3.813 \(\int \frac{(B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx\)
Optimal. Leaf size=299 \[ \frac{b \left (-15 a^2 b^3 B+5 a^3 b^2 C+12 a^4 b B-6 a^5 C-2 a b^4 C+6 b^5 B\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{\left (-11 a^2 b^2 B+5 a^3 b C+2 a^4 B-2 a b^3 C+6 b^4 B\right ) \tan (c+d x)}{2 a^3 d \left (a^2-b^2\right )^2}+\frac{b \left (6 a^2 b B-4 a^3 C+a b^2 C-3 b^3 B\right ) \tan (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{b (b B-a C) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac{(3 b B-a C) \tanh ^{-1}(\sin (c+d x))}{a^4 d} \]
[Out]
(b*(12*a^4*b*B - 15*a^2*b^3*B + 6*b^5*B - 6*a^5*C + 5*a^3*b^2*C - 2*a*b^4*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*b*B - a*C)*ArcTanh[Sin[c + d*x]])/(a^4*d) + ((2*a
^4*B - 11*a^2*b^2*B + 6*b^4*B + 5*a^3*b*C - 2*a*b^3*C)*Tan[c + d*x])/(2*a^3*(a^2 - b^2)^2*d) + (b*(b*B - a*C)*
Tan[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) + (b*(6*a^2*b*B - 3*b^3*B - 4*a^3*C + a*b^2*C)*Tan[c
+ d*x])/(2*a^2*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x]))
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Rubi [A] time = 1.80346, antiderivative size = 299, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used =
{3029, 3000, 3055, 3001, 3770, 2659, 205} \[ \frac{b \left (-15 a^2 b^3 B+5 a^3 b^2 C+12 a^4 b B-6 a^5 C-2 a b^4 C+6 b^5 B\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{\left (-11 a^2 b^2 B+5 a^3 b C+2 a^4 B-2 a b^3 C+6 b^4 B\right ) \tan (c+d x)}{2 a^3 d \left (a^2-b^2\right )^2}+\frac{b \left (6 a^2 b B-4 a^3 C+a b^2 C-3 b^3 B\right ) \tan (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{b (b B-a C) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac{(3 b B-a C) \tanh ^{-1}(\sin (c+d x))}{a^4 d} \]
Antiderivative was successfully verified.
[In]
Int[((B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + b*Cos[c + d*x])^3,x]
[Out]
(b*(12*a^4*b*B - 15*a^2*b^3*B + 6*b^5*B - 6*a^5*C + 5*a^3*b^2*C - 2*a*b^4*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*b*B - a*C)*ArcTanh[Sin[c + d*x]])/(a^4*d) + ((2*a
^4*B - 11*a^2*b^2*B + 6*b^4*B + 5*a^3*b*C - 2*a*b^3*C)*Tan[c + d*x])/(2*a^3*(a^2 - b^2)^2*d) + (b*(b*B - a*C)*
Tan[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) + (b*(6*a^2*b*B - 3*b^3*B - 4*a^3*C + a*b^2*C)*Tan[c
+ d*x])/(2*a^2*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x]))
Rule 3029
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Rule 3000
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((A*b^2 - a*b*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*
Sin[e + f*x])^(1 + n))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[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[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m
+ n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B)*(m + n + 3)*Sin[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^
2, 0] && RationalQ[m] && 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 (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx &=\int \frac{(B+C \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx\\ &=\frac{b (b B-a C) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{\int \frac{\left (2 a^2 B-3 b^2 B+a b C-2 a (b B-a C) \cos (c+d x)+2 b (b B-a C) \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{b (b B-a C) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{b \left (6 a^2 b B-3 b^3 B-4 a^3 C+a b^2 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 (2 a^4 B-11 a^2 b^2 B+6 b^4 B+5 a^3 b C-2 a b^3 C-a \left (4 a^2 b B-b^3 B-2 a^3 C-a b^2 C\right ) \cos (c+d x)+b \left (6 a^2 b B-3 b^3 B-4 a^3 C+a b^2 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 (2 a^4 B-11 a^2 b^2 B+6 b^4 B+5 a^3 b C-2 a b^3 C\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{b (b B-a C) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{b \left (6 a^2 b B-3 b^3 B-4 a^3 C+a b^2 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 (-2 \left (a^2-b^2\right )^2 (3 b B-a C)+a b \left (6 a^2 b B-3 b^3 B-4 a^3 C+a b^2 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 (2 a^4 B-11 a^2 b^2 B+6 b^4 B+5 a^3 b C-2 a b^3 C\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{b (b B-a C) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{b \left (6 a^2 b B-3 b^3 B-4 a^3 C+a b^2 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 b B-a C) \int \sec (c+d x) \, dx}{a^4}+\frac{\left (b \left (12 a^4 b B-15 a^2 b^3 B+6 b^5 B-6 a^5 C+5 a^3 b^2 C-2 a b^4 C\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 a^4 \left (a^2-b^2\right )^2}\\ &=-\frac{(3 b B-a C) \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac{\left (2 a^4 B-11 a^2 b^2 B+6 b^4 B+5 a^3 b C-2 a b^3 C\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{b (b B-a C) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{b \left (6 a^2 b B-3 b^3 B-4 a^3 C+a b^2 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 (b \left (12 a^4 b B-15 a^2 b^3 B+6 b^5 B-6 a^5 C+5 a^3 b^2 C-2 a b^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^4 \left (a^2-b^2\right )^2 d}\\ &=\frac{b \left (12 a^4 b B-15 a^2 b^3 B+6 b^5 B-6 a^5 C+5 a^3 b^2 C-2 a b^4 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 b B-a C) \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac{\left (2 a^4 B-11 a^2 b^2 B+6 b^4 B+5 a^3 b C-2 a b^3 C\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{b (b B-a C) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{b \left (6 a^2 b B-3 b^3 B-4 a^3 C+a b^2 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 [A] time = 5.85741, size = 352, normalized size = 1.18 \[ \frac{\frac{a b^2 \left (-7 a^2 b B+5 a^3 C-2 a b^2 C+4 b^3 B\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}+\frac{a^2 b^2 (a C-b B) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}-\frac{2 b \left (-15 a^2 b^3 B+5 a^3 b^2 C+12 a^4 b B-6 a^5 C-2 a b^4 C+6 b^5 B\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}+2 (3 b B-a C) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 (a C-3 b B) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{2 a B \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{2 a B \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}}{2 a^4 d} \]
Antiderivative was successfully verified.
[In]
Integrate[((B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + b*Cos[c + d*x])^3,x]
[Out]
((-2*b*(12*a^4*b*B - 15*a^2*b^3*B + 6*b^5*B - 6*a^5*C + 5*a^3*b^2*C - 2*a*b^4*C)*ArcTanh[((a - b)*Tan[(c + d*x
)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) + 2*(3*b*B - a*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*(-3
*b*B + a*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (2*a*B*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d
*x)/2]) + (2*a*B*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + (a^2*b^2*(-(b*B) + a*C)*Sin[c + d*x
])/((a - b)*(a + b)*(a + b*Cos[c + d*x])^2) + (a*b^2*(-7*a^2*b*B + 4*b^3*B + 5*a^3*C - 2*a*b^2*C)*Sin[c + d*x]
)/((a - b)^2*(a + b)^2*(a + b*Cos[c + d*x])))/(2*a^4*d)
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Maple [B] time = 0.091, size = 1358, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c))^3,x)
[Out]
-1/d/a^3*B/(tan(1/2*d*x+1/2*c)-1)+3/d/a^4*ln(tan(1/2*d*x+1/2*c)-1)*b*B-1/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)*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)^3*B-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)^3*B+4/d*
b^5/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/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*B+6/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+1/d*b^
3/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*C-2/d*b^4
/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*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-b)^2*tan(1/2*d*x+1/2*c)*B+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-b)^2*tan(1/2*d*x+1/2*c)*B+4/d*b^5/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/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*B+6/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-b)^2*tan(1/2*d*x+1/2*c)*b^2*C-1/d*b^3/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-b)^2*tan(1/2*d*x+1/2*c)*C-2/d*b^4/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-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))*B-15/d*b^4/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))*B+6/d*b^6/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))*B-6/d*b*a/(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+5/d*b^3/a/(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-2/d*b^5/a^3/(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-1/d/a^3*B/(tan(1/2*d*x+1/2*c)+1)-3/d/a^4*ln(tan(1/2*d*x+1/2*c)+1)*b*B+1/d/a^3*ln(tan(1/2*d*x+1
/2*c)+1)*C
________________________________________________________________________________________
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((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c))^3,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [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((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c))^3,x, algorithm="fricas")
[Out]
Timed out
________________________________________________________________________________________
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((B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**3/(a+b*cos(d*x+c))**3,x)
[Out]
Timed out
________________________________________________________________________________________
Giac [B] time = 1.44892, size = 775, normalized size = 2.59 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c))^3,x, algorithm="giac")
[Out]
((6*C*a^5*b - 12*B*a^4*b^2 - 5*C*a^3*b^3 + 15*B*a^2*b^4 + 2*C*a*b^5 - 6*B*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)) + (6*C*a^4*b^2*tan(1/2*d*x + 1/2*c)^3 - 8*B*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 -
5*C*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 + 7*B*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 - 3*C*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 +
5*B*a*b^5*tan(1/2*d*x + 1/2*c)^3 + 2*C*a*b^5*tan(1/2*d*x + 1/2*c)^3 - 4*B*b^6*tan(1/2*d*x + 1/2*c)^3 + 6*C*a^
4*b^2*tan(1/2*d*x + 1/2*c) - 8*B*a^3*b^3*tan(1/2*d*x + 1/2*c) + 5*C*a^3*b^3*tan(1/2*d*x + 1/2*c) - 7*B*a^2*b^4
*tan(1/2*d*x + 1/2*c) - 3*C*a^2*b^4*tan(1/2*d*x + 1/2*c) + 5*B*a*b^5*tan(1/2*d*x + 1/2*c) - 2*C*a*b^5*tan(1/2*
d*x + 1/2*c) + 4*B*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) + (C*a - 3*B*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - (C*a - 3*B*b)*log(abs(ta
n(1/2*d*x + 1/2*c) - 1))/a^4 - 2*B*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^3))/d