3.577 \(\int \frac{(A+C \cos ^2(c+d x)) \sec ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=335 \[ \frac{2 b^2 \left (5 a^2 A b^2-2 a^2 b^2 C+3 a^4 C-4 A b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{\left (-a^2 b^2 (7 A-6 C)+a^4 (-(2 A+3 C))+12 A b^4\right ) \tan (c+d x)}{3 a^4 d \left (a^2-b^2\right )}-\frac{b \left (a^2 (A+2 C)+4 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{a^5 d}-\frac{\left (4 A b^2-a^2 (A-3 C)\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d \left (a^2-b^2\right )}+\frac{b \left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x) \sec (c+d x)}{a^3 d \left (a^2-b^2\right )}+\frac{\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]

[Out]

(2*b^2*(5*a^2*A*b^2 - 4*A*b^4 + 3*a^4*C - 2*a^2*b^2*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^
5*(a - b)^(3/2)*(a + b)^(3/2)*d) - (b*(4*A*b^2 + a^2*(A + 2*C))*ArcTanh[Sin[c + d*x]])/(a^5*d) - ((12*A*b^4 -
a^2*b^2*(7*A - 6*C) - a^4*(2*A + 3*C))*Tan[c + d*x])/(3*a^4*(a^2 - b^2)*d) + (b*(2*A*b^2 - a^2*(A - C))*Sec[c
+ d*x]*Tan[c + d*x])/(a^3*(a^2 - b^2)*d) - ((4*A*b^2 - a^2*(A - 3*C))*Sec[c + d*x]^2*Tan[c + d*x])/(3*a^2*(a^2
- b^2)*d) + ((A*b^2 + a^2*C)*Sec[c + d*x]^2*Tan[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.46962, antiderivative size = 335, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{2 b^2 \left (5 a^2 A b^2-2 a^2 b^2 C+3 a^4 C-4 A b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{\left (-a^2 b^2 (7 A-6 C)+a^4 (-(2 A+3 C))+12 A b^4\right ) \tan (c+d x)}{3 a^4 d \left (a^2-b^2\right )}-\frac{b \left (a^2 (A+2 C)+4 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{a^5 d}-\frac{\left (4 A b^2-a^2 (A-3 C)\right ) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d \left (a^2-b^2\right )}+\frac{b \left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x) \sec (c+d x)}{a^3 d \left (a^2-b^2\right )}+\frac{\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^2*(5*a^2*A*b^2 - 4*A*b^4 + 3*a^4*C - 2*a^2*b^2*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^
5*(a - b)^(3/2)*(a + b)^(3/2)*d) - (b*(4*A*b^2 + a^2*(A + 2*C))*ArcTanh[Sin[c + d*x]])/(a^5*d) - ((12*A*b^4 -
a^2*b^2*(7*A - 6*C) - a^4*(2*A + 3*C))*Tan[c + d*x])/(3*a^4*(a^2 - b^2)*d) + (b*(2*A*b^2 - a^2*(A - C))*Sec[c
+ d*x]*Tan[c + d*x])/(a^3*(a^2 - b^2)*d) - ((4*A*b^2 - a^2*(A - 3*C))*Sec[c + d*x]^2*Tan[c + d*x])/(3*a^2*(a^2
- b^2)*d) + ((A*b^2 + a^2*C)*Sec[c + d*x]^2*Tan[c + d*x])/(a*(a^2 - b^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 ^4(c+d x)}{(a+b \cos (c+d x))^2} \, dx &=\frac{\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\left (-4 A b^2+a^2 (A-3 C)-a b (A+C) \cos (c+d x)+3 \left (A b^2+a^2 C\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac{\left (4 A b^2-a^2 (A-3 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\left (6 b \left (2 A b^2-a^2 (A-C)\right )+a \left (A b^2+a^2 (2 A+3 C)\right ) \cos (c+d x)-2 b \left (4 A b^2-a^2 (A-3 C)\right ) \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx}{3 a^2 \left (a^2-b^2\right )}\\ &=\frac{b \left (2 A b^2-a^2 (A-C)\right ) \sec (c+d x) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac{\left (4 A b^2-a^2 (A-3 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\left (-2 \left (12 A b^4-a^2 b^2 (7 A-6 C)-a^4 (2 A+3 C)\right )-2 a b \left (2 A b^2+a^2 (A+3 C)\right ) \cos (c+d x)+6 b^2 \left (2 A b^2-a^2 (A-C)\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{6 a^3 \left (a^2-b^2\right )}\\ &=-\frac{\left (12 A b^4-a^2 b^2 (7 A-6 C)-a^4 (2 A+3 C)\right ) \tan (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}+\frac{b \left (2 A b^2-a^2 (A-C)\right ) \sec (c+d x) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac{\left (4 A b^2-a^2 (A-3 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\left (-6 b \left (a^2-b^2\right ) \left (4 A b^2+a^2 (A+2 C)\right )+6 a b^2 \left (2 A b^2-a^2 (A-C)\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{6 a^4 \left (a^2-b^2\right )}\\ &=-\frac{\left (12 A b^4-a^2 b^2 (7 A-6 C)-a^4 (2 A+3 C)\right ) \tan (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}+\frac{b \left (2 A b^2-a^2 (A-C)\right ) \sec (c+d x) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac{\left (4 A b^2-a^2 (A-3 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\left (b^2 \left (4 A b^4-a^2 b^2 (5 A-2 C)-3 a^4 C\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{a^5 \left (a^2-b^2\right )}-\frac{\left (b \left (4 A b^2+a^2 (A+2 C)\right )\right ) \int \sec (c+d x) \, dx}{a^5}\\ &=-\frac{b \left (4 A b^2+a^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{a^5 d}-\frac{\left (12 A b^4-a^2 b^2 (7 A-6 C)-a^4 (2 A+3 C)\right ) \tan (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}+\frac{b \left (2 A b^2-a^2 (A-C)\right ) \sec (c+d x) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac{\left (4 A b^2-a^2 (A-3 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\left (2 b^2 \left (4 A b^4-a^2 b^2 (5 A-2 C)-3 a^4 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^5 \left (a^2-b^2\right ) d}\\ &=\frac{2 b^2 \left (5 a^2 A b^2-4 A b^4+3 a^4 C-2 a^2 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^5 (a-b)^{3/2} (a+b)^{3/2} d}-\frac{b \left (4 A b^2+a^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{a^5 d}-\frac{\left (12 A b^4-a^2 b^2 (7 A-6 C)-a^4 (2 A+3 C)\right ) \tan (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}+\frac{b \left (2 A b^2-a^2 (A-C)\right ) \sec (c+d x) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac{\left (4 A b^2-a^2 (A-3 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 6.2806, size = 593, normalized size = 1.77 \[ \frac{2 a^2 A \sin \left (\frac{1}{2} (c+d x)\right )+3 a^2 C \sin \left (\frac{1}{2} (c+d x)\right )+9 A b^2 \sin \left (\frac{1}{2} (c+d x)\right )}{3 a^4 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{2 a^2 A \sin \left (\frac{1}{2} (c+d x)\right )+3 a^2 C \sin \left (\frac{1}{2} (c+d x)\right )+9 A b^2 \sin \left (\frac{1}{2} (c+d x)\right )}{3 a^4 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{-a^2 b^3 C \sin (c+d x)-A b^5 \sin (c+d x)}{a^4 d (a-b) (a+b) (a+b \cos (c+d x))}-\frac{2 b^2 \left (5 a^2 A b^2-2 a^2 b^2 C+3 a^4 C-4 A b^4\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{a^5 d \left (a^2-b^2\right ) \sqrt{b^2-a^2}}+\frac{\left (a^2 A b+2 a^2 b C+4 A b^3\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{a^5 d}+\frac{\left (-a^2 A b-2 a^2 b C-4 A b^3\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^5 d}+\frac{A (a-6 b)}{12 a^3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{A (a-6 b)}{12 a^3 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{A \sin \left (\frac{1}{2} (c+d x)\right )}{6 a^2 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{A \sin \left (\frac{1}{2} (c+d x)\right )}{6 a^2 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*b^2*(5*a^2*A*b^2 - 4*A*b^4 + 3*a^4*C - 2*a^2*b^2*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/
(a^5*(a^2 - b^2)*Sqrt[-a^2 + b^2]*d) + ((a^2*A*b + 4*A*b^3 + 2*a^2*b*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2
]])/(a^5*d) + ((-(a^2*A*b) - 4*A*b^3 - 2*a^2*b*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/(a^5*d) + (A*(a -
6*b))/(12*a^3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (A*Sin[(c + d*x)/2])/(6*a^2*d*(Cos[(c + d*x)/2] - S
in[(c + d*x)/2])^3) + (A*Sin[(c + d*x)/2])/(6*a^2*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) - (A*(a - 6*b))/(
12*a^3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (2*a^2*A*Sin[(c + d*x)/2] + 9*A*b^2*Sin[(c + d*x)/2] + 3*a
^2*C*Sin[(c + d*x)/2])/(3*a^4*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (2*a^2*A*Sin[(c + d*x)/2] + 9*A*b^2*S
in[(c + d*x)/2] + 3*a^2*C*Sin[(c + d*x)/2])/(3*a^4*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (-(A*b^5*Sin[c +
d*x]) - a^2*b^3*C*Sin[c + d*x])/(a^4*(a - b)*(a + b)*d*(a + b*Cos[c + d*x]))

________________________________________________________________________________________

Maple [B]  time = 0.089, size = 830, normalized size = 2.5 \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)^4/(a+b*cos(d*x+c))^2,x)

[Out]

-1/3/d/a^2*A/(tan(1/2*d*x+1/2*c)+1)^3-1/d/a^2/(tan(1/2*d*x+1/2*c)+1)*C-1/3/d/a^2*A/(tan(1/2*d*x+1/2*c)-1)^3-1/
d/a^2/(tan(1/2*d*x+1/2*c)-1)*C-1/2/d/a^2*A/(tan(1/2*d*x+1/2*c)-1)^2+1/2/d/a^2*A/(tan(1/2*d*x+1/2*c)+1)^2-1/d/a
^2*A/(tan(1/2*d*x+1/2*c)-1)-1/d/a^2*A/(tan(1/2*d*x+1/2*c)+1)+2/d*b/a^3*ln(tan(1/2*d*x+1/2*c)-1)*C+1/d*A/a^3/(t
an(1/2*d*x+1/2*c)+1)^2*b-4/d*b^3/a^5*ln(tan(1/2*d*x+1/2*c)+1)*A-2/d*b/a^3*ln(tan(1/2*d*x+1/2*c)+1)*C-3/d/a^4/(
tan(1/2*d*x+1/2*c)-1)*A*b^2-1/d*A/a^3/(tan(1/2*d*x+1/2*c)-1)^2*b+4/d*b^3/a^5*ln(tan(1/2*d*x+1/2*c)-1)*A-3/d/a^
4/(tan(1/2*d*x+1/2*c)+1)*A*b^2-1/d*A/a^3/(tan(1/2*d*x+1/2*c)-1)*b-1/d*A/a^3/(tan(1/2*d*x+1/2*c)+1)*b-1/d*A*b/a
^3*ln(tan(1/2*d*x+1/2*c)+1)+1/d*A*b/a^3*ln(tan(1/2*d*x+1/2*c)-1)+6/d*b^2/a/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arc
tan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C-4/d*b^4/a^3/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctan((a-b)*t
an(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C-8/d*b^6/a^5/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+
1/2*c)/((a+b)*(a-b))^(1/2))*A+10/d/a^3/(a+b)/(a-b)/((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^4-2/d*b^5/a^4/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+
b)*A-2/d*b^3/a^2/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)*C

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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)^4/(a+b*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 70.9184, size = 2917, normalized size = 8.71 \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)^4/(a+b*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

[-1/6*(3*((3*C*a^4*b^3 + (5*A - 2*C)*a^2*b^5 - 4*A*b^7)*cos(d*x + c)^4 + (3*C*a^5*b^2 + (5*A - 2*C)*a^3*b^4 -
4*A*a*b^6)*cos(d*x + c)^3)*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)) +
3*(((A + 2*C)*a^6*b^2 + 2*(A - 2*C)*a^4*b^4 - (7*A - 2*C)*a^2*b^6 + 4*A*b^8)*cos(d*x + c)^4 + ((A + 2*C)*a^7*b
+ 2*(A - 2*C)*a^5*b^3 - (7*A - 2*C)*a^3*b^5 + 4*A*a*b^7)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 3*(((A + 2*C
)*a^6*b^2 + 2*(A - 2*C)*a^4*b^4 - (7*A - 2*C)*a^2*b^6 + 4*A*b^8)*cos(d*x + c)^4 + ((A + 2*C)*a^7*b + 2*(A - 2*
C)*a^5*b^3 - (7*A - 2*C)*a^3*b^5 + 4*A*a*b^7)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 2*(A*a^8 - 2*A*a^6*b^2
+ A*a^4*b^4 + ((2*A + 3*C)*a^7*b + (5*A - 9*C)*a^5*b^3 - (19*A - 6*C)*a^3*b^5 + 12*A*a*b^7)*cos(d*x + c)^3 + (
(2*A + 3*C)*a^8 + 2*(A - 3*C)*a^6*b^2 - (10*A - 3*C)*a^4*b^4 + 6*A*a^2*b^6)*cos(d*x + c)^2 - 2*(A*a^7*b - 2*A*
a^5*b^3 + A*a^3*b^5)*cos(d*x + c))*sin(d*x + c))/((a^9*b - 2*a^7*b^3 + a^5*b^5)*d*cos(d*x + c)^4 + (a^10 - 2*a
^8*b^2 + a^6*b^4)*d*cos(d*x + c)^3), 1/6*(6*((3*C*a^4*b^3 + (5*A - 2*C)*a^2*b^5 - 4*A*b^7)*cos(d*x + c)^4 + (3
*C*a^5*b^2 + (5*A - 2*C)*a^3*b^4 - 4*A*a*b^6)*cos(d*x + c)^3)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sq
rt(a^2 - b^2)*sin(d*x + c))) - 3*(((A + 2*C)*a^6*b^2 + 2*(A - 2*C)*a^4*b^4 - (7*A - 2*C)*a^2*b^6 + 4*A*b^8)*co
s(d*x + c)^4 + ((A + 2*C)*a^7*b + 2*(A - 2*C)*a^5*b^3 - (7*A - 2*C)*a^3*b^5 + 4*A*a*b^7)*cos(d*x + c)^3)*log(s
in(d*x + c) + 1) + 3*(((A + 2*C)*a^6*b^2 + 2*(A - 2*C)*a^4*b^4 - (7*A - 2*C)*a^2*b^6 + 4*A*b^8)*cos(d*x + c)^4
+ ((A + 2*C)*a^7*b + 2*(A - 2*C)*a^5*b^3 - (7*A - 2*C)*a^3*b^5 + 4*A*a*b^7)*cos(d*x + c)^3)*log(-sin(d*x + c)
+ 1) + 2*(A*a^8 - 2*A*a^6*b^2 + A*a^4*b^4 + ((2*A + 3*C)*a^7*b + (5*A - 9*C)*a^5*b^3 - (19*A - 6*C)*a^3*b^5 +
12*A*a*b^7)*cos(d*x + c)^3 + ((2*A + 3*C)*a^8 + 2*(A - 3*C)*a^6*b^2 - (10*A - 3*C)*a^4*b^4 + 6*A*a^2*b^6)*cos
(d*x + c)^2 - 2*(A*a^7*b - 2*A*a^5*b^3 + A*a^3*b^5)*cos(d*x + c))*sin(d*x + c))/((a^9*b - 2*a^7*b^3 + a^5*b^5)
*d*cos(d*x + c)^4 + (a^10 - 2*a^8*b^2 + a^6*b^4)*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((A+C*cos(d*x+c)**2)*sec(d*x+c)**4/(a+b*cos(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.69331, size = 652, normalized size = 1.95 \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)^4/(a+b*cos(d*x+c))^2,x, algorithm="giac")

[Out]

-1/3*(6*(3*C*a^4*b^2 + 5*A*a^2*b^4 - 2*C*a^2*b^4 - 4*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^7 - a^5*b^2)*sqrt(a^2 - b^2)
) + 6*(C*a^2*b^3*tan(1/2*d*x + 1/2*c) + A*b^5*tan(1/2*d*x + 1/2*c))/((a^6 - a^4*b^2)*(a*tan(1/2*d*x + 1/2*c)^2
- b*tan(1/2*d*x + 1/2*c)^2 + a + b)) + 3*(A*a^2*b + 2*C*a^2*b + 4*A*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a
^5 - 3*(A*a^2*b + 2*C*a^2*b + 4*A*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^5 + 2*(3*A*a^2*tan(1/2*d*x + 1/2*c
)^5 + 3*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*A*a*b*tan(1/2*d*x + 1/2*c)^5 + 9*A*b^2*tan(1/2*d*x + 1/2*c)^5 - 2*A*a
^2*tan(1/2*d*x + 1/2*c)^3 - 6*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 18*A*b^2*tan(1/2*d*x + 1/2*c)^3 + 3*A*a^2*tan(1/2
*d*x + 1/2*c) + 3*C*a^2*tan(1/2*d*x + 1/2*c) - 3*A*a*b*tan(1/2*d*x + 1/2*c) + 9*A*b^2*tan(1/2*d*x + 1/2*c))/((
tan(1/2*d*x + 1/2*c)^2 - 1)^3*a^4))/d