\(\int (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx\) [65]

Optimal result

Integrand size = 23, antiderivative size = 111 \[ \int (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=a^3 A x+\frac {a^3 (7 A+5 B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {5 a^3 (A+B) \tan (c+d x)}{2 d}+\frac {a B (a+a \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {(3 A+5 B) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{6 d} \]

[Out]

a^3*A*x+1/2*a^3*(7*A+5*B)*arctanh(sin(d*x+c))/d+5/2*a^3*(A+B)*tan(d*x+c)/d+1/3*a*B*(a+a*sec(d*x+c))^2*tan(d*x+
c)/d+1/6*(3*A+5*B)*(a^3+a^3*sec(d*x+c))*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4002, 3999, 3852, 8, 3855} \[ \int (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {a^3 (7 A+5 B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {5 a^3 (A+B) \tan (c+d x)}{2 d}+\frac {(3 A+5 B) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{6 d}+a^3 A x+\frac {a B \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d} \]

[In]

Int[(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x]),x]

[Out]

a^3*A*x + (a^3*(7*A + 5*B)*ArcTanh[Sin[c + d*x]])/(2*d) + (5*a^3*(A + B)*Tan[c + d*x])/(2*d) + (a*B*(a + a*Sec
[c + d*x])^2*Tan[c + d*x])/(3*d) + ((3*A + 5*B)*(a^3 + a^3*Sec[c + d*x])*Tan[c + d*x])/(6*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3855

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

Rule 3999

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[a*c*x, x]
+ (Dist[b*d, Int[Csc[e + f*x]^2, x], x] + Dist[b*c + a*d, Int[Csc[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rule 4002

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[(-b)
*d*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m - 1)/(f*m)), x] + Dist[1/m, Int[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*c
*m + (b*c*m + a*d*(2*m - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &&
GtQ[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rubi steps \begin{align*} \text {integral}& = \frac {a B (a+a \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {1}{3} \int (a+a \sec (c+d x))^2 (3 a A+a (3 A+5 B) \sec (c+d x)) \, dx \\ & = \frac {a B (a+a \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {(3 A+5 B) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{6 d}+\frac {1}{6} \int (a+a \sec (c+d x)) \left (6 a^2 A+15 a^2 (A+B) \sec (c+d x)\right ) \, dx \\ & = a^3 A x+\frac {a B (a+a \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {(3 A+5 B) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{6 d}+\frac {1}{2} \left (5 a^3 (A+B)\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{2} \left (a^3 (7 A+5 B)\right ) \int \sec (c+d x) \, dx \\ & = a^3 A x+\frac {a^3 (7 A+5 B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a B (a+a \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {(3 A+5 B) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{6 d}-\frac {\left (5 a^3 (A+B)\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d} \\ & = a^3 A x+\frac {a^3 (7 A+5 B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {5 a^3 (A+B) \tan (c+d x)}{2 d}+\frac {a B (a+a \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {(3 A+5 B) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{6 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.07 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.63 \[ \int (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {a^3 \left (6 A d x+3 (7 A+5 B) \text {arctanh}(\sin (c+d x))+3 (6 A+8 B+(A+3 B) \sec (c+d x)) \tan (c+d x)+2 B \tan ^3(c+d x)\right )}{6 d} \]

[In]

Integrate[(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x]),x]

[Out]

(a^3*(6*A*d*x + 3*(7*A + 5*B)*ArcTanh[Sin[c + d*x]] + 3*(6*A + 8*B + (A + 3*B)*Sec[c + d*x])*Tan[c + d*x] + 2*
B*Tan[c + d*x]^3))/(6*d)

Maple [A] (verified)

Time = 3.64 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.22

method	result	size
parts	\(a^{3} A x +\frac {\left (a^{3} A +3 B \,a^{3}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (3 a^{3} A +B \,a^{3}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (3 a^{3} A +3 B \,a^{3}\right ) \tan \left (d x +c \right )}{d}-\frac {B \,a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(135\)
parallelrisch	\(\frac {\left (-\frac {21 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A +\frac {5 B}{7}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+\frac {21 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A +\frac {5 B}{7}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+A x d \cos \left (3 d x +3 c \right )+\left (A +3 B \right ) \sin \left (2 d x +2 c \right )+\left (3 A +\frac {11 B}{3}\right ) \sin \left (3 d x +3 c \right )+3 A x d \cos \left (d x +c \right )+3 \sin \left (d x +c \right ) \left (A +\frac {5 B}{3}\right )\right ) a^{3}}{d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(172\)
derivativedivides	\(\frac {a^{3} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 a^{3} A \tan \left (d x +c \right )+3 B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B \,a^{3} \tan \left (d x +c \right )+a^{3} A \left (d x +c \right )+B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(176\)
default	\(\frac {a^{3} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 a^{3} A \tan \left (d x +c \right )+3 B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B \,a^{3} \tan \left (d x +c \right )+a^{3} A \left (d x +c \right )+B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(176\)
norman	\(\frac {a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-a^{3} A x +3 a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\frac {5 a^{3} \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {a^{3} \left (7 A +11 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 a^{3} \left (9 A +10 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-\frac {a^{3} \left (7 A +5 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{3} \left (7 A +5 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(205\)
risch	\(a^{3} A x -\frac {i a^{3} \left (3 A \,{\mathrm e}^{5 i \left (d x +c \right )}+9 B \,{\mathrm e}^{5 i \left (d x +c \right )}-18 A \,{\mathrm e}^{4 i \left (d x +c \right )}-18 B \,{\mathrm e}^{4 i \left (d x +c \right )}-36 A \,{\mathrm e}^{2 i \left (d x +c \right )}-48 B \,{\mathrm e}^{2 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )} A -9 B \,{\mathrm e}^{i \left (d x +c \right )}-18 A -22 B \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}+\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}\)	\(221\)

[In]

int((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.27 \[ \int (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {12 \, A a^{3} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (7 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (7 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (9 \, A + 11 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 3 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right ) + 2 \, B a^{3}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

[In]

integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/12*(12*A*a^3*d*x*cos(d*x + c)^3 + 3*(7*A + 5*B)*a^3*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(7*A + 5*B)*a^3
*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*(9*A + 11*B)*a^3*cos(d*x + c)^2 + 3*(A + 3*B)*a^3*cos(d*x + c) +
2*B*a^3)*sin(d*x + c))/(d*cos(d*x + c)^3)

Sympy [F]

\[ \int (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=a^{3} \left (\int A\, dx + \int 3 A \sec {\left (c + d x \right )}\, dx + \int 3 A \sec ^{2}{\left (c + d x \right )}\, dx + \int A \sec ^{3}{\left (c + d x \right )}\, dx + \int B \sec {\left (c + d x \right )}\, dx + \int 3 B \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 B \sec ^{3}{\left (c + d x \right )}\, dx + \int B \sec ^{4}{\left (c + d x \right )}\, dx\right ) \]

[In]

integrate((a+a*sec(d*x+c))**3*(A+B*sec(d*x+c)),x)

[Out]

a**3*(Integral(A, x) + Integral(3*A*sec(c + d*x), x) + Integral(3*A*sec(c + d*x)**2, x) + Integral(A*sec(c + d
*x)**3, x) + Integral(B*sec(c + d*x), x) + Integral(3*B*sec(c + d*x)**2, x) + Integral(3*B*sec(c + d*x)**3, x)
+ Integral(B*sec(c + d*x)**4, x))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.78 \[ \int (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {12 \, {\left (d x + c\right )} A a^{3} + 4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} - 3 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 9 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 12 \, B a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 36 \, A a^{3} \tan \left (d x + c\right ) + 36 \, B a^{3} \tan \left (d x + c\right )}{12 \, d} \]

[In]

integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(12*(d*x + c)*A*a^3 + 4*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3 - 3*A*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2
- 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 9*B*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(si
n(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 36*A*a^3*log(sec(d*x + c) + tan(d*x + c)) + 12*B*a^3*log(sec(d*x +
c) + tan(d*x + c)) + 36*A*a^3*tan(d*x + c) + 36*B*a^3*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.70 \[ \int (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {6 \, {\left (d x + c\right )} A a^{3} + 3 \, {\left (7 \, A a^{3} + 5 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (7 \, A a^{3} + 5 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 36 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 33 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

[In]

integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/6*(6*(d*x + c)*A*a^3 + 3*(7*A*a^3 + 5*B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(7*A*a^3 + 5*B*a^3)*log(
abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(15*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 15*B*a^3*tan(1/2*d*x + 1/2*c)^5 - 36*A*a
^3*tan(1/2*d*x + 1/2*c)^3 - 40*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 21*A*a^3*tan(1/2*d*x + 1/2*c) + 33*B*a^3*tan(1/2
*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 13.77 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.88 \[ \int (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {2\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {7\,A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {5\,B\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,A\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {11\,B\,a^3\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {3\,B\,a^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {B\,a^3\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3} \]

[In]

int((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^3,x)

[Out]

(2*A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (7*A*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)
))/d + (5*B*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (3*A*a^3*sin(c + d*x))/(d*cos(c + d*x)) + (A
*a^3*sin(c + d*x))/(2*d*cos(c + d*x)^2) + (11*B*a^3*sin(c + d*x))/(3*d*cos(c + d*x)) + (3*B*a^3*sin(c + d*x))/
(2*d*cos(c + d*x)^2) + (B*a^3*sin(c + d*x))/(3*d*cos(c + d*x)^3)