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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {4001, 3791, 3770, 3767, 8, 3768} \[ \frac {a^3 (4 A+3 B) \tan ^3(c+d x)}{12 d}+\frac {a^3 (4 A+3 B) \tan (c+d x)}{d}+\frac {5 a^3 (4 A+3 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {3 a^3 (4 A+3 B) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {B \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 3767

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 3768

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 3770

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

Rule 3791

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 4001

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

________________________________________________________________________________________

Mathematica [B]  time = 1.30, size = 273, normalized size = 2.18 \[ -\frac {a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (120 (4 A+3 B) \cos ^4(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-\sec (c) (-24 (11 A+9 B) \sin (c)+(36 A+69 B) \sin (d x)+36 A \sin (2 c+d x)+280 A \sin (c+2 d x)-72 A \sin (3 c+2 d x)+36 A \sin (2 c+3 d x)+36 A \sin (4 c+3 d x)+88 A \sin (3 c+4 d x)+69 B \sin (2 c+d x)+264 B \sin (c+2 d x)-24 B \sin (3 c+2 d x)+45 B \sin (2 c+3 d x)+45 B \sin (4 c+3 d x)+72 B \sin (3 c+4 d x))\right )}{1536 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/1536*(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*Sec[c + d*x]^4*(120*(4*A + 3*B)*Cos[c + d*x]^4*(Log[Cos[(
c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c]*(-24*(11*A + 9*B)*Sin[c]
+ (36*A + 69*B)*Sin[d*x] + 36*A*Sin[2*c + d*x] + 69*B*Sin[2*c + d*x] + 280*A*Sin[c + 2*d*x] + 264*B*Sin[c + 2*
d*x] - 72*A*Sin[3*c + 2*d*x] - 24*B*Sin[3*c + 2*d*x] + 36*A*Sin[2*c + 3*d*x] + 45*B*Sin[2*c + 3*d*x] + 36*A*Si
n[4*c + 3*d*x] + 45*B*Sin[4*c + 3*d*x] + 88*A*Sin[3*c + 4*d*x] + 72*B*Sin[3*c + 4*d*x])))/d

________________________________________________________________________________________

fricas [A]  time = 0.46, size = 145, normalized size = 1.16 \[ \frac {15 \, {\left (4 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (11 \, A + 9 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} + 9 \, {\left (4 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 8 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right ) + 6 \, B a^{3}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.37, size = 212, normalized size = 1.70 \[ \frac {15 \, {\left (4 \, A a^{3} + 3 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (4 \, A a^{3} + 3 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (60 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 45 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 220 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 165 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 292 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 219 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 132 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 147 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 1.38, size = 188, normalized size = 1.50 \[ \frac {5 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {3 a^{3} B \tan \left (d x +c \right )}{d}+\frac {11 A \,a^{3} \tan \left (d x +c \right )}{3 d}+\frac {15 a^{3} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {15 a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {3 A \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{3} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a^{3} B \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/2/d*A*a^3*ln(sec(d*x+c)+tan(d*x+c))+3/d*a^3*B*tan(d*x+c)+11/3/d*A*a^3*tan(d*x+c)+15/8/d*a^3*B*sec(d*x+c)*tan
(d*x+c)+15/8/d*a^3*B*ln(sec(d*x+c)+tan(d*x+c))+3/2/d*A*a^3*sec(d*x+c)*tan(d*x+c)+1/d*a^3*B*tan(d*x+c)*sec(d*x+
c)^2+1/3/d*A*a^3*tan(d*x+c)*sec(d*x+c)^2+1/4/d*a^3*B*tan(d*x+c)*sec(d*x+c)^3

________________________________________________________________________________________

maxima [B]  time = 0.37, size = 262, normalized size = 2.10 \[ \frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 48 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} - 3 \, B a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 144 \, A a^{3} \tan \left (d x + c\right ) + 48 \, B a^{3} \tan \left (d x + c\right )}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^3 + 48*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3 - 3*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)) - 36*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
)) - 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)) + 48*A*a^3
*log(sec(d*x + c) + tan(d*x + c)) + 144*A*a^3*tan(d*x + c) + 48*B*a^3*tan(d*x + c))/d

________________________________________________________________________________________

mupad [B]  time = 4.47, size = 185, normalized size = 1.48 \[ \frac {\left (-5\,A\,a^3-\frac {15\,B\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {55\,A\,a^3}{3}+\frac {55\,B\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {73\,A\,a^3}{3}-\frac {73\,B\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (11\,A\,a^3+\frac {49\,B\,a^3}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {5\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,A+3\,B\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(11*A*a^3 + (49*B*a^3)/4) - tan(c/2 + (d*x)/2)^7*(5*A*a^3 + (15*B*a^3)/4) + tan(c/2 + (d*x
)/2)^5*((55*A*a^3)/3 + (55*B*a^3)/4) - tan(c/2 + (d*x)/2)^3*((73*A*a^3)/3 + (73*B*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*A + 3*B))/(4*d)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{3} \left (\int A \sec {\left (c + d x \right )}\, dx + \int 3 A \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 A \sec ^{3}{\left (c + d x \right )}\, dx + \int A \sec ^{4}{\left (c + d x \right )}\, dx + \int B \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 B \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 B \sec ^{4}{\left (c + d x \right )}\, dx + \int B \sec ^{5}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________