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

Optimal. Leaf size=108 \[ \frac {4 i a^4 \tan (c+d x)}{d}-\frac {8 a^4 \log (\cos (c+d x))}{d}-8 i a^4 x+\frac {\left (a^2+i a^2 \tan (c+d x)\right )^2}{d}+\frac {a (a+i a \tan (c+d x))^3}{3 d}+\frac {(a+i a \tan (c+d x))^4}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3527, 3478, 3477, 3475} \[ \frac {4 i a^4 \tan (c+d x)}{d}+\frac {\left (a^2+i a^2 \tan (c+d x)\right )^2}{d}-\frac {8 a^4 \log (\cos (c+d x))}{d}-8 i a^4 x+\frac {a (a+i a \tan (c+d x))^3}{3 d}+\frac {(a+i a \tan (c+d x))^4}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*I)*a^4*x - (8*a^4*Log[Cos[c + d*x]])/d + ((4*I)*a^4*Tan[c + d*x])/d + (a*(a + I*a*Tan[c + d*x])^3)/(3*d) +
(a + I*a*Tan[c + d*x])^4/(4*d) + (a^2 + I*a^2*Tan[c + d*x])^2/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 3477

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(a^2 - b^2)*x, x] + (Dist[2*a*b, Int[Tan[c + d
*x], x], x] + Simp[(b^2*Tan[c + d*x])/d, x]) /; FreeQ[{a, b, c, d}, x]

Rule 3478

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Dist[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] && G
tQ[n, 1]

Rule 3527

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] + Dist[(b*c + a*d)/b, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] &&  !LtQ[m, 0]

Rubi steps

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

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Mathematica [B]  time = 1.26, size = 231, normalized size = 2.14 \[ -\frac {i a^4 \sec (c) \sec ^4(c+d x) \left (-38 \sin (c+2 d x)+18 \sin (3 c+2 d x)-14 \sin (3 c+4 d x)+24 d x \cos (3 c+2 d x)-12 i \cos (3 c+2 d x)+6 d x \cos (3 c+4 d x)+6 d x \cos (5 c+4 d x)-12 i \cos (3 c+2 d x) \log \left (\cos ^2(c+d x)\right )+12 \cos (c+2 d x) \left (-i \log \left (\cos ^2(c+d x)\right )+2 d x-i\right )+3 \cos (c) \left (-6 i \log \left (\cos ^2(c+d x)\right )+12 d x-7 i\right )-3 i \cos (3 c+4 d x) \log \left (\cos ^2(c+d x)\right )-3 i \cos (5 c+4 d x) \log \left (\cos ^2(c+d x)\right )+42 \sin (c)\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/12*I)*a^4*Sec[c]*Sec[c + d*x]^4*((-12*I)*Cos[3*c + 2*d*x] + 24*d*x*Cos[3*c + 2*d*x] + 6*d*x*Cos[3*c + 4*d
*x] + 6*d*x*Cos[5*c + 4*d*x] + 12*Cos[c + 2*d*x]*(-I + 2*d*x - I*Log[Cos[c + d*x]^2]) + 3*Cos[c]*(-7*I + 12*d*
x - (6*I)*Log[Cos[c + d*x]^2]) - (12*I)*Cos[3*c + 2*d*x]*Log[Cos[c + d*x]^2] - (3*I)*Cos[3*c + 4*d*x]*Log[Cos[
c + d*x]^2] - (3*I)*Cos[5*c + 4*d*x]*Log[Cos[c + d*x]^2] + 42*Sin[c] - 38*Sin[c + 2*d*x] + 18*Sin[3*c + 2*d*x]
- 14*Sin[3*c + 4*d*x]))/d

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fricas [A]  time = 0.42, size = 174, normalized size = 1.61 \[ -\frac {4 \, {\left (30 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 63 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 50 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 14 \, a^{4} + 6 \, {\left (a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3*(30*a^4*e^(6*I*d*x + 6*I*c) + 63*a^4*e^(4*I*d*x + 4*I*c) + 50*a^4*e^(2*I*d*x + 2*I*c) + 14*a^4 + 6*(a^4*e
^(8*I*d*x + 8*I*c) + 4*a^4*e^(6*I*d*x + 6*I*c) + 6*a^4*e^(4*I*d*x + 4*I*c) + 4*a^4*e^(2*I*d*x + 2*I*c) + a^4)*
log(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d
*e^(2*I*d*x + 2*I*c) + d)

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giac [B]  time = 1.33, size = 222, normalized size = 2.06 \[ -\frac {4 \, {\left (6 \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 24 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 36 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 24 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 30 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 63 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 50 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 6 \, a^{4} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 14 \, a^{4}\right )}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3*(6*a^4*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 24*a^4*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I
*c) + 1) + 36*a^4*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 24*a^4*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x
+ 2*I*c) + 1) + 30*a^4*e^(6*I*d*x + 6*I*c) + 63*a^4*e^(4*I*d*x + 4*I*c) + 50*a^4*e^(2*I*d*x + 2*I*c) + 6*a^4*
log(e^(2*I*d*x + 2*I*c) + 1) + 14*a^4)/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I
*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)

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maple [A]  time = 0.02, size = 101, normalized size = 0.94 \[ \frac {8 i a^{4} \tan \left (d x +c \right )}{d}+\frac {a^{4} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {4 i a^{4} \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {7 a^{4} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {4 a^{4} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {8 i a^{4} \arctan \left (\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*I/d*a^4*tan(d*x+c)+1/4*a^4*tan(d*x+c)^4/d-4/3*I/d*a^4*tan(d*x+c)^3-7/2*a^4*tan(d*x+c)^2/d+4/d*a^4*ln(1+tan(d
*x+c)^2)-8*I/d*a^4*arctan(tan(d*x+c))

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maxima [A]  time = 0.69, size = 82, normalized size = 0.76 \[ \frac {3 \, a^{4} \tan \left (d x + c\right )^{4} - 16 i \, a^{4} \tan \left (d x + c\right )^{3} - 42 \, a^{4} \tan \left (d x + c\right )^{2} - 96 i \, {\left (d x + c\right )} a^{4} + 48 \, a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 96 i \, a^{4} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a^4*tan(d*x + c)^4 - 16*I*a^4*tan(d*x + c)^3 - 42*a^4*tan(d*x + c)^2 - 96*I*(d*x + c)*a^4 + 48*a^4*log
(tan(d*x + c)^2 + 1) + 96*I*a^4*tan(d*x + c))/d

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mupad [B]  time = 3.72, size = 72, normalized size = 0.67 \[ \frac {8\,a^4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )-\frac {7\,a^4\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}+\frac {a^4\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4}+a^4\,\mathrm {tan}\left (c+d\,x\right )\,8{}\mathrm {i}-\frac {a^4\,{\mathrm {tan}\left (c+d\,x\right )}^3\,4{}\mathrm {i}}{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8*a^4*log(tan(c + d*x) + 1i) + a^4*tan(c + d*x)*8i - (7*a^4*tan(c + d*x)^2)/2 - (a^4*tan(c + d*x)^3*4i)/3 + (
a^4*tan(c + d*x)^4)/4)/d

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sympy [B]  time = 0.88, size = 185, normalized size = 1.71 \[ - \frac {8 a^{4} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {- 120 i a^{4} e^{6 i c} e^{6 i d x} - 252 i a^{4} e^{4 i c} e^{4 i d x} - 200 i a^{4} e^{2 i c} e^{2 i d x} - 56 i a^{4}}{3 i d e^{8 i c} e^{8 i d x} + 12 i d e^{6 i c} e^{6 i d x} + 18 i d e^{4 i c} e^{4 i d x} + 12 i d e^{2 i c} e^{2 i d x} + 3 i d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*a**4*log(exp(2*I*d*x) + exp(-2*I*c))/d + (-120*I*a**4*exp(6*I*c)*exp(6*I*d*x) - 252*I*a**4*exp(4*I*c)*exp(4
*I*d*x) - 200*I*a**4*exp(2*I*c)*exp(2*I*d*x) - 56*I*a**4)/(3*I*d*exp(8*I*c)*exp(8*I*d*x) + 12*I*d*exp(6*I*c)*e
xp(6*I*d*x) + 18*I*d*exp(4*I*c)*exp(4*I*d*x) + 12*I*d*exp(2*I*c)*exp(2*I*d*x) + 3*I*d)

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