3.785 \(\int \sec (c+d x) (a+b \sec (c+d x))^3 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=252 \[ \frac {\left (-3 a^2 C+15 a b B+16 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{60 b d}+\frac {\left (4 a^3 C+12 a^2 b B+9 a b^2 C+3 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (-6 a^3 C+30 a^2 b B+71 a b^2 C+45 b^3 B\right ) \tan (c+d x) \sec (c+d x)}{120 d}+\frac {\left (-3 a^4 C+15 a^3 b B+52 a^2 b^2 C+60 a b^3 B+16 b^4 C\right ) \tan (c+d x)}{30 b d}+\frac {(5 b B-a C) \tan (c+d x) (a+b \sec (c+d x))^3}{20 b d}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^4}{5 b d} \]

[Out]

1/8*(12*B*a^2*b+3*B*b^3+4*C*a^3+9*C*a*b^2)*arctanh(sin(d*x+c))/d+1/30*(15*B*a^3*b+60*B*a*b^3-3*C*a^4+52*C*a^2*
b^2+16*C*b^4)*tan(d*x+c)/b/d+1/120*(30*B*a^2*b+45*B*b^3-6*C*a^3+71*C*a*b^2)*sec(d*x+c)*tan(d*x+c)/d+1/60*(15*B
*a*b-3*C*a^2+16*C*b^2)*(a+b*sec(d*x+c))^2*tan(d*x+c)/b/d+1/20*(5*B*b-C*a)*(a+b*sec(d*x+c))^3*tan(d*x+c)/b/d+1/
5*C*(a+b*sec(d*x+c))^4*tan(d*x+c)/b/d

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Rubi [A]  time = 0.50, antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {4072, 4010, 4002, 3997, 3787, 3770, 3767, 8} \[ \frac {\left (52 a^2 b^2 C+15 a^3 b B-3 a^4 C+60 a b^3 B+16 b^4 C\right ) \tan (c+d x)}{30 b d}+\frac {\left (12 a^2 b B+4 a^3 C+9 a b^2 C+3 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (-3 a^2 C+15 a b B+16 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{60 b d}+\frac {\left (30 a^2 b B-6 a^3 C+71 a b^2 C+45 b^3 B\right ) \tan (c+d x) \sec (c+d x)}{120 d}+\frac {(5 b B-a C) \tan (c+d x) (a+b \sec (c+d x))^3}{20 b d}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^4}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((12*a^2*b*B + 3*b^3*B + 4*a^3*C + 9*a*b^2*C)*ArcTanh[Sin[c + d*x]])/(8*d) + ((15*a^3*b*B + 60*a*b^3*B - 3*a^4
*C + 52*a^2*b^2*C + 16*b^4*C)*Tan[c + d*x])/(30*b*d) + ((30*a^2*b*B + 45*b^3*B - 6*a^3*C + 71*a*b^2*C)*Sec[c +
d*x]*Tan[c + d*x])/(120*d) + ((15*a*b*B - 3*a^2*C + 16*b^2*C)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(60*b*d) +
((5*b*B - a*C)*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(20*b*d) + (C*(a + b*Sec[c + d*x])^4*Tan[c + d*x])/(5*b*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 3770

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

Rule 3787

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 3997

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*C
sc[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 4002

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[1/(m + 1), Int[Csc[e + f*x
]*(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1))*Csc[e + f*x], x], x], x] /; Fr
eeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]

Rule 4010

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)), I
nt[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] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rule 4072

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

Rubi steps

\begin {align*} \int \sec (c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sec ^2(c+d x) (a+b \sec (c+d x))^3 (B+C \sec (c+d x)) \, dx\\ &=\frac {C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^3 (4 b C+(5 b B-a C) \sec (c+d x)) \, dx}{5 b}\\ &=\frac {(5 b B-a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac {C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (b (15 b B+13 a C)+\left (15 a b B-3 a^2 C+16 b^2 C\right ) \sec (c+d x)\right ) \, dx}{20 b}\\ &=\frac {\left (15 a b B-3 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}+\frac {(5 b B-a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac {C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x)) \left (b \left (75 a b B+33 a^2 C+32 b^2 C\right )+\left (30 a^2 b B+45 b^3 B-6 a^3 C+71 a b^2 C\right ) \sec (c+d x)\right ) \, dx}{60 b}\\ &=\frac {\left (30 a^2 b B+45 b^3 B-6 a^3 C+71 a b^2 C\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac {\left (15 a b B-3 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}+\frac {(5 b B-a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac {C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) \left (15 b \left (12 a^2 b B+3 b^3 B+4 a^3 C+9 a b^2 C\right )+4 \left (15 a^3 b B+60 a b^3 B-3 a^4 C+52 a^2 b^2 C+16 b^4 C\right ) \sec (c+d x)\right ) \, dx}{120 b}\\ &=\frac {\left (30 a^2 b B+45 b^3 B-6 a^3 C+71 a b^2 C\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac {\left (15 a b B-3 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}+\frac {(5 b B-a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac {C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac {1}{8} \left (12 a^2 b B+3 b^3 B+4 a^3 C+9 a b^2 C\right ) \int \sec (c+d x) \, dx+\frac {\left (15 a^3 b B+60 a b^3 B-3 a^4 C+52 a^2 b^2 C+16 b^4 C\right ) \int \sec ^2(c+d x) \, dx}{30 b}\\ &=\frac {\left (12 a^2 b B+3 b^3 B+4 a^3 C+9 a b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (30 a^2 b B+45 b^3 B-6 a^3 C+71 a b^2 C\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac {\left (15 a b B-3 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}+\frac {(5 b B-a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac {C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}-\frac {\left (15 a^3 b B+60 a b^3 B-3 a^4 C+52 a^2 b^2 C+16 b^4 C\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{30 b d}\\ &=\frac {\left (12 a^2 b B+3 b^3 B+4 a^3 C+9 a b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (15 a^3 b B+60 a b^3 B-3 a^4 C+52 a^2 b^2 C+16 b^4 C\right ) \tan (c+d x)}{30 b d}+\frac {\left (30 a^2 b B+45 b^3 B-6 a^3 C+71 a b^2 C\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac {\left (15 a b B-3 a^2 C+16 b^2 C\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}+\frac {(5 b B-a C) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac {C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}\\ \end {align*}

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Mathematica [A]  time = 6.15, size = 211, normalized size = 0.84 \[ \frac {a^3 B \tan (c+d x)}{d}+\frac {a^2 (a C+3 b B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^2 (a C+3 b B) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {b^2 (3 a C+b B) \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {3 b^2 (3 a C+b B) \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )}{8 d}+\frac {a b (a C+b B) \left (\tan ^3(c+d x)+3 \tan (c+d x)\right )}{d}+\frac {b^3 C \left (3 \tan ^5(c+d x)+10 \tan ^3(c+d x)+15 \tan (c+d x)\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(3*b*B + a*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (a^3*B*Tan[c + d*x])/d + (a^2*(3*b*B + a*C)*Sec[c + d*x]*Tan
[c + d*x])/(2*d) + (b^2*(b*B + 3*a*C)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (3*b^2*(b*B + 3*a*C)*(ArcTanh[Sin[c
+ d*x]] + Sec[c + d*x]*Tan[c + d*x]))/(8*d) + (a*b*(b*B + a*C)*(3*Tan[c + d*x] + Tan[c + d*x]^3))/d + (b^3*C*
(15*Tan[c + d*x] + 10*Tan[c + d*x]^3 + 3*Tan[c + d*x]^5))/(15*d)

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fricas [A]  time = 0.76, size = 249, normalized size = 0.99 \[ \frac {15 \, {\left (4 \, C a^{3} + 12 \, B a^{2} b + 9 \, C a b^{2} + 3 \, B b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, C a^{3} + 12 \, B a^{2} b + 9 \, C a b^{2} + 3 \, B b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (15 \, B a^{3} + 30 \, C a^{2} b + 30 \, B a b^{2} + 8 \, C b^{3}\right )} \cos \left (d x + c\right )^{4} + 24 \, C b^{3} + 15 \, {\left (4 \, C a^{3} + 12 \, B a^{2} b + 9 \, C a b^{2} + 3 \, B b^{3}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (15 \, C a^{2} b + 15 \, B a b^{2} + 4 \, C b^{3}\right )} \cos \left (d x + c\right )^{2} + 30 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(4*C*a^3 + 12*B*a^2*b + 9*C*a*b^2 + 3*B*b^3)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(4*C*a^3 + 12
*B*a^2*b + 9*C*a*b^2 + 3*B*b^3)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(8*(15*B*a^3 + 30*C*a^2*b + 30*B*a*b
^2 + 8*C*b^3)*cos(d*x + c)^4 + 24*C*b^3 + 15*(4*C*a^3 + 12*B*a^2*b + 9*C*a*b^2 + 3*B*b^3)*cos(d*x + c)^3 + 8*(
15*C*a^2*b + 15*B*a*b^2 + 4*C*b^3)*cos(d*x + c)^2 + 30*(3*C*a*b^2 + B*b^3)*cos(d*x + c))*sin(d*x + c))/(d*cos(
d*x + c)^5)

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giac [B]  time = 0.36, size = 722, normalized size = 2.87 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(4*C*a^3 + 12*B*a^2*b + 9*C*a*b^2 + 3*B*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*C*a^3 + 12*B
*a^2*b + 9*C*a*b^2 + 3*B*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(120*B*a^3*tan(1/2*d*x + 1/2*c)^9 - 60*C*
a^3*tan(1/2*d*x + 1/2*c)^9 - 180*B*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 360*C*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 360*B*a
*b^2*tan(1/2*d*x + 1/2*c)^9 - 225*C*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 75*B*b^3*tan(1/2*d*x + 1/2*c)^9 + 120*C*b^3
*tan(1/2*d*x + 1/2*c)^9 - 480*B*a^3*tan(1/2*d*x + 1/2*c)^7 + 120*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 360*B*a^2*b*ta
n(1/2*d*x + 1/2*c)^7 - 960*C*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 960*B*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 90*C*a*b^2*ta
n(1/2*d*x + 1/2*c)^7 + 30*B*b^3*tan(1/2*d*x + 1/2*c)^7 - 160*C*b^3*tan(1/2*d*x + 1/2*c)^7 + 720*B*a^3*tan(1/2*
d*x + 1/2*c)^5 + 1200*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 1200*B*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 464*C*b^3*tan(1/2
*d*x + 1/2*c)^5 - 480*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 120*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 360*B*a^2*b*tan(1/2*d*
x + 1/2*c)^3 - 960*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 960*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 90*C*a*b^2*tan(1/2*d*
x + 1/2*c)^3 - 30*B*b^3*tan(1/2*d*x + 1/2*c)^3 - 160*C*b^3*tan(1/2*d*x + 1/2*c)^3 + 120*B*a^3*tan(1/2*d*x + 1/
2*c) + 60*C*a^3*tan(1/2*d*x + 1/2*c) + 180*B*a^2*b*tan(1/2*d*x + 1/2*c) + 360*C*a^2*b*tan(1/2*d*x + 1/2*c) + 3
60*B*a*b^2*tan(1/2*d*x + 1/2*c) + 225*C*a*b^2*tan(1/2*d*x + 1/2*c) + 75*B*b^3*tan(1/2*d*x + 1/2*c) + 120*C*b^3
*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d

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maple [A]  time = 2.04, size = 382, normalized size = 1.52 \[ \frac {a^{3} B \tan \left (d x +c \right )}{d}+\frac {C \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {3 a^{2} b B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {3 a^{2} b B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 C \,a^{2} b \tan \left (d x +c \right )}{d}+\frac {C \,a^{2} b \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {2 B a \,b^{2} \tan \left (d x +c \right )}{d}+\frac {B a \,b^{2} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {3 C a \,b^{2} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {9 C a \,b^{2} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {9 C a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {b^{3} B \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 b^{3} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 b^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {8 b^{3} C \tan \left (d x +c \right )}{15 d}+\frac {b^{3} C \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {4 b^{3} C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*a^3*B*tan(d*x+c)+1/2/d*C*a^3*sec(d*x+c)*tan(d*x+c)+1/2/d*C*a^3*ln(sec(d*x+c)+tan(d*x+c))+3/2/d*a^2*b*B*sec
(d*x+c)*tan(d*x+c)+3/2/d*a^2*b*B*ln(sec(d*x+c)+tan(d*x+c))+2/d*C*a^2*b*tan(d*x+c)+1/d*C*a^2*b*tan(d*x+c)*sec(d
*x+c)^2+2/d*B*a*b^2*tan(d*x+c)+1/d*B*a*b^2*tan(d*x+c)*sec(d*x+c)^2+3/4/d*C*a*b^2*tan(d*x+c)*sec(d*x+c)^3+9/8/d
*C*a*b^2*sec(d*x+c)*tan(d*x+c)+9/8/d*C*a*b^2*ln(sec(d*x+c)+tan(d*x+c))+1/4/d*b^3*B*tan(d*x+c)*sec(d*x+c)^3+3/8
/d*b^3*B*sec(d*x+c)*tan(d*x+c)+3/8/d*b^3*B*ln(sec(d*x+c)+tan(d*x+c))+8/15/d*b^3*C*tan(d*x+c)+1/5/d*b^3*C*tan(d
*x+c)*sec(d*x+c)^4+4/15/d*b^3*C*tan(d*x+c)*sec(d*x+c)^2

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maxima [A]  time = 0.33, size = 341, normalized size = 1.35 \[ \frac {240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} b + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a b^{2} + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C b^{3} - 45 \, C a b^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, B b^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, B a^{2} b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, B a^{3} \tan \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(240*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2*b + 240*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a*b^2 + 16*(3*t
an(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C*b^3 - 45*C*a*b^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)) - 15*B*b^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(si
n(d*x + c) - 1)) - 60*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)) - 180*B*a^2*b*(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^3*tan(d*x + c))/d

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mupad [B]  time = 7.73, size = 470, normalized size = 1.87 \[ \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {C\,a^3}{2}+\frac {3\,B\,a^2\,b}{2}+\frac {9\,C\,a\,b^2}{8}+\frac {3\,B\,b^3}{8}\right )}{2\,C\,a^3+6\,B\,a^2\,b+\frac {9\,C\,a\,b^2}{2}+\frac {3\,B\,b^3}{2}}\right )\,\left (C\,a^3+3\,B\,a^2\,b+\frac {9\,C\,a\,b^2}{4}+\frac {3\,B\,b^3}{4}\right )}{d}-\frac {\left (2\,B\,a^3-\frac {5\,B\,b^3}{4}-C\,a^3+2\,C\,b^3+6\,B\,a\,b^2-3\,B\,a^2\,b-\frac {15\,C\,a\,b^2}{4}+6\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {B\,b^3}{2}-8\,B\,a^3+2\,C\,a^3-\frac {8\,C\,b^3}{3}-16\,B\,a\,b^2+6\,B\,a^2\,b+\frac {3\,C\,a\,b^2}{2}-16\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (12\,B\,a^3+20\,C\,a^2\,b+20\,B\,a\,b^2+\frac {116\,C\,b^3}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-8\,B\,a^3-\frac {B\,b^3}{2}-2\,C\,a^3-\frac {8\,C\,b^3}{3}-16\,B\,a\,b^2-6\,B\,a^2\,b-\frac {3\,C\,a\,b^2}{2}-16\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,B\,a^3+\frac {5\,B\,b^3}{4}+C\,a^3+2\,C\,b^3+6\,B\,a\,b^2+3\,B\,a^2\,b+\frac {15\,C\,a\,b^2}{4}+6\,C\,a^2\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atanh((4*tan(c/2 + (d*x)/2)*((3*B*b^3)/8 + (C*a^3)/2 + (3*B*a^2*b)/2 + (9*C*a*b^2)/8))/((3*B*b^3)/2 + 2*C*a^3
+ 6*B*a^2*b + (9*C*a*b^2)/2))*((3*B*b^3)/4 + C*a^3 + 3*B*a^2*b + (9*C*a*b^2)/4))/d - (tan(c/2 + (d*x)/2)*(2*B
*a^3 + (5*B*b^3)/4 + C*a^3 + 2*C*b^3 + 6*B*a*b^2 + 3*B*a^2*b + (15*C*a*b^2)/4 + 6*C*a^2*b) + tan(c/2 + (d*x)/2
)^5*(12*B*a^3 + (116*C*b^3)/15 + 20*B*a*b^2 + 20*C*a^2*b) + tan(c/2 + (d*x)/2)^9*(2*B*a^3 - (5*B*b^3)/4 - C*a^
3 + 2*C*b^3 + 6*B*a*b^2 - 3*B*a^2*b - (15*C*a*b^2)/4 + 6*C*a^2*b) - tan(c/2 + (d*x)/2)^3*(8*B*a^3 + (B*b^3)/2
+ 2*C*a^3 + (8*C*b^3)/3 + 16*B*a*b^2 + 6*B*a^2*b + (3*C*a*b^2)/2 + 16*C*a^2*b) - tan(c/2 + (d*x)/2)^7*(8*B*a^3
- (B*b^3)/2 - 2*C*a^3 + (8*C*b^3)/3 + 16*B*a*b^2 - 6*B*a^2*b - (3*C*a*b^2)/2 + 16*C*a^2*b))/(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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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