3.583 \(\int \frac{(A+C \cos ^2(c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx\)
Optimal. Leaf size=275 \[ -\frac{\left (-a^4 b^2 (12 A+C)+15 a^2 A b^4-2 a^6 C-6 A b^6\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\left (11 a^2 A b^2+a^4 (-(2 A-3 C))-6 A b^4\right ) \tan (c+d x)}{2 a^3 d \left (a^2-b^2\right )^2}-\frac{\left (-a^2 b^2 (6 A+C)-2 a^4 C+3 A b^4\right ) \tan (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{\left (a^2 C+A b^2\right ) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac{3 A b \tanh ^{-1}(\sin (c+d x))}{a^4 d} \]
[Out]
-(((15*a^2*A*b^4 - 6*A*b^6 - 2*a^6*C - a^4*b^2*(12*A + C))*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])
/(a^4*(a - b)^(5/2)*(a + b)^(5/2)*d)) - (3*A*b*ArcTanh[Sin[c + d*x]])/(a^4*d) - ((11*a^2*A*b^2 - 6*A*b^4 - a^4
*(2*A - 3*C))*Tan[c + d*x])/(2*a^3*(a^2 - b^2)^2*d) + ((A*b^2 + a^2*C)*Tan[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b
*Cos[c + d*x])^2) - ((3*A*b^4 - 2*a^4*C - a^2*b^2*(6*A + C))*Tan[c + d*x])/(2*a^2*(a^2 - b^2)^2*d*(a + b*Cos[c
+ d*x]))
________________________________________________________________________________________
Rubi [A] time = 1.18937, antiderivative size = 275, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used =
{3056, 3055, 3001, 3770, 2659, 205} \[ -\frac{\left (-a^4 b^2 (12 A+C)+15 a^2 A b^4-2 a^6 C-6 A b^6\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\left (11 a^2 A b^2+a^4 (-(2 A-3 C))-6 A b^4\right ) \tan (c+d x)}{2 a^3 d \left (a^2-b^2\right )^2}-\frac{\left (-a^2 b^2 (6 A+C)-2 a^4 C+3 A b^4\right ) \tan (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{\left (a^2 C+A b^2\right ) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac{3 A b \tanh ^{-1}(\sin (c+d x))}{a^4 d} \]
Antiderivative was successfully verified.
[In]
Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + b*Cos[c + d*x])^3,x]
[Out]
-(((15*a^2*A*b^4 - 6*A*b^6 - 2*a^6*C - a^4*b^2*(12*A + C))*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])
/(a^4*(a - b)^(5/2)*(a + b)^(5/2)*d)) - (3*A*b*ArcTanh[Sin[c + d*x]])/(a^4*d) - ((11*a^2*A*b^2 - 6*A*b^4 - a^4
*(2*A - 3*C))*Tan[c + d*x])/(2*a^3*(a^2 - b^2)^2*d) + ((A*b^2 + a^2*C)*Tan[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b
*Cos[c + d*x])^2) - ((3*A*b^4 - 2*a^4*C - a^2*b^2*(6*A + C))*Tan[c + d*x])/(2*a^2*(a^2 - b^2)^2*d*(a + b*Cos[c
+ d*x]))
Rule 3056
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Rule 3055
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))
Rule 3001
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Rule 3770
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Rule 2659
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]
Rule 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rubi steps
\begin{align*} \int \frac{\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx &=\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{\int \frac{\left (-3 A b^2+a^2 (2 A-C)-2 a b (A+C) \cos (c+d x)+2 \left (A b^2+a^2 C\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\left (3 A b^4-2 a^4 C-a^2 b^2 (6 A+C)\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\int \frac{\left (-11 a^2 A b^2+6 A b^4+a^4 (2 A-3 C)+a b \left (A b^2-a^2 (4 A+3 C)\right ) \cos (c+d x)-\left (3 A b^4-2 a^4 C-a^2 b^2 (6 A+C)\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (11 a^2 A b^2-6 A b^4-a^4 (2 A-3 C)\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\left (3 A b^4-2 a^4 C-a^2 b^2 (6 A+C)\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\int \frac{\left (-6 A b \left (a^2-b^2\right )^2-a \left (3 A b^4-2 a^4 C-a^2 b^2 (6 A+C)\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (11 a^2 A b^2-6 A b^4-a^4 (2 A-3 C)\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\left (3 A b^4-2 a^4 C-a^2 b^2 (6 A+C)\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{(3 A b) \int \sec (c+d x) \, dx}{a^4}-\frac{\left (15 a^2 A b^4-6 A b^6-2 a^6 C-a^4 b^2 (12 A+C)\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 a^4 \left (a^2-b^2\right )^2}\\ &=-\frac{3 A b \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac{\left (11 a^2 A b^2-6 A b^4-a^4 (2 A-3 C)\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\left (3 A b^4-2 a^4 C-a^2 b^2 (6 A+C)\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\left (15 a^2 A b^4-6 A b^6-2 a^6 C-a^4 b^2 (12 A+C)\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^4 \left (a^2-b^2\right )^2 d}\\ &=\frac{\left (12 a^4 A b^2-15 a^2 A b^4+6 A b^6+2 a^6 C+a^4 b^2 C\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 (a-b)^{5/2} (a+b)^{5/2} d}-\frac{3 A b \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac{\left (11 a^2 A b^2-6 A b^4-a^4 (2 A-3 C)\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\left (3 A b^4-2 a^4 C-a^2 b^2 (6 A+C)\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}\\ \end{align*}
Mathematica [B] time = 6.33256, size = 649, normalized size = 2.36 \[ \frac{\cos ^2(c+d x) \left (A \sec ^2(c+d x)+C\right ) \left (-a^2 b C \sin (c+d x)-A b^3 \sin (c+d x)\right )}{a^2 d (a-b) (a+b) (a+b \cos (c+d x))^2 (2 A+C \cos (2 c+2 d x)+C)}+\frac{\cos ^2(c+d x) \left (A \sec ^2(c+d x)+C\right ) \left (-7 a^2 A b^3 \sin (c+d x)-3 a^4 b C \sin (c+d x)+4 A b^5 \sin (c+d x)\right )}{a^3 d (a-b)^2 (a+b)^2 (a+b \cos (c+d x)) (2 A+C \cos (2 c+2 d x)+C)}-\frac{2 \left (12 a^4 A b^2-15 a^2 A b^4+a^4 b^2 C+2 a^6 C+6 A b^6\right ) \cos ^2(c+d x) \left (A \sec ^2(c+d x)+C\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{a^4 d \left (a^2-b^2\right )^2 \sqrt{b^2-a^2} (2 A+C \cos (2 c+2 d x)+C)}+\frac{6 A b \cos ^2(c+d x) \left (A \sec ^2(c+d x)+C\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{a^4 d (2 A+C \cos (2 c+2 d x)+C)}-\frac{6 A b \cos ^2(c+d x) \left (A \sec ^2(c+d x)+C\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^4 d (2 A+C \cos (2 c+2 d x)+C)}+\frac{2 A \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^2(c+d x) \left (A \sec ^2(c+d x)+C\right )}{a^3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) (2 A+C \cos (2 c+2 d x)+C)}+\frac{2 A \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^2(c+d x) \left (A \sec ^2(c+d x)+C\right )}{a^3 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) (2 A+C \cos (2 c+2 d x)+C)} \]
Antiderivative was successfully verified.
[In]
Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + b*Cos[c + d*x])^3,x]
[Out]
(-2*(12*a^4*A*b^2 - 15*a^2*A*b^4 + 6*A*b^6 + 2*a^6*C + a^4*b^2*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2
+ b^2]]*Cos[c + d*x]^2*(C + A*Sec[c + d*x]^2))/(a^4*(a^2 - b^2)^2*Sqrt[-a^2 + b^2]*d*(2*A + C + C*Cos[2*c + 2
*d*x])) + (6*A*b*Cos[c + d*x]^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(C + A*Sec[c + d*x]^2))/(a^4*d*(2*A +
C + C*Cos[2*c + 2*d*x])) - (6*A*b*Cos[c + d*x]^2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(C + A*Sec[c + d*x]
^2))/(a^4*d*(2*A + C + C*Cos[2*c + 2*d*x])) + (2*A*Cos[c + d*x]^2*(C + A*Sec[c + d*x]^2)*Sin[(c + d*x)/2])/(a^
3*d*(2*A + C + C*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (2*A*Cos[c + d*x]^2*(C + A*Sec[c +
d*x]^2)*Sin[(c + d*x)/2])/(a^3*d*(2*A + C + C*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (Cos
[c + d*x]^2*(C + A*Sec[c + d*x]^2)*(-(A*b^3*Sin[c + d*x]) - a^2*b*C*Sin[c + d*x]))/(a^2*(a - b)*(a + b)*d*(a +
b*Cos[c + d*x])^2*(2*A + C + C*Cos[2*c + 2*d*x])) + (Cos[c + d*x]^2*(C + A*Sec[c + d*x]^2)*(-7*a^2*A*b^3*Sin[
c + d*x] + 4*A*b^5*Sin[c + d*x] - 3*a^4*b*C*Sin[c + d*x]))/(a^3*(a - b)^2*(a + b)^2*d*(a + b*Cos[c + d*x])*(2*
A + C + C*Cos[2*c + 2*d*x]))
________________________________________________________________________________________
Maple [B] time = 0.085, size = 1129, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x)
[Out]
-8/d/a/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A*b^3-
1/d/a^2/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A*b^4
+4/d/a^3/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*b^5/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*
A-4/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*a*b*C-1
/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*b^2*C-8/d/
a/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*b^3/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)*A+1/d/a^2
/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*b^4/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)*A+4/d/a^3/
(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*b^5/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)*A-4/d/(a*ta
n(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)*a*b*C+1/d/(a*tan(1/2
*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*b^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)*C+12/d*b^2/(a^4-2*a^2
*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A-15/d/a^2/(a^4-2*a^2*b^2+b
^4)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A*b^4+6/d/a^4/(a^4-2*a^2*b^2+b^4)
/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A*b^6+2/d/(a^4-2*a^2*b^2+b^4)/((a+b)
*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C*a^2+1/d/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))
^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*b^2*C-1/d/a^3*A/(tan(1/2*d*x+1/2*c)-1)+3/d*A*b/a^4
*ln(tan(1/2*d*x+1/2*c)-1)-1/d/a^3*A/(tan(1/2*d*x+1/2*c)+1)-3/d*A*b/a^4*ln(tan(1/2*d*x+1/2*c)+1)
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 82.808, size = 3401, normalized size = 12.37 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm="fricas")
[Out]
[-1/4*(((2*C*a^6*b^2 + (12*A + C)*a^4*b^4 - 15*A*a^2*b^6 + 6*A*b^8)*cos(d*x + c)^3 + 2*(2*C*a^7*b + (12*A + C)
*a^5*b^3 - 15*A*a^3*b^5 + 6*A*a*b^7)*cos(d*x + c)^2 + (2*C*a^8 + (12*A + C)*a^6*b^2 - 15*A*a^4*b^4 + 6*A*a^2*b
^6)*cos(d*x + c))*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 + 2*sqrt(-a^2 + b^2)
*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) + 6*((A*a^6
*b^3 - 3*A*a^4*b^5 + 3*A*a^2*b^7 - A*b^9)*cos(d*x + c)^3 + 2*(A*a^7*b^2 - 3*A*a^5*b^4 + 3*A*a^3*b^6 - A*a*b^8)
*cos(d*x + c)^2 + (A*a^8*b - 3*A*a^6*b^3 + 3*A*a^4*b^5 - A*a^2*b^7)*cos(d*x + c))*log(sin(d*x + c) + 1) - 6*((
A*a^6*b^3 - 3*A*a^4*b^5 + 3*A*a^2*b^7 - A*b^9)*cos(d*x + c)^3 + 2*(A*a^7*b^2 - 3*A*a^5*b^4 + 3*A*a^3*b^6 - A*a
*b^8)*cos(d*x + c)^2 + (A*a^8*b - 3*A*a^6*b^3 + 3*A*a^4*b^5 - A*a^2*b^7)*cos(d*x + c))*log(-sin(d*x + c) + 1)
- 2*(2*A*a^9 - 6*A*a^7*b^2 + 6*A*a^5*b^4 - 2*A*a^3*b^6 + ((2*A - 3*C)*a^7*b^2 - (13*A - 3*C)*a^5*b^4 + 17*A*a^
3*b^6 - 6*A*a*b^8)*cos(d*x + c)^2 + (4*(A - C)*a^8*b - 5*(4*A - C)*a^6*b^3 + (25*A - C)*a^4*b^5 - 9*A*a^2*b^7)
*cos(d*x + c))*sin(d*x + c))/((a^10*b^2 - 3*a^8*b^4 + 3*a^6*b^6 - a^4*b^8)*d*cos(d*x + c)^3 + 2*(a^11*b - 3*a^
9*b^3 + 3*a^7*b^5 - a^5*b^7)*d*cos(d*x + c)^2 + (a^12 - 3*a^10*b^2 + 3*a^8*b^4 - a^6*b^6)*d*cos(d*x + c)), 1/2
*(((2*C*a^6*b^2 + (12*A + C)*a^4*b^4 - 15*A*a^2*b^6 + 6*A*b^8)*cos(d*x + c)^3 + 2*(2*C*a^7*b + (12*A + C)*a^5*
b^3 - 15*A*a^3*b^5 + 6*A*a*b^7)*cos(d*x + c)^2 + (2*C*a^8 + (12*A + C)*a^6*b^2 - 15*A*a^4*b^4 + 6*A*a^2*b^6)*c
os(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - 3*((A*a^6*b^3 - 3*
A*a^4*b^5 + 3*A*a^2*b^7 - A*b^9)*cos(d*x + c)^3 + 2*(A*a^7*b^2 - 3*A*a^5*b^4 + 3*A*a^3*b^6 - A*a*b^8)*cos(d*x
+ c)^2 + (A*a^8*b - 3*A*a^6*b^3 + 3*A*a^4*b^5 - A*a^2*b^7)*cos(d*x + c))*log(sin(d*x + c) + 1) + 3*((A*a^6*b^3
- 3*A*a^4*b^5 + 3*A*a^2*b^7 - A*b^9)*cos(d*x + c)^3 + 2*(A*a^7*b^2 - 3*A*a^5*b^4 + 3*A*a^3*b^6 - A*a*b^8)*cos
(d*x + c)^2 + (A*a^8*b - 3*A*a^6*b^3 + 3*A*a^4*b^5 - A*a^2*b^7)*cos(d*x + c))*log(-sin(d*x + c) + 1) + (2*A*a^
9 - 6*A*a^7*b^2 + 6*A*a^5*b^4 - 2*A*a^3*b^6 + ((2*A - 3*C)*a^7*b^2 - (13*A - 3*C)*a^5*b^4 + 17*A*a^3*b^6 - 6*A
*a*b^8)*cos(d*x + c)^2 + (4*(A - C)*a^8*b - 5*(4*A - C)*a^6*b^3 + (25*A - C)*a^4*b^5 - 9*A*a^2*b^7)*cos(d*x +
c))*sin(d*x + c))/((a^10*b^2 - 3*a^8*b^4 + 3*a^6*b^6 - a^4*b^8)*d*cos(d*x + c)^3 + 2*(a^11*b - 3*a^9*b^3 + 3*a
^7*b^5 - a^5*b^7)*d*cos(d*x + c)^2 + (a^12 - 3*a^10*b^2 + 3*a^8*b^4 - a^6*b^6)*d*cos(d*x + c))]
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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((A+C*cos(d*x+c)**2)*sec(d*x+c)**2/(a+b*cos(d*x+c))**3,x)
[Out]
Timed out
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Giac [B] time = 1.63396, size = 698, normalized size = 2.54 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm="giac")
[Out]
-((2*C*a^6 + 12*A*a^4*b^2 + C*a^4*b^2 - 15*A*a^2*b^4 + 6*A*b^6)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2
*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/((a^8 - 2*a^6*b^2 + a^4*b^4)
*sqrt(a^2 - b^2)) + 3*A*b*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 3*A*b*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^
4 + (4*C*a^5*b*tan(1/2*d*x + 1/2*c)^3 - 3*C*a^4*b^2*tan(1/2*d*x + 1/2*c)^3 + 8*A*a^3*b^3*tan(1/2*d*x + 1/2*c)^
3 - C*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 - 7*A*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 - 5*A*a*b^5*tan(1/2*d*x + 1/2*c)^3 +
4*A*b^6*tan(1/2*d*x + 1/2*c)^3 + 4*C*a^5*b*tan(1/2*d*x + 1/2*c) + 3*C*a^4*b^2*tan(1/2*d*x + 1/2*c) + 8*A*a^3*
b^3*tan(1/2*d*x + 1/2*c) - C*a^3*b^3*tan(1/2*d*x + 1/2*c) + 7*A*a^2*b^4*tan(1/2*d*x + 1/2*c) - 5*A*a*b^5*tan(1
/2*d*x + 1/2*c) - 4*A*b^6*tan(1/2*d*x + 1/2*c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan
(1/2*d*x + 1/2*c)^2 + a + b)^2) + 2*A*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^3))/d