3.13 \(\int \tan ^4(c+d x) (a+i a \tan (c+d x))^2 \, dx\)

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

[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 = 0.14, 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 = {3543, 3528, 3525, 3475} \[ -\frac {a^2 \tan ^5(c+d x)}{5 d}+\frac {i a^2 \tan ^4(c+d x)}{2 d}+\frac {2 a^2 \tan ^3(c+d x)}{3 d}-\frac {i a^2 \tan ^2(c+d x)}{d}-\frac {2 a^2 \tan (c+d x)}{d}-\frac {2 i a^2 \log (\cos (c+d x))}{d}+2 a^2 x \]

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 3475

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

Rule 3525

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 3528

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 3543

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 = 0.30, size = 108, normalized size = 0.96 \[ \frac {2 a^2 \tan ^{-1}(\tan (c+d x))}{d}-\frac {a^2 \tan ^5(c+d x)}{5 d}+\frac {2 a^2 \tan ^3(c+d x)}{3 d}-\frac {2 a^2 \tan (c+d x)}{d}-\frac {i a^2 \left (-\tan ^4(c+d x)+2 \tan ^2(c+d x)+4 \log (\cos (c+d x))\right )}{2 d} \]

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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fricas [B]  time = 0.43, size = 216, normalized size = 1.93 \[ \frac {-270 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 600 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 740 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 400 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 86 i \, a^{2} + {\left (-30 i \, a^{2} e^{\left (10 i \, d x + 10 i \, c\right )} - 150 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 300 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 300 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 150 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 30 i \, a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}{15 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

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]

1/15*(-270*I*a^2*e^(8*I*d*x + 8*I*c) - 600*I*a^2*e^(6*I*d*x + 6*I*c) - 740*I*a^2*e^(4*I*d*x + 4*I*c) - 400*I*a
^2*e^(2*I*d*x + 2*I*c) - 86*I*a^2 + (-30*I*a^2*e^(10*I*d*x + 10*I*c) - 150*I*a^2*e^(8*I*d*x + 8*I*c) - 300*I*a
^2*e^(6*I*d*x + 6*I*c) - 300*I*a^2*e^(4*I*d*x + 4*I*c) - 150*I*a^2*e^(2*I*d*x + 2*I*c) - 30*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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giac [B]  time = 2.94, size = 274, normalized size = 2.45 \[ \frac {-30 i \, a^{2} e^{\left (10 i \, d x + 10 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 150 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 300 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 300 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 150 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 270 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 600 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 740 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 400 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 30 i \, a^{2} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 86 i \, a^{2}}{15 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

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]

1/15*(-30*I*a^2*e^(10*I*d*x + 10*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 150*I*a^2*e^(8*I*d*x + 8*I*c)*log(e^(2*I*
d*x + 2*I*c) + 1) - 300*I*a^2*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 300*I*a^2*e^(4*I*d*x + 4*I*c)
*log(e^(2*I*d*x + 2*I*c) + 1) - 150*I*a^2*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 270*I*a^2*e^(8*I*
d*x + 8*I*c) - 600*I*a^2*e^(6*I*d*x + 6*I*c) - 740*I*a^2*e^(4*I*d*x + 4*I*c) - 400*I*a^2*e^(2*I*d*x + 2*I*c) -
30*I*a^2*log(e^(2*I*d*x + 2*I*c) + 1) - 86*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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maple [A]  time = 0.02, size = 117, normalized size = 1.04 \[ -\frac {2 a^{2} \tan \left (d x +c \right )}{d}-\frac {a^{2} \left (\tan ^{5}\left (d x +c \right )\right )}{5 d}+\frac {i a^{2} \left (\tan ^{4}\left (d x +c \right )\right )}{2 d}+\frac {2 a^{2} \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {i a^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{d}+\frac {i a^{2} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}+\frac {2 a^{2} \arctan \left (\tan \left (d x +c \right )\right )}{d} \]

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)

[Out]

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

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maxima [A]  time = 0.81, size = 95, normalized size = 0.85 \[ -\frac {6 \, a^{2} \tan \left (d x + c\right )^{5} - 15 i \, a^{2} \tan \left (d x + c\right )^{4} - 20 \, a^{2} \tan \left (d x + c\right )^{3} + 30 i \, a^{2} \tan \left (d x + c\right )^{2} - 60 \, {\left (d x + c\right )} a^{2} - 30 i \, a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 60 \, a^{2} \tan \left (d x + c\right )}{30 \, d} \]

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

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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sympy [B]  time = 0.60, size = 219, normalized size = 1.96 \[ - \frac {2 i a^{2} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {270 i a^{2} e^{8 i c} e^{8 i d x} + 600 i a^{2} e^{6 i c} e^{6 i d x} + 740 i a^{2} e^{4 i c} e^{4 i d x} + 400 i a^{2} e^{2 i c} e^{2 i d x} + 86 i a^{2}}{- 15 d e^{10 i c} e^{10 i d x} - 75 d e^{8 i c} e^{8 i d x} - 150 d e^{6 i c} e^{6 i d x} - 150 d e^{4 i c} e^{4 i d x} - 75 d e^{2 i c} e^{2 i d x} - 15 d} \]

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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