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

Optimal. Leaf size=200 \[ a^4 A x+\frac {b^2 \left (26 a^2 B+32 a A b+9 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{24 d}+\frac {b \left (19 a^3 B+34 a^2 A b+16 a b^2 B+4 A b^3\right ) \tan (c+d x)}{6 d}+\frac {\left (8 a^4 B+32 a^3 A b+24 a^2 b^2 B+16 a A b^3+3 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b (7 a B+4 A b) \tan (c+d x) (a+b \sec (c+d x))^2}{12 d}+\frac {b B \tan (c+d x) (a+b \sec (c+d x))^3}{4 d} \]

[Out]

a^4*A*x+1/8*(32*A*a^3*b+16*A*a*b^3+8*B*a^4+24*B*a^2*b^2+3*B*b^4)*arctanh(sin(d*x+c))/d+1/6*b*(34*A*a^2*b+4*A*b
^3+19*B*a^3+16*B*a*b^2)*tan(d*x+c)/d+1/24*b^2*(32*A*a*b+26*B*a^2+9*B*b^2)*sec(d*x+c)*tan(d*x+c)/d+1/12*b*(4*A*
b+7*B*a)*(a+b*sec(d*x+c))^2*tan(d*x+c)/d+1/4*b*B*(a+b*sec(d*x+c))^3*tan(d*x+c)/d

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Rubi [A]  time = 0.33, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3918, 4056, 4048, 3770, 3767, 8} \[ \frac {b \left (34 a^2 A b+19 a^3 B+16 a b^2 B+4 A b^3\right ) \tan (c+d x)}{6 d}+\frac {\left (32 a^3 A b+24 a^2 b^2 B+8 a^4 B+16 a A b^3+3 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b^2 \left (26 a^2 B+32 a A b+9 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{24 d}+a^4 A x+\frac {b (7 a B+4 A b) \tan (c+d x) (a+b \sec (c+d x))^2}{12 d}+\frac {b B \tan (c+d x) (a+b \sec (c+d x))^3}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^4*A*x + ((32*a^3*A*b + 16*a*A*b^3 + 8*a^4*B + 24*a^2*b^2*B + 3*b^4*B)*ArcTanh[Sin[c + d*x]])/(8*d) + (b*(34*
a^2*A*b + 4*A*b^3 + 19*a^3*B + 16*a*b^2*B)*Tan[c + d*x])/(6*d) + (b^2*(32*a*A*b + 26*a^2*B + 9*b^2*B)*Sec[c +
d*x]*Tan[c + d*x])/(24*d) + (b*(4*A*b + 7*a*B)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(12*d) + (b*B*(a + b*Sec[c
+ d*x])^3*Tan[c + d*x])/(4*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 3918

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> -Simp[(b*
d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1))/(f*m), x] + Dist[1/m, Int[(a + b*Csc[e + f*x])^(m - 2)*Simp[a^2*c
*m + (b^2*d*(m - 1) + 2*a*b*c*m + a^2*d*m)*Csc[e + f*x] + b*(b*c*m + a*d*(2*m - 1))*Csc[e + f*x]^2, x], x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && GtQ[m, 1] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 4048

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

Rule 4056

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

Rubi steps

\begin {align*} \int (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx &=\frac {b B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{4} \int (a+b \sec (c+d x))^2 \left (4 a^2 A+\left (8 a A b+4 a^2 B+3 b^2 B\right ) \sec (c+d x)+b (4 A b+7 a B) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {b (4 A b+7 a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac {b B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{12} \int (a+b \sec (c+d x)) \left (12 a^3 A+\left (36 a^2 A b+8 A b^3+12 a^3 B+23 a b^2 B\right ) \sec (c+d x)+b \left (32 a A b+26 a^2 B+9 b^2 B\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {b^2 \left (32 a A b+26 a^2 B+9 b^2 B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {b (4 A b+7 a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac {b B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{24} \int \left (24 a^4 A+3 \left (32 a^3 A b+16 a A b^3+8 a^4 B+24 a^2 b^2 B+3 b^4 B\right ) \sec (c+d x)+4 b \left (34 a^2 A b+4 A b^3+19 a^3 B+16 a b^2 B\right ) \sec ^2(c+d x)\right ) \, dx\\ &=a^4 A x+\frac {b^2 \left (32 a A b+26 a^2 B+9 b^2 B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {b (4 A b+7 a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac {b B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {1}{6} \left (b \left (34 a^2 A b+4 A b^3+19 a^3 B+16 a b^2 B\right )\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{8} \left (32 a^3 A b+16 a A b^3+8 a^4 B+24 a^2 b^2 B+3 b^4 B\right ) \int \sec (c+d x) \, dx\\ &=a^4 A x+\frac {\left (32 a^3 A b+16 a A b^3+8 a^4 B+24 a^2 b^2 B+3 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b^2 \left (32 a A b+26 a^2 B+9 b^2 B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {b (4 A b+7 a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac {b B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac {\left (b \left (34 a^2 A b+4 A b^3+19 a^3 B+16 a b^2 B\right )\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 d}\\ &=a^4 A x+\frac {\left (32 a^3 A b+16 a A b^3+8 a^4 B+24 a^2 b^2 B+3 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b \left (34 a^2 A b+4 A b^3+19 a^3 B+16 a b^2 B\right ) \tan (c+d x)}{6 d}+\frac {b^2 \left (32 a A b+26 a^2 B+9 b^2 B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {b (4 A b+7 a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac {b B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 1.03, size = 160, normalized size = 0.80 \[ \frac {24 a^4 A d x+3 b \tan (c+d x) \left (b \left (24 a^2 B+16 a A b+3 b^2 B\right ) \sec (c+d x)+8 \left (4 a^3 B+6 a^2 A b+4 a b^2 B+A b^3\right )+2 b^3 B \sec ^3(c+d x)\right )+3 \left (8 a^4 B+32 a^3 A b+24 a^2 b^2 B+16 a A b^3+3 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))+8 b^3 (4 a B+A b) \tan ^3(c+d x)}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(24*a^4*A*d*x + 3*(32*a^3*A*b + 16*a*A*b^3 + 8*a^4*B + 24*a^2*b^2*B + 3*b^4*B)*ArcTanh[Sin[c + d*x]] + 3*b*(8*
(6*a^2*A*b + A*b^3 + 4*a^3*B + 4*a*b^2*B) + b*(16*a*A*b + 24*a^2*B + 3*b^2*B)*Sec[c + d*x] + 2*b^3*B*Sec[c + d
*x]^3)*Tan[c + d*x] + 8*b^3*(A*b + 4*a*B)*Tan[c + d*x]^3)/(24*d)

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fricas [A]  time = 0.49, size = 250, normalized size = 1.25 \[ \frac {48 \, A a^{4} d x \cos \left (d x + c\right )^{4} + 3 \, {\left (8 \, B a^{4} + 32 \, A a^{3} b + 24 \, B a^{2} b^{2} + 16 \, A a b^{3} + 3 \, B b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (8 \, B a^{4} + 32 \, A a^{3} b + 24 \, B a^{2} b^{2} + 16 \, A a b^{3} + 3 \, B b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6 \, B b^{4} + 16 \, {\left (6 \, B a^{3} b + 9 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (24 \, B a^{2} b^{2} + 16 \, A a b^{3} + 3 \, B b^{4}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (4 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(48*A*a^4*d*x*cos(d*x + c)^4 + 3*(8*B*a^4 + 32*A*a^3*b + 24*B*a^2*b^2 + 16*A*a*b^3 + 3*B*b^4)*cos(d*x + c
)^4*log(sin(d*x + c) + 1) - 3*(8*B*a^4 + 32*A*a^3*b + 24*B*a^2*b^2 + 16*A*a*b^3 + 3*B*b^4)*cos(d*x + c)^4*log(
-sin(d*x + c) + 1) + 2*(6*B*b^4 + 16*(6*B*a^3*b + 9*A*a^2*b^2 + 4*B*a*b^3 + A*b^4)*cos(d*x + c)^3 + 3*(24*B*a^
2*b^2 + 16*A*a*b^3 + 3*B*b^4)*cos(d*x + c)^2 + 8*(4*B*a*b^3 + A*b^4)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x +
c)^4)

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giac [B]  time = 2.39, size = 635, normalized size = 3.18 \[ \frac {24 \, {\left (d x + c\right )} A a^{4} + 3 \, {\left (8 \, B a^{4} + 32 \, A a^{3} b + 24 \, B a^{2} b^{2} + 16 \, A a b^{3} + 3 \, B b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (8 \, B a^{4} + 32 \, A a^{3} b + 24 \, B a^{2} b^{2} + 16 \, A a b^{3} + 3 \, B b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (96 \, B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 144 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 72 \, B a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 48 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 96 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 288 \, B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 432 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, B a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 48 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 160 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, B b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 288 \, B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 432 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 72 \, B a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 48 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 160 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, B b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 96 \, B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 144 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 72 \, B a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 48 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 96 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, B b^{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 )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(24*(d*x + c)*A*a^4 + 3*(8*B*a^4 + 32*A*a^3*b + 24*B*a^2*b^2 + 16*A*a*b^3 + 3*B*b^4)*log(abs(tan(1/2*d*x
+ 1/2*c) + 1)) - 3*(8*B*a^4 + 32*A*a^3*b + 24*B*a^2*b^2 + 16*A*a*b^3 + 3*B*b^4)*log(abs(tan(1/2*d*x + 1/2*c) -
1)) - 2*(96*B*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 144*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 72*B*a^2*b^2*tan(1/2*d*x
+ 1/2*c)^7 - 48*A*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 96*B*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 24*A*b^4*tan(1/2*d*x + 1/
2*c)^7 - 15*B*b^4*tan(1/2*d*x + 1/2*c)^7 - 288*B*a^3*b*tan(1/2*d*x + 1/2*c)^5 - 432*A*a^2*b^2*tan(1/2*d*x + 1/
2*c)^5 + 72*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 48*A*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 160*B*a*b^3*tan(1/2*d*x + 1
/2*c)^5 - 40*A*b^4*tan(1/2*d*x + 1/2*c)^5 - 9*B*b^4*tan(1/2*d*x + 1/2*c)^5 + 288*B*a^3*b*tan(1/2*d*x + 1/2*c)^
3 + 432*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 72*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 48*A*a*b^3*tan(1/2*d*x + 1/2*
c)^3 + 160*B*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 40*A*b^4*tan(1/2*d*x + 1/2*c)^3 - 9*B*b^4*tan(1/2*d*x + 1/2*c)^3 -
96*B*a^3*b*tan(1/2*d*x + 1/2*c) - 144*A*a^2*b^2*tan(1/2*d*x + 1/2*c) - 72*B*a^2*b^2*tan(1/2*d*x + 1/2*c) - 48
*A*a*b^3*tan(1/2*d*x + 1/2*c) - 96*B*a*b^3*tan(1/2*d*x + 1/2*c) - 24*A*b^4*tan(1/2*d*x + 1/2*c) - 15*B*b^4*tan
(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

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maple [A]  time = 1.38, size = 338, normalized size = 1.69 \[ A \,a^{4} x +\frac {A \,a^{4} c}{d}+\frac {a^{4} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 A \,a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 B \,a^{3} b \tan \left (d x +c \right )}{d}+\frac {6 A \,a^{2} b^{2} \tan \left (d x +c \right )}{d}+\frac {3 a^{2} b^{2} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {3 a^{2} b^{2} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {2 a A \,b^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {2 a A \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {8 B a \,b^{3} \tan \left (d x +c \right )}{3 d}+\frac {4 B a \,b^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {2 A \,b^{4} \tan \left (d x +c \right )}{3 d}+\frac {A \,b^{4} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {B \,b^{4} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 B \,b^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 B \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.84, size = 303, normalized size = 1.52 \[ \frac {48 \, {\left (d x + c\right )} A a^{4} + 64 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a b^{3} + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{4} - 3 \, B b^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, B a^{2} b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 48 \, A a b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, B a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 192 \, A a^{3} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 192 \, B a^{3} b \tan \left (d x + c\right ) + 288 \, A a^{2} b^{2} \tan \left (d x + c\right )}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(48*(d*x + c)*A*a^4 + 64*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a*b^3 + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))
*A*b^4 - 3*B*b^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)) - 72*B*a^2*b^2*(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*b^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*B*a^4*log(sec(d*x + c) + tan(d*x + c)) + 192*A*a^3*b*log(sec(d*x + c) + tan(d*x + c))
+ 192*B*a^3*b*tan(d*x + c) + 288*A*a^2*b^2*tan(d*x + c))/d

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mupad [B]  time = 5.01, size = 1969, normalized size = 9.84 \[ \text {result too large to display} \]

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,x)

[Out]

(9*B*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + (27*B*b^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))
)/8 + 4*A*b^4*sin(2*c + 2*d*x) + A*b^4*sin(4*c + 4*d*x) + (9*B*b^4*sin(3*c + 3*d*x))/8 + 9*A*a^4*atan((64*A^2*
a^8*sin(c/2 + (d*x)/2) + 64*B^2*a^8*sin(c/2 + (d*x)/2) + 9*B^2*b^8*sin(c/2 + (d*x)/2) + 256*A^2*a^2*b^6*sin(c/
2 + (d*x)/2) + 1024*A^2*a^4*b^4*sin(c/2 + (d*x)/2) + 1024*A^2*a^6*b^2*sin(c/2 + (d*x)/2) + 144*B^2*a^2*b^6*sin
(c/2 + (d*x)/2) + 624*B^2*a^4*b^4*sin(c/2 + (d*x)/2) + 384*B^2*a^6*b^2*sin(c/2 + (d*x)/2) + 960*A*B*a^3*b^5*si
n(c/2 + (d*x)/2) + 1792*A*B*a^5*b^3*sin(c/2 + (d*x)/2) + 96*A*B*a*b^7*sin(c/2 + (d*x)/2) + 512*A*B*a^7*b*sin(c
/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(64*A^2*a^8 + 64*B^2*a^8 + 9*B^2*b^8 + 256*A^2*a^2*b^6 + 1024*A^2*a^4*b^4 +
1024*A^2*a^6*b^2 + 144*B^2*a^2*b^6 + 624*B^2*a^4*b^4 + 384*B^2*a^6*b^2 + 96*A*B*a*b^7 + 512*A*B*a^7*b + 960*A
*B*a^3*b^5 + 1792*A*B*a^5*b^3))) + (33*B*b^4*sin(c + d*x))/8 + 12*A*a^4*cos(2*c + 2*d*x)*atan((64*A^2*a^8*sin(
c/2 + (d*x)/2) + 64*B^2*a^8*sin(c/2 + (d*x)/2) + 9*B^2*b^8*sin(c/2 + (d*x)/2) + 256*A^2*a^2*b^6*sin(c/2 + (d*x
)/2) + 1024*A^2*a^4*b^4*sin(c/2 + (d*x)/2) + 1024*A^2*a^6*b^2*sin(c/2 + (d*x)/2) + 144*B^2*a^2*b^6*sin(c/2 + (
d*x)/2) + 624*B^2*a^4*b^4*sin(c/2 + (d*x)/2) + 384*B^2*a^6*b^2*sin(c/2 + (d*x)/2) + 960*A*B*a^3*b^5*sin(c/2 +
(d*x)/2) + 1792*A*B*a^5*b^3*sin(c/2 + (d*x)/2) + 96*A*B*a*b^7*sin(c/2 + (d*x)/2) + 512*A*B*a^7*b*sin(c/2 + (d*
x)/2))/(cos(c/2 + (d*x)/2)*(64*A^2*a^8 + 64*B^2*a^8 + 9*B^2*b^8 + 256*A^2*a^2*b^6 + 1024*A^2*a^4*b^4 + 1024*A^
2*a^6*b^2 + 144*B^2*a^2*b^6 + 624*B^2*a^4*b^4 + 384*B^2*a^6*b^2 + 96*A*B*a*b^7 + 512*A*B*a^7*b + 960*A*B*a^3*b
^5 + 1792*A*B*a^5*b^3))) + 3*A*a^4*cos(4*c + 4*d*x)*atan((64*A^2*a^8*sin(c/2 + (d*x)/2) + 64*B^2*a^8*sin(c/2 +
(d*x)/2) + 9*B^2*b^8*sin(c/2 + (d*x)/2) + 256*A^2*a^2*b^6*sin(c/2 + (d*x)/2) + 1024*A^2*a^4*b^4*sin(c/2 + (d*
x)/2) + 1024*A^2*a^6*b^2*sin(c/2 + (d*x)/2) + 144*B^2*a^2*b^6*sin(c/2 + (d*x)/2) + 624*B^2*a^4*b^4*sin(c/2 + (
d*x)/2) + 384*B^2*a^6*b^2*sin(c/2 + (d*x)/2) + 960*A*B*a^3*b^5*sin(c/2 + (d*x)/2) + 1792*A*B*a^5*b^3*sin(c/2 +
(d*x)/2) + 96*A*B*a*b^7*sin(c/2 + (d*x)/2) + 512*A*B*a^7*b*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(64*A^2*a^
8 + 64*B^2*a^8 + 9*B^2*b^8 + 256*A^2*a^2*b^6 + 1024*A^2*a^4*b^4 + 1024*A^2*a^6*b^2 + 144*B^2*a^2*b^6 + 624*B^2
*a^4*b^4 + 384*B^2*a^6*b^2 + 96*A*B*a*b^7 + 512*A*B*a^7*b + 960*A*B*a^3*b^5 + 1792*A*B*a^5*b^3))) + 6*A*a*b^3*
sin(c + d*x) + 18*A*a*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 36*A*a^3*b*atanh(sin(c/2 + (d*x)/2)/c
os(c/2 + (d*x)/2)) + 6*A*a*b^3*sin(3*c + 3*d*x) + 16*B*a*b^3*sin(2*c + 2*d*x) + 12*B*a^3*b*sin(2*c + 2*d*x) +
4*B*a*b^3*sin(4*c + 4*d*x) + 6*B*a^3*b*sin(4*c + 4*d*x) + 9*B*a^2*b^2*sin(c + d*x) + 12*B*a^4*atanh(sin(c/2 +
(d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x) + 3*B*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4*c +
4*d*x) + 27*B*a^2*b^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + (9*B*b^4*atanh(sin(c/2 + (d*x)/2)/cos(c/
2 + (d*x)/2))*cos(2*c + 2*d*x))/2 + (9*B*b^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x))/8
+ 18*A*a^2*b^2*sin(2*c + 2*d*x) + 9*A*a^2*b^2*sin(4*c + 4*d*x) + 9*B*a^2*b^2*sin(3*c + 3*d*x) + 24*A*a*b^3*ata
nh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x) + 48*A*a^3*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*
x)/2))*cos(2*c + 2*d*x) + 6*A*a*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x) + 12*A*a^3*b
*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x) + 36*B*a^2*b^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2
+ (d*x)/2))*cos(2*c + 2*d*x) + 9*B*a^2*b^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x))/(12
*d*(cos(2*c + 2*d*x)/2 + cos(4*c + 4*d*x)/8 + 3/8))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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, x)

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