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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.17, antiderivative size = 86, normalized size of antiderivative = 1.00, 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 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]

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 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 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]

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 = 1.96, 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)

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 137, normalized size = 1.59 \[ \frac {10 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 8 \, a^{4} + 7 \, {\left (a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + {\left (a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \]

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)

________________________________________________________________________________________

giac [A]  time = 2.54, size = 157, normalized size = 1.83 \[ \frac {14 \, a^{4} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 32 \, a^{4} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) + 14 \, a^{4} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) + 2 \, a^{4} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - \frac {21 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 16 i \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 46 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 16 i \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21 \, a^{4}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

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*(14*a^4*log(tan(1/2*d*x + 1/2*c) + 1) - 32*a^4*log(tan(1/2*d*x + 1/2*c) + I) + 14*a^4*log(tan(1/2*d*x + 1/
2*c) - 1) + 2*a^4*log(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

________________________________________________________________________________________

maple [A]  time = 0.35, size = 79, normalized size = 0.92 \[ \frac {a^{4} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {7 a^{4} \ln \left (\cos \left (d x +c \right )\right )}{d}+8 i a^{4} x -\frac {4 i a^{4} \tan \left (d x +c \right )}{d}+\frac {8 i a^{4} c}{d}+\frac {a^{4} \ln \left (\sin \left (d x +c \right )\right )}{d} \]

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

________________________________________________________________________________________

maxima [A]  time = 0.52, size = 67, normalized size = 0.78 \[ \frac {a^{4} \tan \left (d x + c\right )^{2} + 16 i \, {\left (d x + c\right )} a^{4} - 8 \, a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, a^{4} \log \left (\tan \left (d x + c\right )\right ) - 8 i \, a^{4} \tan \left (d x + c\right )}{2 \, d} \]

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

________________________________________________________________________________________

mupad [B]  time = 3.80, size = 64, normalized size = 0.74 \[ \frac {a^4\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,d}-\frac {8\,a^4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{d}+\frac {a^4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{d}-\frac {a^4\,\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotInvertible} \]

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]

Exception raised: NotInvertible

________________________________________________________________________________________