\(\int \sec ^2(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx\) [63]

Optimal result

Integrand size = 31, antiderivative size = 163 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {a^3 (15 A+13 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^3 (15 A+13 B) \tan (c+d x)}{5 d}+\frac {3 a^3 (15 A+13 B) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {(5 A-B) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac {B (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac {a^3 (15 A+13 B) \tan ^3(c+d x)}{60 d} \]

[Out]

1/8*a^3*(15*A+13*B)*arctanh(sin(d*x+c))/d+1/5*a^3*(15*A+13*B)*tan(d*x+c)/d+3/40*a^3*(15*A+13*B)*sec(d*x+c)*tan
(d*x+c)/d+1/20*(5*A-B)*(a+a*sec(d*x+c))^3*tan(d*x+c)/d+1/5*B*(a+a*sec(d*x+c))^4*tan(d*x+c)/a/d+1/60*a^3*(15*A+
13*B)*tan(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4095, 4086, 3876, 3855, 3852, 8, 3853} \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {a^3 (15 A+13 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^3 (15 A+13 B) \tan ^3(c+d x)}{60 d}+\frac {a^3 (15 A+13 B) \tan (c+d x)}{5 d}+\frac {3 a^3 (15 A+13 B) \tan (c+d x) \sec (c+d x)}{40 d}+\frac {(5 A-B) \tan (c+d x) (a \sec (c+d x)+a)^3}{20 d}+\frac {B \tan (c+d x) (a \sec (c+d x)+a)^4}{5 a d} \]

[In]

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

[Out]

(a^3*(15*A + 13*B)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^3*(15*A + 13*B)*Tan[c + d*x])/(5*d) + (3*a^3*(15*A + 13*B
)*Sec[c + d*x]*Tan[c + d*x])/(40*d) + ((5*A - B)*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(20*d) + (B*(a + a*Sec[c
+ d*x])^4*Tan[c + d*x])/(5*a*d) + (a^3*(15*A + 13*B)*Tan[c + d*x]^3)/(60*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)]

Rule 4095

Int[csc[(e_.) + (f_.)*(x_)]^2*(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 + 1)/(b*f*(m + 2))), x] + Dist[1/(b*(m + 2)),
Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Fr
eeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {B (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac {\int \sec (c+d x) (a+a \sec (c+d x))^3 (4 a B+a (5 A-B) \sec (c+d x)) \, dx}{5 a} \\ & = \frac {(5 A-B) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac {B (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac {1}{20} (15 A+13 B) \int \sec (c+d x) (a+a \sec (c+d x))^3 \, dx \\ & = \frac {(5 A-B) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac {B (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac {1}{20} (15 A+13 B) \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 {(5 A-B) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac {B (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac {1}{20} \left (a^3 (15 A+13 B)\right ) \int \sec (c+d x) \, dx+\frac {1}{20} \left (a^3 (15 A+13 B)\right ) \int \sec ^4(c+d x) \, dx+\frac {1}{20} \left (3 a^3 (15 A+13 B)\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{20} \left (3 a^3 (15 A+13 B)\right ) \int \sec ^3(c+d x) \, dx \\ & = \frac {a^3 (15 A+13 B) \text {arctanh}(\sin (c+d x))}{20 d}+\frac {3 a^3 (15 A+13 B) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {(5 A-B) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac {B (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac {1}{40} \left (3 a^3 (15 A+13 B)\right ) \int \sec (c+d x) \, dx-\frac {\left (a^3 (15 A+13 B)\right ) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{20 d}-\frac {\left (3 a^3 (15 A+13 B)\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{20 d} \\ & = \frac {a^3 (15 A+13 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^3 (15 A+13 B) \tan (c+d x)}{5 d}+\frac {3 a^3 (15 A+13 B) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {(5 A-B) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac {B (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac {a^3 (15 A+13 B) \tan ^3(c+d x)}{60 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.12 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.63 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {a^3 \left (15 (15 A+13 B) \text {arctanh}(\sin (c+d x))+\left (8 (45 A+38 B)+15 (15 A+13 B) \sec (c+d x)+8 (15 A+19 B) \sec ^2(c+d x)+30 (A+3 B) \sec ^3(c+d x)+24 B \sec ^4(c+d x)\right ) \tan (c+d x)\right )}{120 d} \]

[In]

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

[Out]

(a^3*(15*(15*A + 13*B)*ArcTanh[Sin[c + d*x]] + (8*(45*A + 38*B) + 15*(15*A + 13*B)*Sec[c + d*x] + 8*(15*A + 19
*B)*Sec[c + d*x]^2 + 30*(A + 3*B)*Sec[c + d*x]^3 + 24*B*Sec[c + d*x]^4)*Tan[c + d*x]))/(120*d)

Maple [A] (verified)

Time = 5.02 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.20

method	result	size
parts	\(\frac {\left (a^{3} A +3 B \,a^{3}\right ) \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {\left (3 a^{3} A +B \,a^{3}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {\left (3 a^{3} A +3 B \,a^{3}\right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}-\frac {B \,a^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {a^{3} A \tan \left (d x +c \right )}{d}\)	\(196\)
norman	\(\frac {-\frac {32 a^{3} \left (15 A +13 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {7 a^{3} \left (15 A +13 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}-\frac {a^{3} \left (15 A +13 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {a^{3} \left (49 A +51 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a^{3} \left (183 A +133 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}-\frac {a^{3} \left (15 A +13 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a^{3} \left (15 A +13 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(201\)
parallelrisch	\(\frac {10 \left (-\frac {15 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (A +\frac {13 B}{15}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8}+\frac {15 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (A +\frac {13 B}{15}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8}+\left (\frac {19 A}{20}+\frac {5 B}{4}\right ) \sin \left (2 d x +2 c \right )+\left (\frac {13 A}{10}+\frac {19 B}{15}\right ) \sin \left (3 d x +3 c \right )+\left (\frac {3 A}{8}+\frac {13 B}{40}\right ) \sin \left (4 d x +4 c \right )+\left (\frac {3 A}{10}+\frac {19 B}{75}\right ) \sin \left (5 d x +5 c \right )+\sin \left (d x +c \right ) \left (A +\frac {4 B}{3}\right )\right ) a^{3}}{d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\)	\(217\)
derivativedivides	\(\frac {a^{3} A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,a^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )-3 a^{3} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 B \,a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{3} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 B \,a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a^{3} A \tan \left (d x +c \right )+B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(271\)
default	\(\frac {a^{3} A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,a^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )-3 a^{3} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 B \,a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{3} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 B \,a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a^{3} A \tan \left (d x +c \right )+B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(271\)
risch	\(-\frac {i a^{3} \left (225 A \,{\mathrm e}^{9 i \left (d x +c \right )}+195 B \,{\mathrm e}^{9 i \left (d x +c \right )}-120 A \,{\mathrm e}^{8 i \left (d x +c \right )}+570 A \,{\mathrm e}^{7 i \left (d x +c \right )}+750 B \,{\mathrm e}^{7 i \left (d x +c \right )}-1200 A \,{\mathrm e}^{6 i \left (d x +c \right )}-720 B \,{\mathrm e}^{6 i \left (d x +c \right )}-2400 A \,{\mathrm e}^{4 i \left (d x +c \right )}-2320 B \,{\mathrm e}^{4 i \left (d x +c \right )}-570 A \,{\mathrm e}^{3 i \left (d x +c \right )}-750 B \,{\mathrm e}^{3 i \left (d x +c \right )}-1680 A \,{\mathrm e}^{2 i \left (d x +c \right )}-1520 B \,{\mathrm e}^{2 i \left (d x +c \right )}-225 \,{\mathrm e}^{i \left (d x +c \right )} A -195 B \,{\mathrm e}^{i \left (d x +c \right )}-360 A -304 B \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{8 d}+\frac {13 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{8 d}-\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{8 d}-\frac {13 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{8 d}\)	\(299\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.01 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {15 \, {\left (15 \, A + 13 \, B\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (15 \, A + 13 \, B\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (45 \, A + 38 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} + 15 \, {\left (15 \, A + 13 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} + 8 \, {\left (15 \, A + 19 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 30 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right ) + 24 \, B a^{3}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

[In]

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

[Out]

1/240*(15*(15*A + 13*B)*a^3*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(15*A + 13*B)*a^3*cos(d*x + c)^5*log(-si
n(d*x + c) + 1) + 2*(8*(45*A + 38*B)*a^3*cos(d*x + c)^4 + 15*(15*A + 13*B)*a^3*cos(d*x + c)^3 + 8*(15*A + 19*B
)*a^3*cos(d*x + c)^2 + 30*(A + 3*B)*a^3*cos(d*x + c) + 24*B*a^3)*sin(d*x + c))/(d*cos(d*x + c)^5)

Sympy [F]

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

[In]

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

[Out]

a**3*(Integral(A*sec(c + d*x)**2, x) + Integral(3*A*sec(c + d*x)**3, x) + Integral(3*A*sec(c + d*x)**4, x) + I
ntegral(A*sec(c + d*x)**5, x) + Integral(B*sec(c + d*x)**3, x) + Integral(3*B*sec(c + d*x)**4, x) + Integral(3
*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. 337 vs. \(2 (151) = 302\).

Time = 0.22 (sec) , antiderivative size = 337, normalized size of antiderivative = 2.07 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} B a^{3} + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} - 15 \, A a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 45 \, B a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, A a^{3} \tan \left (d x + c\right )}{240 \, d} \]

[In]

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

[Out]

1/240*(240*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^3 + 16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c
))*B*a^3 + 240*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3 - 15*A*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)) - 45*B*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)) - 180*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)) -
60*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)) + 240*A*a^3*ta
n(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.51 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {15 \, {\left (15 \, A a^{3} + 13 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (15 \, A a^{3} + 13 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (225 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 195 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 1050 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 910 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1920 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1664 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 1830 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1330 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 735 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 765 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

[In]

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

[Out]

1/120*(15*(15*A*a^3 + 13*B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(15*A*a^3 + 13*B*a^3)*log(abs(tan(1/2*
d*x + 1/2*c) - 1)) - 2*(225*A*a^3*tan(1/2*d*x + 1/2*c)^9 + 195*B*a^3*tan(1/2*d*x + 1/2*c)^9 - 1050*A*a^3*tan(1
/2*d*x + 1/2*c)^7 - 910*B*a^3*tan(1/2*d*x + 1/2*c)^7 + 1920*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 1664*B*a^3*tan(1/2*
d*x + 1/2*c)^5 - 1830*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 1330*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 735*A*a^3*tan(1/2*d*x
+ 1/2*c) + 765*B*a^3*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 = 16.11 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.37 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (15\,A+13\,B\right )}{4\,d}-\frac {\left (\frac {15\,A\,a^3}{4}+\frac {13\,B\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {35\,A\,a^3}{2}-\frac {91\,B\,a^3}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (32\,A\,a^3+\frac {416\,B\,a^3}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {61\,A\,a^3}{2}-\frac {133\,B\,a^3}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {49\,A\,a^3}{4}+\frac {51\,B\,a^3}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[In]

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

[Out]

(a^3*atanh(tan(c/2 + (d*x)/2))*(15*A + 13*B))/(4*d) - (tan(c/2 + (d*x)/2)*((49*A*a^3)/4 + (51*B*a^3)/4) + tan(
c/2 + (d*x)/2)^9*((15*A*a^3)/4 + (13*B*a^3)/4) - tan(c/2 + (d*x)/2)^7*((35*A*a^3)/2 + (91*B*a^3)/6) - tan(c/2
+ (d*x)/2)^3*((61*A*a^3)/2 + (133*B*a^3)/6) + tan(c/2 + (d*x)/2)^5*(32*A*a^3 + (416*B*a^3)/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)
^10 - 1))