3.1.86 \(\int \frac {\sin ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx\) [86]

Optimal. Leaf size=117 \[ \frac {\left (8 a^2-4 a b+3 b^2\right ) x}{8 b^3}-\frac {a^{5/2} \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{b^3 \sqrt {a+b} d}+\frac {(4 a-3 b) \cos (c+d x) \sin (c+d x)}{8 b^2 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b d} \]

[Out]

1/8*(8*a^2-4*a*b+3*b^2)*x/b^3+1/8*(4*a-3*b)*cos(d*x+c)*sin(d*x+c)/b^2/d-1/4*cos(d*x+c)*sin(d*x+c)^3/b/d-a^(5/2
)*arctan((a+b)^(1/2)*tan(d*x+c)/a^(1/2))/b^3/d/(a+b)^(1/2)

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Rubi [A]
time = 0.14, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3266, 481, 592, 536, 209, 211} \begin {gather*} -\frac {a^{5/2} \text {ArcTan}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{b^3 d \sqrt {a+b}}+\frac {x \left (8 a^2-4 a b+3 b^2\right )}{8 b^3}+\frac {(4 a-3 b) \sin (c+d x) \cos (c+d x)}{8 b^2 d}-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^6/(a + b*Sin[c + d*x]^2),x]

[Out]

((8*a^2 - 4*a*b + 3*b^2)*x)/(8*b^3) - (a^(5/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(b^3*Sqrt[a + b]*d)
+ ((4*a - 3*b)*Cos[c + d*x]*Sin[c + d*x])/(8*b^2*d) - (Cos[c + d*x]*Sin[c + d*x]^3)/(4*b*d)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 481

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-a)*e^(
2*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*n*(b*c - a*d)*(p + 1))), x] + Dist[e^
(2*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1)
+ (a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0]
&& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 536

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 592

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[g^(n - 1)*(b*e - a*f)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*n*(b*c -
a*d)*(p + 1))), x] - Dist[g^n/(b*n*(b*c - a*d)*(p + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*S
imp[c*(b*e - a*f)*(m - n + 1) + (d*(b*e - a*f)*(m + n*q + 1) - b*n*(c*f - d*e)*(p + 1))*x^n, x], x], x] /; Fre
eQ[{a, b, c, d, e, f, g, q}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, 0]

Rule 3266

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[x^m*((a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p +
1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {\sin ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx &=\frac {\text {Subst}\left (\int \frac {x^6}{\left (1+x^2\right )^3 \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b d}+\frac {\text {Subst}\left (\int \frac {x^2 \left (3 a+(-a+3 b) x^2\right )}{\left (1+x^2\right )^2 \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{4 b d}\\ &=\frac {(4 a-3 b) \cos (c+d x) \sin (c+d x)}{8 b^2 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b d}-\frac {\text {Subst}\left (\int \frac {a (4 a-3 b)+\left (-4 a^2+a b-3 b^2\right ) x^2}{\left (1+x^2\right ) \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{8 b^2 d}\\ &=\frac {(4 a-3 b) \cos (c+d x) \sin (c+d x)}{8 b^2 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b d}-\frac {a^3 \text {Subst}\left (\int \frac {1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{b^3 d}+\frac {\left (8 a^2-4 a b+3 b^2\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{8 b^3 d}\\ &=\frac {\left (8 a^2-4 a b+3 b^2\right ) x}{8 b^3}-\frac {a^{5/2} \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{b^3 \sqrt {a+b} d}+\frac {(4 a-3 b) \cos (c+d x) \sin (c+d x)}{8 b^2 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b d}\\ \end {align*}

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Mathematica [A]
time = 0.32, size = 95, normalized size = 0.81 \begin {gather*} \frac {4 \left (8 a^2-4 a b+3 b^2\right ) (c+d x)-\frac {32 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a+b}}+8 (a-b) b \sin (2 (c+d x))+b^2 \sin (4 (c+d x))}{32 b^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^6/(a + b*Sin[c + d*x]^2),x]

[Out]

(4*(8*a^2 - 4*a*b + 3*b^2)*(c + d*x) - (32*a^(5/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/Sqrt[a + b] + 8
*(a - b)*b*Sin[2*(c + d*x)] + b^2*Sin[4*(c + d*x)])/(32*b^3*d)

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Maple [A]
time = 0.30, size = 118, normalized size = 1.01

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (\frac {1}{2} a b -\frac {5}{8} b^{2}\right ) \left (\tan ^{3}\left (d x +c \right )\right )+\left (\frac {1}{2} a b -\frac {3}{8} b^{2}\right ) \tan \left (d x +c \right )}{\left (\tan ^{2}\left (d x +c \right )+1\right )^{2}}+\frac {\left (8 a^{2}-4 a b +3 b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{8}}{b^{3}}-\frac {a^{3} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{b^{3} \sqrt {a \left (a +b \right )}}}{d}\)	\(118\)
default	\(\frac {\frac {\frac {\left (\frac {1}{2} a b -\frac {5}{8} b^{2}\right ) \left (\tan ^{3}\left (d x +c \right )\right )+\left (\frac {1}{2} a b -\frac {3}{8} b^{2}\right ) \tan \left (d x +c \right )}{\left (\tan ^{2}\left (d x +c \right )+1\right )^{2}}+\frac {\left (8 a^{2}-4 a b +3 b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{8}}{b^{3}}-\frac {a^{3} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{b^{3} \sqrt {a \left (a +b \right )}}}{d}\)	\(118\)
risch	\(\frac {x \,a^{2}}{b^{3}}-\frac {a x}{2 b^{2}}+\frac {3 x}{8 b}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} a}{8 b^{2} d}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 b d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} a}{8 b^{2} d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 b d}-\frac {\sqrt {-a \left (a +b \right )}\, a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{2 \left (a +b \right ) d \,b^{3}}+\frac {\sqrt {-a \left (a +b \right )}\, a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{2 \left (a +b \right ) d \,b^{3}}+\frac {\sin \left (4 d x +4 c \right )}{32 b d}\)	\(227\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^6/(a+sin(d*x+c)^2*b),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.54, size = 128, normalized size = 1.09 \begin {gather*} -\frac {\frac {8 \, a^{3} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} b^{3}} - \frac {{\left (4 \, a - 5 \, b\right )} \tan \left (d x + c\right )^{3} + {\left (4 \, a - 3 \, b\right )} \tan \left (d x + c\right )}{b^{2} \tan \left (d x + c\right )^{4} + 2 \, b^{2} \tan \left (d x + c\right )^{2} + b^{2}} - \frac {{\left (8 \, a^{2} - 4 \, a b + 3 \, b^{2}\right )} {\left (d x + c\right )}}{b^{3}}}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^6/(a+b*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/8*(8*a^3*arctan((a + b)*tan(d*x + c)/sqrt((a + b)*a))/(sqrt((a + b)*a)*b^3) - ((4*a - 5*b)*tan(d*x + c)^3 +
(4*a - 3*b)*tan(d*x + c))/(b^2*tan(d*x + c)^4 + 2*b^2*tan(d*x + c)^2 + b^2) - (8*a^2 - 4*a*b + 3*b^2)*(d*x +
c)/b^3)/d

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Fricas [A]
time = 0.47, size = 372, normalized size = 3.18 \begin {gather*} \left [\frac {2 \, a^{2} \sqrt {-\frac {a}{a + b}} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {-\frac {a}{a + b}} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) + {\left (8 \, a^{2} - 4 \, a b + 3 \, b^{2}\right )} d x + {\left (2 \, b^{2} \cos \left (d x + c\right )^{3} + {\left (4 \, a b - 5 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, b^{3} d}, \frac {4 \, a^{2} \sqrt {\frac {a}{a + b}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {\frac {a}{a + b}}}{2 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) + {\left (8 \, a^{2} - 4 \, a b + 3 \, b^{2}\right )} d x + {\left (2 \, b^{2} \cos \left (d x + c\right )^{3} + {\left (4 \, a b - 5 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, b^{3} d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^6/(a+b*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/8*(2*a^2*sqrt(-a/(a + b))*log(((8*a^2 + 8*a*b + b^2)*cos(d*x + c)^4 - 2*(4*a^2 + 5*a*b + b^2)*cos(d*x + c)^
2 + 4*((2*a^2 + 3*a*b + b^2)*cos(d*x + c)^3 - (a^2 + 2*a*b + b^2)*cos(d*x + c))*sqrt(-a/(a + b))*sin(d*x + c)
+ a^2 + 2*a*b + b^2)/(b^2*cos(d*x + c)^4 - 2*(a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2)) + (8*a^2 - 4*a*b
+ 3*b^2)*d*x + (2*b^2*cos(d*x + c)^3 + (4*a*b - 5*b^2)*cos(d*x + c))*sin(d*x + c))/(b^3*d), 1/8*(4*a^2*sqrt(a
/(a + b))*arctan(1/2*((2*a + b)*cos(d*x + c)^2 - a - b)*sqrt(a/(a + b))/(a*cos(d*x + c)*sin(d*x + c))) + (8*a^
2 - 4*a*b + 3*b^2)*d*x + (2*b^2*cos(d*x + c)^3 + (4*a*b - 5*b^2)*cos(d*x + c))*sin(d*x + c))/(b^3*d)]

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Sympy [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(sin(d*x+c)**6/(a+b*sin(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]
time = 0.48, size = 157, normalized size = 1.34 \begin {gather*} -\frac {\frac {8 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )} a^{3}}{\sqrt {a^{2} + a b} b^{3}} - \frac {{\left (8 \, a^{2} - 4 \, a b + 3 \, b^{2}\right )} {\left (d x + c\right )}}{b^{3}} - \frac {4 \, a \tan \left (d x + c\right )^{3} - 5 \, b \tan \left (d x + c\right )^{3} + 4 \, a \tan \left (d x + c\right ) - 3 \, b \tan \left (d x + c\right )}{{\left (\tan \left (d x + c\right )^{2} + 1\right )}^{2} b^{2}}}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^6/(a+b*sin(d*x+c)^2),x, algorithm="giac")

[Out]

-1/8*(8*(pi*floor((d*x + c)/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(d*x + c) + b*tan(d*x + c))/sqrt(a^2 + a*b
)))*a^3/(sqrt(a^2 + a*b)*b^3) - (8*a^2 - 4*a*b + 3*b^2)*(d*x + c)/b^3 - (4*a*tan(d*x + c)^3 - 5*b*tan(d*x + c)
^3 + 4*a*tan(d*x + c) - 3*b*tan(d*x + c))/((tan(d*x + c)^2 + 1)^2*b^2))/d

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Mupad [B]
time = 14.82, size = 1892, normalized size = 16.17 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^6/(a + b*sin(c + d*x)^2),x)

[Out]

((tan(c + d*x)*(4*a - 3*b))/(8*b^2) + (tan(c + d*x)^3*(4*a - 5*b))/(8*b^2))/(d*(2*tan(c + d*x)^2 + tan(c + d*x
)^4 + 1)) - (atan((((((((3*a*b^9)/2 + a^2*b^8 - (5*a^3*b^7)/2 - 2*a^4*b^6)/(2*b^6) - (tan(c + d*x)*(-a^5*(a +
b))^(1/2)*(1024*a*b^8 + 256*b^9 + 1280*a^2*b^7 + 512*a^3*b^6))/(128*b^4*(a*b^3 + b^4)))*(-a^5*(a + b))^(1/2))/
(2*(a*b^3 + b^4)) - (tan(c + d*x)*(3*a*b^6 + 192*a^6*b + 128*a^7 + 9*b^7 + 19*a^2*b^5 + 65*a^3*b^4 + 40*a^4*b^
3 + 64*a^5*b^2))/(64*b^4))*(-a^5*(a + b))^(1/2)*1i)/(a*b^3 + b^4) - ((((((3*a*b^9)/2 + a^2*b^8 - (5*a^3*b^7)/2
- 2*a^4*b^6)/(2*b^6) + (tan(c + d*x)*(-a^5*(a + b))^(1/2)*(1024*a*b^8 + 256*b^9 + 1280*a^2*b^7 + 512*a^3*b^6)
)/(128*b^4*(a*b^3 + b^4)))*(-a^5*(a + b))^(1/2))/(2*(a*b^3 + b^4)) + (tan(c + d*x)*(3*a*b^6 + 192*a^6*b + 128*
a^7 + 9*b^7 + 19*a^2*b^5 + 65*a^3*b^4 + 40*a^4*b^3 + 64*a^5*b^2))/(64*b^4))*(-a^5*(a + b))^(1/2)*1i)/(a*b^3 +
b^4))/(((((((3*a*b^9)/2 + a^2*b^8 - (5*a^3*b^7)/2 - 2*a^4*b^6)/(2*b^6) - (tan(c + d*x)*(-a^5*(a + b))^(1/2)*(1
024*a*b^8 + 256*b^9 + 1280*a^2*b^7 + 512*a^3*b^6))/(128*b^4*(a*b^3 + b^4)))*(-a^5*(a + b))^(1/2))/(2*(a*b^3 +
b^4)) - (tan(c + d*x)*(3*a*b^6 + 192*a^6*b + 128*a^7 + 9*b^7 + 19*a^2*b^5 + 65*a^3*b^4 + 40*a^4*b^3 + 64*a^5*b
^2))/(64*b^4))*(-a^5*(a + b))^(1/2))/(a*b^3 + b^4) - ((a^7*b)/4 + a^8 + (9*a^3*b^5)/32 - (3*a^4*b^4)/16 + (25*
a^5*b^3)/32 + (a^6*b^2)/2)/b^6 + ((((((3*a*b^9)/2 + a^2*b^8 - (5*a^3*b^7)/2 - 2*a^4*b^6)/(2*b^6) + (tan(c + d*
x)*(-a^5*(a + b))^(1/2)*(1024*a*b^8 + 256*b^9 + 1280*a^2*b^7 + 512*a^3*b^6))/(128*b^4*(a*b^3 + b^4)))*(-a^5*(a
+ b))^(1/2))/(2*(a*b^3 + b^4)) + (tan(c + d*x)*(3*a*b^6 + 192*a^6*b + 128*a^7 + 9*b^7 + 19*a^2*b^5 + 65*a^3*b
^4 + 40*a^4*b^3 + 64*a^5*b^2))/(64*b^4))*(-a^5*(a + b))^(1/2))/(a*b^3 + b^4)))*(-a^5*(a + b))^(1/2)*1i)/(d*(a*
b^3 + b^4)) - (atan((((((((3*a*b^9)/2 + a^2*b^8 - (5*a^3*b^7)/2 - 2*a^4*b^6)/b^6 - (tan(c + d*x)*(a^2*8i - a*b
*4i + b^2*3i)*(1024*a*b^8 + 256*b^9 + 1280*a^2*b^7 + 512*a^3*b^6))/(512*b^7))*(a^2*8i - a*b*4i + b^2*3i))/(16*
b^3) - (tan(c + d*x)*(3*a*b^6 + 192*a^6*b + 128*a^7 + 9*b^7 + 19*a^2*b^5 + 65*a^3*b^4 + 40*a^4*b^3 + 64*a^5*b^
2))/(32*b^4))*(a^2*8i - a*b*4i + b^2*3i)*1i)/(16*b^3) - ((((((3*a*b^9)/2 + a^2*b^8 - (5*a^3*b^7)/2 - 2*a^4*b^6
)/b^6 + (tan(c + d*x)*(a^2*8i - a*b*4i + b^2*3i)*(1024*a*b^8 + 256*b^9 + 1280*a^2*b^7 + 512*a^3*b^6))/(512*b^7
))*(a^2*8i - a*b*4i + b^2*3i))/(16*b^3) + (tan(c + d*x)*(3*a*b^6 + 192*a^6*b + 128*a^7 + 9*b^7 + 19*a^2*b^5 +
65*a^3*b^4 + 40*a^4*b^3 + 64*a^5*b^2))/(32*b^4))*(a^2*8i - a*b*4i + b^2*3i)*1i)/(16*b^3))/(((((((3*a*b^9)/2 +
a^2*b^8 - (5*a^3*b^7)/2 - 2*a^4*b^6)/b^6 - (tan(c + d*x)*(a^2*8i - a*b*4i + b^2*3i)*(1024*a*b^8 + 256*b^9 + 12
80*a^2*b^7 + 512*a^3*b^6))/(512*b^7))*(a^2*8i - a*b*4i + b^2*3i))/(16*b^3) - (tan(c + d*x)*(3*a*b^6 + 192*a^6*
b + 128*a^7 + 9*b^7 + 19*a^2*b^5 + 65*a^3*b^4 + 40*a^4*b^3 + 64*a^5*b^2))/(32*b^4))*(a^2*8i - a*b*4i + b^2*3i)
)/(16*b^3) - ((a^7*b)/4 + a^8 + (9*a^3*b^5)/32 - (3*a^4*b^4)/16 + (25*a^5*b^3)/32 + (a^6*b^2)/2)/b^6 + ((((((3
*a*b^9)/2 + a^2*b^8 - (5*a^3*b^7)/2 - 2*a^4*b^6)/b^6 + (tan(c + d*x)*(a^2*8i - a*b*4i + b^2*3i)*(1024*a*b^8 +
256*b^9 + 1280*a^2*b^7 + 512*a^3*b^6))/(512*b^7))*(a^2*8i - a*b*4i + b^2*3i))/(16*b^3) + (tan(c + d*x)*(3*a*b^
6 + 192*a^6*b + 128*a^7 + 9*b^7 + 19*a^2*b^5 + 65*a^3*b^4 + 40*a^4*b^3 + 64*a^5*b^2))/(32*b^4))*(a^2*8i - a*b*
4i + b^2*3i))/(16*b^3)))*(a^2*8i - a*b*4i + b^2*3i)*1i)/(8*b^3*d)

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