3.590 \(\int \frac{(A+C \cos ^2(c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^4} \, dx\)
Optimal. Leaf size=301 \[ -\frac{b \left (-a^4 b^2 (8 A-C)+7 a^2 A b^4+4 a^6 (2 A+C)-2 A b^6\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{7/2} (a+b)^{7/2}}-\frac{\left (-13 a^4 b^2 (2 A+C)+17 a^2 A b^4-2 a^6 C-6 A b^6\right ) \sin (c+d x)}{6 a^3 d \left (a^2-b^2\right )^3 (a+b \cos (c+d x))}-\frac{\left (-a^2 b^2 (8 A+3 C)-2 a^4 C+3 A b^4\right ) \sin (c+d x)}{6 a^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+\frac{\left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a^4 d} \]
[Out]
-((b*(7*a^2*A*b^4 - 2*A*b^6 - a^4*b^2*(8*A - C) + 4*a^6*(2*A + C))*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[
a + b]])/(a^4*(a - b)^(7/2)*(a + b)^(7/2)*d)) + (A*ArcTanh[Sin[c + d*x]])/(a^4*d) + ((A*b^2 + a^2*C)*Sin[c + d
*x])/(3*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^3) - ((3*A*b^4 - 2*a^4*C - a^2*b^2*(8*A + 3*C))*Sin[c + d*x])/(6*
a^2*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x])^2) - ((17*a^2*A*b^4 - 6*A*b^6 - 2*a^6*C - 13*a^4*b^2*(2*A + C))*Sin[c
+ d*x])/(6*a^3*(a^2 - b^2)^3*d*(a + b*Cos[c + d*x]))
________________________________________________________________________________________
Rubi [A] time = 1.30167, antiderivative size = 301, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used =
{3056, 3055, 3001, 3770, 2659, 205} \[ -\frac{b \left (-a^4 b^2 (8 A-C)+7 a^2 A b^4+4 a^6 (2 A+C)-2 A b^6\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{7/2} (a+b)^{7/2}}-\frac{\left (-13 a^4 b^2 (2 A+C)+17 a^2 A b^4-2 a^6 C-6 A b^6\right ) \sin (c+d x)}{6 a^3 d \left (a^2-b^2\right )^3 (a+b \cos (c+d x))}-\frac{\left (-a^2 b^2 (8 A+3 C)-2 a^4 C+3 A b^4\right ) \sin (c+d x)}{6 a^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+\frac{\left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a^4 d} \]
Antiderivative was successfully verified.
[In]
Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x])^4,x]
[Out]
-((b*(7*a^2*A*b^4 - 2*A*b^6 - a^4*b^2*(8*A - C) + 4*a^6*(2*A + C))*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[
a + b]])/(a^4*(a - b)^(7/2)*(a + b)^(7/2)*d)) + (A*ArcTanh[Sin[c + d*x]])/(a^4*d) + ((A*b^2 + a^2*C)*Sin[c + d
*x])/(3*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^3) - ((3*A*b^4 - 2*a^4*C - a^2*b^2*(8*A + 3*C))*Sin[c + d*x])/(6*
a^2*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x])^2) - ((17*a^2*A*b^4 - 6*A*b^6 - 2*a^6*C - 13*a^4*b^2*(2*A + C))*Sin[c
+ d*x])/(6*a^3*(a^2 - b^2)^3*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 (c+d x)}{(a+b \cos (c+d x))^4} \, dx &=\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac{\int \frac{\left (3 A \left (a^2-b^2\right )-3 a b (A+C) \cos (c+d x)+2 \left (A b^2+a^2 C\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx}{3 a \left (a^2-b^2\right )}\\ &=\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac{\left (3 A b^4-2 a^4 C-a^2 b^2 (8 A+3 C)\right ) \sin (c+d x)}{6 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac{\int \frac{\left (6 A \left (a^2-b^2\right )^2+2 a b \left (A b^2-a^2 (6 A+5 C)\right ) \cos (c+d x)-\left (3 A b^4-2 a^4 C-a^2 b^2 (8 A+3 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{6 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac{\left (3 A b^4-2 a^4 C-a^2 b^2 (8 A+3 C)\right ) \sin (c+d x)}{6 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}-\frac{\left (17 a^2 A b^4-6 A b^6-2 a^6 C-13 a^4 b^2 (2 A+C)\right ) \sin (c+d x)}{6 a^3 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}+\frac{\int \frac{\left (6 A \left (a^2-b^2\right )^3-3 a b \left (A b^4-a^2 b^2 (2 A-C)+a^4 (6 A+4 C)\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{6 a^3 \left (a^2-b^2\right )^3}\\ &=\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac{\left (3 A b^4-2 a^4 C-a^2 b^2 (8 A+3 C)\right ) \sin (c+d x)}{6 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}-\frac{\left (17 a^2 A b^4-6 A b^6-2 a^6 C-13 a^4 b^2 (2 A+C)\right ) \sin (c+d x)}{6 a^3 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}+\frac{A \int \sec (c+d x) \, dx}{a^4}-\frac{\left (b \left (7 a^2 A b^4-2 A b^6-a^4 b^2 (8 A-C)+4 a^6 (2 A+C)\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 a^4 \left (a^2-b^2\right )^3}\\ &=\frac{A \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac{\left (3 A b^4-2 a^4 C-a^2 b^2 (8 A+3 C)\right ) \sin (c+d x)}{6 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}-\frac{\left (17 a^2 A b^4-6 A b^6-2 a^6 C-13 a^4 b^2 (2 A+C)\right ) \sin (c+d x)}{6 a^3 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}-\frac{\left (b \left (7 a^2 A b^4-2 A b^6-a^4 b^2 (8 A-C)+4 a^6 (2 A+C)\right )\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 )^3 d}\\ &=-\frac{b \left (8 a^6 A-8 a^4 A b^2+7 a^2 A b^4-2 A b^6+4 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)^{7/2} (a+b)^{7/2} d}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac{\left (3 A b^4-2 a^4 C-a^2 b^2 (8 A+3 C)\right ) \sin (c+d x)}{6 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}-\frac{\left (17 a^2 A b^4-6 A b^6-2 a^6 C-13 a^4 b^2 (2 A+C)\right ) \sin (c+d x)}{6 a^3 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}\\ \end{align*}
Mathematica [C] time = 5.24793, size = 498, normalized size = 1.65 \[ \frac{\cos (c+d x) (A \sec (c+d x)+C \cos (c+d x)) \left (\frac{6 b (\sin (c)+i \cos (c)) \left (a^4 b^2 (C-8 A)+7 a^2 A b^4+4 a^6 (2 A+C)-2 A b^6\right ) \tan ^{-1}\left (\frac{(\sin (c)+i \cos (c)) \left (\tan \left (\frac{d x}{2}\right ) (b \cos (c)-a)+b \sin (c)\right )}{\sqrt{-\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2}}\right )}{a^4 \left (a^2-b^2\right )^3 \sqrt{-\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2}}+\frac{2 \sec (c) \left (a^2 C+A b^2\right ) (b \sin (d x)-a \sin (c))}{a b \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}+\frac{\sec (c) \sin (d x) \left (13 a^4 b^2 (2 A+C)-17 a^2 A b^4+2 a^6 C+6 A b^6\right )-3 a b \tan (c) \left (a^2 b^2 (C-2 A)+a^4 (6 A+4 C)+A b^4\right )}{\left (a^3-a b^2\right )^3 (a+b \cos (c+d x))}+\frac{\sec (c) \sin (d x) \left (a^2 b^2 (8 A+3 C)+2 a^4 C-3 A b^4\right )+a b \tan (c) \left (A b^2-a^2 (6 A+5 C)\right )}{a^2 \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}-\frac{6 A \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{a^4}+\frac{6 A \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^4}\right )}{3 d (2 A+C \cos (2 (c+d x))+C)} \]
Antiderivative was successfully verified.
[In]
Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x])^4,x]
[Out]
(Cos[c + d*x]*(C*Cos[c + d*x] + A*Sec[c + d*x])*((-6*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/a^4 + (6*A*Lo
g[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/a^4 + (6*b*(7*a^2*A*b^4 - 2*A*b^6 + a^4*b^2*(-8*A + C) + 4*a^6*(2*A +
C))*ArcTan[((I*Cos[c] + Sin[c])*(b*Sin[c] + (-a + b*Cos[c])*Tan[(d*x)/2]))/Sqrt[-((a^2 - b^2)*(Cos[c] - I*Sin[
c])^2)]]*(I*Cos[c] + Sin[c]))/(a^4*(a^2 - b^2)^3*Sqrt[-((a^2 - b^2)*(Cos[c] - I*Sin[c])^2)]) + (2*(A*b^2 + a^2
*C)*Sec[c]*(-(a*Sin[c]) + b*Sin[d*x]))/(a*b*(a^2 - b^2)*(a + b*Cos[c + d*x])^3) + ((-17*a^2*A*b^4 + 6*A*b^6 +
2*a^6*C + 13*a^4*b^2*(2*A + C))*Sec[c]*Sin[d*x] - 3*a*b*(A*b^4 + a^2*b^2*(-2*A + C) + a^4*(6*A + 4*C))*Tan[c])
/((a^3 - a*b^2)^3*(a + b*Cos[c + d*x])) + ((-3*A*b^4 + 2*a^4*C + a^2*b^2*(8*A + 3*C))*Sec[c]*Sin[d*x] + a*b*(A
*b^2 - a^2*(6*A + 5*C))*Tan[c])/(a^2*(a^2 - b^2)^2*(a + b*Cos[c + d*x])^2)))/(3*d*(2*A + C + C*Cos[2*(c + d*x)
]))
________________________________________________________________________________________
Maple [B] time = 0.082, size = 2337, normalized size = 7.8 \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)/(a+b*cos(d*x+c))^4,x)
[Out]
1/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5
*C*b^3-1/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1
/2*c)*C*b^3+12/d*b^2/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3*a/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*t
an(1/2*d*x+1/2*c)*A+24/d*b^2/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3*a/(a^2+2*a*b+b^2)/(a^2-2*a*
b+b^2)*tan(1/2*d*x+1/2*c)^3*A+12/d*b^2/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3*a/(a-b)/(a^3+3*a^
2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5*A+8/d*b^3/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a+b)*(a-b))^(1/2)*arctan((a-b)
*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A-1/d*b^3/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a+b)*(a-b))^(1/2)*arctan((a
-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C+1/d/a^4*A*ln(tan(1/2*d*x+1/2*c)+1)-1/d/a^4*A*ln(tan(1/2*d*x+1/2*
c)-1)+6/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/
2*c)^5*C*a*b^2+6/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1
/2*d*x+1/2*c)*C*a*b^2+28/3/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3*a/(a^2-2*a*b+b^2)/(a^2+2*a*
b+b^2)*tan(1/2*d*x+1/2*c)^3*b^2*C+2/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3*a^3/(a+b)/(a^3-3*a
^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)*C-4/d*b/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a+b)*(a-b))^(1/2)*arctan((a-b)*ta
n(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C*a^2-8/d*b/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a+b)*(a-b))^(1/2)*arctan((a-
b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*a^2*A+2/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3*a^3
/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5*C+4/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b
)^3*a^3/(a^2+2*a*b+b^2)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*C-4/d*b^3/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2
*c)^2*b+a+b)^3/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)*A+4/d*b^3/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*
x+1/2*c)^2*b+a+b)^3/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5*A-2/d*b/(a*tan(1/2*d*x+1/2*c)^2-tan(1
/2*d*x+1/2*c)^2*b+a+b)^3/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)*C*a^2+2/d*b/(a*tan(1/2*d*x+1/2*c)^
2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5*C*a^2-7/d/a^2*b^5/(a^6-3*
a^4*b^2+3*a^2*b^4-b^6)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A+2/d/a^4*b^7/
(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A+2/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)^3/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5*
A*b^6-44/3/d/a/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a^2+2*a*b+b^2)/(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)^3/(a^2+2*a*b+b^2)/(a^2-2*a*b+b^2
)*tan(1/2*d*x+1/2*c)^3*A*b^6-6/d/a/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a+b)/(a^3-3*a^2*b+3*
a*b^2-b^3)*tan(1/2*d*x+1/2*c)*A*b^4+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)^3/(a+b)/(a^3-3
*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)*A*b^5+2/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)^3/(a+
b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)*A*b^6-6/d/a/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b
)^3/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5*A*b^4-1/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)^3/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5*A*b^5
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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)/(a+b*cos(d*x+c))^4,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 159.947, size = 4764, normalized size = 15.83 \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)/(a+b*cos(d*x+c))^4,x, algorithm="fricas")
[Out]
[-1/12*(3*(4*(2*A + C)*a^9*b - (8*A - C)*a^7*b^3 + 7*A*a^5*b^5 - 2*A*a^3*b^7 + (4*(2*A + C)*a^6*b^4 - (8*A - C
)*a^4*b^6 + 7*A*a^2*b^8 - 2*A*b^10)*cos(d*x + c)^3 + 3*(4*(2*A + C)*a^7*b^3 - (8*A - C)*a^5*b^5 + 7*A*a^3*b^7
- 2*A*a*b^9)*cos(d*x + c)^2 + 3*(4*(2*A + C)*a^8*b^2 - (8*A - C)*a^6*b^4 + 7*A*a^4*b^6 - 2*A*a^2*b^8)*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^11 - 4*A*a^9*b
^2 + 6*A*a^7*b^4 - 4*A*a^5*b^6 + A*a^3*b^8 + (A*a^8*b^3 - 4*A*a^6*b^5 + 6*A*a^4*b^7 - 4*A*a^2*b^9 + A*b^11)*co
s(d*x + c)^3 + 3*(A*a^9*b^2 - 4*A*a^7*b^4 + 6*A*a^5*b^6 - 4*A*a^3*b^8 + A*a*b^10)*cos(d*x + c)^2 + 3*(A*a^10*b
- 4*A*a^8*b^3 + 6*A*a^6*b^5 - 4*A*a^4*b^7 + A*a^2*b^9)*cos(d*x + c))*log(sin(d*x + c) + 1) + 6*(A*a^11 - 4*A*
a^9*b^2 + 6*A*a^7*b^4 - 4*A*a^5*b^6 + A*a^3*b^8 + (A*a^8*b^3 - 4*A*a^6*b^5 + 6*A*a^4*b^7 - 4*A*a^2*b^9 + A*b^1
1)*cos(d*x + c)^3 + 3*(A*a^9*b^2 - 4*A*a^7*b^4 + 6*A*a^5*b^6 - 4*A*a^3*b^8 + A*a*b^10)*cos(d*x + c)^2 + 3*(A*a
^10*b - 4*A*a^8*b^3 + 6*A*a^6*b^5 - 4*A*a^4*b^7 + A*a^2*b^9)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(6*C*a^1
1 + 4*(9*A + C)*a^9*b^2 - (68*A + 11*C)*a^7*b^4 + (43*A + C)*a^5*b^6 - 11*A*a^3*b^8 + (2*C*a^9*b^2 + (26*A + 1
1*C)*a^7*b^4 - (43*A + 13*C)*a^5*b^6 + 23*A*a^3*b^8 - 6*A*a*b^10)*cos(d*x + c)^2 + 3*(2*C*a^10*b + (20*A + 7*C
)*a^8*b^3 - 5*(7*A + 2*C)*a^6*b^5 + (20*A + C)*a^4*b^7 - 5*A*a^2*b^9)*cos(d*x + c))*sin(d*x + c))/((a^12*b^3 -
4*a^10*b^5 + 6*a^8*b^7 - 4*a^6*b^9 + a^4*b^11)*d*cos(d*x + c)^3 + 3*(a^13*b^2 - 4*a^11*b^4 + 6*a^9*b^6 - 4*a^
7*b^8 + a^5*b^10)*d*cos(d*x + c)^2 + 3*(a^14*b - 4*a^12*b^3 + 6*a^10*b^5 - 4*a^8*b^7 + a^6*b^9)*d*cos(d*x + c)
+ (a^15 - 4*a^13*b^2 + 6*a^11*b^4 - 4*a^9*b^6 + a^7*b^8)*d), -1/6*(3*(4*(2*A + C)*a^9*b - (8*A - C)*a^7*b^3 +
7*A*a^5*b^5 - 2*A*a^3*b^7 + (4*(2*A + C)*a^6*b^4 - (8*A - C)*a^4*b^6 + 7*A*a^2*b^8 - 2*A*b^10)*cos(d*x + c)^3
+ 3*(4*(2*A + C)*a^7*b^3 - (8*A - C)*a^5*b^5 + 7*A*a^3*b^7 - 2*A*a*b^9)*cos(d*x + c)^2 + 3*(4*(2*A + C)*a^8*b
^2 - (8*A - C)*a^6*b^4 + 7*A*a^4*b^6 - 2*A*a^2*b^8)*cos(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^11 - 4*A*a^9*b^2 + 6*A*a^7*b^4 - 4*A*a^5*b^6 + A*a^3*b^8 + (A*a^8*b^
3 - 4*A*a^6*b^5 + 6*A*a^4*b^7 - 4*A*a^2*b^9 + A*b^11)*cos(d*x + c)^3 + 3*(A*a^9*b^2 - 4*A*a^7*b^4 + 6*A*a^5*b^
6 - 4*A*a^3*b^8 + A*a*b^10)*cos(d*x + c)^2 + 3*(A*a^10*b - 4*A*a^8*b^3 + 6*A*a^6*b^5 - 4*A*a^4*b^7 + A*a^2*b^9
)*cos(d*x + c))*log(sin(d*x + c) + 1) + 3*(A*a^11 - 4*A*a^9*b^2 + 6*A*a^7*b^4 - 4*A*a^5*b^6 + A*a^3*b^8 + (A*a
^8*b^3 - 4*A*a^6*b^5 + 6*A*a^4*b^7 - 4*A*a^2*b^9 + A*b^11)*cos(d*x + c)^3 + 3*(A*a^9*b^2 - 4*A*a^7*b^4 + 6*A*a
^5*b^6 - 4*A*a^3*b^8 + A*a*b^10)*cos(d*x + c)^2 + 3*(A*a^10*b - 4*A*a^8*b^3 + 6*A*a^6*b^5 - 4*A*a^4*b^7 + A*a^
2*b^9)*cos(d*x + c))*log(-sin(d*x + c) + 1) - (6*C*a^11 + 4*(9*A + C)*a^9*b^2 - (68*A + 11*C)*a^7*b^4 + (43*A
+ C)*a^5*b^6 - 11*A*a^3*b^8 + (2*C*a^9*b^2 + (26*A + 11*C)*a^7*b^4 - (43*A + 13*C)*a^5*b^6 + 23*A*a^3*b^8 - 6*
A*a*b^10)*cos(d*x + c)^2 + 3*(2*C*a^10*b + (20*A + 7*C)*a^8*b^3 - 5*(7*A + 2*C)*a^6*b^5 + (20*A + C)*a^4*b^7 -
5*A*a^2*b^9)*cos(d*x + c))*sin(d*x + c))/((a^12*b^3 - 4*a^10*b^5 + 6*a^8*b^7 - 4*a^6*b^9 + a^4*b^11)*d*cos(d*
x + c)^3 + 3*(a^13*b^2 - 4*a^11*b^4 + 6*a^9*b^6 - 4*a^7*b^8 + a^5*b^10)*d*cos(d*x + c)^2 + 3*(a^14*b - 4*a^12*
b^3 + 6*a^10*b^5 - 4*a^8*b^7 + a^6*b^9)*d*cos(d*x + c) + (a^15 - 4*a^13*b^2 + 6*a^11*b^4 - 4*a^9*b^6 + a^7*b^8
)*d)]
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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)/(a+b*cos(d*x+c))**4,x)
[Out]
Timed out
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Giac [B] time = 1.4977, size = 1172, normalized size = 3.89 \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)/(a+b*cos(d*x+c))^4,x, algorithm="giac")
[Out]
-1/3*(3*(8*A*a^6*b + 4*C*a^6*b - 8*A*a^4*b^3 + C*a^4*b^3 + 7*A*a^2*b^5 - 2*A*b^7)*(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^10 - 3*a
^8*b^2 + 3*a^6*b^4 - a^4*b^6)*sqrt(a^2 - b^2)) - 3*A*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 + 3*A*log(abs(tan(
1/2*d*x + 1/2*c) - 1))/a^4 - (6*C*a^8*tan(1/2*d*x + 1/2*c)^5 - 6*C*a^7*b*tan(1/2*d*x + 1/2*c)^5 + 36*A*a^6*b^2
*tan(1/2*d*x + 1/2*c)^5 + 12*C*a^6*b^2*tan(1/2*d*x + 1/2*c)^5 - 60*A*a^5*b^3*tan(1/2*d*x + 1/2*c)^5 - 27*C*a^5
*b^3*tan(1/2*d*x + 1/2*c)^5 - 6*A*a^4*b^4*tan(1/2*d*x + 1/2*c)^5 + 12*C*a^4*b^4*tan(1/2*d*x + 1/2*c)^5 + 45*A*
a^3*b^5*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^3*b^5*tan(1/2*d*x + 1/2*c)^5 - 6*A*a^2*b^6*tan(1/2*d*x + 1/2*c)^5 - 15*
A*a*b^7*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^8*tan(1/2*d*x + 1/2*c)^5 + 12*C*a^8*tan(1/2*d*x + 1/2*c)^3 + 72*A*a^6*b
^2*tan(1/2*d*x + 1/2*c)^3 + 16*C*a^6*b^2*tan(1/2*d*x + 1/2*c)^3 - 116*A*a^4*b^4*tan(1/2*d*x + 1/2*c)^3 - 28*C*
a^4*b^4*tan(1/2*d*x + 1/2*c)^3 + 56*A*a^2*b^6*tan(1/2*d*x + 1/2*c)^3 - 12*A*b^8*tan(1/2*d*x + 1/2*c)^3 + 6*C*a
^8*tan(1/2*d*x + 1/2*c) + 6*C*a^7*b*tan(1/2*d*x + 1/2*c) + 36*A*a^6*b^2*tan(1/2*d*x + 1/2*c) + 12*C*a^6*b^2*ta
n(1/2*d*x + 1/2*c) + 60*A*a^5*b^3*tan(1/2*d*x + 1/2*c) + 27*C*a^5*b^3*tan(1/2*d*x + 1/2*c) - 6*A*a^4*b^4*tan(1
/2*d*x + 1/2*c) + 12*C*a^4*b^4*tan(1/2*d*x + 1/2*c) - 45*A*a^3*b^5*tan(1/2*d*x + 1/2*c) - 3*C*a^3*b^5*tan(1/2*
d*x + 1/2*c) - 6*A*a^2*b^6*tan(1/2*d*x + 1/2*c) + 15*A*a*b^7*tan(1/2*d*x + 1/2*c) + 6*A*b^8*tan(1/2*d*x + 1/2*
c))/((a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^3))
/d