3.1.57 \(\int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) [57]

Optimal. Leaf size=124 \[ \frac {15 i x}{4 a^2}-\frac {4 \log (\cos (c+d x))}{a^2 d}-\frac {15 i \tan (c+d x)}{4 a^2 d}-\frac {2 \tan ^2(c+d x)}{a^2 d}+\frac {5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

15/4*I*x/a^2-4*ln(cos(d*x+c))/a^2/d-15/4*I*tan(d*x+c)/a^2/d-2*tan(d*x+c)^2/a^2/d+5/4*I*tan(d*x+c)^3/a^2/d/(1+I
*tan(d*x+c))-1/4*tan(d*x+c)^4/d/(a+I*a*tan(d*x+c))^2

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Rubi [A]
time = 0.14, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3639, 3676, 3609, 3606, 3556} \begin {gather*} \frac {5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac {2 \tan ^2(c+d x)}{a^2 d}-\frac {15 i \tan (c+d x)}{4 a^2 d}-\frac {4 \log (\cos (c+d x))}{a^2 d}+\frac {15 i x}{4 a^2}-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^5/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(((15*I)/4)*x)/a^2 - (4*Log[Cos[c + d*x]])/(a^2*d) - (((15*I)/4)*Tan[c + d*x])/(a^2*d) - (2*Tan[c + d*x]^2)/(a
^2*d) + (((5*I)/4)*Tan[c + d*x]^3)/(a^2*d*(1 + I*Tan[c + d*x])) - Tan[c + d*x]^4/(4*d*(a + I*a*Tan[c + d*x])^2
)

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 3609

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

Rule 3639

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Dist[1/(2*a^2*m), Int[(
a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1)
) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*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] && LtQ[m, 0] && GtQ[n, 1] && (IntegerQ[m] || IntegersQ[2*m
, 2*n])

Rule 3676

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

Rubi steps

\begin {align*} \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {\tan ^3(c+d x) (-4 a+6 i a \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac {5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \tan ^2(c+d x) \left (-30 i a^2-32 a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=-\frac {2 \tan ^2(c+d x)}{a^2 d}+\frac {5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \tan (c+d x) \left (32 a^2-30 i a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=\frac {15 i x}{4 a^2}-\frac {15 i \tan (c+d x)}{4 a^2 d}-\frac {2 \tan ^2(c+d x)}{a^2 d}+\frac {5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {4 \int \tan (c+d x) \, dx}{a^2}\\ &=\frac {15 i x}{4 a^2}-\frac {4 \log (\cos (c+d x))}{a^2 d}-\frac {15 i \tan (c+d x)}{4 a^2 d}-\frac {2 \tan ^2(c+d x)}{a^2 d}+\frac {5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(300\) vs. \(2(124)=248\).
time = 1.82, size = 300, normalized size = 2.42 \begin {gather*} \frac {\sec ^2(c+d x) (\cos (d x)+i \sin (d x))^2 \left (64 i d x-16 \cos (2 d x)-16 \cos (2 c-d x) \sec (c) \sec (c+d x)+16 \cos (2 c+d x) \sec (c) \sec (c+d x)-128 i d x \sin ^2(c)+60 d x \sin (2 c)-i \cos (4 d x) \sin (2 c)+32 i \log \left (\cos ^2(c+d x)\right ) \sin (2 c)+8 i \sec ^2(c+d x) \sin (2 c)+64 \text {ArcTan}(\tan (d x)) (-i \cos (2 c)+\sin (2 c))+16 i \sin (2 d x)-\sin (2 c) \sin (4 d x)-16 i \sec (c) \sec (c+d x) \sin (2 c-d x)+16 i \sec (c) \sec (c+d x) \sin (2 c+d x)-64 d x \tan (c)+\cos (2 c) \left (-60 i d x+\cos (4 d x)+32 \log \left (\cos ^2(c+d x)\right )+8 \sec ^2(c+d x)-i \sin (4 d x)-64 d x \tan (c)\right )\right )}{16 a^2 d (-i+\tan (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^5/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(Sec[c + d*x]^2*(Cos[d*x] + I*Sin[d*x])^2*((64*I)*d*x - 16*Cos[2*d*x] - 16*Cos[2*c - d*x]*Sec[c]*Sec[c + d*x]
+ 16*Cos[2*c + d*x]*Sec[c]*Sec[c + d*x] - (128*I)*d*x*Sin[c]^2 + 60*d*x*Sin[2*c] - I*Cos[4*d*x]*Sin[2*c] + (32
*I)*Log[Cos[c + d*x]^2]*Sin[2*c] + (8*I)*Sec[c + d*x]^2*Sin[2*c] + 64*ArcTan[Tan[d*x]]*((-I)*Cos[2*c] + Sin[2*
c]) + (16*I)*Sin[2*d*x] - Sin[2*c]*Sin[4*d*x] - (16*I)*Sec[c]*Sec[c + d*x]*Sin[2*c - d*x] + (16*I)*Sec[c]*Sec[
c + d*x]*Sin[2*c + d*x] - 64*d*x*Tan[c] + Cos[2*c]*((-60*I)*d*x + Cos[4*d*x] + 32*Log[Cos[c + d*x]^2] + 8*Sec[
c + d*x]^2 - I*Sin[4*d*x] - 64*d*x*Tan[c])))/(16*a^2*d*(-I + Tan[c + d*x])^2)

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Maple [A]
time = 0.11, size = 79, normalized size = 0.64

method	result	size
derivativedivides	\(\frac {-2 i \tan \left (d x +c \right )-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {9 i}{4 \left (\tan \left (d x +c \right )-i\right )}+\frac {1}{4 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {31 \ln \left (\tan \left (d x +c \right )-i\right )}{8}+\frac {\ln \left (\tan \left (d x +c \right )+i\right )}{8}}{d \,a^{2}}\)	\(79\)
default	\(\frac {-2 i \tan \left (d x +c \right )-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {9 i}{4 \left (\tan \left (d x +c \right )-i\right )}+\frac {1}{4 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {31 \ln \left (\tan \left (d x +c \right )-i\right )}{8}+\frac {\ln \left (\tan \left (d x +c \right )+i\right )}{8}}{d \,a^{2}}\)	\(79\)
risch	\(\frac {31 i x}{4 a^{2}}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{a^{2} d}-\frac {{\mathrm e}^{-4 i \left (d x +c \right )}}{16 a^{2} d}+\frac {8 i c}{a^{2} d}+\frac {2 \,{\mathrm e}^{2 i \left (d x +c \right )}+4}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {4 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{2} d}\)	\(104\)
norman	\(\frac {\frac {3}{d a}-\frac {\tan ^{6}\left (d x +c \right )}{2 d a}+\frac {15 i x}{4 a}+\frac {4 \left (\tan ^{2}\left (d x +c \right )\right )}{d a}+\frac {15 i x \left (\tan ^{2}\left (d x +c \right )\right )}{2 a}+\frac {15 i x \left (\tan ^{4}\left (d x +c \right )\right )}{4 a}-\frac {15 i \tan \left (d x +c \right )}{4 d a}-\frac {25 i \left (\tan ^{3}\left (d x +c \right )\right )}{4 d a}-\frac {2 i \left (\tan ^{5}\left (d x +c \right )\right )}{d a}}{a \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {2 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{a^{2} d}\)	\(164\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d/a^2*(-2*I*tan(d*x+c)-1/2*tan(d*x+c)^2-9/4*I/(tan(d*x+c)-I)+1/4/(tan(d*x+c)-I)^2+31/8*ln(tan(d*x+c)-I)+1/8*
ln(tan(d*x+c)+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(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^2,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 = 0.38, size = 151, normalized size = 1.22 \begin {gather*} \frac {124 i \, d x e^{\left (8 i \, d x + 8 i \, c\right )} - 8 \, {\left (-31 i \, d x - 6\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (124 i \, d x + 95\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 64 \, {\left (e^{\left (8 i \, d x + 8 i \, c\right )} + 2 \, e^{\left (6 i \, d x + 6 i \, c\right )} + e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 14 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 1}{16 \, {\left (a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 2 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/16*(124*I*d*x*e^(8*I*d*x + 8*I*c) - 8*(-31*I*d*x - 6)*e^(6*I*d*x + 6*I*c) + (124*I*d*x + 95)*e^(4*I*d*x + 4*
I*c) - 64*(e^(8*I*d*x + 8*I*c) + 2*e^(6*I*d*x + 6*I*c) + e^(4*I*d*x + 4*I*c))*log(e^(2*I*d*x + 2*I*c) + 1) + 1
4*e^(2*I*d*x + 2*I*c) - 1)/(a^2*d*e^(8*I*d*x + 8*I*c) + 2*a^2*d*e^(6*I*d*x + 6*I*c) + a^2*d*e^(4*I*d*x + 4*I*c
))

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Sympy [A]
time = 1.00, size = 214, normalized size = 1.73 \begin {gather*} \frac {2 e^{2 i c} e^{2 i d x} + 4}{a^{2} d e^{4 i c} e^{4 i d x} + 2 a^{2} d e^{2 i c} e^{2 i d x} + a^{2} d} + \begin {cases} \frac {\left (16 a^{2} d e^{4 i c} e^{- 2 i d x} - a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{16 a^{4} d^{2}} & \text {for}\: a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac {\left (31 i e^{4 i c} - 8 i e^{2 i c} + i\right ) e^{- 4 i c}}{4 a^{2}} - \frac {31 i}{4 a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {31 i x}{4 a^{2}} - \frac {4 \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**5/(a+I*a*tan(d*x+c))**2,x)

[Out]

(2*exp(2*I*c)*exp(2*I*d*x) + 4)/(a**2*d*exp(4*I*c)*exp(4*I*d*x) + 2*a**2*d*exp(2*I*c)*exp(2*I*d*x) + a**2*d) +
Piecewise(((16*a**2*d*exp(4*I*c)*exp(-2*I*d*x) - a**2*d*exp(2*I*c)*exp(-4*I*d*x))*exp(-6*I*c)/(16*a**4*d**2),
Ne(a**4*d**2*exp(6*I*c), 0)), (x*((31*I*exp(4*I*c) - 8*I*exp(2*I*c) + I)*exp(-4*I*c)/(4*a**2) - 31*I/(4*a**2)
), True)) + 31*I*x/(4*a**2) - 4*log(exp(2*I*d*x) + exp(-2*I*c))/(a**2*d)

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Giac [A]
time = 1.65, size = 98, normalized size = 0.79 \begin {gather*} \frac {\frac {2 \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac {62 \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} - \frac {8 \, {\left (a^{2} \tan \left (d x + c\right )^{2} + 4 i \, a^{2} \tan \left (d x + c\right )\right )}}{a^{4}} - \frac {93 \, \tan \left (d x + c\right )^{2} - 150 i \, \tan \left (d x + c\right ) - 61}{a^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/16*(2*log(tan(d*x + c) + I)/a^2 + 62*log(tan(d*x + c) - I)/a^2 - 8*(a^2*tan(d*x + c)^2 + 4*I*a^2*tan(d*x + c
))/a^4 - (93*tan(d*x + c)^2 - 150*I*tan(d*x + c) - 61)/(a^2*(tan(d*x + c) - I)^2))/d

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Mupad [B]
time = 4.03, size = 114, normalized size = 0.92 \begin {gather*} \frac {31\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{8\,a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{8\,a^2\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}}{a^2\,d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,a^2\,d}+\frac {\frac {9\,\mathrm {tan}\left (c+d\,x\right )}{4\,a^2}-\frac {2{}\mathrm {i}}{a^2}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^5/(a + a*tan(c + d*x)*1i)^2,x)

[Out]

(31*log(tan(c + d*x) - 1i))/(8*a^2*d) + log(tan(c + d*x) + 1i)/(8*a^2*d) - (tan(c + d*x)*2i)/(a^2*d) - tan(c +
d*x)^2/(2*a^2*d) + ((9*tan(c + d*x))/(4*a^2) - 2i/a^2)/(d*(2*tan(c + d*x) + tan(c + d*x)^2*1i - 1i))

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