3.7.46 \(\int \sec ^2(c+d x) (a+b \sec (c+d x))^2 (A+C \sec ^2(c+d x)) \, dx\) [646]

Optimal. Leaf size=226 \[ \frac {a b (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {\left (a^4 C+2 a^2 b^2 (5 A+3 C)+2 b^4 (5 A+4 C)\right ) \tan (c+d x)}{15 b^2 d}+\frac {a \left (20 A b^2+2 a^2 C+13 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{60 b d}+\frac {\left (a^2 C+2 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{30 b^2 d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d} \]

[Out]

1/4*a*b*(4*A+3*C)*arctanh(sin(d*x+c))/d+1/15*(a^4*C+2*a^2*b^2*(5*A+3*C)+2*b^4*(5*A+4*C))*tan(d*x+c)/b^2/d+1/60
*a*(20*A*b^2+2*C*a^2+13*C*b^2)*sec(d*x+c)*tan(d*x+c)/b/d+1/30*(a^2*C+2*b^2*(5*A+4*C))*(a+b*sec(d*x+c))^2*tan(d
*x+c)/b^2/d-1/10*a*C*(a+b*sec(d*x+c))^3*tan(d*x+c)/b^2/d+1/5*C*sec(d*x+c)*(a+b*sec(d*x+c))^3*tan(d*x+c)/b/d

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Rubi [A]
time = 0.35, antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {4178, 4167, 4087, 4082, 3872, 3855, 3852, 8} \begin {gather*} \frac {\left (a^2 C+2 b^2 (5 A+4 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{30 b^2 d}+\frac {a \left (2 a^2 C+20 A b^2+13 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{60 b d}+\frac {\left (a^4 C+2 a^2 b^2 (5 A+3 C)+2 b^4 (5 A+4 C)\right ) \tan (c+d x)}{15 b^2 d}+\frac {a b (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{4 d}-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^3}{10 b^2 d}+\frac {C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^3}{5 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b*(4*A + 3*C)*ArcTanh[Sin[c + d*x]])/(4*d) + ((a^4*C + 2*a^2*b^2*(5*A + 3*C) + 2*b^4*(5*A + 4*C))*Tan[c + d
*x])/(15*b^2*d) + (a*(20*A*b^2 + 2*a^2*C + 13*b^2*C)*Sec[c + d*x]*Tan[c + d*x])/(60*b*d) + ((a^2*C + 2*b^2*(5*
A + 4*C))*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(30*b^2*d) - (a*C*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(10*b^2*
d) + (C*Sec[c + d*x]*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(5*b*d)

Rule 8

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

Rule 3852

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 3855

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

Rule 3872

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 4082

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
*Csc[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 4087

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] /;
FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]

Rule 4167

Int[csc[(e_.) + (f_.)*(x_)]*((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 + 1)/(b*f*(m
+ 2))), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) + (b*
B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 4178

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))
^(m_), x_Symbol] :> Simp[(-C)*Csc[e + f*x]*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Dis
t[1/(b*(m + 3)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m + 2) + A*(m + 3))*Csc[e + f*x] - 2
*a*C*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (a C+b (5 A+4 C) \sec (c+d x)-2 a C \sec ^2(c+d x)\right ) \, dx}{5 b}\\ &=-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (-2 a b C+2 \left (a^2 C+2 b^2 (5 A+4 C)\right ) \sec (c+d x)\right ) \, dx}{20 b^2}\\ &=\frac {\left (a^2 C+2 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{30 b^2 d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x)) \left (2 b \left (20 A b^2-a^2 C+16 b^2 C\right )+2 a \left (20 A b^2+2 a^2 C+13 b^2 C\right ) \sec (c+d x)\right ) \, dx}{60 b^2}\\ &=\frac {a \left (20 A b^2+2 a^2 C+13 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{60 b d}+\frac {\left (a^2 C+2 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{30 b^2 d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) \left (30 a b^3 (4 A+3 C)+8 \left (a^4 C+2 a^2 b^2 (5 A+3 C)+2 b^4 (5 A+4 C)\right ) \sec (c+d x)\right ) \, dx}{120 b^2}\\ &=\frac {a \left (20 A b^2+2 a^2 C+13 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{60 b d}+\frac {\left (a^2 C+2 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{30 b^2 d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}+\frac {1}{4} (a b (4 A+3 C)) \int \sec (c+d x) \, dx+\frac {\left (a^4 C+2 a^2 b^2 (5 A+3 C)+2 b^4 (5 A+4 C)\right ) \int \sec ^2(c+d x) \, dx}{15 b^2}\\ &=\frac {a b (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {a \left (20 A b^2+2 a^2 C+13 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{60 b d}+\frac {\left (a^2 C+2 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{30 b^2 d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}-\frac {\left (a^4 C+2 a^2 b^2 (5 A+3 C)+2 b^4 (5 A+4 C)\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 b^2 d}\\ &=\frac {a b (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {\left (a^4 C+2 a^2 b^2 (5 A+3 C)+2 b^4 (5 A+4 C)\right ) \tan (c+d x)}{15 b^2 d}+\frac {a \left (20 A b^2+2 a^2 C+13 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{60 b d}+\frac {\left (a^2 C+2 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{30 b^2 d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{10 b^2 d}+\frac {C \sec (c+d x) (a+b \sec (c+d x))^3 \tan (c+d x)}{5 b d}\\ \end {align*}

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Mathematica [A]
time = 2.33, size = 275, normalized size = 1.22 \begin {gather*} -\frac {\left (C+A \cos ^2(c+d x)\right ) \sec ^5(c+d x) \left (60 a b (4 A+3 C) \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 (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-\left (90 a^2 A+100 A b^2+100 a^2 C+128 b^2 C+15 a b (12 A+17 C) \cos (c+d x)+24 \left (5 a^2 (A+C)+b^2 (5 A+4 C)\right ) \cos (2 (c+d x))+60 a A b \cos (3 (c+d x))+45 a b C \cos (3 (c+d x))+30 a^2 A \cos (4 (c+d x))+20 A b^2 \cos (4 (c+d x))+20 a^2 C \cos (4 (c+d x))+16 b^2 C \cos (4 (c+d x))\right ) \sin (c+d x)\right )}{120 d (A+2 C+A \cos (2 (c+d x)))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/120*((C + A*Cos[c + d*x]^2)*Sec[c + d*x]^5*(60*a*b*(4*A + 3*C)*Cos[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]]) - (90*a^2*A + 100*A*b^2 + 100*a^2*C + 128*b^2*C + 15*
a*b*(12*A + 17*C)*Cos[c + d*x] + 24*(5*a^2*(A + C) + b^2*(5*A + 4*C))*Cos[2*(c + d*x)] + 60*a*A*b*Cos[3*(c + d
*x)] + 45*a*b*C*Cos[3*(c + d*x)] + 30*a^2*A*Cos[4*(c + d*x)] + 20*A*b^2*Cos[4*(c + d*x)] + 20*a^2*C*Cos[4*(c +
d*x)] + 16*b^2*C*Cos[4*(c + d*x)])*Sin[c + d*x]))/(d*(A + 2*C + A*Cos[2*(c + d*x)]))

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Maple [A]
time = 0.11, size = 184, normalized size = 0.81

method	result	size
derivativedivides	\(\frac {-A \,b^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )-b^{2} C \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+2 a A b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 a b C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+A \,a^{2} \tan \left (d x +c \right )-a^{2} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(184\)
default	\(\frac {-A \,b^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )-b^{2} C \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+2 a A b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 a b C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+A \,a^{2} \tan \left (d x +c \right )-a^{2} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(184\)
norman	\(\frac {-\frac {4 \left (45 A \,a^{2}+25 A \,b^{2}+25 a^{2} C +29 b^{2} C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}-\frac {\left (4 A \,a^{2}-4 a A b +4 A \,b^{2}+4 a^{2} C -5 a b C +4 b^{2} C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {\left (4 A \,a^{2}+4 a A b +4 A \,b^{2}+4 a^{2} C +5 a b C +4 b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {\left (24 A \,a^{2}-12 a A b +16 A \,b^{2}+16 a^{2} C -3 a b C +8 b^{2} C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {\left (24 A \,a^{2}+12 a A b +16 A \,b^{2}+16 a^{2} C +3 a b C +8 b^{2} C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}-\frac {b a \left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}+\frac {b a \left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}\)	\(314\)
risch	\(-\frac {i \left (60 a A b \,{\mathrm e}^{9 i \left (d x +c \right )}+45 C a b \,{\mathrm e}^{9 i \left (d x +c \right )}-60 A \,a^{2} {\mathrm e}^{8 i \left (d x +c \right )}+120 A a b \,{\mathrm e}^{7 i \left (d x +c \right )}+210 C a b \,{\mathrm e}^{7 i \left (d x +c \right )}-240 A \,a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-120 A \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-120 C \,a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-360 A \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-280 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-280 C \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-320 C \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-120 a A b \,{\mathrm e}^{3 i \left (d x +c \right )}-210 C a b \,{\mathrm e}^{3 i \left (d x +c \right )}-240 A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-200 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-200 C \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-160 C \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-60 a A b \,{\mathrm e}^{i \left (d x +c \right )}-45 C b a \,{\mathrm e}^{i \left (d x +c \right )}-60 A \,a^{2}-40 A \,b^{2}-40 a^{2} C -32 b^{2} C \right )}{30 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {3 b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{4 d}+\frac {b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {3 b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{4 d}\)	\(421\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 216, normalized size = 0.96 \begin {gather*} \frac {40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} + 40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{2} + 8 \, {\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^{2} - 15 \, C a b {\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 \, A a 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 )} + 120 \, A a^{2} \tan \left (d x + c\right )}{120 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(40*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2 + 40*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*b^2 + 8*(3*tan(d*x
+ c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C*b^2 - 15*C*a*b*(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*A*a*b*(2*sin(d*x + c
)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 120*A*a^2*tan(d*x + c))/d

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Fricas [A]
time = 3.82, size = 180, normalized size = 0.80 \begin {gather*} \frac {15 \, {\left (4 \, A + 3 \, C\right )} a b \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, A + 3 \, C\right )} a b \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (15 \, {\left (4 \, A + 3 \, C\right )} a b \cos \left (d x + c\right )^{3} + 4 \, {\left (5 \, {\left (3 \, A + 2 \, C\right )} a^{2} + 2 \, {\left (5 \, A + 4 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} + 30 \, C a b \cos \left (d x + c\right ) + 12 \, C b^{2} + 4 \, {\left (5 \, C a^{2} + {\left (5 \, A + 4 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{120 \, d \cos \left (d x + c\right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(4*A + 3*C)*a*b*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(4*A + 3*C)*a*b*cos(d*x + c)^5*log(-sin(d*
x + c) + 1) + 2*(15*(4*A + 3*C)*a*b*cos(d*x + c)^3 + 4*(5*(3*A + 2*C)*a^2 + 2*(5*A + 4*C)*b^2)*cos(d*x + c)^4
+ 30*C*a*b*cos(d*x + c) + 12*C*b^2 + 4*(5*C*a^2 + (5*A + 4*C)*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x +
c)^5)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sec ^{2}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 532 vs. \(2 (214) = 428\).
time = 0.50, size = 532, normalized size = 2.35 \begin {gather*} \frac {15 \, {\left (4 \, A a b + 3 \, C a b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (4 \, A a b + 3 \, C a b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (60 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 240 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 160 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 160 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 80 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 360 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 200 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 200 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 232 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 240 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 80 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 60 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, C b^{2} \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 )}^{5}}}{60 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(15*(4*A*a*b + 3*C*a*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*A*a*b + 3*C*a*b)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) - 2*(60*A*a^2*tan(1/2*d*x + 1/2*c)^9 + 60*C*a^2*tan(1/2*d*x + 1/2*c)^9 - 60*A*a*b*tan(1/2*d*x +
1/2*c)^9 - 75*C*a*b*tan(1/2*d*x + 1/2*c)^9 + 60*A*b^2*tan(1/2*d*x + 1/2*c)^9 + 60*C*b^2*tan(1/2*d*x + 1/2*c)^9
- 240*A*a^2*tan(1/2*d*x + 1/2*c)^7 - 160*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 120*A*a*b*tan(1/2*d*x + 1/2*c)^7 + 30
*C*a*b*tan(1/2*d*x + 1/2*c)^7 - 160*A*b^2*tan(1/2*d*x + 1/2*c)^7 - 80*C*b^2*tan(1/2*d*x + 1/2*c)^7 + 360*A*a^2
*tan(1/2*d*x + 1/2*c)^5 + 200*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 200*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 232*C*b^2*tan(
1/2*d*x + 1/2*c)^5 - 240*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 160*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 120*A*a*b*tan(1/2*d
*x + 1/2*c)^3 - 30*C*a*b*tan(1/2*d*x + 1/2*c)^3 - 160*A*b^2*tan(1/2*d*x + 1/2*c)^3 - 80*C*b^2*tan(1/2*d*x + 1/
2*c)^3 + 60*A*a^2*tan(1/2*d*x + 1/2*c) + 60*C*a^2*tan(1/2*d*x + 1/2*c) + 60*A*a*b*tan(1/2*d*x + 1/2*c) + 75*C*
a*b*tan(1/2*d*x + 1/2*c) + 60*A*b^2*tan(1/2*d*x + 1/2*c) + 60*C*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c
)^2 - 1)^5)/d

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Mupad [B]
time = 7.55, size = 322, normalized size = 1.42 \begin {gather*} \frac {a\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,A+3\,C\right )}{2\,d}-\frac {\left (2\,A\,a^2+2\,A\,b^2+2\,C\,a^2+2\,C\,b^2-2\,A\,a\,b-\frac {5\,C\,a\,b}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (4\,A\,a\,b-\frac {16\,A\,b^2}{3}-\frac {16\,C\,a^2}{3}-\frac {8\,C\,b^2}{3}-8\,A\,a^2+C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (12\,A\,a^2+\frac {20\,A\,b^2}{3}+\frac {20\,C\,a^2}{3}+\frac {116\,C\,b^2}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-8\,A\,a^2-\frac {16\,A\,b^2}{3}-\frac {16\,C\,a^2}{3}-\frac {8\,C\,b^2}{3}-4\,A\,a\,b-C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^2+2\,A\,b^2+2\,C\,a^2+2\,C\,b^2+2\,A\,a\,b+\frac {5\,C\,a\,b}{2}\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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