\(\int \sec (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx\) [73]

Optimal result

Integrand size = 29, antiderivative size = 159 \[ \int \sec (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {7 a^4 (5 A+4 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {8 a^4 (5 A+4 B) \tan (c+d x)}{5 d}+\frac {27 a^4 (5 A+4 B) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {a^4 (5 A+4 B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {B (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {4 a^4 (5 A+4 B) \tan ^3(c+d x)}{15 d} \]

[Out]

7/8*a^4*(5*A+4*B)*arctanh(sin(d*x+c))/d+8/5*a^4*(5*A+4*B)*tan(d*x+c)/d+27/40*a^4*(5*A+4*B)*sec(d*x+c)*tan(d*x+
c)/d+1/20*a^4*(5*A+4*B)*sec(d*x+c)^3*tan(d*x+c)/d+1/5*B*(a+a*sec(d*x+c))^4*tan(d*x+c)/d+4/15*a^4*(5*A+4*B)*tan
(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {4086, 3876, 3855, 3852, 8, 3853} \[ \int \sec (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {7 a^4 (5 A+4 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {4 a^4 (5 A+4 B) \tan ^3(c+d x)}{15 d}+\frac {8 a^4 (5 A+4 B) \tan (c+d x)}{5 d}+\frac {a^4 (5 A+4 B) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac {27 a^4 (5 A+4 B) \tan (c+d x) \sec (c+d x)}{40 d}+\frac {B \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d} \]

[In]

Int[Sec[c + d*x]*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]

[Out]

(7*a^4*(5*A + 4*B)*ArcTanh[Sin[c + d*x]])/(8*d) + (8*a^4*(5*A + 4*B)*Tan[c + d*x])/(5*d) + (27*a^4*(5*A + 4*B)
*Sec[c + d*x]*Tan[c + d*x])/(40*d) + (a^4*(5*A + 4*B)*Sec[c + d*x]^3*Tan[c + d*x])/(20*d) + (B*(a + a*Sec[c +
d*x])^4*Tan[c + d*x])/(5*d) + (4*a^4*(5*A + 4*B)*Tan[c + d*x]^3)/(15*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 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 4086

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

Rubi steps \begin{align*} \text {integral}& = \frac {B (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {1}{5} (5 A+4 B) \int \sec (c+d x) (a+a \sec (c+d x))^4 \, dx \\ & = \frac {B (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {1}{5} (5 A+4 B) \int \left (a^4 \sec (c+d x)+4 a^4 \sec ^2(c+d x)+6 a^4 \sec ^3(c+d x)+4 a^4 \sec ^4(c+d x)+a^4 \sec ^5(c+d x)\right ) \, dx \\ & = \frac {B (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {1}{5} \left (a^4 (5 A+4 B)\right ) \int \sec (c+d x) \, dx+\frac {1}{5} \left (a^4 (5 A+4 B)\right ) \int \sec ^5(c+d x) \, dx+\frac {1}{5} \left (4 a^4 (5 A+4 B)\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{5} \left (4 a^4 (5 A+4 B)\right ) \int \sec ^4(c+d x) \, dx+\frac {1}{5} \left (6 a^4 (5 A+4 B)\right ) \int \sec ^3(c+d x) \, dx \\ & = \frac {a^4 (5 A+4 B) \text {arctanh}(\sin (c+d x))}{5 d}+\frac {3 a^4 (5 A+4 B) \sec (c+d x) \tan (c+d x)}{5 d}+\frac {a^4 (5 A+4 B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {B (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {1}{20} \left (3 a^4 (5 A+4 B)\right ) \int \sec ^3(c+d x) \, dx+\frac {1}{5} \left (3 a^4 (5 A+4 B)\right ) \int \sec (c+d x) \, dx-\frac {\left (4 a^4 (5 A+4 B)\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{5 d}-\frac {\left (4 a^4 (5 A+4 B)\right ) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d} \\ & = \frac {4 a^4 (5 A+4 B) \text {arctanh}(\sin (c+d x))}{5 d}+\frac {8 a^4 (5 A+4 B) \tan (c+d x)}{5 d}+\frac {27 a^4 (5 A+4 B) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {a^4 (5 A+4 B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {B (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {4 a^4 (5 A+4 B) \tan ^3(c+d x)}{15 d}+\frac {1}{40} \left (3 a^4 (5 A+4 B)\right ) \int \sec (c+d x) \, dx \\ & = \frac {7 a^4 (5 A+4 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {8 a^4 (5 A+4 B) \tan (c+d x)}{5 d}+\frac {27 a^4 (5 A+4 B) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {a^4 (5 A+4 B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {B (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {4 a^4 (5 A+4 B) \tan ^3(c+d x)}{15 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.02 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.36 \[ \int \sec (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {35 a^4 A \text {arctanh}(\sin (c+d x))}{8 d}+\frac {7 a^4 B \text {arctanh}(\sin (c+d x))}{2 d}+\frac {8 a^4 A \tan (c+d x)}{d}+\frac {8 a^4 B \tan (c+d x)}{d}+\frac {27 a^4 A \sec (c+d x) \tan (c+d x)}{8 d}+\frac {7 a^4 B \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^4 A \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a^4 B \sec ^3(c+d x) \tan (c+d x)}{d}+\frac {4 a^4 A \tan ^3(c+d x)}{3 d}+\frac {8 a^4 B \tan ^3(c+d x)}{3 d}+\frac {a^4 B \tan ^5(c+d x)}{5 d} \]

[In]

Integrate[Sec[c + d*x]*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]

[Out]

(35*a^4*A*ArcTanh[Sin[c + d*x]])/(8*d) + (7*a^4*B*ArcTanh[Sin[c + d*x]])/(2*d) + (8*a^4*A*Tan[c + d*x])/d + (8
*a^4*B*Tan[c + d*x])/d + (27*a^4*A*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (7*a^4*B*Sec[c + d*x]*Tan[c + d*x])/(2*d
) + (a^4*A*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (a^4*B*Sec[c + d*x]^3*Tan[c + d*x])/d + (4*a^4*A*Tan[c + d*x]^
3)/(3*d) + (8*a^4*B*Tan[c + d*x]^3)/(3*d) + (a^4*B*Tan[c + d*x]^5)/(5*d)

Maple [A] (verified)

Time = 5.38 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.26

method	result	size
norman	\(\frac {\frac {79 a^{4} \left (5 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}-\frac {224 a^{4} \left (5 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {49 a^{4} \left (5 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}-\frac {7 a^{4} \left (5 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {a^{4} \left (93 A +100 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}-\frac {7 a^{4} \left (5 A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {7 a^{4} \left (5 A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(201\)
parallelrisch	\(\frac {56 \left (-\frac {75 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (A +\frac {4 B}{5}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{32}+\frac {75 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (A +\frac {4 B}{5}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{32}+\left (\frac {93 A}{112}+\frac {33 B}{28}\right ) \sin \left (2 d x +2 c \right )+\left (\frac {19 A}{14}+\frac {11 B}{8}\right ) \sin \left (3 d x +3 c \right )+\left (\frac {81 A}{224}+\frac {3 B}{8}\right ) \sin \left (4 d x +4 c \right )+\left (\frac {5 A}{14}+\frac {83 B}{280}\right ) \sin \left (5 d x +5 c \right )+\sin \left (d x +c \right ) \left (A +\frac {5 B}{4}\right )\right ) a^{4}}{3 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\)	\(217\)
parts	\(\frac {\left (a^{4} A +4 B \,a^{4}\right ) \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {\left (4 a^{4} A +B \,a^{4}\right ) \tan \left (d x +c \right )}{d}-\frac {\left (4 a^{4} A +6 B \,a^{4}\right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (6 a^{4} A +4 B \,a^{4}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {B \,a^{4} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right ) a^{4}}{d}\)	\(227\)
derivativedivides	\(\frac {a^{4} A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,a^{4} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )-4 a^{4} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+4 B \,a^{4} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+6 a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-6 B \,a^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+4 a^{4} A \tan \left (d x +c \right )+4 B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{4} \tan \left (d x +c \right )}{d}\)	\(303\)
default	\(\frac {a^{4} A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,a^{4} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )-4 a^{4} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+4 B \,a^{4} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+6 a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-6 B \,a^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+4 a^{4} A \tan \left (d x +c \right )+4 B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{4} \tan \left (d x +c \right )}{d}\)	\(303\)
risch	\(-\frac {i a^{4} \left (405 A \,{\mathrm e}^{9 i \left (d x +c \right )}+420 B \,{\mathrm e}^{9 i \left (d x +c \right )}-480 A \,{\mathrm e}^{8 i \left (d x +c \right )}-120 B \,{\mathrm e}^{8 i \left (d x +c \right )}+930 A \,{\mathrm e}^{7 i \left (d x +c \right )}+1320 B \,{\mathrm e}^{7 i \left (d x +c \right )}-2880 A \,{\mathrm e}^{6 i \left (d x +c \right )}-1920 B \,{\mathrm e}^{6 i \left (d x +c \right )}-5120 A \,{\mathrm e}^{4 i \left (d x +c \right )}-4720 B \,{\mathrm e}^{4 i \left (d x +c \right )}-930 A \,{\mathrm e}^{3 i \left (d x +c \right )}-1320 B \,{\mathrm e}^{3 i \left (d x +c \right )}-3520 A \,{\mathrm e}^{2 i \left (d x +c \right )}-3200 B \,{\mathrm e}^{2 i \left (d x +c \right )}-405 \,{\mathrm e}^{i \left (d x +c \right )} A -420 B \,{\mathrm e}^{i \left (d x +c \right )}-800 A -664 B \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {35 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{8 d}-\frac {7 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}+\frac {35 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{8 d}+\frac {7 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}\)	\(311\)

[In]

int(sec(d*x+c)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

(79/6*a^4*(5*A+4*B)/d*tan(1/2*d*x+1/2*c)^3-224/15*a^4*(5*A+4*B)/d*tan(1/2*d*x+1/2*c)^5+49/6*a^4*(5*A+4*B)/d*ta
n(1/2*d*x+1/2*c)^7-7/4*a^4*(5*A+4*B)/d*tan(1/2*d*x+1/2*c)^9-1/4*a^4*(93*A+100*B)/d*tan(1/2*d*x+1/2*c))/(-1+tan
(1/2*d*x+1/2*c)^2)^5-7/8*a^4*(5*A+4*B)/d*ln(tan(1/2*d*x+1/2*c)-1)+7/8*a^4*(5*A+4*B)/d*ln(tan(1/2*d*x+1/2*c)+1)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.04 \[ \int \sec (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {105 \, {\left (5 \, A + 4 \, B\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left (5 \, A + 4 \, B\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (100 \, A + 83 \, B\right )} a^{4} \cos \left (d x + c\right )^{4} + 15 \, {\left (27 \, A + 28 \, B\right )} a^{4} \cos \left (d x + c\right )^{3} + 16 \, {\left (10 \, A + 17 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 30 \, {\left (A + 4 \, B\right )} a^{4} \cos \left (d x + c\right ) + 24 \, B a^{4}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

[In]

integrate(sec(d*x+c)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/240*(105*(5*A + 4*B)*a^4*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 105*(5*A + 4*B)*a^4*cos(d*x + c)^5*log(-sin(
d*x + c) + 1) + 2*(8*(100*A + 83*B)*a^4*cos(d*x + c)^4 + 15*(27*A + 28*B)*a^4*cos(d*x + c)^3 + 16*(10*A + 17*B
)*a^4*cos(d*x + c)^2 + 30*(A + 4*B)*a^4*cos(d*x + c) + 24*B*a^4)*sin(d*x + c))/(d*cos(d*x + c)^5)

Sympy [F]

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

[In]

integrate(sec(d*x+c)*(a+a*sec(d*x+c))**4*(A+B*sec(d*x+c)),x)

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 369 vs. \(2 (147) = 294\).

Time = 0.22 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.32 \[ \int \sec (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {320 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} B a^{4} + 480 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{4} - 15 \, A a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, B a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 360 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 240 \, B a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, A a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 960 \, A a^{4} \tan \left (d x + c\right ) + 240 \, B a^{4} \tan \left (d x + c\right )}{240 \, d} \]

[In]

integrate(sec(d*x+c)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/240*(320*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^4 + 16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c
))*B*a^4 + 480*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^4 - 15*A*a^4*(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*B*a^4*(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)) - 360*A*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) -
240*B*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 240*A*a^4*l
og(sec(d*x + c) + tan(d*x + c)) + 960*A*a^4*tan(d*x + c) + 240*B*a^4*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.55 \[ \int \sec (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {105 \, {\left (5 \, A a^{4} + 4 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 105 \, {\left (5 \, A a^{4} + 4 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (525 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 420 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 2450 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1960 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 4480 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3584 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3950 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3160 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1395 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1500 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

[In]

integrate(sec(d*x+c)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/120*(105*(5*A*a^4 + 4*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 105*(5*A*a^4 + 4*B*a^4)*log(abs(tan(1/2*d*
x + 1/2*c) - 1)) - 2*(525*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 420*B*a^4*tan(1/2*d*x + 1/2*c)^9 - 2450*A*a^4*tan(1/2
*d*x + 1/2*c)^7 - 1960*B*a^4*tan(1/2*d*x + 1/2*c)^7 + 4480*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 3584*B*a^4*tan(1/2*d
*x + 1/2*c)^5 - 3950*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 3160*B*a^4*tan(1/2*d*x + 1/2*c)^3 + 1395*A*a^4*tan(1/2*d*x
+ 1/2*c) + 1500*B*a^4*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 = 16.08 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.41 \[ \int \sec (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {7\,a^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (5\,A+4\,B\right )}{4\,d}-\frac {\left (\frac {35\,A\,a^4}{4}+7\,B\,a^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {245\,A\,a^4}{6}-\frac {98\,B\,a^4}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {224\,A\,a^4}{3}+\frac {896\,B\,a^4}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {395\,A\,a^4}{6}-\frac {158\,B\,a^4}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {93\,A\,a^4}{4}+25\,B\,a^4\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[In]

int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^4)/cos(c + d*x),x)

[Out]

(7*a^4*atanh(tan(c/2 + (d*x)/2))*(5*A + 4*B))/(4*d) - (tan(c/2 + (d*x)/2)*((93*A*a^4)/4 + 25*B*a^4) + tan(c/2
+ (d*x)/2)^9*((35*A*a^4)/4 + 7*B*a^4) - tan(c/2 + (d*x)/2)^7*((245*A*a^4)/6 + (98*B*a^4)/3) - tan(c/2 + (d*x)/
2)^3*((395*A*a^4)/6 + (158*B*a^4)/3) + tan(c/2 + (d*x)/2)^5*((224*A*a^4)/3 + (896*B*a^4)/15))/(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)^1
0 - 1))