\(\int \cot ^5(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx\) [239]

Optimal result

Integrand size = 29, antiderivative size = 108 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=(A b+a B) x+\frac {(A b+a B) \cot (c+d x)}{d}+\frac {(a A-b B) \cot ^2(c+d x)}{2 d}-\frac {(A b+a B) \cot ^3(c+d x)}{3 d}-\frac {a A \cot ^4(c+d x)}{4 d}+\frac {(a A-b B) \log (\sin (c+d x))}{d} \]

[Out]

(A*b+B*a)*x+(A*b+B*a)*cot(d*x+c)/d+1/2*(A*a-B*b)*cot(d*x+c)^2/d-1/3*(A*b+B*a)*cot(d*x+c)^3/d-1/4*a*A*cot(d*x+c
)^4/d+(A*a-B*b)*ln(sin(d*x+c))/d

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {3672, 3610, 3612, 3556} \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=-\frac {(a B+A b) \cot ^3(c+d x)}{3 d}+\frac {(a A-b B) \cot ^2(c+d x)}{2 d}+\frac {(a B+A b) \cot (c+d x)}{d}+\frac {(a A-b B) \log (\sin (c+d x))}{d}+x (a B+A b)-\frac {a A \cot ^4(c+d x)}{4 d} \]

[In]

Int[Cot[c + d*x]^5*(a + b*Tan[c + d*x])*(A + B*Tan[c + d*x]),x]

[Out]

(A*b + a*B)*x + ((A*b + a*B)*Cot[c + d*x])/d + ((a*A - b*B)*Cot[c + d*x]^2)/(2*d) - ((A*b + a*B)*Cot[c + d*x]^
3)/(3*d) - (a*A*Cot[c + d*x]^4)/(4*d) + ((a*A - b*B)*Log[Sin[c + d*x]])/d

Rule 3556

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

Rule 3610

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 3612

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 3672

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

Rubi steps \begin{align*} \text {integral}& = -\frac {a A \cot ^4(c+d x)}{4 d}+\int \cot ^4(c+d x) (A b+a B-(a A-b B) \tan (c+d x)) \, dx \\ & = -\frac {(A b+a B) \cot ^3(c+d x)}{3 d}-\frac {a A \cot ^4(c+d x)}{4 d}+\int \cot ^3(c+d x) (-a A+b B-(A b+a B) \tan (c+d x)) \, dx \\ & = \frac {(a A-b B) \cot ^2(c+d x)}{2 d}-\frac {(A b+a B) \cot ^3(c+d x)}{3 d}-\frac {a A \cot ^4(c+d x)}{4 d}+\int \cot ^2(c+d x) (-A b-a B+(a A-b B) \tan (c+d x)) \, dx \\ & = \frac {(A b+a B) \cot (c+d x)}{d}+\frac {(a A-b B) \cot ^2(c+d x)}{2 d}-\frac {(A b+a B) \cot ^3(c+d x)}{3 d}-\frac {a A \cot ^4(c+d x)}{4 d}+\int \cot (c+d x) (a A-b B+(A b+a B) \tan (c+d x)) \, dx \\ & = (A b+a B) x+\frac {(A b+a B) \cot (c+d x)}{d}+\frac {(a A-b B) \cot ^2(c+d x)}{2 d}-\frac {(A b+a B) \cot ^3(c+d x)}{3 d}-\frac {a A \cot ^4(c+d x)}{4 d}+(a A-b B) \int \cot (c+d x) \, dx \\ & = (A b+a B) x+\frac {(A b+a B) \cot (c+d x)}{d}+\frac {(a A-b B) \cot ^2(c+d x)}{2 d}-\frac {(A b+a B) \cot ^3(c+d x)}{3 d}-\frac {a A \cot ^4(c+d x)}{4 d}+\frac {(a A-b B) \log (\sin (c+d x))}{d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 1.25 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.93 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=-\frac {4 (A b+a B) \cot ^3(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\tan ^2(c+d x)\right )+3 \left ((-2 a A+2 b B) \cot ^2(c+d x)+a A \cot ^4(c+d x)-4 (a A-b B) (\log (\cos (c+d x))+\log (\tan (c+d x)))\right )}{12 d} \]

[In]

Integrate[Cot[c + d*x]^5*(a + b*Tan[c + d*x])*(A + B*Tan[c + d*x]),x]

[Out]

-1/12*(4*(A*b + a*B)*Cot[c + d*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, -Tan[c + d*x]^2] + 3*((-2*a*A + 2*b*B)*Co
t[c + d*x]^2 + a*A*Cot[c + d*x]^4 - 4*(a*A - b*B)*(Log[Cos[c + d*x]] + Log[Tan[c + d*x]])))/d

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.15

method	result	size
derivativedivides	\(\frac {\frac {\left (-a A +B b \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (A b +B a \right ) \arctan \left (\tan \left (d x +c \right )\right )-\frac {-A b -B a}{\tan \left (d x +c \right )}-\frac {A b +B a}{3 \tan \left (d x +c \right )^{3}}-\frac {-a A +B b}{2 \tan \left (d x +c \right )^{2}}+\left (a A -B b \right ) \ln \left (\tan \left (d x +c \right )\right )-\frac {a A}{4 \tan \left (d x +c \right )^{4}}}{d}\)	\(124\)
default	\(\frac {\frac {\left (-a A +B b \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (A b +B a \right ) \arctan \left (\tan \left (d x +c \right )\right )-\frac {-A b -B a}{\tan \left (d x +c \right )}-\frac {A b +B a}{3 \tan \left (d x +c \right )^{3}}-\frac {-a A +B b}{2 \tan \left (d x +c \right )^{2}}+\left (a A -B b \right ) \ln \left (\tan \left (d x +c \right )\right )-\frac {a A}{4 \tan \left (d x +c \right )^{4}}}{d}\)	\(124\)
norman	\(\frac {\frac {\left (A b +B a \right ) \left (\tan ^{3}\left (d x +c \right )\right )}{d}+\left (A b +B a \right ) x \left (\tan ^{4}\left (d x +c \right )\right )-\frac {a A}{4 d}-\frac {\left (A b +B a \right ) \tan \left (d x +c \right )}{3 d}+\frac {\left (a A -B b \right ) \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}}{\tan \left (d x +c \right )^{4}}+\frac {\left (a A -B b \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {\left (a A -B b \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(137\)
parallelrisch	\(\frac {-3 A \left (\cot ^{4}\left (d x +c \right )\right ) a -4 A b \left (\cot ^{3}\left (d x +c \right )\right )-4 B a \left (\cot ^{3}\left (d x +c \right )\right )+6 A \left (\cot ^{2}\left (d x +c \right )\right ) a +12 A b d x -6 B b \left (\cot ^{2}\left (d x +c \right )\right )+12 B x a d +12 A \cot \left (d x +c \right ) b +12 a A \ln \left (\tan \left (d x +c \right )\right )-6 A \ln \left (\sec ^{2}\left (d x +c \right )\right ) a +12 B \cot \left (d x +c \right ) a -12 B \ln \left (\tan \left (d x +c \right )\right ) b +6 B \ln \left (\sec ^{2}\left (d x +c \right )\right ) b}{12 d}\)	\(147\)
risch	\(A b x +B a x -i A a x +i B b x -\frac {2 i a A c}{d}+\frac {2 i B b c}{d}-\frac {2 \left (-6 i A b \,{\mathrm e}^{6 i \left (d x +c \right )}-6 i B a \,{\mathrm e}^{6 i \left (d x +c \right )}+6 A a \,{\mathrm e}^{6 i \left (d x +c \right )}-3 B b \,{\mathrm e}^{6 i \left (d x +c \right )}+12 i A b \,{\mathrm e}^{4 i \left (d x +c \right )}+12 i B a \,{\mathrm e}^{4 i \left (d x +c \right )}-6 A a \,{\mathrm e}^{4 i \left (d x +c \right )}+6 B b \,{\mathrm e}^{4 i \left (d x +c \right )}-10 i A b \,{\mathrm e}^{2 i \left (d x +c \right )}-10 i B a \,{\mathrm e}^{2 i \left (d x +c \right )}+6 A a \,{\mathrm e}^{2 i \left (d x +c \right )}-3 B b \,{\mathrm e}^{2 i \left (d x +c \right )}+4 i A b +4 i a B \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {a A \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B b}{d}\)	\(268\)

[In]

int(cot(d*x+c)^5*(a+b*tan(d*x+c))*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(1/2*(-A*a+B*b)*ln(1+tan(d*x+c)^2)+(A*b+B*a)*arctan(tan(d*x+c))-(-A*b-B*a)/tan(d*x+c)-1/3*(A*b+B*a)/tan(d*
x+c)^3-1/2*(-A*a+B*b)/tan(d*x+c)^2+(A*a-B*b)*ln(tan(d*x+c))-1/4*a*A/tan(d*x+c)^4)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.28 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {6 \, {\left (A a - B b\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{4} + 3 \, {\left (4 \, {\left (B a + A b\right )} d x + 3 \, A a - 2 \, B b\right )} \tan \left (d x + c\right )^{4} + 12 \, {\left (B a + A b\right )} \tan \left (d x + c\right )^{3} + 6 \, {\left (A a - B b\right )} \tan \left (d x + c\right )^{2} - 3 \, A a - 4 \, {\left (B a + A b\right )} \tan \left (d x + c\right )}{12 \, d \tan \left (d x + c\right )^{4}} \]

[In]

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/12*(6*(A*a - B*b)*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^4 + 3*(4*(B*a + A*b)*d*x + 3*A*a - 2
*B*b)*tan(d*x + c)^4 + 12*(B*a + A*b)*tan(d*x + c)^3 + 6*(A*a - B*b)*tan(d*x + c)^2 - 3*A*a - 4*(B*a + A*b)*ta
n(d*x + c))/(d*tan(d*x + c)^4)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (95) = 190\).

Time = 1.12 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.94 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\begin {cases} \tilde {\infty } A a x & \text {for}\: c = 0 \wedge d = 0 \\x \left (A + B \tan {\left (c \right )}\right ) \left (a + b \tan {\left (c \right )}\right ) \cot ^{5}{\left (c \right )} & \text {for}\: d = 0 \\\tilde {\infty } A a x & \text {for}\: c = - d x \\- \frac {A a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {A a \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + \frac {A a}{2 d \tan ^{2}{\left (c + d x \right )}} - \frac {A a}{4 d \tan ^{4}{\left (c + d x \right )}} + A b x + \frac {A b}{d \tan {\left (c + d x \right )}} - \frac {A b}{3 d \tan ^{3}{\left (c + d x \right )}} + B a x + \frac {B a}{d \tan {\left (c + d x \right )}} - \frac {B a}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac {B b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {B b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {B b}{2 d \tan ^{2}{\left (c + d x \right )}} & \text {otherwise} \end {cases} \]

[In]

integrate(cot(d*x+c)**5*(a+b*tan(d*x+c))*(A+B*tan(d*x+c)),x)

[Out]

Piecewise((zoo*A*a*x, Eq(c, 0) & Eq(d, 0)), (x*(A + B*tan(c))*(a + b*tan(c))*cot(c)**5, Eq(d, 0)), (zoo*A*a*x,
Eq(c, -d*x)), (-A*a*log(tan(c + d*x)**2 + 1)/(2*d) + A*a*log(tan(c + d*x))/d + A*a/(2*d*tan(c + d*x)**2) - A*
a/(4*d*tan(c + d*x)**4) + A*b*x + A*b/(d*tan(c + d*x)) - A*b/(3*d*tan(c + d*x)**3) + B*a*x + B*a/(d*tan(c + d*
x)) - B*a/(3*d*tan(c + d*x)**3) + B*b*log(tan(c + d*x)**2 + 1)/(2*d) - B*b*log(tan(c + d*x))/d - B*b/(2*d*tan(
c + d*x)**2), True))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.13 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {12 \, {\left (B a + A b\right )} {\left (d x + c\right )} - 6 \, {\left (A a - B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 12 \, {\left (A a - B b\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {12 \, {\left (B a + A b\right )} \tan \left (d x + c\right )^{3} + 6 \, {\left (A a - B b\right )} \tan \left (d x + c\right )^{2} - 3 \, A a - 4 \, {\left (B a + A b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4}}}{12 \, d} \]

[In]

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(12*(B*a + A*b)*(d*x + c) - 6*(A*a - B*b)*log(tan(d*x + c)^2 + 1) + 12*(A*a - B*b)*log(tan(d*x + c)) + (1
2*(B*a + A*b)*tan(d*x + c)^3 + 6*(A*a - B*b)*tan(d*x + c)^2 - 3*A*a - 4*(B*a + A*b)*tan(d*x + c))/tan(d*x + c)
^4)/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (102) = 204\).

Time = 0.90 (sec) , antiderivative size = 299, normalized size of antiderivative = 2.77 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=-\frac {3 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 8 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 120 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 192 \, {\left (B a + A b\right )} {\left (d x + c\right )} + 192 \, {\left (A a - B b\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 192 \, {\left (A a - B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {400 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 400 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

[In]

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/192*(3*A*a*tan(1/2*d*x + 1/2*c)^4 - 8*B*a*tan(1/2*d*x + 1/2*c)^3 - 8*A*b*tan(1/2*d*x + 1/2*c)^3 - 36*A*a*ta
n(1/2*d*x + 1/2*c)^2 + 24*B*b*tan(1/2*d*x + 1/2*c)^2 + 120*B*a*tan(1/2*d*x + 1/2*c) + 120*A*b*tan(1/2*d*x + 1/
2*c) - 192*(B*a + A*b)*(d*x + c) + 192*(A*a - B*b)*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 192*(A*a - B*b)*log(abs(t
an(1/2*d*x + 1/2*c))) + (400*A*a*tan(1/2*d*x + 1/2*c)^4 - 400*B*b*tan(1/2*d*x + 1/2*c)^4 - 120*B*a*tan(1/2*d*x
+ 1/2*c)^3 - 120*A*b*tan(1/2*d*x + 1/2*c)^3 - 36*A*a*tan(1/2*d*x + 1/2*c)^2 + 24*B*b*tan(1/2*d*x + 1/2*c)^2 +
8*B*a*tan(1/2*d*x + 1/2*c) + 8*A*b*tan(1/2*d*x + 1/2*c) + 3*A*a)/tan(1/2*d*x + 1/2*c)^4)/d

Mupad [B] (verification not implemented)

Time = 7.87 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.34 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (A\,a-B\,b\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^4\,\left (\left (-A\,b-B\,a\right )\,{\mathrm {tan}\left (c+d\,x\right )}^3+\left (\frac {B\,b}{2}-\frac {A\,a}{2}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^2+\left (\frac {A\,b}{3}+\frac {B\,a}{3}\right )\,\mathrm {tan}\left (c+d\,x\right )+\frac {A\,a}{4}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,1{}\mathrm {i}\right )\,\left (a+b\,1{}\mathrm {i}\right )}{2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )\,\left (b+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d} \]

[In]

int(cot(c + d*x)^5*(A + B*tan(c + d*x))*(a + b*tan(c + d*x)),x)

[Out]

(log(tan(c + d*x))*(A*a - B*b))/d - (cot(c + d*x)^4*((A*a)/4 + tan(c + d*x)*((A*b)/3 + (B*a)/3) - tan(c + d*x)
^3*(A*b + B*a) - tan(c + d*x)^2*((A*a)/2 - (B*b)/2)))/d - (log(tan(c + d*x) - 1i)*(A + B*1i)*(a + b*1i))/(2*d)
+ (log(tan(c + d*x) + 1i)*(A - B*1i)*(a*1i + b)*1i)/(2*d)