3.1457 \(\int \sin (c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx\)

Optimal. Leaf size=197 \[ \frac{9 a^2 b \tan (c+d x)}{2 d}-\frac{3 a^2 b \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac{9}{2} a^2 b x+\frac{a^3 \cos (c+d x)}{d}+\frac{a^3 \sec (c+d x)}{d}-\frac{a b^2 \cos ^3(c+d x)}{d}+\frac{6 a b^2 \cos (c+d x)}{d}+\frac{3 a b^2 \sec (c+d x)}{d}+\frac{15 b^3 \tan (c+d x)}{8 d}-\frac{b^3 \sin ^4(c+d x) \tan (c+d x)}{4 d}-\frac{5 b^3 \sin ^2(c+d x) \tan (c+d x)}{8 d}-\frac{15 b^3 x}{8} \]

[Out]

(-9*a^2*b*x)/2 - (15*b^3*x)/8 + (a^3*Cos[c + d*x])/d + (6*a*b^2*Cos[c + d*x])/d - (a*b^2*Cos[c + d*x]^3)/d + (
a^3*Sec[c + d*x])/d + (3*a*b^2*Sec[c + d*x])/d + (9*a^2*b*Tan[c + d*x])/(2*d) + (15*b^3*Tan[c + d*x])/(8*d) -
(3*a^2*b*Sin[c + d*x]^2*Tan[c + d*x])/(2*d) - (5*b^3*Sin[c + d*x]^2*Tan[c + d*x])/(8*d) - (b^3*Sin[c + d*x]^4*
Tan[c + d*x])/(4*d)

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Rubi [A]  time = 0.260817, antiderivative size = 197, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {2912, 2590, 14, 2591, 288, 321, 203, 270} \[ \frac{9 a^2 b \tan (c+d x)}{2 d}-\frac{3 a^2 b \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac{9}{2} a^2 b x+\frac{a^3 \cos (c+d x)}{d}+\frac{a^3 \sec (c+d x)}{d}-\frac{a b^2 \cos ^3(c+d x)}{d}+\frac{6 a b^2 \cos (c+d x)}{d}+\frac{3 a b^2 \sec (c+d x)}{d}+\frac{15 b^3 \tan (c+d x)}{8 d}-\frac{b^3 \sin ^4(c+d x) \tan (c+d x)}{4 d}-\frac{5 b^3 \sin ^2(c+d x) \tan (c+d x)}{8 d}-\frac{15 b^3 x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*a^2*b*x)/2 - (15*b^3*x)/8 + (a^3*Cos[c + d*x])/d + (6*a*b^2*Cos[c + d*x])/d - (a*b^2*Cos[c + d*x]^3)/d + (
a^3*Sec[c + d*x])/d + (3*a*b^2*Sec[c + d*x])/d + (9*a^2*b*Tan[c + d*x])/(2*d) + (15*b^3*Tan[c + d*x])/(8*d) -
(3*a^2*b*Sin[c + d*x]^2*Tan[c + d*x])/(2*d) - (5*b^3*Sin[c + d*x]^2*Tan[c + d*x])/(8*d) - (b^3*Sin[c + d*x]^4*
Tan[c + d*x])/(4*d)

Rule 2912

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[m] && (GtQ[m, 0] || IntegerQ[n])

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2591

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[(b*ff)/f, Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, (b*Tan[e + f*x])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rule 288

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

Rule 321

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

Rule 203

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

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \sin (c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx &=\int \left (a^3 \sin (c+d x) \tan ^2(c+d x)+3 a^2 b \sin ^2(c+d x) \tan ^2(c+d x)+3 a b^2 \sin ^3(c+d x) \tan ^2(c+d x)+b^3 \sin ^4(c+d x) \tan ^2(c+d x)\right ) \, dx\\ &=a^3 \int \sin (c+d x) \tan ^2(c+d x) \, dx+\left (3 a^2 b\right ) \int \sin ^2(c+d x) \tan ^2(c+d x) \, dx+\left (3 a b^2\right ) \int \sin ^3(c+d x) \tan ^2(c+d x) \, dx+b^3 \int \sin ^4(c+d x) \tan ^2(c+d x) \, dx\\ &=-\frac{a^3 \operatorname{Subst}\left (\int \frac{1-x^2}{x^2} \, dx,x,\cos (c+d x)\right )}{d}+\frac{\left (3 a^2 b\right ) \operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{d}-\frac{\left (3 a b^2\right ) \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{x^2} \, dx,x,\cos (c+d x)\right )}{d}+\frac{b^3 \operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^2\right )^3} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{3 a^2 b \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac{b^3 \sin ^4(c+d x) \tan (c+d x)}{4 d}-\frac{a^3 \operatorname{Subst}\left (\int \left (-1+\frac{1}{x^2}\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac{\left (9 a^2 b\right ) \operatorname{Subst}\left (\int \frac{x^2}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac{\left (3 a b^2\right ) \operatorname{Subst}\left (\int \left (-2+\frac{1}{x^2}+x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac{\left (5 b^3\right ) \operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{4 d}\\ &=\frac{a^3 \cos (c+d x)}{d}+\frac{6 a b^2 \cos (c+d x)}{d}-\frac{a b^2 \cos ^3(c+d x)}{d}+\frac{a^3 \sec (c+d x)}{d}+\frac{3 a b^2 \sec (c+d x)}{d}+\frac{9 a^2 b \tan (c+d x)}{2 d}-\frac{3 a^2 b \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac{5 b^3 \sin ^2(c+d x) \tan (c+d x)}{8 d}-\frac{b^3 \sin ^4(c+d x) \tan (c+d x)}{4 d}-\frac{\left (9 a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{\left (15 b^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{1+x^2} \, dx,x,\tan (c+d x)\right )}{8 d}\\ &=-\frac{9}{2} a^2 b x+\frac{a^3 \cos (c+d x)}{d}+\frac{6 a b^2 \cos (c+d x)}{d}-\frac{a b^2 \cos ^3(c+d x)}{d}+\frac{a^3 \sec (c+d x)}{d}+\frac{3 a b^2 \sec (c+d x)}{d}+\frac{9 a^2 b \tan (c+d x)}{2 d}+\frac{15 b^3 \tan (c+d x)}{8 d}-\frac{3 a^2 b \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac{5 b^3 \sin ^2(c+d x) \tan (c+d x)}{8 d}-\frac{b^3 \sin ^4(c+d x) \tan (c+d x)}{4 d}-\frac{\left (15 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{8 d}\\ &=-\frac{9}{2} a^2 b x-\frac{15 b^3 x}{8}+\frac{a^3 \cos (c+d x)}{d}+\frac{6 a b^2 \cos (c+d x)}{d}-\frac{a b^2 \cos ^3(c+d x)}{d}+\frac{a^3 \sec (c+d x)}{d}+\frac{3 a b^2 \sec (c+d x)}{d}+\frac{9 a^2 b \tan (c+d x)}{2 d}+\frac{15 b^3 \tan (c+d x)}{8 d}-\frac{3 a^2 b \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac{5 b^3 \sin ^2(c+d x) \tan (c+d x)}{8 d}-\frac{b^3 \sin ^4(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.759607, size = 147, normalized size = 0.75 \[ \frac{\sec (c+d x) \left (-24 b \left (12 a^2+5 b^2\right ) (c+d x) \cos (c+d x)+32 \left (a^3+5 a b^2\right ) \cos (2 (c+d x))+216 a^2 b \sin (c+d x)+24 a^2 b \sin (3 (c+d x))+96 a^3-8 a b^2 \cos (4 (c+d x))+360 a b^2+80 b^3 \sin (c+d x)+15 b^3 \sin (3 (c+d x))-b^3 \sin (5 (c+d x))\right )}{64 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]*(96*a^3 + 360*a*b^2 - 24*b*(12*a^2 + 5*b^2)*(c + d*x)*Cos[c + d*x] + 32*(a^3 + 5*a*b^2)*Cos[2*(c
+ d*x)] - 8*a*b^2*Cos[4*(c + d*x)] + 216*a^2*b*Sin[c + d*x] + 80*b^3*Sin[c + d*x] + 24*a^2*b*Sin[3*(c + d*x)]
+ 15*b^3*Sin[3*(c + d*x)] - b^3*Sin[5*(c + d*x)]))/(64*d)

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Maple [A]  time = 0.059, size = 214, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{\cos \left ( dx+c \right ) }}+ \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) \right ) +3\,{a}^{2}b \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{\cos \left ( dx+c \right ) }}+ \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{3}+3/2\,\sin \left ( dx+c \right ) \right ) \cos \left ( dx+c \right ) -3/2\,dx-3/2\,c \right ) +3\,a{b}^{2} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{\cos \left ( dx+c \right ) }}+ \left ( 8/3+ \left ( \sin \left ( dx+c \right ) \right ) ^{4}+4/3\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) \right ) +{b}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{\cos \left ( dx+c \right ) }}+ \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\sin \left ( dx+c \right ) }{8}} \right ) \cos \left ( dx+c \right ) -{\frac{15\,dx}{8}}-{\frac{15\,c}{8}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*sin(d*x+c)^3*(a+b*sin(d*x+c))^3,x)

[Out]

1/d*(a^3*(sin(d*x+c)^4/cos(d*x+c)+(2+sin(d*x+c)^2)*cos(d*x+c))+3*a^2*b*(sin(d*x+c)^5/cos(d*x+c)+(sin(d*x+c)^3+
3/2*sin(d*x+c))*cos(d*x+c)-3/2*d*x-3/2*c)+3*a*b^2*(sin(d*x+c)^6/cos(d*x+c)+(8/3+sin(d*x+c)^4+4/3*sin(d*x+c)^2)
*cos(d*x+c))+b^3*(sin(d*x+c)^7/cos(d*x+c)+(sin(d*x+c)^5+5/4*sin(d*x+c)^3+15/8*sin(d*x+c))*cos(d*x+c)-15/8*d*x-
15/8*c))

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Maxima [A]  time = 1.49203, size = 221, normalized size = 1.12 \begin{align*} -\frac{12 \,{\left (3 \, d x + 3 \, c - \frac{\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a^{2} b + 8 \,{\left (\cos \left (d x + c\right )^{3} - \frac{3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} a b^{2} +{\left (15 \, d x + 15 \, c - \frac{9 \, \tan \left (d x + c\right )^{3} + 7 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1} - 8 \, \tan \left (d x + c\right )\right )} b^{3} - 8 \, a^{3}{\left (\frac{1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(12*(3*d*x + 3*c - tan(d*x + c)/(tan(d*x + c)^2 + 1) - 2*tan(d*x + c))*a^2*b + 8*(cos(d*x + c)^3 - 3/cos(
d*x + c) - 6*cos(d*x + c))*a*b^2 + (15*d*x + 15*c - (9*tan(d*x + c)^3 + 7*tan(d*x + c))/(tan(d*x + c)^4 + 2*ta
n(d*x + c)^2 + 1) - 8*tan(d*x + c))*b^3 - 8*a^3*(1/cos(d*x + c) + cos(d*x + c)))/d

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Fricas [A]  time = 1.67019, size = 319, normalized size = 1.62 \begin{align*} -\frac{8 \, a b^{2} \cos \left (d x + c\right )^{4} + 3 \,{\left (12 \, a^{2} b + 5 \, b^{3}\right )} d x \cos \left (d x + c\right ) - 8 \, a^{3} - 24 \, a b^{2} - 8 \,{\left (a^{3} + 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} +{\left (2 \, b^{3} \cos \left (d x + c\right )^{4} - 24 \, a^{2} b - 8 \, b^{3} - 3 \,{\left (4 \, a^{2} b + 3 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{8 \, d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(8*a*b^2*cos(d*x + c)^4 + 3*(12*a^2*b + 5*b^3)*d*x*cos(d*x + c) - 8*a^3 - 24*a*b^2 - 8*(a^3 + 6*a*b^2)*co
s(d*x + c)^2 + (2*b^3*cos(d*x + c)^4 - 24*a^2*b - 8*b^3 - 3*(4*a^2*b + 3*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(d
*cos(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*sin(d*x+c)**3*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.22656, size = 454, normalized size = 2.3 \begin{align*} -\frac{3 \,{\left (12 \, a^{2} b + 5 \, b^{3}\right )}{\left (d x + c\right )} + \frac{16 \,{\left (3 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{3} + 3 \, a b^{2}\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1} + \frac{2 \,{\left (12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 7 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 8 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 24 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 15 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 24 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 120 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 15 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 24 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 136 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 7 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 8 \, a^{3} - 40 \, a b^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{4}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(3*(12*a^2*b + 5*b^3)*(d*x + c) + 16*(3*a^2*b*tan(1/2*d*x + 1/2*c) + b^3*tan(1/2*d*x + 1/2*c) + a^3 + 3*a
*b^2)/(tan(1/2*d*x + 1/2*c)^2 - 1) + 2*(12*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 7*b^3*tan(1/2*d*x + 1/2*c)^7 - 8*a^3
*tan(1/2*d*x + 1/2*c)^6 - 24*a*b^2*tan(1/2*d*x + 1/2*c)^6 + 12*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 15*b^3*tan(1/2*d
*x + 1/2*c)^5 - 24*a^3*tan(1/2*d*x + 1/2*c)^4 - 120*a*b^2*tan(1/2*d*x + 1/2*c)^4 - 12*a^2*b*tan(1/2*d*x + 1/2*
c)^3 - 15*b^3*tan(1/2*d*x + 1/2*c)^3 - 24*a^3*tan(1/2*d*x + 1/2*c)^2 - 136*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 12*a
^2*b*tan(1/2*d*x + 1/2*c) - 7*b^3*tan(1/2*d*x + 1/2*c) - 8*a^3 - 40*a*b^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d