3.4.24 \(\int (a+a \sec (c+d x))^3 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [324]

Optimal. Leaf size=125 \[ \frac {5 a^3 (4 B+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 (4 B+3 C) \tan (c+d x)}{d}+\frac {3 a^3 (4 B+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {a^3 (4 B+3 C) \tan ^3(c+d x)}{12 d} \]

[Out]

5/8*a^3*(4*B+3*C)*arctanh(sin(d*x+c))/d+a^3*(4*B+3*C)*tan(d*x+c)/d+3/8*a^3*(4*B+3*C)*sec(d*x+c)*tan(d*x+c)/d+1
/4*C*(a+a*sec(d*x+c))^3*tan(d*x+c)/d+1/12*a^3*(4*B+3*C)*tan(d*x+c)^3/d

________________________________________________________________________________________

Rubi [A]
time = 0.10, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {4139, 12, 3876, 3855, 3852, 8, 3853} \begin {gather*} \frac {a^3 (4 B+3 C) \tan ^3(c+d x)}{12 d}+\frac {a^3 (4 B+3 C) \tan (c+d x)}{d}+\frac {5 a^3 (4 B+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {3 a^3 (4 B+3 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(5*a^3*(4*B + 3*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^3*(4*B + 3*C)*Tan[c + d*x])/d + (3*a^3*(4*B + 3*C)*Sec[c
+ d*x]*Tan[c + d*x])/(8*d) + (C*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(4*d) + (a^3*(4*B + 3*C)*Tan[c + d*x]^3)/
(12*d)

Rule 8

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

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

Rule 3876

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[n]

Rule 4139

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_.), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[1/(b*(m + 1)
), Int[(a + b*Csc[e + f*x])^m*Simp[A*b*(m + 1) + (a*C*m + b*B*(m + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b
, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {\int a (4 B+3 C) \sec (c+d x) (a+a \sec (c+d x))^3 \, dx}{4 a}\\ &=\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{4} (4 B+3 C) \int \sec (c+d x) (a+a \sec (c+d x))^3 \, dx\\ &=\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{4} (4 B+3 C) \int \left (a^3 \sec (c+d x)+3 a^3 \sec ^2(c+d x)+3 a^3 \sec ^3(c+d x)+a^3 \sec ^4(c+d x)\right ) \, dx\\ &=\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{4} \left (a^3 (4 B+3 C)\right ) \int \sec (c+d x) \, dx+\frac {1}{4} \left (a^3 (4 B+3 C)\right ) \int \sec ^4(c+d x) \, dx+\frac {1}{4} \left (3 a^3 (4 B+3 C)\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{4} \left (3 a^3 (4 B+3 C)\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac {a^3 (4 B+3 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {3 a^3 (4 B+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{8} \left (3 a^3 (4 B+3 C)\right ) \int \sec (c+d x) \, dx-\frac {\left (a^3 (4 B+3 C)\right ) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{4 d}-\frac {\left (3 a^3 (4 B+3 C)\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{4 d}\\ &=\frac {5 a^3 (4 B+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 (4 B+3 C) \tan (c+d x)}{d}+\frac {3 a^3 (4 B+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {a^3 (4 B+3 C) \tan ^3(c+d x)}{12 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.50, size = 81, normalized size = 0.65 \begin {gather*} \frac {a^3 \left (15 (4 B+3 C) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (96 (B+C)+9 (4 B+5 C) \sec (c+d x)+6 C \sec ^3(c+d x)+8 (B+3 C) \tan ^2(c+d x)\right )\right )}{24 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^3*(15*(4*B + 3*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(96*(B + C) + 9*(4*B + 5*C)*Sec[c + d*x] + 6*C*Sec[c
+ d*x]^3 + 8*(B + 3*C)*Tan[c + d*x]^2)))/(24*d)

________________________________________________________________________________________

Maple [A]
time = 0.84, size = 219, normalized size = 1.75

method	result	size
norman	\(\frac {\frac {a^{3} \left (49 C +44 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {73 \left (4 B +3 C \right ) a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {55 \left (4 B +3 C \right ) a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}-\frac {5 \left (4 B +3 C \right ) a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {5 \left (4 B +3 C \right ) a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {5 \left (4 B +3 C \right ) a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(175\)
derivativedivides	\(\frac {-a^{3} B \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a^{3} C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{3} B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 a^{3} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 a^{3} B \tan \left (d x +c \right )+3 a^{3} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{3} C \tan \left (d x +c \right )}{d}\)	\(219\)
default	\(\frac {-a^{3} B \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a^{3} C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{3} B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 a^{3} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 a^{3} B \tan \left (d x +c \right )+3 a^{3} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{3} C \tan \left (d x +c \right )}{d}\)	\(219\)
risch	\(-\frac {i a^{3} \left (36 B \,{\mathrm e}^{7 i \left (d x +c \right )}+45 C \,{\mathrm e}^{7 i \left (d x +c \right )}-72 B \,{\mathrm e}^{6 i \left (d x +c \right )}-24 C \,{\mathrm e}^{6 i \left (d x +c \right )}+36 B \,{\mathrm e}^{5 i \left (d x +c \right )}+69 C \,{\mathrm e}^{5 i \left (d x +c \right )}-264 B \,{\mathrm e}^{4 i \left (d x +c \right )}-216 C \,{\mathrm e}^{4 i \left (d x +c \right )}-36 B \,{\mathrm e}^{3 i \left (d x +c \right )}-69 C \,{\mathrm e}^{3 i \left (d x +c \right )}-280 B \,{\mathrm e}^{2 i \left (d x +c \right )}-264 C \,{\mathrm e}^{2 i \left (d x +c \right )}-36 B \,{\mathrm e}^{i \left (d x +c \right )}-45 C \,{\mathrm e}^{i \left (d x +c \right )}-88 B -72 C \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}-\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}+\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}+\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}\)	\(287\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/d*(-a^3*B*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+a^3*C*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(se
c(d*x+c)+tan(d*x+c)))+3*a^3*B*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))-3*a^3*C*(-2/3-1/3*sec(
d*x+c)^2)*tan(d*x+c)+3*a^3*B*tan(d*x+c)+3*a^3*C*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+a^3*
B*ln(sec(d*x+c)+tan(d*x+c))+a^3*C*tan(d*x+c))

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 262 vs. \(2 (117) = 234\).
time = 0.28, size = 262, normalized size = 2.10 \begin {gather*} \frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} + 48 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 3 \, C a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, 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 \, C 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 )} + 48 \, B a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 144 \, B a^{3} \tan \left (d x + c\right ) + 48 \, C a^{3} \tan \left (d x + c\right )}{48 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3 + 48*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3 - 3*C*a^3*(2*(3*
sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin
(d*x + c) - 1)) - 36*B*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
)) - 36*C*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)) + 48*B*a^3
*log(sec(d*x + c) + tan(d*x + c)) + 144*B*a^3*tan(d*x + c) + 48*C*a^3*tan(d*x + c))/d

________________________________________________________________________________________

Fricas [A]
time = 2.50, size = 145, normalized size = 1.16 \begin {gather*} \frac {15 \, {\left (4 \, B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (11 \, B + 9 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 9 \, {\left (4 \, B + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 8 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + 6 \, C a^{3}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(15*(4*B + 3*C)*a^3*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 15*(4*B + 3*C)*a^3*cos(d*x + c)^4*log(-sin(d*x
+ c) + 1) + 2*(8*(11*B + 9*C)*a^3*cos(d*x + c)^3 + 9*(4*B + 5*C)*a^3*cos(d*x + c)^2 + 8*(B + 3*C)*a^3*cos(d*x
+ c) + 6*C*a^3)*sin(d*x + c))/(d*cos(d*x + c)^4)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{3} \left (\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 + \int C \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 C \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 C \sec ^{4}{\left (c + d x \right )}\, dx + \int C \sec ^{5}{\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**3*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

a**3*(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) + Inte
gral(B*sec(c + d*x)**4, x) + Integral(C*sec(c + d*x)**2, x) + Integral(3*C*sec(c + d*x)**3, x) + Integral(3*C*
sec(c + d*x)**4, x) + Integral(C*sec(c + d*x)**5, x))

________________________________________________________________________________________

Giac [A]
time = 0.54, size = 212, normalized size = 1.70 \begin {gather*} \frac {15 \, {\left (4 \, B a^{3} + 3 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (4 \, B a^{3} + 3 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (60 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 45 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 220 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 165 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 292 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 219 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 132 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 147 \, C 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 )}^{4}}}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(15*(4*B*a^3 + 3*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*B*a^3 + 3*C*a^3)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) - 2*(60*B*a^3*tan(1/2*d*x + 1/2*c)^7 + 45*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 220*B*a^3*tan(1/2*d*x +
1/2*c)^5 - 165*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 292*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 219*C*a^3*tan(1/2*d*x + 1/2*
c)^3 - 132*B*a^3*tan(1/2*d*x + 1/2*c) - 147*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

________________________________________________________________________________________

Mupad [B]
time = 5.42, size = 185, normalized size = 1.48 \begin {gather*} \frac {\left (-5\,B\,a^3-\frac {15\,C\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {55\,B\,a^3}{3}+\frac {55\,C\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {73\,B\,a^3}{3}-\frac {73\,C\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (11\,B\,a^3+\frac {49\,C\,a^3}{4}\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 )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {5\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,B+3\,C\right )}{4\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^3,x)

[Out]

(tan(c/2 + (d*x)/2)*(11*B*a^3 + (49*C*a^3)/4) - tan(c/2 + (d*x)/2)^7*(5*B*a^3 + (15*C*a^3)/4) + tan(c/2 + (d*x
)/2)^5*((55*B*a^3)/3 + (55*C*a^3)/4) - tan(c/2 + (d*x)/2)^3*((73*B*a^3)/3 + (73*C*a^3)/4))/(d*(6*tan(c/2 + (d*
x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) + (5*a^3*atanh(tan(c/2
+ (d*x)/2))*(4*B + 3*C))/(4*d)

________________________________________________________________________________________