3.1.13 \(\int \tan ^4(c+d x) (a+i a \tan (c+d x))^2 \, dx\) [13]

Optimal. Leaf size=112 \[ 2 a^2 x-\frac {2 i a^2 \log (\cos (c+d x))}{d}-\frac {2 a^2 \tan (c+d x)}{d}-\frac {i a^2 \tan ^2(c+d x)}{d}+\frac {2 a^2 \tan ^3(c+d x)}{3 d}+\frac {i a^2 \tan ^4(c+d x)}{2 d}-\frac {a^2 \tan ^5(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3624, 3609, 3606, 3556} \begin {gather*} -\frac {a^2 \tan ^5(c+d x)}{5 d}+\frac {i a^2 \tan ^4(c+d x)}{2 d}+\frac {2 a^2 \tan ^3(c+d x)}{3 d}-\frac {i a^2 \tan ^2(c+d x)}{d}-\frac {2 a^2 \tan (c+d x)}{d}-\frac {2 i a^2 \log (\cos (c+d x))}{d}+2 a^2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^4*(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

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 3624

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

Rubi steps

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

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Mathematica [A]
time = 0.33, size = 108, normalized size = 0.96 \begin {gather*} \frac {2 a^2 \text {ArcTan}(\tan (c+d x))}{d}-\frac {2 a^2 \tan (c+d x)}{d}+\frac {2 a^2 \tan ^3(c+d x)}{3 d}-\frac {a^2 \tan ^5(c+d x)}{5 d}-\frac {i a^2 \left (4 \log (\cos (c+d x))+2 \tan ^2(c+d x)-\tan ^4(c+d x)\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^4*(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

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Maple [A]
time = 0.05, size = 82, normalized size = 0.73

method	result	size
derivativedivides	\(\frac {a^{2} \left (-2 \tan \left (d x +c \right )-\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {i \left (\tan ^{4}\left (d x +c \right )\right )}{2}+\frac {2 \left (\tan ^{3}\left (d x +c \right )\right )}{3}-i \left (\tan ^{2}\left (d x +c \right )\right )+i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+2 \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(82\)
default	\(\frac {a^{2} \left (-2 \tan \left (d x +c \right )-\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {i \left (\tan ^{4}\left (d x +c \right )\right )}{2}+\frac {2 \left (\tan ^{3}\left (d x +c \right )\right )}{3}-i \left (\tan ^{2}\left (d x +c \right )\right )+i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+2 \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(82\)
risch	\(-\frac {4 a^{2} c}{d}-\frac {2 i a^{2} \left (135 \,{\mathrm e}^{8 i \left (d x +c \right )}+300 \,{\mathrm e}^{6 i \left (d x +c \right )}+370 \,{\mathrm e}^{4 i \left (d x +c \right )}+200 \,{\mathrm e}^{2 i \left (d x +c \right )}+43\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {2 i a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(100\)
norman	\(2 a^{2} x -\frac {2 a^{2} \tan \left (d x +c \right )}{d}+\frac {2 a^{2} \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {a^{2} \left (\tan ^{5}\left (d x +c \right )\right )}{5 d}-\frac {i a^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{d}+\frac {i a^{2} \left (\tan ^{4}\left (d x +c \right )\right )}{2 d}+\frac {i a^{2} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^4*(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.50, size = 95, normalized size = 0.85 \begin {gather*} -\frac {6 \, a^{2} \tan \left (d x + c\right )^{5} - 15 i \, a^{2} \tan \left (d x + c\right )^{4} - 20 \, a^{2} \tan \left (d x + c\right )^{3} + 30 i \, a^{2} \tan \left (d x + c\right )^{2} - 60 \, {\left (d x + c\right )} a^{2} - 30 i \, a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 60 \, a^{2} \tan \left (d x + c\right )}{30 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/30*(6*a^2*tan(d*x + c)^5 - 15*I*a^2*tan(d*x + c)^4 - 20*a^2*tan(d*x + c)^3 + 30*I*a^2*tan(d*x + c)^2 - 60*(
d*x + c)*a^2 - 30*I*a^2*log(tan(d*x + c)^2 + 1) + 60*a^2*tan(d*x + c))/d

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (100) = 200\).
time = 0.43, size = 217, normalized size = 1.94 \begin {gather*} -\frac {2 \, {\left (135 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + 300 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 370 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 200 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 43 i \, a^{2} + 15 \, {\left (i \, a^{2} e^{\left (10 i \, d x + 10 i \, c\right )} + 5 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + 10 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 10 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 5 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{15 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-2/15*(135*I*a^2*e^(8*I*d*x + 8*I*c) + 300*I*a^2*e^(6*I*d*x + 6*I*c) + 370*I*a^2*e^(4*I*d*x + 4*I*c) + 200*I*a
^2*e^(2*I*d*x + 2*I*c) + 43*I*a^2 + 15*(I*a^2*e^(10*I*d*x + 10*I*c) + 5*I*a^2*e^(8*I*d*x + 8*I*c) + 10*I*a^2*e
^(6*I*d*x + 6*I*c) + 10*I*a^2*e^(4*I*d*x + 4*I*c) + 5*I*a^2*e^(2*I*d*x + 2*I*c) + I*a^2)*log(e^(2*I*d*x + 2*I*
c) + 1))/(d*e^(10*I*d*x + 10*I*c) + 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) + 10*d*e^(4*I*d*x + 4*I
*c) + 5*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (100) = 200\).
time = 0.32, size = 219, normalized size = 1.96 \begin {gather*} - \frac {2 i a^{2} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {- 270 i a^{2} e^{8 i c} e^{8 i d x} - 600 i a^{2} e^{6 i c} e^{6 i d x} - 740 i a^{2} e^{4 i c} e^{4 i d x} - 400 i a^{2} e^{2 i c} e^{2 i d x} - 86 i a^{2}}{15 d e^{10 i c} e^{10 i d x} + 75 d e^{8 i c} e^{8 i d x} + 150 d e^{6 i c} e^{6 i d x} + 150 d e^{4 i c} e^{4 i d x} + 75 d e^{2 i c} e^{2 i d x} + 15 d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**4*(a+I*a*tan(d*x+c))**2,x)

[Out]

-2*I*a**2*log(exp(2*I*d*x) + exp(-2*I*c))/d + (-270*I*a**2*exp(8*I*c)*exp(8*I*d*x) - 600*I*a**2*exp(6*I*c)*exp
(6*I*d*x) - 740*I*a**2*exp(4*I*c)*exp(4*I*d*x) - 400*I*a**2*exp(2*I*c)*exp(2*I*d*x) - 86*I*a**2)/(15*d*exp(10*
I*c)*exp(10*I*d*x) + 75*d*exp(8*I*c)*exp(8*I*d*x) + 150*d*exp(6*I*c)*exp(6*I*d*x) + 150*d*exp(4*I*c)*exp(4*I*d
*x) + 75*d*exp(2*I*c)*exp(2*I*d*x) + 15*d)

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 274 vs. \(2 (100) = 200\).
time = 1.14, size = 274, normalized size = 2.45 \begin {gather*} -\frac {2 \, {\left (15 i \, a^{2} e^{\left (10 i \, d x + 10 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 75 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 150 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 150 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 75 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 135 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + 300 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 370 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 200 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 15 i \, a^{2} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 43 i \, a^{2}\right )}}{15 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-2/15*(15*I*a^2*e^(10*I*d*x + 10*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 75*I*a^2*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d
*x + 2*I*c) + 1) + 150*I*a^2*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 150*I*a^2*e^(4*I*d*x + 4*I*c)*
log(e^(2*I*d*x + 2*I*c) + 1) + 75*I*a^2*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 135*I*a^2*e^(8*I*d*
x + 8*I*c) + 300*I*a^2*e^(6*I*d*x + 6*I*c) + 370*I*a^2*e^(4*I*d*x + 4*I*c) + 200*I*a^2*e^(2*I*d*x + 2*I*c) + 1
5*I*a^2*log(e^(2*I*d*x + 2*I*c) + 1) + 43*I*a^2)/(d*e^(10*I*d*x + 10*I*c) + 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(
6*I*d*x + 6*I*c) + 10*d*e^(4*I*d*x + 4*I*c) + 5*d*e^(2*I*d*x + 2*I*c) + d)

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Mupad [B]
time = 3.70, size = 86, normalized size = 0.77 \begin {gather*} \frac {\frac {2\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3}-2\,a^2\,\mathrm {tan}\left (c+d\,x\right )-\frac {a^2\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5}+a^2\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,2{}\mathrm {i}-a^2\,{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+\frac {a^2\,{\mathrm {tan}\left (c+d\,x\right )}^4\,1{}\mathrm {i}}{2}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^4*(a + a*tan(c + d*x)*1i)^2,x)

[Out]

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

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