3.302 \(\int \sec ^2(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=334 \[ \frac {\left (-4 a^2 B+24 a A b+25 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{120 b d}+\frac {\left (-4 a^3 B+24 a^2 A b+53 a b^2 B+32 A b^3\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{120 b d}+\frac {\left (8 a^4 B+32 a^3 A b+36 a^2 b^2 B+24 a A b^3+5 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {\left (-8 a^4 B+48 a^3 A b+178 a^2 b^2 B+232 a A b^3+75 b^4 B\right ) \tan (c+d x) \sec (c+d x)}{240 d}+\frac {\left (-4 a^5 B+24 a^4 A b+121 a^3 b^2 B+224 a^2 A b^3+128 a b^4 B+32 A b^5\right ) \tan (c+d x)}{60 b d}+\frac {(6 A b-a B) \tan (c+d x) (a+b \sec (c+d x))^4}{30 b d}+\frac {B \tan (c+d x) (a+b \sec (c+d x))^5}{6 b d} \]

[Out]

1/16*(32*A*a^3*b+24*A*a*b^3+8*B*a^4+36*B*a^2*b^2+5*B*b^4)*arctanh(sin(d*x+c))/d+1/60*(24*A*a^4*b+224*A*a^2*b^3
+32*A*b^5-4*B*a^5+121*B*a^3*b^2+128*B*a*b^4)*tan(d*x+c)/b/d+1/240*(48*A*a^3*b+232*A*a*b^3-8*B*a^4+178*B*a^2*b^
2+75*B*b^4)*sec(d*x+c)*tan(d*x+c)/d+1/120*(24*A*a^2*b+32*A*b^3-4*B*a^3+53*B*a*b^2)*(a+b*sec(d*x+c))^2*tan(d*x+
c)/b/d+1/120*(24*A*a*b-4*B*a^2+25*B*b^2)*(a+b*sec(d*x+c))^3*tan(d*x+c)/b/d+1/30*(6*A*b-B*a)*(a+b*sec(d*x+c))^4
*tan(d*x+c)/b/d+1/6*B*(a+b*sec(d*x+c))^5*tan(d*x+c)/b/d

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Rubi [A]  time = 0.71, antiderivative size = 334, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4010, 4002, 3997, 3787, 3770, 3767, 8} \[ \frac {\left (224 a^2 A b^3+24 a^4 A b+121 a^3 b^2 B-4 a^5 B+128 a b^4 B+32 A b^5\right ) \tan (c+d x)}{60 b d}+\frac {\left (32 a^3 A b+36 a^2 b^2 B+8 a^4 B+24 a A b^3+5 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {\left (-4 a^2 B+24 a A b+25 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{120 b d}+\frac {\left (24 a^2 A b-4 a^3 B+53 a b^2 B+32 A b^3\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{120 b d}+\frac {\left (48 a^3 A b+178 a^2 b^2 B-8 a^4 B+232 a A b^3+75 b^4 B\right ) \tan (c+d x) \sec (c+d x)}{240 d}+\frac {(6 A b-a B) \tan (c+d x) (a+b \sec (c+d x))^4}{30 b d}+\frac {B \tan (c+d x) (a+b \sec (c+d x))^5}{6 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]

[Out]

((32*a^3*A*b + 24*a*A*b^3 + 8*a^4*B + 36*a^2*b^2*B + 5*b^4*B)*ArcTanh[Sin[c + d*x]])/(16*d) + ((24*a^4*A*b + 2
24*a^2*A*b^3 + 32*A*b^5 - 4*a^5*B + 121*a^3*b^2*B + 128*a*b^4*B)*Tan[c + d*x])/(60*b*d) + ((48*a^3*A*b + 232*a
*A*b^3 - 8*a^4*B + 178*a^2*b^2*B + 75*b^4*B)*Sec[c + d*x]*Tan[c + d*x])/(240*d) + ((24*a^2*A*b + 32*A*b^3 - 4*
a^3*B + 53*a*b^2*B)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(120*b*d) + ((24*a*A*b - 4*a^2*B + 25*b^2*B)*(a + b*S
ec[c + d*x])^3*Tan[c + d*x])/(120*b*d) + ((6*A*b - a*B)*(a + b*Sec[c + d*x])^4*Tan[c + d*x])/(30*b*d) + (B*(a
+ b*Sec[c + d*x])^5*Tan[c + d*x])/(6*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]

Rubi steps

\begin {align*} \int \sec ^2(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx &=\frac {B (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^4 (5 b B+(6 A b-a B) \sec (c+d x)) \, dx}{6 b}\\ &=\frac {(6 A b-a B) (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac {B (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^3 \left (3 b (8 A b+7 a B)+\left (24 a A b-4 a^2 B+25 b^2 B\right ) \sec (c+d x)\right ) \, dx}{30 b}\\ &=\frac {\left (24 a A b-4 a^2 B+25 b^2 B\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}+\frac {(6 A b-a B) (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac {B (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (3 b \left (56 a A b+24 a^2 B+25 b^2 B\right )+3 \left (24 a^2 A b+32 A b^3-4 a^3 B+53 a b^2 B\right ) \sec (c+d x)\right ) \, dx}{120 b}\\ &=\frac {\left (24 a^2 A b+32 A b^3-4 a^3 B+53 a b^2 B\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}+\frac {\left (24 a A b-4 a^2 B+25 b^2 B\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}+\frac {(6 A b-a B) (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac {B (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x)) \left (3 b \left (216 a^2 A b+64 A b^3+64 a^3 B+181 a b^2 B\right )+3 \left (48 a^3 A b+232 a A b^3-8 a^4 B+178 a^2 b^2 B+75 b^4 B\right ) \sec (c+d x)\right ) \, dx}{360 b}\\ &=\frac {\left (48 a^3 A b+232 a A b^3-8 a^4 B+178 a^2 b^2 B+75 b^4 B\right ) \sec (c+d x) \tan (c+d x)}{240 d}+\frac {\left (24 a^2 A b+32 A b^3-4 a^3 B+53 a b^2 B\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}+\frac {\left (24 a A b-4 a^2 B+25 b^2 B\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}+\frac {(6 A b-a B) (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac {B (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac {\int \sec (c+d x) \left (45 b \left (32 a^3 A b+24 a A b^3+8 a^4 B+36 a^2 b^2 B+5 b^4 B\right )+12 \left (24 a^4 A b+224 a^2 A b^3+32 A b^5-4 a^5 B+121 a^3 b^2 B+128 a b^4 B\right ) \sec (c+d x)\right ) \, dx}{720 b}\\ &=\frac {\left (48 a^3 A b+232 a A b^3-8 a^4 B+178 a^2 b^2 B+75 b^4 B\right ) \sec (c+d x) \tan (c+d x)}{240 d}+\frac {\left (24 a^2 A b+32 A b^3-4 a^3 B+53 a b^2 B\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}+\frac {\left (24 a A b-4 a^2 B+25 b^2 B\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}+\frac {(6 A b-a B) (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac {B (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac {1}{16} \left (32 a^3 A b+24 a A b^3+8 a^4 B+36 a^2 b^2 B+5 b^4 B\right ) \int \sec (c+d x) \, dx+\frac {\left (24 a^4 A b+224 a^2 A b^3+32 A b^5-4 a^5 B+121 a^3 b^2 B+128 a b^4 B\right ) \int \sec ^2(c+d x) \, dx}{60 b}\\ &=\frac {\left (32 a^3 A b+24 a A b^3+8 a^4 B+36 a^2 b^2 B+5 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {\left (48 a^3 A b+232 a A b^3-8 a^4 B+178 a^2 b^2 B+75 b^4 B\right ) \sec (c+d x) \tan (c+d x)}{240 d}+\frac {\left (24 a^2 A b+32 A b^3-4 a^3 B+53 a b^2 B\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}+\frac {\left (24 a A b-4 a^2 B+25 b^2 B\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}+\frac {(6 A b-a B) (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac {B (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}-\frac {\left (24 a^4 A b+224 a^2 A b^3+32 A b^5-4 a^5 B+121 a^3 b^2 B+128 a b^4 B\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{60 b d}\\ &=\frac {\left (32 a^3 A b+24 a A b^3+8 a^4 B+36 a^2 b^2 B+5 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {\left (24 a^4 A b+224 a^2 A b^3+32 A b^5-4 a^5 B+121 a^3 b^2 B+128 a b^4 B\right ) \tan (c+d x)}{60 b d}+\frac {\left (48 a^3 A b+232 a A b^3-8 a^4 B+178 a^2 b^2 B+75 b^4 B\right ) \sec (c+d x) \tan (c+d x)}{240 d}+\frac {\left (24 a^2 A b+32 A b^3-4 a^3 B+53 a b^2 B\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}+\frac {\left (24 a A b-4 a^2 B+25 b^2 B\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}+\frac {(6 A b-a B) (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac {B (a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}\\ \end {align*}

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Mathematica [A]  time = 2.85, size = 244, normalized size = 0.73 \[ \frac {15 \left (8 a^4 B+32 a^3 A b+36 a^2 b^2 B+24 a A b^3+5 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (10 b^2 \left (36 a^2 B+24 a A b+5 b^2 B\right ) \sec ^3(c+d x)+160 b \left (2 a^3 B+3 a^2 A b+4 a b^2 B+A b^3\right ) \tan ^2(c+d x)+15 \left (8 a^4 B+32 a^3 A b+36 a^2 b^2 B+24 a A b^3+5 b^4 B\right ) \sec (c+d x)+240 \left (a^4 A+4 a^3 b B+6 a^2 A b^2+4 a b^3 B+A b^4\right )+48 b^3 (4 a B+A b) \tan ^4(c+d x)+40 b^4 B \sec ^5(c+d x)\right )}{240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]

[Out]

(15*(32*a^3*A*b + 24*a*A*b^3 + 8*a^4*B + 36*a^2*b^2*B + 5*b^4*B)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(240*(a^
4*A + 6*a^2*A*b^2 + A*b^4 + 4*a^3*b*B + 4*a*b^3*B) + 15*(32*a^3*A*b + 24*a*A*b^3 + 8*a^4*B + 36*a^2*b^2*B + 5*
b^4*B)*Sec[c + d*x] + 10*b^2*(24*a*A*b + 36*a^2*B + 5*b^2*B)*Sec[c + d*x]^3 + 40*b^4*B*Sec[c + d*x]^5 + 160*b*
(3*a^2*A*b + A*b^3 + 2*a^3*B + 4*a*b^2*B)*Tan[c + d*x]^2 + 48*b^3*(A*b + 4*a*B)*Tan[c + d*x]^4))/(240*d)

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fricas [A]  time = 0.49, size = 327, normalized size = 0.98 \[ \frac {15 \, {\left (8 \, B a^{4} + 32 \, A a^{3} b + 36 \, B a^{2} b^{2} + 24 \, A a b^{3} + 5 \, B b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (8 \, B a^{4} + 32 \, A a^{3} b + 36 \, B a^{2} b^{2} + 24 \, A a b^{3} + 5 \, B b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (15 \, A a^{4} + 40 \, B a^{3} b + 60 \, A a^{2} b^{2} + 32 \, B a b^{3} + 8 \, A b^{4}\right )} \cos \left (d x + c\right )^{5} + 40 \, B b^{4} + 15 \, {\left (8 \, B a^{4} + 32 \, A a^{3} b + 36 \, B a^{2} b^{2} + 24 \, A a b^{3} + 5 \, B b^{4}\right )} \cos \left (d x + c\right )^{4} + 32 \, {\left (10 \, B a^{3} b + 15 \, A a^{2} b^{2} + 8 \, B a b^{3} + 2 \, A b^{4}\right )} \cos \left (d x + c\right )^{3} + 10 \, {\left (36 \, B a^{2} b^{2} + 24 \, A a b^{3} + 5 \, B b^{4}\right )} \cos \left (d x + c\right )^{2} + 48 \, {\left (4 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/480*(15*(8*B*a^4 + 32*A*a^3*b + 36*B*a^2*b^2 + 24*A*a*b^3 + 5*B*b^4)*cos(d*x + c)^6*log(sin(d*x + c) + 1) -
15*(8*B*a^4 + 32*A*a^3*b + 36*B*a^2*b^2 + 24*A*a*b^3 + 5*B*b^4)*cos(d*x + c)^6*log(-sin(d*x + c) + 1) + 2*(16*
(15*A*a^4 + 40*B*a^3*b + 60*A*a^2*b^2 + 32*B*a*b^3 + 8*A*b^4)*cos(d*x + c)^5 + 40*B*b^4 + 15*(8*B*a^4 + 32*A*a
^3*b + 36*B*a^2*b^2 + 24*A*a*b^3 + 5*B*b^4)*cos(d*x + c)^4 + 32*(10*B*a^3*b + 15*A*a^2*b^2 + 8*B*a*b^3 + 2*A*b
^4)*cos(d*x + c)^3 + 10*(36*B*a^2*b^2 + 24*A*a*b^3 + 5*B*b^4)*cos(d*x + c)^2 + 48*(4*B*a*b^3 + A*b^4)*cos(d*x
+ c))*sin(d*x + c))/(d*cos(d*x + c)^6)

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giac [B]  time = 2.28, size = 1186, normalized size = 3.55 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/240*(15*(8*B*a^4 + 32*A*a^3*b + 36*B*a^2*b^2 + 24*A*a*b^3 + 5*B*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15
*(8*B*a^4 + 32*A*a^3*b + 36*B*a^2*b^2 + 24*A*a*b^3 + 5*B*b^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(240*A*a^
4*tan(1/2*d*x + 1/2*c)^11 - 120*B*a^4*tan(1/2*d*x + 1/2*c)^11 - 480*A*a^3*b*tan(1/2*d*x + 1/2*c)^11 + 960*B*a^
3*b*tan(1/2*d*x + 1/2*c)^11 + 1440*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^11 - 900*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^11 -
600*A*a*b^3*tan(1/2*d*x + 1/2*c)^11 + 960*B*a*b^3*tan(1/2*d*x + 1/2*c)^11 + 240*A*b^4*tan(1/2*d*x + 1/2*c)^11
- 165*B*b^4*tan(1/2*d*x + 1/2*c)^11 - 1200*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 360*B*a^4*tan(1/2*d*x + 1/2*c)^9 +
1440*A*a^3*b*tan(1/2*d*x + 1/2*c)^9 - 3520*B*a^3*b*tan(1/2*d*x + 1/2*c)^9 - 5280*A*a^2*b^2*tan(1/2*d*x + 1/2*c
)^9 + 1260*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 + 840*A*a*b^3*tan(1/2*d*x + 1/2*c)^9 - 2240*B*a*b^3*tan(1/2*d*x +
1/2*c)^9 - 560*A*b^4*tan(1/2*d*x + 1/2*c)^9 - 25*B*b^4*tan(1/2*d*x + 1/2*c)^9 + 2400*A*a^4*tan(1/2*d*x + 1/2*c
)^7 - 240*B*a^4*tan(1/2*d*x + 1/2*c)^7 - 960*A*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 5760*B*a^3*b*tan(1/2*d*x + 1/2*c
)^7 + 8640*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 360*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 240*A*a*b^3*tan(1/2*d*x +
1/2*c)^7 + 4992*B*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 1248*A*b^4*tan(1/2*d*x + 1/2*c)^7 - 450*B*b^4*tan(1/2*d*x +
1/2*c)^7 - 2400*A*a^4*tan(1/2*d*x + 1/2*c)^5 - 240*B*a^4*tan(1/2*d*x + 1/2*c)^5 - 960*A*a^3*b*tan(1/2*d*x + 1/
2*c)^5 - 5760*B*a^3*b*tan(1/2*d*x + 1/2*c)^5 - 8640*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 360*B*a^2*b^2*tan(1/2*d
*x + 1/2*c)^5 - 240*A*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 4992*B*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 1248*A*b^4*tan(1/2*
d*x + 1/2*c)^5 - 450*B*b^4*tan(1/2*d*x + 1/2*c)^5 + 1200*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 360*B*a^4*tan(1/2*d*x
+ 1/2*c)^3 + 1440*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 3520*B*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 5280*A*a^2*b^2*tan(1/
2*d*x + 1/2*c)^3 + 1260*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 840*A*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 2240*B*a*b^3*t
an(1/2*d*x + 1/2*c)^3 + 560*A*b^4*tan(1/2*d*x + 1/2*c)^3 - 25*B*b^4*tan(1/2*d*x + 1/2*c)^3 - 240*A*a^4*tan(1/2
*d*x + 1/2*c) - 120*B*a^4*tan(1/2*d*x + 1/2*c) - 480*A*a^3*b*tan(1/2*d*x + 1/2*c) - 960*B*a^3*b*tan(1/2*d*x +
1/2*c) - 1440*A*a^2*b^2*tan(1/2*d*x + 1/2*c) - 900*B*a^2*b^2*tan(1/2*d*x + 1/2*c) - 600*A*a*b^3*tan(1/2*d*x +
1/2*c) - 960*B*a*b^3*tan(1/2*d*x + 1/2*c) - 240*A*b^4*tan(1/2*d*x + 1/2*c) - 165*B*b^4*tan(1/2*d*x + 1/2*c))/(
tan(1/2*d*x + 1/2*c)^2 - 1)^6)/d

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maple [A]  time = 1.85, size = 550, normalized size = 1.65 \[ \frac {A \,b^{4} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {4 A \,b^{4} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {A \,a^{4} \tan \left (d x +c \right )}{d}+\frac {a^{4} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {9 a^{2} b^{2} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {3 a A \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {5 B \,b^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {32 B a \,b^{3} \tan \left (d x +c \right )}{15 d}+\frac {8 B \,a^{3} b \tan \left (d x +c \right )}{3 d}+\frac {4 A \,a^{2} b^{2} \tan \left (d x +c \right )}{d}+\frac {5 B \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {8 A \,b^{4} \tan \left (d x +c \right )}{15 d}+\frac {5 B \,b^{4} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{24 d}+\frac {B \,b^{4} \tan \left (d x +c \right ) \left (\sec ^{5}\left (d x +c \right )\right )}{6 d}+\frac {4 B a \,b^{3} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {3 a A \,b^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {2 A \,a^{3} b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {9 a^{2} b^{2} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{4 d}+\frac {16 B a \,b^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {4 B \,a^{3} b \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {3 a^{2} b^{2} B \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{2 d}+\frac {a A \,b^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{d}+\frac {2 A \,a^{2} b^{2} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {a^{4} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {2 A \,a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x)

[Out]

1/d*A*a^4*tan(d*x+c)+1/2/d*a^4*B*ln(sec(d*x+c)+tan(d*x+c))+9/4/d*a^2*b^2*B*ln(sec(d*x+c)+tan(d*x+c))+3/2/d*a*A
*b^3*ln(sec(d*x+c)+tan(d*x+c))+1/6/d*B*b^4*tan(d*x+c)*sec(d*x+c)^5+5/24/d*B*b^4*tan(d*x+c)*sec(d*x+c)^3+5/16/d
*B*b^4*sec(d*x+c)*tan(d*x+c)+32/15/d*B*a*b^3*tan(d*x+c)+8/3/d*B*a^3*b*tan(d*x+c)+4/d*A*a^2*b^2*tan(d*x+c)+5/16
/d*B*b^4*ln(sec(d*x+c)+tan(d*x+c))+8/15/d*A*b^4*tan(d*x+c)+3/2/d*a*A*b^3*sec(d*x+c)*tan(d*x+c)+4/5/d*B*a*b^3*t
an(d*x+c)*sec(d*x+c)^4+16/15/d*B*a*b^3*tan(d*x+c)*sec(d*x+c)^2+4/3/d*B*a^3*b*tan(d*x+c)*sec(d*x+c)^2+2/d*A*a^3
*b*sec(d*x+c)*tan(d*x+c)+3/2/d*a^2*b^2*B*tan(d*x+c)*sec(d*x+c)^3+9/4/d*a^2*b^2*B*sec(d*x+c)*tan(d*x+c)+1/d*a*A
*b^3*tan(d*x+c)*sec(d*x+c)^3+2/d*A*a^2*b^2*tan(d*x+c)*sec(d*x+c)^2+1/2/d*a^4*B*sec(d*x+c)*tan(d*x+c)+2/d*A*a^3
*b*ln(sec(d*x+c)+tan(d*x+c))+1/5/d*A*b^4*tan(d*x+c)*sec(d*x+c)^4+4/15/d*A*b^4*tan(d*x+c)*sec(d*x+c)^2

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maxima [A]  time = 0.76, size = 474, normalized size = 1.42 \[ \frac {640 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} b + 960 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} b^{2} + 128 \, {\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 b^{3} + 32 \, {\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 )} A b^{4} - 5 \, B b^{4} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, B a^{2} 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 )} - 120 \, A a 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 )} - 120 \, 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 )} - 480 \, A a^{3} 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 )} + 480 \, A a^{4} \tan \left (d x + c\right )}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/480*(640*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3*b + 960*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^2*b^2 + 128*(
3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*B*a*b^3 + 32*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 1
5*tan(d*x + c))*A*b^4 - 5*B*b^4*(2*(15*sin(d*x + c)^5 - 40*sin(d*x + c)^3 + 33*sin(d*x + c))/(sin(d*x + c)^6 -
3*sin(d*x + c)^4 + 3*sin(d*x + c)^2 - 1) - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)) - 180*B*a^2*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)) - 120*A*a*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(sin(d*x + c) - 1)) - 120*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)) - 480*A*a^3*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(
d*x + c) + 1) + log(sin(d*x + c) - 1)) + 480*A*a^4*tan(d*x + c))/d

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mupad [B]  time = 5.74, size = 709, normalized size = 2.12 \[ \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B\,a^4}{2}+2\,A\,a^3\,b+\frac {9\,B\,a^2\,b^2}{4}+\frac {3\,A\,a\,b^3}{2}+\frac {5\,B\,b^4}{16}\right )}{2\,B\,a^4+8\,A\,a^3\,b+9\,B\,a^2\,b^2+6\,A\,a\,b^3+\frac {5\,B\,b^4}{4}}\right )\,\left (B\,a^4+4\,A\,a^3\,b+\frac {9\,B\,a^2\,b^2}{2}+3\,A\,a\,b^3+\frac {5\,B\,b^4}{8}\right )}{d}+\frac {\left (B\,a^4-2\,A\,b^4-2\,A\,a^4+\frac {11\,B\,b^4}{8}-12\,A\,a^2\,b^2+\frac {15\,B\,a^2\,b^2}{2}+5\,A\,a\,b^3+4\,A\,a^3\,b-8\,B\,a\,b^3-8\,B\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (10\,A\,a^4+\frac {14\,A\,b^4}{3}-3\,B\,a^4+\frac {5\,B\,b^4}{24}+44\,A\,a^2\,b^2-\frac {21\,B\,a^2\,b^2}{2}-7\,A\,a\,b^3-12\,A\,a^3\,b+\frac {56\,B\,a\,b^3}{3}+\frac {88\,B\,a^3\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (2\,B\,a^4-\frac {52\,A\,b^4}{5}-20\,A\,a^4+\frac {15\,B\,b^4}{4}-72\,A\,a^2\,b^2+3\,B\,a^2\,b^2+2\,A\,a\,b^3+8\,A\,a^3\,b-\frac {208\,B\,a\,b^3}{5}-48\,B\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (20\,A\,a^4+\frac {52\,A\,b^4}{5}+2\,B\,a^4+\frac {15\,B\,b^4}{4}+72\,A\,a^2\,b^2+3\,B\,a^2\,b^2+2\,A\,a\,b^3+8\,A\,a^3\,b+\frac {208\,B\,a\,b^3}{5}+48\,B\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {5\,B\,b^4}{24}-\frac {14\,A\,b^4}{3}-3\,B\,a^4-10\,A\,a^4-44\,A\,a^2\,b^2-\frac {21\,B\,a^2\,b^2}{2}-7\,A\,a\,b^3-12\,A\,a^3\,b-\frac {56\,B\,a\,b^3}{3}-\frac {88\,B\,a^3\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^4+2\,A\,b^4+B\,a^4+\frac {11\,B\,b^4}{8}+12\,A\,a^2\,b^2+\frac {15\,B\,a^2\,b^2}{2}+5\,A\,a\,b^3+4\,A\,a^3\,b+8\,B\,a\,b^3+8\,B\,a^3\,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 )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\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(((A + B/cos(c + d*x))*(a + b/cos(c + d*x))^4)/cos(c + d*x)^2,x)

[Out]

(atanh((4*tan(c/2 + (d*x)/2)*((B*a^4)/2 + (5*B*b^4)/16 + (9*B*a^2*b^2)/4 + (3*A*a*b^3)/2 + 2*A*a^3*b))/(2*B*a^
4 + (5*B*b^4)/4 + 9*B*a^2*b^2 + 6*A*a*b^3 + 8*A*a^3*b))*(B*a^4 + (5*B*b^4)/8 + (9*B*a^2*b^2)/2 + 3*A*a*b^3 + 4
*A*a^3*b))/d + (tan(c/2 + (d*x)/2)*(2*A*a^4 + 2*A*b^4 + B*a^4 + (11*B*b^4)/8 + 12*A*a^2*b^2 + (15*B*a^2*b^2)/2
+ 5*A*a*b^3 + 4*A*a^3*b + 8*B*a*b^3 + 8*B*a^3*b) - tan(c/2 + (d*x)/2)^11*(2*A*a^4 + 2*A*b^4 - B*a^4 - (11*B*b
^4)/8 + 12*A*a^2*b^2 - (15*B*a^2*b^2)/2 - 5*A*a*b^3 - 4*A*a^3*b + 8*B*a*b^3 + 8*B*a^3*b) - tan(c/2 + (d*x)/2)^
3*(10*A*a^4 + (14*A*b^4)/3 + 3*B*a^4 - (5*B*b^4)/24 + 44*A*a^2*b^2 + (21*B*a^2*b^2)/2 + 7*A*a*b^3 + 12*A*a^3*b
+ (56*B*a*b^3)/3 + (88*B*a^3*b)/3) + tan(c/2 + (d*x)/2)^9*(10*A*a^4 + (14*A*b^4)/3 - 3*B*a^4 + (5*B*b^4)/24 +
44*A*a^2*b^2 - (21*B*a^2*b^2)/2 - 7*A*a*b^3 - 12*A*a^3*b + (56*B*a*b^3)/3 + (88*B*a^3*b)/3) + tan(c/2 + (d*x)
/2)^5*(20*A*a^4 + (52*A*b^4)/5 + 2*B*a^4 + (15*B*b^4)/4 + 72*A*a^2*b^2 + 3*B*a^2*b^2 + 2*A*a*b^3 + 8*A*a^3*b +
(208*B*a*b^3)/5 + 48*B*a^3*b) - tan(c/2 + (d*x)/2)^7*(20*A*a^4 + (52*A*b^4)/5 - 2*B*a^4 - (15*B*b^4)/4 + 72*A
*a^2*b^2 - 3*B*a^2*b^2 - 2*A*a*b^3 - 8*A*a^3*b + (208*B*a*b^3)/5 + 48*B*a^3*b))/(d*(15*tan(c/2 + (d*x)/2)^4 -
6*tan(c/2 + (d*x)/2)^2 - 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c/2 + (d*x)/2)^8 - 6*tan(c/2 + (d*x)/2)^10 + tan(c/2
+ (d*x)/2)^12 + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(a+b*sec(d*x+c))**4*(A+B*sec(d*x+c)),x)

[Out]

Integral((A + B*sec(c + d*x))*(a + b*sec(c + d*x))**4*sec(c + d*x)**2, x)

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