 190, normalized size = 1.64 \[ \begin {cases} \frac {\left (98304 i a^{12} d^{3} e^{18 i c} e^{- 2 i d x} + 73728 i a^{12} d^{3} e^{16 i c} e^{- 4 i d x} + 32768 i a^{12} d^{3} e^{14 i c} e^{- 6 i d x} + 6144 i a^{12} d^{3} e^{12 i c} e^{- 8 i d x}\right ) e^{- 20 i c}}{786432 a^{16} d^{4}} & \text {for}\: 786432 a^{16} d^{4} e^{20 i c} \neq 0 \\x \left (\frac {\left (e^{8 i c} + 4 e^{6 i c} + 6 e^{4 i c} + 4 e^{2 i c} + 1\right ) e^{- 8 i c}}{16 a^{4}} - \frac {1}{16 a^{4}}\right ) & \text {otherwise} \end {cases} + \frac {x}{16 a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))**4,x)

[Out]

Piecewise(((98304*I*a**12*d**3*exp(18*I*c)*exp(-2*I*d*x) + 73728*I*a**12*d**3*exp(16*I*c)*exp(-4*I*d*x) + 3276
8*I*a**12*d**3*exp(14*I*c)*exp(-6*I*d*x) + 6144*I*a**12*d**3*exp(12*I*c)*exp(-8*I*d*x))*exp(-20*I*c)/(786432*a
**16*d**4), Ne(786432*a**16*d**4*exp(20*I*c), 0)), (x*((exp(8*I*c) + 4*exp(6*I*c) + 6*exp(4*I*c) + 4*exp(2*I*c
) + 1)*exp(-8*I*c)/(16*a**4) - 1/(16*a**4)), True)) + x/(16*a**4)

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