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

Optimal. Leaf size=195 \[ -\frac{b \left (6 a^3 A-17 a^2 b B-12 a A b^2-2 b^3 B\right ) \tan (c+d x)}{3 d}+\frac{b \left (12 a^2 A b+8 a^3 B+4 a b^2 B+A b^3\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b^2 \left (6 a^2 A-8 a b B-3 A b^2\right ) \tan (c+d x) \sec (c+d x)}{6 d}+a^3 x (a B+4 A b)-\frac{b (3 a A-b B) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac{a A \sin (c+d x) (a+b \sec (c+d x))^3}{d} \]

[Out]

a^3*(4*A*b + a*B)*x + (b*(12*a^2*A*b + A*b^3 + 8*a^3*B + 4*a*b^2*B)*ArcTanh[Sin[c + d*x]])/(2*d) + (a*A*(a + b
*Sec[c + d*x])^3*Sin[c + d*x])/d - (b*(6*a^3*A - 12*a*A*b^2 - 17*a^2*b*B - 2*b^3*B)*Tan[c + d*x])/(3*d) - (b^2
*(6*a^2*A - 3*A*b^2 - 8*a*b*B)*Sec[c + d*x]*Tan[c + d*x])/(6*d) - (b*(3*a*A - b*B)*(a + b*Sec[c + d*x])^2*Tan[
c + d*x])/(3*d)

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Rubi [A]  time = 0.368302, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {4025, 4056, 4048, 3770, 3767, 8} \[ -\frac{b \left (6 a^3 A-17 a^2 b B-12 a A b^2-2 b^3 B\right ) \tan (c+d x)}{3 d}+\frac{b \left (12 a^2 A b+8 a^3 B+4 a b^2 B+A b^3\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b^2 \left (6 a^2 A-8 a b B-3 A b^2\right ) \tan (c+d x) \sec (c+d x)}{6 d}+a^3 x (a B+4 A b)-\frac{b (3 a A-b B) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac{a A \sin (c+d x) (a+b \sec (c+d x))^3}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^3*(4*A*b + a*B)*x + (b*(12*a^2*A*b + A*b^3 + 8*a^3*B + 4*a*b^2*B)*ArcTanh[Sin[c + d*x]])/(2*d) + (a*A*(a + b
*Sec[c + d*x])^3*Sin[c + d*x])/d - (b*(6*a^3*A - 12*a*A*b^2 - 17*a^2*b*B - 2*b^3*B)*Tan[c + d*x])/(3*d) - (b^2
*(6*a^2*A - 3*A*b^2 - 8*a*b*B)*Sec[c + d*x]*Tan[c + d*x])/(6*d) - (b*(3*a*A - b*B)*(a + b*Sec[c + d*x])^2*Tan[
c + d*x])/(3*d)

Rule 4025

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*n), x]
+ Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^(n + 1)*Simp[a*(a*B*n - A*b*(m - n - 1)) + (
2*a*b*B*n + A*(b^2*n + a^2*(1 + n)))*Csc[e + f*x] + b*(b*B*n + a*A*(m + n))*Csc[e + f*x]^2, x], x], x] /; Free
Q[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LeQ[n, -1]

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]

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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, 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 8

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

Rubi steps

\begin{align*} \int \cos (c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx &=\frac{a A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}-\int (a+b \sec (c+d x))^2 \left (-a (4 A b+a B)-b (A b+2 a B) \sec (c+d x)+b (3 a A-b B) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}-\frac{b (3 a A-b B) (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}-\frac{1}{3} \int (a+b \sec (c+d x)) \left (-3 a^2 (4 A b+a B)-b \left (9 a A b+9 a^2 B+2 b^2 B\right ) \sec (c+d x)+b \left (6 a^2 A-3 A b^2-8 a b B\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}-\frac{b^2 \left (6 a^2 A-3 A b^2-8 a b B\right ) \sec (c+d x) \tan (c+d x)}{6 d}-\frac{b (3 a A-b B) (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}-\frac{1}{6} \int \left (-6 a^3 (4 A b+a B)-3 b \left (12 a^2 A b+A b^3+8 a^3 B+4 a b^2 B\right ) \sec (c+d x)+2 b \left (6 a^3 A-12 a A b^2-17 a^2 b B-2 b^3 B\right ) \sec ^2(c+d x)\right ) \, dx\\ &=a^3 (4 A b+a B) x+\frac{a A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}-\frac{b^2 \left (6 a^2 A-3 A b^2-8 a b B\right ) \sec (c+d x) \tan (c+d x)}{6 d}-\frac{b (3 a A-b B) (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{1}{2} \left (b \left (12 a^2 A b+A b^3+8 a^3 B+4 a b^2 B\right )\right ) \int \sec (c+d x) \, dx-\frac{1}{3} \left (b \left (6 a^3 A-12 a A b^2-17 a^2 b B-2 b^3 B\right )\right ) \int \sec ^2(c+d x) \, dx\\ &=a^3 (4 A b+a B) x+\frac{b \left (12 a^2 A b+A b^3+8 a^3 B+4 a b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}-\frac{b^2 \left (6 a^2 A-3 A b^2-8 a b B\right ) \sec (c+d x) \tan (c+d x)}{6 d}-\frac{b (3 a A-b B) (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{\left (b \left (6 a^3 A-12 a A b^2-17 a^2 b B-2 b^3 B\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=a^3 (4 A b+a B) x+\frac{b \left (12 a^2 A b+A b^3+8 a^3 B+4 a b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a A (a+b \sec (c+d x))^3 \sin (c+d x)}{d}-\frac{b \left (6 a^3 A-12 a A b^2-17 a^2 b B-2 b^3 B\right ) \tan (c+d x)}{3 d}-\frac{b^2 \left (6 a^2 A-3 A b^2-8 a b B\right ) \sec (c+d x) \tan (c+d x)}{6 d}-\frac{b (3 a A-b B) (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}\\ \end{align*}

Mathematica [B]  time = 6.28756, size = 1051, normalized size = 5.39 \[ \frac{\left (-A b^4-4 a B b^3-12 a^2 A b^2-8 a^3 B b\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \cos ^5(c+d x)}{2 d (b+a \cos (c+d x))^4 (B+A \cos (c+d x))}+\frac{\left (A b^4+4 a B b^3+12 a^2 A b^2+8 a^3 B b\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \cos ^5(c+d x)}{2 d (b+a \cos (c+d x))^4 (B+A \cos (c+d x))}+\frac{a^3 (4 A b+a B) (c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \cos ^5(c+d x)}{d (b+a \cos (c+d x))^4 (B+A \cos (c+d x))}+\frac{b^4 B (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^5(c+d x)}{6 d (b+a \cos (c+d x))^4 (B+A \cos (c+d x)) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{2 (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \left (B \sin \left (\frac{1}{2} (c+d x)\right ) b^4+6 a A \sin \left (\frac{1}{2} (c+d x)\right ) b^3+9 a^2 B \sin \left (\frac{1}{2} (c+d x)\right ) b^2\right ) \cos ^5(c+d x)}{3 d (b+a \cos (c+d x))^4 (B+A \cos (c+d x)) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{2 (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \left (B \sin \left (\frac{1}{2} (c+d x)\right ) b^4+6 a A \sin \left (\frac{1}{2} (c+d x)\right ) b^3+9 a^2 B \sin \left (\frac{1}{2} (c+d x)\right ) b^2\right ) \cos ^5(c+d x)}{3 d (b+a \cos (c+d x))^4 (B+A \cos (c+d x)) \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{a^4 A (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \sin (c+d x) \cos ^5(c+d x)}{d (b+a \cos (c+d x))^4 (B+A \cos (c+d x))}+\frac{\left (3 A b^4+B b^4+12 a B b^3\right ) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \cos ^5(c+d x)}{12 d (b+a \cos (c+d x))^4 (B+A \cos (c+d x)) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{\left (-3 A b^4-B b^4-12 a B b^3\right ) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \cos ^5(c+d x)}{12 d (b+a \cos (c+d x))^4 (B+A \cos (c+d x)) \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{b^4 B (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^5(c+d x)}{6 d (b+a \cos (c+d x))^4 (B+A \cos (c+d x)) \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(4*A*b + a*B)*(c + d*x)*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x]))/(d*(b + a*Cos[c + d*x
])^4*(B + A*Cos[c + d*x])) + ((-12*a^2*A*b^2 - A*b^4 - 8*a^3*b*B - 4*a*b^3*B)*Cos[c + d*x]^5*Log[Cos[(c + d*x)
/2] - Sin[(c + d*x)/2]]*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x]))/(2*d*(b + a*Cos[c + d*x])^4*(B + A*Cos[c
+ d*x])) + ((12*a^2*A*b^2 + A*b^4 + 8*a^3*b*B + 4*a*b^3*B)*Cos[c + d*x]^5*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)
/2]]*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x]))/(2*d*(b + a*Cos[c + d*x])^4*(B + A*Cos[c + d*x])) + ((3*A*b^
4 + 12*a*b^3*B + b^4*B)*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x]))/(12*d*(b + a*Cos[c + d*x])
^4*(B + A*Cos[c + d*x])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (b^4*B*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^
4*(A + B*Sec[c + d*x])*Sin[(c + d*x)/2])/(6*d*(b + a*Cos[c + d*x])^4*(B + A*Cos[c + d*x])*(Cos[(c + d*x)/2] -
Sin[(c + d*x)/2])^3) + (b^4*B*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x])*Sin[(c + d*x)/2])/(6*
d*(b + a*Cos[c + d*x])^4*(B + A*Cos[c + d*x])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) + ((-3*A*b^4 - 12*a*b^3
*B - b^4*B)*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x]))/(12*d*(b + a*Cos[c + d*x])^4*(B + A*Co
s[c + d*x])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (2*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^4*(A + B*Sec[c +
d*x])*(6*a*A*b^3*Sin[(c + d*x)/2] + 9*a^2*b^2*B*Sin[(c + d*x)/2] + b^4*B*Sin[(c + d*x)/2]))/(3*d*(b + a*Cos[c
+ d*x])^4*(B + A*Cos[c + d*x])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (2*Cos[c + d*x]^5*(a + b*Sec[c + d*x]
)^4*(A + B*Sec[c + d*x])*(6*a*A*b^3*Sin[(c + d*x)/2] + 9*a^2*b^2*B*Sin[(c + d*x)/2] + b^4*B*Sin[(c + d*x)/2]))
/(3*d*(b + a*Cos[c + d*x])^4*(B + A*Cos[c + d*x])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (a^4*A*Cos[c + d*x]
^5*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x])*Sin[c + d*x])/(d*(b + a*Cos[c + d*x])^4*(B + A*Cos[c + d*x]))

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Maple [A]  time = 0.063, size = 262, normalized size = 1.3 \begin{align*}{\frac{A{a}^{4}\sin \left ( dx+c \right ) }{d}}+B{a}^{4}x+{\frac{B{a}^{4}c}{d}}+4\,A{a}^{3}bx+4\,{\frac{A{a}^{3}bc}{d}}+4\,{\frac{B{a}^{3}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+6\,{\frac{A{a}^{2}{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+6\,{\frac{B{a}^{2}{b}^{2}\tan \left ( dx+c \right ) }{d}}+4\,{\frac{Aa{b}^{3}\tan \left ( dx+c \right ) }{d}}+2\,{\frac{Ba{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+2\,{\frac{Ba{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{A{b}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{A{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,B{b}^{4}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{B{b}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.989914, size = 331, normalized size = 1.7 \begin{align*} \frac{12 \,{\left (d x + c\right )} B a^{4} + 48 \,{\left (d x + c\right )} A a^{3} b + 4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B b^{4} - 12 \, B 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 )} - 3 \, A b^{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 )} + 24 \, B a^{3} b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, A a^{2} b^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{4} \sin \left (d x + c\right ) + 72 \, B a^{2} b^{2} \tan \left (d x + c\right ) + 48 \, A a b^{3} \tan \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*(d*x + c)*B*a^4 + 48*(d*x + c)*A*a^3*b + 4*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*b^4 - 12*B*a*b^3*(2*si
n(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 3*A*b^4*(2*sin(d*x + c)/(si
n(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 24*B*a^3*b*(log(sin(d*x + c) + 1) - log(s
in(d*x + c) - 1)) + 36*A*a^2*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*A*a^4*sin(d*x + c) + 72*
B*a^2*b^2*tan(d*x + c) + 48*A*a*b^3*tan(d*x + c))/d

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Fricas [A]  time = 0.573885, size = 524, normalized size = 2.69 \begin{align*} \frac{12 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} d x \cos \left (d x + c\right )^{3} + 3 \,{\left (8 \, B a^{3} b + 12 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (8 \, B a^{3} b + 12 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (6 \, A a^{4} \cos \left (d x + c\right )^{3} + 2 \, B b^{4} + 4 \,{\left (9 \, B a^{2} b^{2} + 6 \, A a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (4 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*(B*a^4 + 4*A*a^3*b)*d*x*cos(d*x + c)^3 + 3*(8*B*a^3*b + 12*A*a^2*b^2 + 4*B*a*b^3 + A*b^4)*cos(d*x + c
)^3*log(sin(d*x + c) + 1) - 3*(8*B*a^3*b + 12*A*a^2*b^2 + 4*B*a*b^3 + A*b^4)*cos(d*x + c)^3*log(-sin(d*x + c)
+ 1) + 2*(6*A*a^4*cos(d*x + c)^3 + 2*B*b^4 + 4*(9*B*a^2*b^2 + 6*A*a*b^3 + B*b^4)*cos(d*x + c)^2 + 3*(4*B*a*b^3
+ A*b^4)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.2449, size = 522, normalized size = 2.68 \begin{align*} \frac{\frac{12 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1} + 6 \,{\left (B a^{4} + 4 \, A a^{3} b\right )}{\left (d x + c\right )} + 3 \,{\left (8 \, B a^{3} b + 12 \, A a^{2} b^{2} + 4 \, B a b^{3} + A 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^{3} b + 12 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (36 \, B a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 24 \, A a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, B a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, A b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, B b^{4} \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 )^{3} - 48 \, A a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, B b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 36 \, B a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 24 \, A a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, B a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, A b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, 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 )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(12*A*a^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 6*(B*a^4 + 4*A*a^3*b)*(d*x + c) + 3*(8*B*a^3
*b + 12*A*a^2*b^2 + 4*B*a*b^3 + A*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(8*B*a^3*b + 12*A*a^2*b^2 + 4*B*
a*b^3 + A*b^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(36*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 24*A*a*b^3*tan(1/
2*d*x + 1/2*c)^5 - 12*B*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 3*A*b^4*tan(1/2*d*x + 1/2*c)^5 + 6*B*b^4*tan(1/2*d*x +
1/2*c)^5 - 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 - 4*B*b^4*tan(1/2*d*x + 1/2
*c)^3 + 36*B*a^2*b^2*tan(1/2*d*x + 1/2*c) + 24*A*a*b^3*tan(1/2*d*x + 1/2*c) + 12*B*a*b^3*tan(1/2*d*x + 1/2*c)
+ 3*A*b^4*tan(1/2*d*x + 1/2*c) + 6*B*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d