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

Optimal. Leaf size=86 \[ -\frac{\left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac{3 \left (a^4+i a^4 \tan (c+d x)\right )}{d}+\frac{a^4 \log (\sin (c+d x))}{d}+\frac{7 a^4 \log (\cos (c+d x))}{d}+8 i a^4 x \]

[Out]

(8*I)*a^4*x + (7*a^4*Log[Cos[c + d*x]])/d + (a^4*Log[Sin[c + d*x]])/d - (a^2 + I*a^2*Tan[c + d*x])^2/(2*d) - (
3*(a^4 + I*a^4*Tan[c + d*x]))/d

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Rubi [A]  time = 0.173431, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {3556, 3594, 3589, 3475, 3531} \[ -\frac{\left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac{3 \left (a^4+i a^4 \tan (c+d x)\right )}{d}+\frac{a^4 \log (\sin (c+d x))}{d}+\frac{7 a^4 \log (\cos (c+d x))}{d}+8 i a^4 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*I)*a^4*x + (7*a^4*Log[Cos[c + d*x]])/d + (a^4*Log[Sin[c + d*x]])/d - (a^2 + I*a^2*Tan[c + d*x])^2/(2*d) - (
3*(a^4 + I*a^4*Tan[c + d*x]))/d

Rule 3556

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a
^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3594

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

Rule 3589

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

Rule 3475

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

Rule 3531

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

Rubi steps

\begin{align*} \int \cot (c+d x) (a+i a \tan (c+d x))^4 \, dx &=-\frac{\left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}+\frac{1}{2} a \int \cot (c+d x) (a+i a \tan (c+d x))^2 (2 a+6 i a \tan (c+d x)) \, dx\\ &=-\frac{\left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac{3 \left (a^4+i a^4 \tan (c+d x)\right )}{d}+\frac{1}{2} a \int \cot (c+d x) (a+i a \tan (c+d x)) \left (2 a^2+14 i a^2 \tan (c+d x)\right ) \, dx\\ &=-\frac{\left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac{3 \left (a^4+i a^4 \tan (c+d x)\right )}{d}+\frac{1}{2} a \int \cot (c+d x) \left (2 a^3+16 i a^3 \tan (c+d x)\right ) \, dx-\left (7 a^4\right ) \int \tan (c+d x) \, dx\\ &=8 i a^4 x+\frac{7 a^4 \log (\cos (c+d x))}{d}-\frac{\left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac{3 \left (a^4+i a^4 \tan (c+d x)\right )}{d}+a^4 \int \cot (c+d x) \, dx\\ &=8 i a^4 x+\frac{7 a^4 \log (\cos (c+d x))}{d}+\frac{a^4 \log (\sin (c+d x))}{d}-\frac{\left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac{3 \left (a^4+i a^4 \tan (c+d x)\right )}{d}\\ \end{align*}

Mathematica [A]  time = 1.82937, size = 159, normalized size = 1.85 \[ \frac{a^4 \sec (c) \sec ^2(c+d x) \left (-16 i \sin (c+2 d x)+16 i d x \cos (3 c+2 d x)+7 \cos (3 c+2 d x) \log \left (\cos ^2(c+d x)\right )+\cos (c+2 d x) \left (\log \left (\sin ^2(c+d x)\right )+7 \log \left (\cos ^2(c+d x)\right )+16 i d x\right )+2 \cos (c) \left (\log \left (\sin ^2(c+d x)\right )+7 \log \left (\cos ^2(c+d x)\right )+16 i d x+2\right )+\cos (3 c+2 d x) \log \left (\sin ^2(c+d x)\right )+16 i \sin (c)\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*Sec[c]*Sec[c + d*x]^2*((16*I)*d*x*Cos[3*c + 2*d*x] + 7*Cos[3*c + 2*d*x]*Log[Cos[c + d*x]^2] + Cos[3*c + 2
*d*x]*Log[Sin[c + d*x]^2] + Cos[c + 2*d*x]*((16*I)*d*x + 7*Log[Cos[c + d*x]^2] + Log[Sin[c + d*x]^2]) + 2*Cos[
c]*(2 + (16*I)*d*x + 7*Log[Cos[c + d*x]^2] + Log[Sin[c + d*x]^2]) + (16*I)*Sin[c] - (16*I)*Sin[c + 2*d*x]))/(8
*d)

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Maple [A]  time = 0.051, size = 79, normalized size = 0.9 \begin{align*}{\frac{{a}^{4} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+7\,{\frac{{a}^{4}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+8\,i{a}^{4}x-{\frac{4\,i{a}^{4}\tan \left ( dx+c \right ) }{d}}+{\frac{8\,i{a}^{4}c}{d}}+{\frac{{a}^{4}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.64394, size = 90, normalized size = 1.05 \begin{align*} \frac{a^{4} \tan \left (d x + c\right )^{2} + 16 i \,{\left (d x + c\right )} a^{4} - 8 \, a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, a^{4} \log \left (\tan \left (d x + c\right )\right ) - 8 i \, a^{4} \tan \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.27655, size = 373, normalized size = 4.34 \begin{align*} \frac{10 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 8 \, a^{4} + 7 \,{\left (a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, 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 ) +{\left (a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, 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 )}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(10*a^4*e^(2*I*d*x + 2*I*c) + 8*a^4 + 7*(a^4*e^(4*I*d*x + 4*I*c) + 2*a^4*e^(2*I*d*x + 2*I*c) + a^4)*log(e^(2*I
*d*x + 2*I*c) + 1) + (a^4*e^(4*I*d*x + 4*I*c) + 2*a^4*e^(2*I*d*x + 2*I*c) + a^4)*log(e^(2*I*d*x + 2*I*c) - 1))
/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [A]  time = 2.53156, size = 112, normalized size = 1.3 \begin{align*} \frac{a^{4} \left (\log{\left (e^{2 i d x} - e^{- 2 i c} \right )} + 7 \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}\right )}{d} + \frac{\frac{10 a^{4} e^{- 2 i c} e^{2 i d x}}{d} + \frac{8 a^{4} e^{- 4 i c}}{d}}{e^{4 i d x} + 2 e^{- 2 i c} e^{2 i d x} + e^{- 4 i c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**4*(log(exp(2*I*d*x) - exp(-2*I*c)) + 7*log(exp(2*I*d*x) + exp(-2*I*c)))/d + (10*a**4*exp(-2*I*c)*exp(2*I*d*
x)/d + 8*a**4*exp(-4*I*c)/d)/(exp(4*I*d*x) + 2*exp(-2*I*c)*exp(2*I*d*x) + exp(-4*I*c))

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Giac [A]  time = 1.45249, size = 215, normalized size = 2.5 \begin{align*} -\frac{32 \, a^{4} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) - 14 \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 14 \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - 2 \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + \frac{21 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 16 i \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 46 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 16 i \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 21 \, a^{4}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(32*a^4*log(tan(1/2*d*x + 1/2*c) + I) - 14*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 14*a^4*log(abs(tan(1/
2*d*x + 1/2*c) - 1)) - 2*a^4*log(abs(tan(1/2*d*x + 1/2*c))) + (21*a^4*tan(1/2*d*x + 1/2*c)^4 - 16*I*a^4*tan(1/
2*d*x + 1/2*c)^3 - 46*a^4*tan(1/2*d*x + 1/2*c)^2 + 16*I*a^4*tan(1/2*d*x + 1/2*c) + 21*a^4)/(tan(1/2*d*x + 1/2*
c)^2 - 1)^2)/d