3.2.60 \(\int (a+b \sin (c+d x))^3 \tan ^3(c+d x) \, dx\) [160]

Optimal. Leaf size=150 \[ \frac {(a+b)^2 (2 a+5 b) \log (1-\sin (c+d x))}{4 d}+\frac {(2 a-5 b) (a-b)^2 \log (1+\sin (c+d x))}{4 d}+\frac {b \left (6 a^2+5 b^2\right ) \sin (c+d x)}{2 d}+\frac {3 a b^2 \sin ^2(c+d x)}{2 d}+\frac {b^3 \sin ^3(c+d x)}{3 d}+\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^3}{2 d} \]

[Out]

1/4*(a+b)^2*(2*a+5*b)*ln(1-sin(d*x+c))/d+1/4*(2*a-5*b)*(a-b)^2*ln(1+sin(d*x+c))/d+1/2*b*(6*a^2+5*b^2)*sin(d*x+
c)/d+3/2*a*b^2*sin(d*x+c)^2/d+1/3*b^3*sin(d*x+c)^3/d+1/2*sec(d*x+c)^2*(a+b*sin(d*x+c))^3/d

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Rubi [A]
time = 0.16, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2800, 1659, 1643, 647, 31} \begin {gather*} \frac {b \left (6 a^2+5 b^2\right ) \sin (c+d x)}{2 d}+\frac {3 a b^2 \sin ^2(c+d x)}{2 d}+\frac {(a+b)^2 (2 a+5 b) \log (1-\sin (c+d x))}{4 d}+\frac {(2 a-5 b) (a-b)^2 \log (\sin (c+d x)+1)}{4 d}+\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^3}{2 d}+\frac {b^3 \sin ^3(c+d x)}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b)^2*(2*a + 5*b)*Log[1 - Sin[c + d*x]])/(4*d) + ((2*a - 5*b)*(a - b)^2*Log[1 + Sin[c + d*x]])/(4*d) + (b
*(6*a^2 + 5*b^2)*Sin[c + d*x])/(2*d) + (3*a*b^2*Sin[c + d*x]^2)/(2*d) + (b^3*Sin[c + d*x]^3)/(3*d) + (Sec[c +
d*x]^2*(a + b*Sin[c + d*x])^3)/(2*d)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 647

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[e/2 + c*(d/(2*q)),
Int[1/(-q + c*x), x], x] + Dist[e/2 - c*(d/(2*q)), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[(-a)*c]

Rule 1643

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1659

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a + c*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + c
*x^2, x], x, 1]}, Simp[(d + e*x)^m*(a + c*x^2)^(p + 1)*((a*g - c*f*x)/(2*a*c*(p + 1))), x] + Dist[1/(2*a*c*(p
+ 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*c*(p + 1)*(d + e*x)*Q - a*e*g*m + c*d*f*(2*p
+ 3) + c*e*f*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && GtQ[m, 0] &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rule 2800

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps

\begin {align*} \int (a+b \sin (c+d x))^3 \tan ^3(c+d x) \, dx &=\frac {\text {Subst}\left (\int \frac {x^3 (a+x)^3}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^3}{2 d}+\frac {\text {Subst}\left (\int \frac {(a+x)^2 \left (-3 b^4-2 a b^2 x-2 b^2 x^2\right )}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{2 b^2 d}\\ &=\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^3}{2 d}+\frac {\text {Subst}\left (\int \left (6 a^2 b^2+5 b^4+6 a b^2 x+2 b^2 x^2-\frac {9 a^2 b^4+5 b^6+2 a b^2 \left (a^2+6 b^2\right ) x}{b^2-x^2}\right ) \, dx,x,b \sin (c+d x)\right )}{2 b^2 d}\\ &=\frac {b \left (6 a^2+5 b^2\right ) \sin (c+d x)}{2 d}+\frac {3 a b^2 \sin ^2(c+d x)}{2 d}+\frac {b^3 \sin ^3(c+d x)}{3 d}+\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^3}{2 d}-\frac {\text {Subst}\left (\int \frac {9 a^2 b^4+5 b^6+2 a b^2 \left (a^2+6 b^2\right ) x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{2 b^2 d}\\ &=\frac {b \left (6 a^2+5 b^2\right ) \sin (c+d x)}{2 d}+\frac {3 a b^2 \sin ^2(c+d x)}{2 d}+\frac {b^3 \sin ^3(c+d x)}{3 d}+\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^3}{2 d}-\frac {\left ((2 a-5 b) (a-b)^2\right ) \text {Subst}\left (\int \frac {1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{4 d}-\frac {\left ((a+b)^2 (2 a+5 b)\right ) \text {Subst}\left (\int \frac {1}{b-x} \, dx,x,b \sin (c+d x)\right )}{4 d}\\ &=\frac {(a+b)^2 (2 a+5 b) \log (1-\sin (c+d x))}{4 d}+\frac {(2 a-5 b) (a-b)^2 \log (1+\sin (c+d x))}{4 d}+\frac {b \left (6 a^2+5 b^2\right ) \sin (c+d x)}{2 d}+\frac {3 a b^2 \sin ^2(c+d x)}{2 d}+\frac {b^3 \sin ^3(c+d x)}{3 d}+\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^3}{2 d}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 141, normalized size = 0.94 \begin {gather*} \frac {3 (a+b)^2 (2 a+5 b) \log (1-\sin (c+d x))+3 (2 a-5 b) (a-b)^2 \log (1+\sin (c+d x))-\frac {3 (a+b)^3}{-1+\sin (c+d x)}+12 b \left (3 a^2+2 b^2\right ) \sin (c+d x)+18 a b^2 \sin ^2(c+d x)+4 b^3 \sin ^3(c+d x)+\frac {3 (a-b)^3}{1+\sin (c+d x)}}{12 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(a + b)^2*(2*a + 5*b)*Log[1 - Sin[c + d*x]] + 3*(2*a - 5*b)*(a - b)^2*Log[1 + Sin[c + d*x]] - (3*(a + b)^3)
/(-1 + Sin[c + d*x]) + 12*b*(3*a^2 + 2*b^2)*Sin[c + d*x] + 18*a*b^2*Sin[c + d*x]^2 + 4*b^3*Sin[c + d*x]^3 + (3
*(a - b)^3)/(1 + Sin[c + d*x]))/(12*d)

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Maple [A]
time = 0.19, size = 206, normalized size = 1.37

method	result	size
derivativedivides	\(\frac {a^{3} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{2} b \left (\frac {\sin ^{5}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a \,b^{2} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )+b^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{2}+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{6}+\frac {5 \sin \left (d x +c \right )}{2}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(206\)
default	\(\frac {a^{3} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{2} b \left (\frac {\sin ^{5}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a \,b^{2} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )+b^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{2}+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{6}+\frac {5 \sin \left (d x +c \right )}{2}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(206\)
risch	\(\frac {i b^{3} {\mathrm e}^{3 i \left (d x +c \right )}}{24 d}-\frac {3 i b \,{\mathrm e}^{i \left (d x +c \right )} a^{2}}{2 d}-\frac {i \left (2 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+6 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-3 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-b^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{d \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )^{2}}-\frac {3 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {2 i a^{3} c}{d}+\frac {3 i b \,{\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 d}-\frac {9 i b^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}-6 i a \,b^{2} x -\frac {3 a \,b^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {i b^{3} {\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}-\frac {12 i a \,b^{2} c}{d}-i a^{3} x +\frac {9 i b^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} b}{2 d}+\frac {6 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a \,b^{2}}{d}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{3}}{2 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} b}{2 d}+\frac {6 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a \,b^{2}}{d}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{3}}{2 d}\)	\(462\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^3*(1/2*tan(d*x+c)^2+ln(cos(d*x+c)))+3*a^2*b*(1/2*sin(d*x+c)^5/cos(d*x+c)^2+1/2*sin(d*x+c)^3+3/2*sin(d*x
+c)-3/2*ln(sec(d*x+c)+tan(d*x+c)))+3*a*b^2*(1/2*sin(d*x+c)^6/cos(d*x+c)^2+1/2*sin(d*x+c)^4+sin(d*x+c)^2+2*ln(c
os(d*x+c)))+b^3*(1/2*sin(d*x+c)^7/cos(d*x+c)^2+1/2*sin(d*x+c)^5+5/6*sin(d*x+c)^3+5/2*sin(d*x+c)-5/2*ln(sec(d*x
+c)+tan(d*x+c))))

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Maxima [A]
time = 0.29, size = 162, normalized size = 1.08 \begin {gather*} \frac {4 \, b^{3} \sin \left (d x + c\right )^{3} + 18 \, a b^{2} \sin \left (d x + c\right )^{2} + 3 \, {\left (2 \, a^{3} - 9 \, a^{2} b + 12 \, a b^{2} - 5 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (2 \, a^{3} + 9 \, a^{2} b + 12 \, a b^{2} + 5 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 12 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \sin \left (d x + c\right ) - \frac {6 \, {\left (a^{3} + 3 \, a b^{2} + {\left (3 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{2} - 1}}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*b^3*sin(d*x + c)^3 + 18*a*b^2*sin(d*x + c)^2 + 3*(2*a^3 - 9*a^2*b + 12*a*b^2 - 5*b^3)*log(sin(d*x + c)
+ 1) + 3*(2*a^3 + 9*a^2*b + 12*a*b^2 + 5*b^3)*log(sin(d*x + c) - 1) + 12*(3*a^2*b + 2*b^3)*sin(d*x + c) - 6*(
a^3 + 3*a*b^2 + (3*a^2*b + b^3)*sin(d*x + c))/(sin(d*x + c)^2 - 1))/d

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Fricas [A]
time = 0.37, size = 194, normalized size = 1.29 \begin {gather*} -\frac {18 \, a b^{2} \cos \left (d x + c\right )^{4} - 9 \, a b^{2} \cos \left (d x + c\right )^{2} - 3 \, {\left (2 \, a^{3} - 9 \, a^{2} b + 12 \, a b^{2} - 5 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, a^{3} + 9 \, a^{2} b + 12 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 6 \, a^{3} - 18 \, a b^{2} + 2 \, {\left (2 \, b^{3} \cos \left (d x + c\right )^{4} - 9 \, a^{2} b - 3 \, b^{3} - 2 \, {\left (9 \, a^{2} b + 7 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(18*a*b^2*cos(d*x + c)^4 - 9*a*b^2*cos(d*x + c)^2 - 3*(2*a^3 - 9*a^2*b + 12*a*b^2 - 5*b^3)*cos(d*x + c)^
2*log(sin(d*x + c) + 1) - 3*(2*a^3 + 9*a^2*b + 12*a*b^2 + 5*b^3)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) - 6*a^3
- 18*a*b^2 + 2*(2*b^3*cos(d*x + c)^4 - 9*a^2*b - 3*b^3 - 2*(9*a^2*b + 7*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(d
*cos(d*x + c)^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sin {\left (c + d x \right )}\right )^{3} \tan ^{3}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*sin(c + d*x))**3*tan(c + d*x)**3, x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Mupad [B]
time = 6.98, size = 366, normalized size = 2.44 \begin {gather*} \frac {\left (9\,a^2\,b+5\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (2\,a^3+12\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\left (12\,a^2\,b+\frac {20\,b^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (6\,a^3+12\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (6\,a^2\,b-\frac {22\,b^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (6\,a^3+12\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (12\,a^2\,b+\frac {20\,b^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a^3+12\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (9\,a^2\,b+5\,b^3\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\,\left (a^3+6\,a\,b^2\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,{\left (a-b\right )}^2\,\left (a-\frac {5\,b}{2}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,{\left (a+b\right )}^2\,\left (a+\frac {5\,b}{2}\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(9*a^2*b + 5*b^3) + tan(c/2 + (d*x)/2)^2*(12*a*b^2 + 2*a^3) + tan(c/2 + (d*x)/2)^4*(12*a*b
^2 + 6*a^3) + tan(c/2 + (d*x)/2)^8*(12*a*b^2 + 2*a^3) + tan(c/2 + (d*x)/2)^6*(12*a*b^2 + 6*a^3) + tan(c/2 + (d
*x)/2)^9*(9*a^2*b + 5*b^3) + tan(c/2 + (d*x)/2)^5*(6*a^2*b - (22*b^3)/3) + tan(c/2 + (d*x)/2)^3*(12*a^2*b + (2
0*b^3)/3) + tan(c/2 + (d*x)/2)^7*(12*a^2*b + (20*b^3)/3))/(d*(tan(c/2 + (d*x)/2)^2 - 2*tan(c/2 + (d*x)/2)^4 -
2*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) - (log(tan(c/2 + (d*x)/2)^2 + 1)*(
6*a*b^2 + a^3))/d + (log(tan(c/2 + (d*x)/2) + 1)*(a - b)^2*(a - (5*b)/2))/d + (log(tan(c/2 + (d*x)/2) - 1)*(a
+ b)^2*(a + (5*b)/2))/d

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