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

Optimal. Leaf size=227 \[ \frac{\left (a^2 b^2 (85 A+56 C)+6 a^4 C+2 b^4 (5 A+4 C)\right ) \tan (c+d x)}{15 d}+\frac{a b \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a b \left (6 a^2 C+40 A b^2+29 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{30 d}+\frac{\left (3 a^2 C+b^2 (5 A+4 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{15 d}+a^4 A x+\frac{a C \tan (c+d x) (a+b \sec (c+d x))^3}{5 d}+\frac{C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d} \]

[Out]

a^4*A*x + (a*b*(4*a^2*(2*A + C) + b^2*(4*A + 3*C))*ArcTanh[Sin[c + d*x]])/(2*d) + ((6*a^4*C + 2*b^4*(5*A + 4*C
) + a^2*b^2*(85*A + 56*C))*Tan[c + d*x])/(15*d) + (a*b*(40*A*b^2 + 6*a^2*C + 29*b^2*C)*Sec[c + d*x]*Tan[c + d*
x])/(30*d) + ((3*a^2*C + b^2*(5*A + 4*C))*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(15*d) + (a*C*(a + b*Sec[c + d*
x])^3*Tan[c + d*x])/(5*d) + (C*(a + b*Sec[c + d*x])^4*Tan[c + d*x])/(5*d)

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Rubi [A]  time = 0.478201, antiderivative size = 227, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {4057, 4056, 4048, 3770, 3767, 8} \[ \frac{\left (a^2 b^2 (85 A+56 C)+6 a^4 C+2 b^4 (5 A+4 C)\right ) \tan (c+d x)}{15 d}+\frac{a b \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a b \left (6 a^2 C+40 A b^2+29 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{30 d}+\frac{\left (3 a^2 C+b^2 (5 A+4 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{15 d}+a^4 A x+\frac{a C \tan (c+d x) (a+b \sec (c+d x))^3}{5 d}+\frac{C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^4*A*x + (a*b*(4*a^2*(2*A + C) + b^2*(4*A + 3*C))*ArcTanh[Sin[c + d*x]])/(2*d) + ((6*a^4*C + 2*b^4*(5*A + 4*C
) + a^2*b^2*(85*A + 56*C))*Tan[c + d*x])/(15*d) + (a*b*(40*A*b^2 + 6*a^2*C + 29*b^2*C)*Sec[c + d*x]*Tan[c + d*
x])/(30*d) + ((3*a^2*C + b^2*(5*A + 4*C))*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(15*d) + (a*C*(a + b*Sec[c + d*
x])^3*Tan[c + d*x])/(5*d) + (C*(a + b*Sec[c + d*x])^4*Tan[c + d*x])/(5*d)

Rule 4057

Int[((A_.) + 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)*Si
mp[a*A*(m + 1) + (A*b*(m + 1) + b*C*m)*Csc[e + f*x] + a*C*m*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A
, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 0]

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 (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{1}{5} \int (a+b \sec (c+d x))^3 \left (5 a A+b (5 A+4 C) \sec (c+d x)+4 a C \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a C (a+b \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac{C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{1}{20} \int (a+b \sec (c+d x))^2 \left (20 a^2 A+4 a b (10 A+7 C) \sec (c+d x)+4 \left (3 a^2 C+b^2 (5 A+4 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{\left (3 a^2 C+b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{15 d}+\frac{a C (a+b \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac{C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{1}{60} \int (a+b \sec (c+d x)) \left (60 a^3 A+4 b \left (9 a^2 (5 A+3 C)+2 b^2 (5 A+4 C)\right ) \sec (c+d x)+4 a \left (40 A b^2+6 a^2 C+29 b^2 C\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a b \left (40 A b^2+6 a^2 C+29 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{30 d}+\frac{\left (3 a^2 C+b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{15 d}+\frac{a C (a+b \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac{C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{1}{120} \int \left (120 a^4 A+60 a b \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \sec (c+d x)+8 \left (6 a^4 C+2 b^4 (5 A+4 C)+a^2 b^2 (85 A+56 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=a^4 A x+\frac{a b \left (40 A b^2+6 a^2 C+29 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{30 d}+\frac{\left (3 a^2 C+b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{15 d}+\frac{a C (a+b \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac{C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{1}{2} \left (a b \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right )\right ) \int \sec (c+d x) \, dx+\frac{1}{15} \left (6 a^4 C+2 b^4 (5 A+4 C)+a^2 b^2 (85 A+56 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=a^4 A x+\frac{a b \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a b \left (40 A b^2+6 a^2 C+29 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{30 d}+\frac{\left (3 a^2 C+b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{15 d}+\frac{a C (a+b \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac{C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}-\frac{\left (6 a^4 C+2 b^4 (5 A+4 C)+a^2 b^2 (85 A+56 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=a^4 A x+\frac{a b \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{\left (6 a^4 C+2 b^4 (5 A+4 C)+a^2 b^2 (85 A+56 C)\right ) \tan (c+d x)}{15 d}+\frac{a b \left (40 A b^2+6 a^2 C+29 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{30 d}+\frac{\left (3 a^2 C+b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{15 d}+\frac{a C (a+b \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac{C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 d}\\ \end{align*}

Mathematica [B]  time = 2.5729, size = 503, normalized size = 2.22 \[ \frac{\sec ^5(c+d x) \left (A \cos ^2(c+d x)+C\right ) \left (-120 a b \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \cos ^5(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+180 a^2 A b^2 \sin (c+d x)+270 a^2 A b^2 \sin (3 (c+d x))+90 a^2 A b^2 \sin (5 (c+d x))+150 a^4 A (c+d x) \cos (c+d x)+75 a^4 A (c+d x) \cos (3 (c+d x))+15 a^4 A c \cos (5 (c+d x))+15 a^4 A d x \cos (5 (c+d x))+240 a^2 b^2 C \sin (c+d x)+300 a^2 b^2 C \sin (3 (c+d x))+60 a^2 b^2 C \sin (5 (c+d x))+120 a^3 b C \sin (2 (c+d x))+60 a^3 b C \sin (4 (c+d x))+30 a^4 C \sin (c+d x)+45 a^4 C \sin (3 (c+d x))+15 a^4 C \sin (5 (c+d x))+120 a A b^3 \sin (2 (c+d x))+60 a A b^3 \sin (4 (c+d x))+210 a b^3 C \sin (2 (c+d x))+45 a b^3 C \sin (4 (c+d x))+40 A b^4 \sin (c+d x)+50 A b^4 \sin (3 (c+d x))+10 A b^4 \sin (5 (c+d x))+80 b^4 C \sin (c+d x)+40 b^4 C \sin (3 (c+d x))+8 b^4 C \sin (5 (c+d x))\right )}{120 d (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((C + A*Cos[c + d*x]^2)*Sec[c + d*x]^5*(150*a^4*A*(c + d*x)*Cos[c + d*x] + 75*a^4*A*(c + d*x)*Cos[3*(c + d*x)]
+ 15*a^4*A*c*Cos[5*(c + d*x)] + 15*a^4*A*d*x*Cos[5*(c + d*x)] - 120*a*b*(4*a^2*(2*A + C) + b^2*(4*A + 3*C))*C
os[c + d*x]^5*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 180*a^2*
A*b^2*Sin[c + d*x] + 40*A*b^4*Sin[c + d*x] + 30*a^4*C*Sin[c + d*x] + 240*a^2*b^2*C*Sin[c + d*x] + 80*b^4*C*Sin
[c + d*x] + 120*a*A*b^3*Sin[2*(c + d*x)] + 120*a^3*b*C*Sin[2*(c + d*x)] + 210*a*b^3*C*Sin[2*(c + d*x)] + 270*a
^2*A*b^2*Sin[3*(c + d*x)] + 50*A*b^4*Sin[3*(c + d*x)] + 45*a^4*C*Sin[3*(c + d*x)] + 300*a^2*b^2*C*Sin[3*(c + d
*x)] + 40*b^4*C*Sin[3*(c + d*x)] + 60*a*A*b^3*Sin[4*(c + d*x)] + 60*a^3*b*C*Sin[4*(c + d*x)] + 45*a*b^3*C*Sin[
4*(c + d*x)] + 90*a^2*A*b^2*Sin[5*(c + d*x)] + 10*A*b^4*Sin[5*(c + d*x)] + 15*a^4*C*Sin[5*(c + d*x)] + 60*a^2*
b^2*C*Sin[5*(c + d*x)] + 8*b^4*C*Sin[5*(c + d*x)]))/(120*d*(A + 2*C + A*Cos[2*(c + d*x)]))

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Maple [A]  time = 0.056, size = 377, normalized size = 1.7 \begin{align*}{a}^{4}Ax+{\frac{A{a}^{4}c}{d}}+{\frac{{a}^{4}C\tan \left ( dx+c \right ) }{d}}+4\,{\frac{A{a}^{3}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{{a}^{3}bC\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+2\,{\frac{{a}^{3}bC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+6\,{\frac{A{a}^{2}{b}^{2}\tan \left ( dx+c \right ) }{d}}+4\,{\frac{C{a}^{2}{b}^{2}\tan \left ( dx+c \right ) }{d}}+2\,{\frac{C{a}^{2}{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+2\,{\frac{Aa{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+2\,{\frac{Aa{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{Ca{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{3\,Ca{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,Ca{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,A{b}^{4}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{A{b}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{8\,C{b}^{4}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{C{b}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,C{b}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^4*A*x+1/d*A*a^4*c+1/d*a^4*C*tan(d*x+c)+4/d*A*a^3*b*ln(sec(d*x+c)+tan(d*x+c))+2/d*a^3*b*C*sec(d*x+c)*tan(d*x+
c)+2/d*a^3*b*C*ln(sec(d*x+c)+tan(d*x+c))+6/d*A*a^2*b^2*tan(d*x+c)+4/d*C*a^2*b^2*tan(d*x+c)+2/d*C*a^2*b^2*tan(d
*x+c)*sec(d*x+c)^2+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))+1/d*C*a*b^3*tan(d*x
+c)*sec(d*x+c)^3+3/2/d*C*a*b^3*sec(d*x+c)*tan(d*x+c)+3/2/d*C*a*b^3*ln(sec(d*x+c)+tan(d*x+c))+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+8/15/d*C*b^4*tan(d*x+c)+1/5/d*C*b^4*tan(d*x+c)*sec(d*x+c)^4+4/15/d*C
*b^4*tan(d*x+c)*sec(d*x+c)^2

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Maxima [A]  time = 0.979258, size = 429, normalized size = 1.89 \begin{align*} \frac{60 \,{\left (d x + c\right )} A a^{4} + 120 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} b^{2} + 20 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{4} + 4 \,{\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^{4} - 15 \, C 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 )} - 60 \, C 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 )} - 60 \, 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 )} + 240 \, A a^{3} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 60 \, C a^{4} \tan \left (d x + c\right ) + 360 \, A a^{2} b^{2} \tan \left (d x + c\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(60*(d*x + c)*A*a^4 + 120*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2*b^2 + 20*(tan(d*x + c)^3 + 3*tan(d*x +
c))*A*b^4 + 4*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C*b^4 - 15*C*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)
) - 60*C*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)) - 60*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)) + 240*A*a^3*b*log(se
c(d*x + c) + tan(d*x + c)) + 60*C*a^4*tan(d*x + c) + 360*A*a^2*b^2*tan(d*x + c))/d

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Fricas [A]  time = 0.582037, size = 610, normalized size = 2.69 \begin{align*} \frac{60 \, A a^{4} d x \cos \left (d x + c\right )^{5} + 15 \,{\left (4 \,{\left (2 \, A + C\right )} a^{3} b +{\left (4 \, A + 3 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (4 \,{\left (2 \, A + C\right )} a^{3} b +{\left (4 \, A + 3 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (30 \, C a b^{3} \cos \left (d x + c\right ) + 6 \, C b^{4} + 2 \,{\left (15 \, C a^{4} + 30 \,{\left (3 \, A + 2 \, C\right )} a^{2} b^{2} + 2 \,{\left (5 \, A + 4 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{4} + 15 \,{\left (4 \, C a^{3} b +{\left (4 \, A + 3 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{3} + 2 \,{\left (30 \, C a^{2} b^{2} +{\left (5 \, A + 4 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(60*A*a^4*d*x*cos(d*x + c)^5 + 15*(4*(2*A + C)*a^3*b + (4*A + 3*C)*a*b^3)*cos(d*x + c)^5*log(sin(d*x + c)
+ 1) - 15*(4*(2*A + C)*a^3*b + (4*A + 3*C)*a*b^3)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(30*C*a*b^3*cos(d
*x + c) + 6*C*b^4 + 2*(15*C*a^4 + 30*(3*A + 2*C)*a^2*b^2 + 2*(5*A + 4*C)*b^4)*cos(d*x + c)^4 + 15*(4*C*a^3*b +
(4*A + 3*C)*a*b^3)*cos(d*x + c)^3 + 2*(30*C*a^2*b^2 + (5*A + 4*C)*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d
*x + c)^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \left (a + b \sec{\left (c + d x \right )}\right )^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.22534, size = 1050, normalized size = 4.63 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(30*(d*x + c)*A*a^4 + 15*(8*A*a^3*b + 4*C*a^3*b + 4*A*a*b^3 + 3*C*a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1
)) - 15*(8*A*a^3*b + 4*C*a^3*b + 4*A*a*b^3 + 3*C*a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(30*C*a^4*tan(1
/2*d*x + 1/2*c)^9 - 60*C*a^3*b*tan(1/2*d*x + 1/2*c)^9 + 180*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 + 180*C*a^2*b^2*t
an(1/2*d*x + 1/2*c)^9 - 60*A*a*b^3*tan(1/2*d*x + 1/2*c)^9 - 75*C*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 30*A*b^4*tan(1
/2*d*x + 1/2*c)^9 + 30*C*b^4*tan(1/2*d*x + 1/2*c)^9 - 120*C*a^4*tan(1/2*d*x + 1/2*c)^7 + 120*C*a^3*b*tan(1/2*d
*x + 1/2*c)^7 - 720*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 480*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 120*A*a*b^3*tan(
1/2*d*x + 1/2*c)^7 + 30*C*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 80*A*b^4*tan(1/2*d*x + 1/2*c)^7 - 40*C*b^4*tan(1/2*d*
x + 1/2*c)^7 + 180*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 1080*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 600*C*a^2*b^2*tan(1/
2*d*x + 1/2*c)^5 + 100*A*b^4*tan(1/2*d*x + 1/2*c)^5 + 116*C*b^4*tan(1/2*d*x + 1/2*c)^5 - 120*C*a^4*tan(1/2*d*x
+ 1/2*c)^3 - 120*C*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 720*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 480*C*a^2*b^2*tan(1/
2*d*x + 1/2*c)^3 - 120*A*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 30*C*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 80*A*b^4*tan(1/2*d
*x + 1/2*c)^3 - 40*C*b^4*tan(1/2*d*x + 1/2*c)^3 + 30*C*a^4*tan(1/2*d*x + 1/2*c) + 60*C*a^3*b*tan(1/2*d*x + 1/2
*c) + 180*A*a^2*b^2*tan(1/2*d*x + 1/2*c) + 180*C*a^2*b^2*tan(1/2*d*x + 1/2*c) + 60*A*a*b^3*tan(1/2*d*x + 1/2*c
) + 75*C*a*b^3*tan(1/2*d*x + 1/2*c) + 30*A*b^4*tan(1/2*d*x + 1/2*c) + 30*C*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*
d*x + 1/2*c)^2 - 1)^5)/d