3.5.81 \(\int \frac {\tan ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx\) [481]

Optimal. Leaf size=129 \[ -\frac {a \left (a^2-3 b^2\right ) x}{\left (a^2+b^2\right )^3}-\frac {b \left (3 a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^3 d}-\frac {a^2}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {2 a b}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))} \]

[Out]

-a*(a^2-3*b^2)*x/(a^2+b^2)^3-b*(3*a^2-b^2)*ln(a*cos(d*x+c)+b*sin(d*x+c))/(a^2+b^2)^3/d-1/2*a^2/b/(a^2+b^2)/d/(
a+b*tan(d*x+c))^2+2*a*b/(a^2+b^2)^2/d/(a+b*tan(d*x+c))

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Rubi [A]
time = 0.13, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3623, 3610, 3612, 3611} \begin {gather*} -\frac {a^2}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {2 a b}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac {b \left (3 a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}-\frac {a x \left (a^2-3 b^2\right )}{\left (a^2+b^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^2/(a + b*Tan[c + d*x])^3,x]

[Out]

-((a*(a^2 - 3*b^2)*x)/(a^2 + b^2)^3) - (b*(3*a^2 - b^2)*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/((a^2 + b^2)^3*d
) - a^2/(2*b*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + (2*a*b)/((a^2 + b^2)^2*d*(a + b*Tan[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 3611

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c/(b*f))
*Log[RemoveContent[a*Cos[e + f*x] + b*Sin[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] && EqQ[a*c + b*d, 0]

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 3623

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

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 4.04, size = 200, normalized size = 1.55 \begin {gather*} -\frac {-\frac {b^2 \tan ^3(c+d x)}{(a+b \tan (c+d x))^2}+\frac {b \tan ^2(c+d x)}{a+b \tan (c+d x)}+a \left (\frac {(i a+b)^3 \log (i-\tan (c+d x))}{\left (a^2+b^2\right )^2}+\frac {i (a+i b) \log (i+\tan (c+d x))}{(a-i b)^2}-\frac {2 b \left (-3 a^2+b^2\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^2}+\frac {2 a (a-b) (a+b)}{b \left (a^2+b^2\right ) (a+b \tan (c+d x))}\right )}{2 a \left (a^2+b^2\right ) d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^2/(a + b*Tan[c + d*x])^3,x]

[Out]

-1/2*(-((b^2*Tan[c + d*x]^3)/(a + b*Tan[c + d*x])^2) + (b*Tan[c + d*x]^2)/(a + b*Tan[c + d*x]) + a*(((I*a + b)
^3*Log[I - Tan[c + d*x]])/(a^2 + b^2)^2 + (I*(a + I*b)*Log[I + Tan[c + d*x]])/(a - I*b)^2 - (2*b*(-3*a^2 + b^2
)*Log[a + b*Tan[c + d*x]])/(a^2 + b^2)^2 + (2*a*(a - b)*(a + b))/(b*(a^2 + b^2)*(a + b*Tan[c + d*x]))))/(a*(a^
2 + b^2)*d)

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Maple [A]
time = 0.15, size = 150, normalized size = 1.16

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (3 a^{2} b -b^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-a^{3}+3 b^{2} a \right ) \arctan \left (\tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{3}}-\frac {a^{2}}{2 \left (a^{2}+b^{2}\right ) b \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {2 a b}{\left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}-\frac {b \left (3 a^{2}-b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{3}}}{d}\)	\(150\)
default	\(\frac {\frac {\frac {\left (3 a^{2} b -b^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-a^{3}+3 b^{2} a \right ) \arctan \left (\tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{3}}-\frac {a^{2}}{2 \left (a^{2}+b^{2}\right ) b \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {2 a b}{\left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}-\frac {b \left (3 a^{2}-b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{3}}}{d}\)	\(150\)
norman	\(\frac {\frac {a \left (-a^{3}+b^{2} a \right )}{2 b \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d}-\frac {\left (a^{2}-3 b^{2}\right ) a^{3} x}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (a^{2}+b^{2}\right )}-\frac {b^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d}-\frac {2 b \left (a^{2}-3 b^{2}\right ) a^{2} x \tan \left (d x +c \right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (a^{2}+b^{2}\right )}-\frac {b^{2} \left (a^{2}-3 b^{2}\right ) a x \left (\tan ^{2}\left (d x +c \right )\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (a^{2}+b^{2}\right )}}{\left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {b \left (3 a^{2}-b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right )}-\frac {b \left (3 a^{2}-b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{d \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right )}\)	\(331\)
risch	\(\frac {x}{3 i b \,a^{2}-i b^{3}-a^{3}+3 b^{2} a}+\frac {6 i a^{2} b x}{a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}}-\frac {2 i b^{3} x}{a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}}+\frac {6 i a^{2} b c}{\left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right ) d}-\frac {2 i b^{3} c}{d \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right )}+\frac {2 i a \left (-i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-i a^{3}+2 i a \,b^{2}+a^{2} b -2 b^{3}\right )}{\left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right )^{2} \left (-i a +b \right )^{2} d \left (i a +b \right )^{3}}-\frac {3 a^{2} b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{\left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right ) d}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{d \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right )}\)	\(417\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^2/(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 256 vs. \(2 (127) = 254\).
time = 0.53, size = 256, normalized size = 1.98 \begin {gather*} -\frac {\frac {2 \, {\left (a^{3} - 3 \, a b^{2}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {2 \, {\left (3 \, a^{2} b - b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {4 \, a b^{3} \tan \left (d x + c\right ) - a^{4} + 3 \, a^{2} b^{2}}{a^{6} b + 2 \, a^{4} b^{3} + a^{2} b^{5} + {\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} \tan \left (d x + c\right )}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2/(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (127) = 254\).
time = 1.08, size = 326, normalized size = 2.53 \begin {gather*} -\frac {3 \, a^{4} b - 3 \, a^{2} b^{3} + 2 \, {\left (a^{5} - 3 \, a^{3} b^{2}\right )} d x - {\left (a^{4} b - 5 \, a^{2} b^{3} - 2 \, {\left (a^{3} b^{2} - 3 \, a b^{4}\right )} d x\right )} \tan \left (d x + c\right )^{2} + {\left (3 \, a^{4} b - a^{2} b^{3} + {\left (3 \, a^{2} b^{3} - b^{5}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{3} b^{2} - a b^{4}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \, {\left (a^{5} - 3 \, a^{3} b^{2} + 2 \, a b^{4} - 2 \, {\left (a^{4} b - 3 \, a^{2} b^{3}\right )} d x\right )} \tan \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} d \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} d \tan \left (d x + c\right ) + {\left (a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6}\right )} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2/(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/2*(3*a^4*b - 3*a^2*b^3 + 2*(a^5 - 3*a^3*b^2)*d*x - (a^4*b - 5*a^2*b^3 - 2*(a^3*b^2 - 3*a*b^4)*d*x)*tan(d*x
+ c)^2 + (3*a^4*b - a^2*b^3 + (3*a^2*b^3 - b^5)*tan(d*x + c)^2 + 2*(3*a^3*b^2 - a*b^4)*tan(d*x + c))*log((b^2*
tan(d*x + c)^2 + 2*a*b*tan(d*x + c) + a^2)/(tan(d*x + c)^2 + 1)) - 2*(a^5 - 3*a^3*b^2 + 2*a*b^4 - 2*(a^4*b - 3
*a^2*b^3)*d*x)*tan(d*x + c))/((a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*d*tan(d*x + c)^2 + 2*(a^7*b + 3*a^5*b^3
+ 3*a^3*b^5 + a*b^7)*d*tan(d*x + c) + (a^8 + 3*a^6*b^2 + 3*a^4*b^4 + a^2*b^6)*d)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: AttributeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**2/(a+b*tan(d*x+c))**3,x)

[Out]

Exception raised: AttributeError >> 'NoneType' object has no attribute 'primitive'

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 263 vs. \(2 (127) = 254\).
time = 0.81, size = 263, normalized size = 2.04 \begin {gather*} -\frac {\frac {2 \, {\left (a^{3} - 3 \, a b^{2}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {2 \, {\left (3 \, a^{2} b^{2} - b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}} - \frac {9 \, a^{2} b^{4} \tan \left (d x + c\right )^{2} - 3 \, b^{6} \tan \left (d x + c\right )^{2} + 22 \, a^{3} b^{3} \tan \left (d x + c\right ) - 2 \, a b^{5} \tan \left (d x + c\right ) - a^{6} + 11 \, a^{4} b^{2}}{{\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2/(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*(2*(a^3 - 3*a*b^2)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - (3*a^2*b - b^3)*log(tan(d*x + c)^2 + 1
)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + 2*(3*a^2*b^2 - b^4)*log(abs(b*tan(d*x + c) + a))/(a^6*b + 3*a^4*b^3 +
3*a^2*b^5 + b^7) - (9*a^2*b^4*tan(d*x + c)^2 - 3*b^6*tan(d*x + c)^2 + 22*a^3*b^3*tan(d*x + c) - 2*a*b^5*tan(d*
x + c) - a^6 + 11*a^4*b^2)/((a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*(b*tan(d*x + c) + a)^2))/d

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Mupad [B]
time = 4.09, size = 224, normalized size = 1.74 \begin {gather*} -\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{2\,d\,\left (-a^3\,1{}\mathrm {i}-3\,a^2\,b+a\,b^2\,3{}\mathrm {i}+b^3\right )}-\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (\frac {3\,b}{{\left (a^2+b^2\right )}^2}-\frac {4\,b^3}{{\left (a^2+b^2\right )}^3}\right )}{d}-\frac {\frac {a^4-3\,a^2\,b^2}{2\,b\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {2\,a\,b^2\,\mathrm {tan}\left (c+d\,x\right )}{a^4+2\,a^2\,b^2+b^4}}{d\,\left (a^2+2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )+b^2\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d\,\left (-a^3-a^2\,b\,3{}\mathrm {i}+3\,a\,b^2+b^3\,1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^2/(a + b*tan(c + d*x))^3,x)

[Out]

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

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