\(\int \sec ^3(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx\) [53]

Optimal result

Integrand size = 31, antiderivative size = 169 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {a^2 (7 A+6 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^2 (10 A+9 B) \tan (c+d x)}{5 d}+\frac {a^2 (7 A+6 B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 (5 A+6 B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {B \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{5 d}+\frac {a^2 (10 A+9 B) \tan ^3(c+d x)}{15 d} \]

[Out]

1/8*a^2*(7*A+6*B)*arctanh(sin(d*x+c))/d+1/5*a^2*(10*A+9*B)*tan(d*x+c)/d+1/8*a^2*(7*A+6*B)*sec(d*x+c)*tan(d*x+c
)/d+1/20*a^2*(5*A+6*B)*sec(d*x+c)^3*tan(d*x+c)/d+1/5*B*sec(d*x+c)^3*(a^2+a^2*sec(d*x+c))*tan(d*x+c)/d+1/15*a^2
*(10*A+9*B)*tan(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4103, 4082, 3872, 3853, 3855, 3852} \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {a^2 (7 A+6 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^2 (10 A+9 B) \tan ^3(c+d x)}{15 d}+\frac {a^2 (10 A+9 B) \tan (c+d x)}{5 d}+\frac {a^2 (5 A+6 B) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac {a^2 (7 A+6 B) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {B \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d} \]

[In]

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

[Out]

(a^2*(7*A + 6*B)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^2*(10*A + 9*B)*Tan[c + d*x])/(5*d) + (a^2*(7*A + 6*B)*Sec[c
+ d*x]*Tan[c + d*x])/(8*d) + (a^2*(5*A + 6*B)*Sec[c + d*x]^3*Tan[c + d*x])/(20*d) + (B*Sec[c + d*x]^3*(a^2 +
a^2*Sec[c + d*x])*Tan[c + d*x])/(5*d) + (a^2*(10*A + 9*B)*Tan[c + d*x]^3)/(15*d)

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 3872

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

Rule 4082

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Dist[1/(n + 1), Int[(d
*Csc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e,
f, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[n, -1]

Rule 4103

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

Rubi steps \begin{align*} \text {integral}& = \frac {B \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{5 d}+\frac {1}{5} \int \sec ^3(c+d x) (a+a \sec (c+d x)) (a (5 A+3 B)+a (5 A+6 B) \sec (c+d x)) \, dx \\ & = \frac {a^2 (5 A+6 B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {B \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{5 d}+\frac {1}{20} \int \sec ^3(c+d x) \left (5 a^2 (7 A+6 B)+4 a^2 (10 A+9 B) \sec (c+d x)\right ) \, dx \\ & = \frac {a^2 (5 A+6 B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {B \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{5 d}+\frac {1}{4} \left (a^2 (7 A+6 B)\right ) \int \sec ^3(c+d x) \, dx+\frac {1}{5} \left (a^2 (10 A+9 B)\right ) \int \sec ^4(c+d x) \, dx \\ & = \frac {a^2 (7 A+6 B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 (5 A+6 B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {B \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{5 d}+\frac {1}{8} \left (a^2 (7 A+6 B)\right ) \int \sec (c+d x) \, dx-\frac {\left (a^2 (10 A+9 B)\right ) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d} \\ & = \frac {a^2 (7 A+6 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^2 (10 A+9 B) \tan (c+d x)}{5 d}+\frac {a^2 (7 A+6 B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 (5 A+6 B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {B \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{5 d}+\frac {a^2 (10 A+9 B) \tan ^3(c+d x)}{15 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.74 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.60 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {a^2 \left (15 (7 A+6 B) \text {arctanh}(\sin (c+d x))+\sec (c+d x) \left (15 (7 A+6 B)+8 (10 A+9 B) (2+\cos (2 (c+d x))) \sec (c+d x)+30 (A+2 B) \sec ^2(c+d x)+24 B \sec ^3(c+d x)\right ) \tan (c+d x)\right )}{120 d} \]

[In]

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

[Out]

(a^2*(15*(7*A + 6*B)*ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*(15*(7*A + 6*B) + 8*(10*A + 9*B)*(2 + Cos[2*(c + d*x
)])*Sec[c + d*x] + 30*(A + 2*B)*Sec[c + d*x]^2 + 24*B*Sec[c + d*x]^3)*Tan[c + d*x]))/(120*d)

Maple [A] (verified)

Time = 4.28 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.02

method	result	size
parts	\(\frac {\left (A \,a^{2}+2 B \,a^{2}\right ) \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {\left (2 A \,a^{2}+B \,a^{2}\right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {B \,a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(173\)
norman	\(\frac {\frac {7 a^{2} \left (7 A +6 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}-\frac {a^{2} \left (7 A +6 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {a^{2} \left (25 A +26 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {8 a^{2} \left (25 A +27 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {a^{2} \left (79 A +54 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}-\frac {a^{2} \left (7 A +6 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a^{2} \left (7 A +6 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(201\)
parallelrisch	\(\frac {16 \left (-\frac {105 \left (A +\frac {6 B}{7}\right ) \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{64}+\frac {105 \left (A +\frac {6 B}{7}\right ) \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{64}+\left (\frac {33 A}{32}+\frac {21 B}{16}\right ) \sin \left (2 d x +2 c \right )+\left (\frac {5 A}{4}+\frac {9 B}{8}\right ) \sin \left (3 d x +3 c \right )+\left (\frac {21 A}{64}+\frac {9 B}{32}\right ) \sin \left (4 d x +4 c \right )+\left (\frac {A}{4}+\frac {9 B}{40}\right ) \sin \left (5 d x +5 c \right )+\sin \left (d x +c \right ) \left (A +\frac {3 B}{2}\right )\right ) a^{2}}{3 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\)	\(217\)
derivativedivides	\(\frac {A \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )-2 A \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 B \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(223\)
default	\(\frac {A \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )-2 A \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 B \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(223\)
risch	\(-\frac {i a^{2} \left (105 A \,{\mathrm e}^{9 i \left (d x +c \right )}+90 B \,{\mathrm e}^{9 i \left (d x +c \right )}+330 A \,{\mathrm e}^{7 i \left (d x +c \right )}+420 B \,{\mathrm e}^{7 i \left (d x +c \right )}-480 A \,{\mathrm e}^{6 i \left (d x +c \right )}-240 B \,{\mathrm e}^{6 i \left (d x +c \right )}-1120 A \,{\mathrm e}^{4 i \left (d x +c \right )}-1200 B \,{\mathrm e}^{4 i \left (d x +c \right )}-330 A \,{\mathrm e}^{3 i \left (d x +c \right )}-420 B \,{\mathrm e}^{3 i \left (d x +c \right )}-800 A \,{\mathrm e}^{2 i \left (d x +c \right )}-720 B \,{\mathrm e}^{2 i \left (d x +c \right )}-105 \,{\mathrm e}^{i \left (d x +c \right )} A -90 B \,{\mathrm e}^{i \left (d x +c \right )}-160 A -144 B \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {7 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{8 d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{4 d}-\frac {7 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{8 d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{4 d}\)	\(287\)

[In]

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

[Out]

(A*a^2+2*B*a^2)/d*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))-(2*A*a^2+B*a^
2)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+A*a^2/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))-B*a^
2/d*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.98 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {15 \, {\left (7 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (7 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (10 \, A + 9 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 15 \, {\left (7 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (10 \, A + 9 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 30 \, {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right ) + 24 \, B a^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

[In]

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

[Out]

1/240*(15*(7*A + 6*B)*a^2*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(7*A + 6*B)*a^2*cos(d*x + c)^5*log(-sin(d*
x + c) + 1) + 2*(16*(10*A + 9*B)*a^2*cos(d*x + c)^4 + 15*(7*A + 6*B)*a^2*cos(d*x + c)^3 + 8*(10*A + 9*B)*a^2*c
os(d*x + c)^2 + 30*(A + 2*B)*a^2*cos(d*x + c) + 24*B*a^2)*sin(d*x + c))/(d*cos(d*x + c)^5)

Sympy [F]

\[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=a^{2} \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 A \sec ^{4}{\left (c + d x \right )}\, dx + \int A \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sec ^{4}{\left (c + d x \right )}\, dx + \int 2 B \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sec ^{6}{\left (c + d x \right )}\, dx\right ) \]

[In]

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

[Out]

a**2*(Integral(A*sec(c + d*x)**3, x) + Integral(2*A*sec(c + d*x)**4, x) + Integral(A*sec(c + d*x)**5, x) + Int
egral(B*sec(c + d*x)**4, x) + Integral(2*B*sec(c + d*x)**5, x) + Integral(B*sec(c + d*x)**6, x))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.64 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} B a^{2} + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} - 15 \, A a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 30 \, B a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]

[In]

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

[Out]

1/240*(160*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^2 + 16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c
))*B*a^2 + 80*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2 - 15*A*a^2*(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)) - 30*B*a^2*(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)) - 60*A*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.46 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {15 \, {\left (7 \, A a^{2} + 6 \, B a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (7 \, A a^{2} + 6 \, B a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (105 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 90 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 490 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 420 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 800 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 864 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 790 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 540 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 375 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 390 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

[In]

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

[Out]

1/120*(15*(7*A*a^2 + 6*B*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(7*A*a^2 + 6*B*a^2)*log(abs(tan(1/2*d*x
+ 1/2*c) - 1)) - 2*(105*A*a^2*tan(1/2*d*x + 1/2*c)^9 + 90*B*a^2*tan(1/2*d*x + 1/2*c)^9 - 490*A*a^2*tan(1/2*d*x
+ 1/2*c)^7 - 420*B*a^2*tan(1/2*d*x + 1/2*c)^7 + 800*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 864*B*a^2*tan(1/2*d*x + 1/
2*c)^5 - 790*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 540*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 375*A*a^2*tan(1/2*d*x + 1/2*c)
+ 390*B*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d

Mupad [B] (verification not implemented)

Time = 16.34 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.33 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (7\,A+6\,B\right )}{4\,d}-\frac {\left (\frac {7\,A\,a^2}{4}+\frac {3\,B\,a^2}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {49\,A\,a^2}{6}-7\,B\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {40\,A\,a^2}{3}+\frac {72\,B\,a^2}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {79\,A\,a^2}{6}-9\,B\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {25\,A\,a^2}{4}+\frac {13\,B\,a^2}{2}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[In]

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

[Out]

(a^2*atanh(tan(c/2 + (d*x)/2))*(7*A + 6*B))/(4*d) - (tan(c/2 + (d*x)/2)*((25*A*a^2)/4 + (13*B*a^2)/2) + tan(c/
2 + (d*x)/2)^9*((7*A*a^2)/4 + (3*B*a^2)/2) - tan(c/2 + (d*x)/2)^7*((49*A*a^2)/6 + 7*B*a^2) - tan(c/2 + (d*x)/2
)^3*((79*A*a^2)/6 + 9*B*a^2) + tan(c/2 + (d*x)/2)^5*((40*A*a^2)/3 + (72*B*a^2)/5))/(d*(5*tan(c/2 + (d*x)/2)^2
- 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))