3.43 \(\int \frac {\cot ^4(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx\)

Optimal. Leaf size=155 \[ -\frac {(5 A+3 i B) \cot ^3(c+d x)}{6 a d}+\frac {(-B+i A) \cot ^2(c+d x)}{a d}+\frac {(5 A+3 i B) \cot (c+d x)}{2 a d}+\frac {2 (-B+i A) \log (\sin (c+d x))}{a d}+\frac {(A+i B) \cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {x (5 A+3 i B)}{2 a} \]

[Out]

1/2*(5*A+3*I*B)*x/a+1/2*(5*A+3*I*B)*cot(d*x+c)/a/d+(I*A-B)*cot(d*x+c)^2/a/d-1/6*(5*A+3*I*B)*cot(d*x+c)^3/a/d+2
*(I*A-B)*ln(sin(d*x+c))/a/d+1/2*(A+I*B)*cot(d*x+c)^3/d/(a+I*a*tan(d*x+c))

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Rubi [A]  time = 0.25, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3596, 3529, 3531, 3475} \[ -\frac {(5 A+3 i B) \cot ^3(c+d x)}{6 a d}+\frac {(-B+i A) \cot ^2(c+d x)}{a d}+\frac {(5 A+3 i B) \cot (c+d x)}{2 a d}+\frac {2 (-B+i A) \log (\sin (c+d x))}{a d}+\frac {(A+i B) \cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {x (5 A+3 i B)}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*A + (3*I)*B)*x)/(2*a) + ((5*A + (3*I)*B)*Cot[c + d*x])/(2*a*d) + ((I*A - B)*Cot[c + d*x]^2)/(a*d) - ((5*A
+ (3*I)*B)*Cot[c + d*x]^3)/(6*a*d) + (2*(I*A - B)*Log[Sin[c + d*x]])/(a*d) + ((A + I*B)*Cot[c + d*x]^3)/(2*d*(
a + I*a*Tan[c + d*x]))

Rule 3475

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

Rule 3529

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

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 3596

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

Rubi steps

\begin {align*} \int \frac {\cot ^4(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx &=\frac {(A+i B) \cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \cot ^4(c+d x) (a (5 A+3 i B)-4 a (i A-B) \tan (c+d x)) \, dx}{2 a^2}\\ &=-\frac {(5 A+3 i B) \cot ^3(c+d x)}{6 a d}+\frac {(A+i B) \cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \cot ^3(c+d x) (-4 a (i A-B)-a (5 A+3 i B) \tan (c+d x)) \, dx}{2 a^2}\\ &=\frac {(i A-B) \cot ^2(c+d x)}{a d}-\frac {(5 A+3 i B) \cot ^3(c+d x)}{6 a d}+\frac {(A+i B) \cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \cot ^2(c+d x) (-a (5 A+3 i B)+4 a (i A-B) \tan (c+d x)) \, dx}{2 a^2}\\ &=\frac {(5 A+3 i B) \cot (c+d x)}{2 a d}+\frac {(i A-B) \cot ^2(c+d x)}{a d}-\frac {(5 A+3 i B) \cot ^3(c+d x)}{6 a d}+\frac {(A+i B) \cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \cot (c+d x) (4 a (i A-B)+a (5 A+3 i B) \tan (c+d x)) \, dx}{2 a^2}\\ &=\frac {(5 A+3 i B) x}{2 a}+\frac {(5 A+3 i B) \cot (c+d x)}{2 a d}+\frac {(i A-B) \cot ^2(c+d x)}{a d}-\frac {(5 A+3 i B) \cot ^3(c+d x)}{6 a d}+\frac {(A+i B) \cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {(2 (i A-B)) \int \cot (c+d x) \, dx}{a}\\ &=\frac {(5 A+3 i B) x}{2 a}+\frac {(5 A+3 i B) \cot (c+d x)}{2 a d}+\frac {(i A-B) \cot ^2(c+d x)}{a d}-\frac {(5 A+3 i B) \cot ^3(c+d x)}{6 a d}+\frac {2 (i A-B) \log (\sin (c+d x))}{a d}+\frac {(A+i B) \cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}\\ \end {align*}

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Mathematica [B]  time = 7.68, size = 1062, normalized size = 6.85 \[ \frac {\csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) (\cos (d x)+i \sin (d x)) \left (\frac {1}{2} i A \cos (c-d x)-\frac {1}{2} i A \cos (c+d x)-\frac {1}{2} A \sin (c-d x)+\frac {1}{2} A \sin (c+d x)\right ) (A+B \tan (c+d x)) \csc ^3(c+d x)}{6 d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac {\csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (-\frac {\cos (c)}{12}-\frac {1}{12} i \sin (c)\right ) (2 A \cos (c)-3 i A \sin (c)+3 B \sin (c)) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x)) \csc ^2(c+d x)}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac {\csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) (\cos (d x)+i \sin (d x)) \left (-\frac {7}{2} i A \cos (c-d x)+\frac {3}{2} B \cos (c-d x)+\frac {7}{2} i A \cos (c+d x)-\frac {3}{2} B \cos (c+d x)+\frac {7}{2} A \sin (c-d x)+\frac {3}{2} i B \sin (c-d x)-\frac {7}{2} A \sin (c+d x)-\frac {3}{2} i B \sin (c+d x)\right ) (A+B \tan (c+d x)) \csc (c+d x)}{6 d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac {\left (A \cos \left (\frac {c}{2}\right )+i B \cos \left (\frac {c}{2}\right )+i A \sin \left (\frac {c}{2}\right )-B \sin \left (\frac {c}{2}\right )\right ) \left (2 \tan ^{-1}(\tan (d x)) \cos \left (\frac {c}{2}\right )+2 i \tan ^{-1}(\tan (d x)) \sin \left (\frac {c}{2}\right )\right ) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x))}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac {\left (A \cos \left (\frac {c}{2}\right )+i B \cos \left (\frac {c}{2}\right )+i A \sin \left (\frac {c}{2}\right )-B \sin \left (\frac {c}{2}\right )\right ) \left (i \cos \left (\frac {c}{2}\right ) \log \left (\sin ^2(c+d x)\right )-\log \left (\sin ^2(c+d x)\right ) \sin \left (\frac {c}{2}\right )\right ) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x))}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac {x (-2 i A \csc (c)+2 B \csc (c)+i (A+i B) \cot (c) (2 \cos (c)+2 i \sin (c))) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x))}{(A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac {(A+i B) \cos (2 d x) \left (\frac {1}{4} i \cos (c)+\frac {\sin (c)}{4}\right ) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x))}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac {(5 A+3 i B) \left (\frac {1}{2} d x \cos (c)+\frac {1}{2} i d x \sin (c)\right ) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x))}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac {(A+i B) \left (\frac {\cos (c)}{4}-\frac {1}{4} i \sin (c)\right ) (\cos (d x)+i \sin (d x)) \sin (2 d x) (A+B \tan (c+d x))}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*Cos[c/2] + I*B*Cos[c/2] + I*A*Sin[c/2] - B*Sin[c/2])*(2*ArcTan[Tan[d*x]]*Cos[c/2] + (2*I)*ArcTan[Tan[d*x]]
*Sin[c/2])*(Cos[d*x] + I*Sin[d*x])*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c +
d*x])) + ((A*Cos[c/2] + I*B*Cos[c/2] + I*A*Sin[c/2] - B*Sin[c/2])*(I*Cos[c/2]*Log[Sin[c + d*x]^2] - Log[Sin[c
+ d*x]^2]*Sin[c/2])*(Cos[d*x] + I*Sin[d*x])*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I
*a*Tan[c + d*x])) + (x*((-2*I)*A*Csc[c] + 2*B*Csc[c] + I*(A + I*B)*Cot[c]*(2*Cos[c] + (2*I)*Sin[c]))*(Cos[d*x]
+ I*Sin[d*x])*(A + B*Tan[c + d*x]))/((A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])) + ((A + I*B)*C
os[2*d*x]*((I/4)*Cos[c] + Sin[c]/4)*(Cos[d*x] + I*Sin[d*x])*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c
+ d*x])*(a + I*a*Tan[c + d*x])) + (Csc[c/2]*Csc[c + d*x]^2*Sec[c/2]*(-1/12*Cos[c] - (I/12)*Sin[c])*(2*A*Cos[c
] - (3*I)*A*Sin[c] + 3*B*Sin[c])*(Cos[d*x] + I*Sin[d*x])*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c +
d*x])*(a + I*a*Tan[c + d*x])) + ((5*A + (3*I)*B)*((d*x*Cos[c])/2 + (I/2)*d*x*Sin[c])*(Cos[d*x] + I*Sin[d*x])*(
A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])) + ((A + I*B)*(Cos[c]/4 - (I/
4)*Sin[c])*(Cos[d*x] + I*Sin[d*x])*Sin[2*d*x]*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a +
I*a*Tan[c + d*x])) + (Csc[c/2]*Csc[c + d*x]^3*Sec[c/2]*(Cos[d*x] + I*Sin[d*x])*((I/2)*A*Cos[c - d*x] - (I/2)*A
*Cos[c + d*x] - (A*Sin[c - d*x])/2 + (A*Sin[c + d*x])/2)*(A + B*Tan[c + d*x]))/(6*d*(A*Cos[c + d*x] + B*Sin[c
+ d*x])*(a + I*a*Tan[c + d*x])) + (Csc[c/2]*Csc[c + d*x]*Sec[c/2]*(Cos[d*x] + I*Sin[d*x])*(((-7*I)/2)*A*Cos[c
- d*x] + (3*B*Cos[c - d*x])/2 + ((7*I)/2)*A*Cos[c + d*x] - (3*B*Cos[c + d*x])/2 + (7*A*Sin[c - d*x])/2 + ((3*I
)/2)*B*Sin[c - d*x] - (7*A*Sin[c + d*x])/2 - ((3*I)/2)*B*Sin[c + d*x])*(A + B*Tan[c + d*x]))/(6*d*(A*Cos[c + d
*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x]))

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fricas [A]  time = 0.63, size = 251, normalized size = 1.62 \[ \frac {6 \, {\left (9 \, A + 7 i \, B\right )} d x e^{\left (8 i \, d x + 8 i \, c\right )} - {\left (18 \, {\left (9 \, A + 7 i \, B\right )} d x - 51 i \, A + 3 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (18 \, {\left (9 \, A + 7 i \, B\right )} d x - 81 i \, A + 33 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (6 \, {\left (9 \, A + 7 i \, B\right )} d x - 65 i \, A + 33 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left ({\left (24 i \, A - 24 \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + {\left (-72 i \, A + 72 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (72 i \, A - 72 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-24 i \, A + 24 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - 3 i \, A + 3 \, B}{12 \, {\left (a d e^{\left (8 i \, d x + 8 i \, c\right )} - 3 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*(9*A + 7*I*B)*d*x*e^(8*I*d*x + 8*I*c) - (18*(9*A + 7*I*B)*d*x - 51*I*A + 3*B)*e^(6*I*d*x + 6*I*c) + (1
8*(9*A + 7*I*B)*d*x - 81*I*A + 33*B)*e^(4*I*d*x + 4*I*c) - (6*(9*A + 7*I*B)*d*x - 65*I*A + 33*B)*e^(2*I*d*x +
2*I*c) + ((24*I*A - 24*B)*e^(8*I*d*x + 8*I*c) + (-72*I*A + 72*B)*e^(6*I*d*x + 6*I*c) + (72*I*A - 72*B)*e^(4*I*
d*x + 4*I*c) + (-24*I*A + 24*B)*e^(2*I*d*x + 2*I*c))*log(e^(2*I*d*x + 2*I*c) - 1) - 3*I*A + 3*B)/(a*d*e^(8*I*d
*x + 8*I*c) - 3*a*d*e^(6*I*d*x + 6*I*c) + 3*a*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))

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giac [A]  time = 1.30, size = 186, normalized size = 1.20 \[ -\frac {\frac {3 \, {\left (9 i \, A - 7 \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a} + \frac {3 \, {\left (-i \, A - B\right )} \log \left (-i \, \tan \left (d x + c\right ) + 1\right )}{a} + \frac {24 \, {\left (-i \, A + B\right )} \log \left (\tan \left (d x + c\right )\right )}{a} + \frac {3 \, {\left (-9 i \, A \tan \left (d x + c\right ) + 7 \, B \tan \left (d x + c\right ) - 11 \, A - 9 i \, B\right )}}{a {\left (\tan \left (d x + c\right ) - i\right )}} + \frac {2 i \, {\left (22 \, A \tan \left (d x + c\right )^{3} + 22 i \, B \tan \left (d x + c\right )^{3} + 12 i \, A \tan \left (d x + c\right )^{2} - 6 \, B \tan \left (d x + c\right )^{2} - 3 \, A \tan \left (d x + c\right ) - 3 i \, B \tan \left (d x + c\right ) - 2 i \, A\right )}}{a \tan \left (d x + c\right )^{3}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*(9*I*A - 7*B)*log(tan(d*x + c) - I)/a + 3*(-I*A - B)*log(-I*tan(d*x + c) + 1)/a + 24*(-I*A + B)*log(t
an(d*x + c))/a + 3*(-9*I*A*tan(d*x + c) + 7*B*tan(d*x + c) - 11*A - 9*I*B)/(a*(tan(d*x + c) - I)) + 2*I*(22*A*
tan(d*x + c)^3 + 22*I*B*tan(d*x + c)^3 + 12*I*A*tan(d*x + c)^2 - 6*B*tan(d*x + c)^2 - 3*A*tan(d*x + c) - 3*I*B
*tan(d*x + c) - 2*I*A)/(a*tan(d*x + c)^3))/d

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maple [A]  time = 0.66, size = 241, normalized size = 1.55 \[ \frac {B \ln \left (\tan \left (d x +c \right )+i\right )}{4 d a}+\frac {i A \ln \left (\tan \left (d x +c \right )+i\right )}{4 d a}+\frac {i A}{2 a d \tan \left (d x +c \right )^{2}}-\frac {B}{2 a d \tan \left (d x +c \right )^{2}}-\frac {A}{3 a d \tan \left (d x +c \right )^{3}}+\frac {2 i A \ln \left (\tan \left (d x +c \right )\right )}{a d}-\frac {2 B \ln \left (\tan \left (d x +c \right )\right )}{a d}+\frac {i B}{a d \tan \left (d x +c \right )}+\frac {2 A}{a d \tan \left (d x +c \right )}+\frac {A}{2 d a \left (\tan \left (d x +c \right )-i\right )}+\frac {i B}{2 d a \left (\tan \left (d x +c \right )-i\right )}-\frac {9 i \ln \left (\tan \left (d x +c \right )-i\right ) A}{4 d a}+\frac {7 \ln \left (\tan \left (d x +c \right )-i\right ) B}{4 d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/d/a*B*ln(tan(d*x+c)+I)+1/4*I/d/a*A*ln(tan(d*x+c)+I)+1/2*I/a/d/tan(d*x+c)^2*A-1/2/a/d/tan(d*x+c)^2*B-1/3/a/
d*A/tan(d*x+c)^3+2*I/a/d*A*ln(tan(d*x+c))-2/a/d*B*ln(tan(d*x+c))+I/a/d/tan(d*x+c)*B+2/a/d*A/tan(d*x+c)+1/2/d/a
/(tan(d*x+c)-I)*A+1/2*I/d/a/(tan(d*x+c)-I)*B-9/4*I/a/d*ln(tan(d*x+c)-I)*A+7/4/d/a*ln(tan(d*x+c)-I)*B

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 6.55, size = 174, normalized size = 1.12 \[ \frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {3\,A}{2\,a}+\frac {B\,1{}\mathrm {i}}{2\,a}\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (-\frac {3\,B}{2\,a}+\frac {A\,5{}\mathrm {i}}{2\,a}\right )-\frac {A}{3\,a}+\mathrm {tan}\left (c+d\,x\right )\,\left (-\frac {B}{2\,a}+\frac {A\,1{}\mathrm {i}}{6\,a}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4\,1{}\mathrm {i}+{\mathrm {tan}\left (c+d\,x\right )}^3\right )}+\frac {2\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-B+A\,1{}\mathrm {i}\right )}{a\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{4\,a\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-7\,B+A\,9{}\mathrm {i}\right )}{4\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c + d*x)^2*((3*A)/(2*a) + (B*1i)/(2*a)) + tan(c + d*x)^3*((A*5i)/(2*a) - (3*B)/(2*a)) - A/(3*a) + tan(c +
d*x)*((A*1i)/(6*a) - B/(2*a)))/(d*(tan(c + d*x)^3 + tan(c + d*x)^4*1i)) + (2*log(tan(c + d*x))*(A*1i - B))/(a
*d) + (log(tan(c + d*x) + 1i)*(A*1i + B))/(4*a*d) - (log(tan(c + d*x) - 1i)*(A*9i - 7*B))/(4*a*d)

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sympy [A]  time = 1.02, size = 262, normalized size = 1.69 \[ \frac {- 12 i A e^{4 i c} e^{4 i d x} - 14 i A + 6 B + \left (18 i A e^{2 i c} - 6 B e^{2 i c}\right ) e^{2 i d x}}{- 3 a d e^{6 i c} e^{6 i d x} + 9 a d e^{4 i c} e^{4 i d x} - 9 a d e^{2 i c} e^{2 i d x} + 3 a d} + \begin {cases} - \frac {\left (- i A + B\right ) e^{- 2 i c} e^{- 2 i d x}}{4 a d} & \text {for}\: 4 a d e^{2 i c} \neq 0 \\x \left (- \frac {9 A + 7 i B}{2 a} - \frac {i \left (9 i A e^{2 i c} + i A - 7 B e^{2 i c} - B\right ) e^{- 2 i c}}{2 a}\right ) & \text {otherwise} \end {cases} - \frac {x \left (- 9 A - 7 i B\right )}{2 a} + \frac {2 i \left (A + i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-12*I*A*exp(4*I*c)*exp(4*I*d*x) - 14*I*A + 6*B + (18*I*A*exp(2*I*c) - 6*B*exp(2*I*c))*exp(2*I*d*x))/(-3*a*d*e
xp(6*I*c)*exp(6*I*d*x) + 9*a*d*exp(4*I*c)*exp(4*I*d*x) - 9*a*d*exp(2*I*c)*exp(2*I*d*x) + 3*a*d) + Piecewise((-
(-I*A + B)*exp(-2*I*c)*exp(-2*I*d*x)/(4*a*d), Ne(4*a*d*exp(2*I*c), 0)), (x*(-(9*A + 7*I*B)/(2*a) - I*(9*I*A*ex
p(2*I*c) + I*A - 7*B*exp(2*I*c) - B)*exp(-2*I*c)/(2*a)), True)) - x*(-9*A - 7*I*B)/(2*a) + 2*I*(A + I*B)*log(e
xp(2*I*d*x) - exp(-2*I*c))/(a*d)

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