\(\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^3} \, dx\) [896]
Optimal result
Integrand size = 42, antiderivative size = 420 \[
\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^3} \, dx=-\frac {\left (8 a^4 B-29 a^2 b^2 B+15 b^4 B+9 a^3 b C-3 a b^3 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}+\frac {\left (11 a^2 b B-5 b^3 B-7 a^3 C+a b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 a^2 \left (a^2-b^2\right )^2 d}-\frac {\left (35 a^4 b B-38 a^2 b^3 B+15 b^5 B-15 a^5 C+6 a^3 b^2 C-3 a b^4 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a^3 (a-b)^2 (a+b)^3 d}+\frac {\left (8 a^4 B-29 a^2 b^2 B+15 b^4 B+9 a^3 b C-3 a b^3 C\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {b (b B-a C) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {b \left (11 a^2 b B-5 b^3 B-7 a^3 C+a b^2 C\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}
\]
[Out]
-1/4*(8*B*a^4-29*B*a^2*b^2+15*B*b^4+9*C*a^3*b-3*C*a*b^3)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellip
ticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/(a^2-b^2)^2/d+1/4*(11*B*a^2*b-5*B*b^3-7*C*a^3+C*a*b^2)*(cos(1/2*d*x+1/2*c
)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/(a^2-b^2)^2/d-1/4*(35*B*a^4*b-38*B*a^2
*b^3+15*B*b^5-15*C*a^5+6*C*a^3*b^2-3*C*a*b^4)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1
/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/a^3/(a-b)^2/(a+b)^3/d+1/4*(8*B*a^4-29*B*a^2*b^2+15*B*b^4+9*C*a^3*b-3*C*a*b^3)
*sin(d*x+c)/a^3/(a^2-b^2)^2/d/cos(d*x+c)^(1/2)+1/2*b*(B*b-C*a)*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))^2/cos
(d*x+c)^(1/2)+1/4*b*(11*B*a^2*b-5*B*b^3-7*C*a^3+C*a*b^2)*sin(d*x+c)/a^2/(a^2-b^2)^2/d/(a+b*cos(d*x+c))/cos(d*x
+c)^(1/2)
Rubi [A] (verified)
Time = 1.63 (sec) , antiderivative size = 420, normalized size of antiderivative = 1.00, number of
steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3108, 3079, 3134, 3138, 2719,
3081, 2720, 2884} \[
\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^3} \, dx=\frac {b (b B-a C) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {\left (-7 a^3 C+11 a^2 b B+a b^2 C-5 b^3 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 a^2 d \left (a^2-b^2\right )^2}+\frac {b \left (-7 a^3 C+11 a^2 b B+a b^2 C-5 b^3 B\right ) \sin (c+d x)}{4 a^2 d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\left (8 a^4 B+9 a^3 b C-29 a^2 b^2 B-3 a b^3 C+15 b^4 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 d \left (a^2-b^2\right )^2}+\frac {\left (8 a^4 B+9 a^3 b C-29 a^2 b^2 B-3 a b^3 C+15 b^4 B\right ) \sin (c+d x)}{4 a^3 d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)}}-\frac {\left (-15 a^5 C+35 a^4 b B+6 a^3 b^2 C-38 a^2 b^3 B-3 a b^4 C+15 b^5 B\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a^3 d (a-b)^2 (a+b)^3}
\]
[In]
Int[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + b*Cos[c + d*x])^3),x]
[Out]
-1/4*((8*a^4*B - 29*a^2*b^2*B + 15*b^4*B + 9*a^3*b*C - 3*a*b^3*C)*EllipticE[(c + d*x)/2, 2])/(a^3*(a^2 - b^2)^
2*d) + ((11*a^2*b*B - 5*b^3*B - 7*a^3*C + a*b^2*C)*EllipticF[(c + d*x)/2, 2])/(4*a^2*(a^2 - b^2)^2*d) - ((35*a
^4*b*B - 38*a^2*b^3*B + 15*b^5*B - 15*a^5*C + 6*a^3*b^2*C - 3*a*b^4*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2,
2])/(4*a^3*(a - b)^2*(a + b)^3*d) + ((8*a^4*B - 29*a^2*b^2*B + 15*b^4*B + 9*a^3*b*C - 3*a*b^3*C)*Sin[c + d*x])
/(4*a^3*(a^2 - b^2)^2*d*Sqrt[Cos[c + d*x]]) + (b*(b*B - a*C)*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*Sqrt[Cos[c + d*x
]]*(a + b*Cos[c + d*x])^2) + (b*(11*a^2*b*B - 5*b^3*B - 7*a^3*C + a*b^2*C)*Sin[c + d*x])/(4*a^2*(a^2 - b^2)^2*
d*Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x]))
Rule 2719
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]
Rule 2720
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]
Rule 2884
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Rule 3079
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)), I
nt[(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 3081
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Rule 3108
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 3134
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] + D
ist[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] &&
LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n]
&& !IntegerQ[m]) || EqQ[a, 0])))
Rule 3138
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]
Rubi steps \begin{align*}
\text {integral}& = \int \frac {B+C \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^3} \, dx \\ & = \frac {b (b B-a C) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {\int \frac {\frac {1}{2} \left (4 a^2 B-5 b^2 B+a b C\right )-2 a (b B-a C) \cos (c+d x)+\frac {3}{2} b (b B-a C) \cos ^2(c+d x)}{\cos ^{\frac {3}{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) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {b \left (11 a^2 b B-5 b^3 B-7 a^3 C+a b^2 C\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}+\frac {\int \frac {\frac {1}{4} \left (8 a^4 B-29 a^2 b^2 B+15 b^4 B+9 a^3 b C-3 a b^3 C\right )-a \left (4 a^2 b B-b^3 B-2 a^3 C-a b^2 C\right ) \cos (c+d x)+\frac {1}{4} b \left (11 a^2 b B-5 b^3 B-7 a^3 C+a b^2 C\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{2 a^2 \left (a^2-b^2\right )^2} \\ & = \frac {\left (8 a^4 B-29 a^2 b^2 B+15 b^4 B+9 a^3 b C-3 a b^3 C\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {b (b B-a C) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {b \left (11 a^2 b B-5 b^3 B-7 a^3 C+a b^2 C\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}+\frac {\int \frac {\frac {1}{8} \left (-24 a^4 b B+33 a^2 b^3 B-15 b^5 B+8 a^5 C-5 a^3 b^2 C+3 a b^4 C\right )-\frac {1}{2} a \left (2 a^4 B-10 a^2 b^2 B+5 b^4 B+4 a^3 b C-a b^3 C\right ) \cos (c+d x)-\frac {1}{8} b \left (8 a^4 B-29 a^2 b^2 B+15 b^4 B+9 a^3 b C-3 a b^3 C\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a^3 \left (a^2-b^2\right )^2} \\ & = \frac {\left (8 a^4 B-29 a^2 b^2 B+15 b^4 B+9 a^3 b C-3 a b^3 C\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {b (b B-a C) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {b \left (11 a^2 b B-5 b^3 B-7 a^3 C+a b^2 C\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\int \frac {\frac {1}{8} b \left (24 a^4 b B-33 a^2 b^3 B+15 b^5 B-8 a^5 C+5 a^3 b^2 C-3 a b^4 C\right )-\frac {1}{8} a b^2 \left (11 a^2 b B-5 b^3 B-7 a^3 C+a b^2 C\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a^3 b \left (a^2-b^2\right )^2}-\frac {\left (8 a^4 B-29 a^2 b^2 B+15 b^4 B+9 a^3 b C-3 a b^3 C\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 a^3 \left (a^2-b^2\right )^2} \\ & = -\frac {\left (8 a^4 B-29 a^2 b^2 B+15 b^4 B+9 a^3 b C-3 a b^3 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}+\frac {\left (8 a^4 B-29 a^2 b^2 B+15 b^4 B+9 a^3 b C-3 a b^3 C\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {b (b B-a C) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {b \left (11 a^2 b B-5 b^3 B-7 a^3 C+a b^2 C\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}+\frac {\left (11 a^2 b B-5 b^3 B-7 a^3 C+a b^2 C\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{8 a^2 \left (a^2-b^2\right )^2}-\frac {\left (35 a^4 b B-38 a^2 b^3 B+15 b^5 B-15 a^5 C+6 a^3 b^2 C-3 a b^4 C\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{8 a^3 \left (a^2-b^2\right )^2} \\ & = -\frac {\left (8 a^4 B-29 a^2 b^2 B+15 b^4 B+9 a^3 b C-3 a b^3 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}+\frac {\left (11 a^2 b B-5 b^3 B-7 a^3 C+a b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 a^2 \left (a^2-b^2\right )^2 d}-\frac {\left (35 a^4 b B-38 a^2 b^3 B+15 b^5 B-15 a^5 C+6 a^3 b^2 C-3 a b^4 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a^3 (a-b)^2 (a+b)^3 d}+\frac {\left (8 a^4 B-29 a^2 b^2 B+15 b^4 B+9 a^3 b C-3 a b^3 C\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {b (b B-a C) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {b \left (11 a^2 b B-5 b^3 B-7 a^3 C+a b^2 C\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \\
\end{align*}
Mathematica [A] (verified)
Time = 5.92 (sec) , antiderivative size = 458, normalized size of antiderivative = 1.09
\[
\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^3} \, dx=\frac {-\frac {\frac {\left (56 a^4 b B-95 a^2 b^3 B+45 b^5 B-16 a^5 C+19 a^3 b^2 C-9 a b^4 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {8 a \left (2 a^4 B-10 a^2 b^2 B+5 b^4 B+4 a^3 b C-a b^3 C\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{b (a+b)}+\frac {\left (8 a^4 B-29 a^2 b^2 B+15 b^4 B+9 a^3 b C-3 a b^3 C\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}+\frac {\sqrt {\cos (c+d x)} \left (2 a b \left (16 a^4 B-47 a^2 b^2 B+25 b^4 B+11 a^3 b C-5 a b^3 C\right ) \sin (c+d x)+b^2 \left (8 a^4 B-29 a^2 b^2 B+15 b^4 B+9 a^3 b C-3 a b^3 C\right ) \sin (2 (c+d x))+16 \left (a^3-a b^2\right )^2 B \tan (c+d x)\right )}{\left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}}{8 a^3 d}
\]
[In]
Integrate[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + b*Cos[c + d*x])^3),x]
[Out]
(-((((56*a^4*b*B - 95*a^2*b^3*B + 45*b^5*B - 16*a^5*C + 19*a^3*b^2*C - 9*a*b^4*C)*EllipticPi[(2*b)/(a + b), (c
+ d*x)/2, 2])/(a + b) + (8*a*(2*a^4*B - 10*a^2*b^2*B + 5*b^4*B + 4*a^3*b*C - a*b^3*C)*((a + b)*EllipticF[(c +
d*x)/2, 2] - a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]))/(b*(a + b)) + ((8*a^4*B - 29*a^2*b^2*B + 15*b^4*B
+ 9*a^3*b*C - 3*a*b^3*C)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt
[Cos[c + d*x]]], -1] + (-2*a^2 + b^2)*EllipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b*S
qrt[Sin[c + d*x]^2]))/((a - b)^2*(a + b)^2)) + (Sqrt[Cos[c + d*x]]*(2*a*b*(16*a^4*B - 47*a^2*b^2*B + 25*b^4*B
+ 11*a^3*b*C - 5*a*b^3*C)*Sin[c + d*x] + b^2*(8*a^4*B - 29*a^2*b^2*B + 15*b^4*B + 9*a^3*b*C - 3*a*b^3*C)*Sin[2
*(c + d*x)] + 16*(a^3 - a*b^2)^2*B*Tan[c + d*x]))/((a^2 - b^2)^2*(a + b*Cos[c + d*x])^2))/(8*a^3*d)
Maple [B] (verified)
Leaf count of result is larger than twice the leaf count of optimal. \(1974\) vs. \(2(480)=960\).
Time = 18.11 (sec) , antiderivative size = 1975, normalized size of antiderivative =
4.70
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method | result | size |
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default |
\(\text {Expression too large to display}\) |
\(1975\) |
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[In]
int((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+cos(d*x+c)*b)^3,x,method=_RETURNVERBOSE)
[Out]
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*B/a^3/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)
^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(sin(1/2
*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-2*B/a^2*b*(-1/a*b
^2/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*b*cos(1/2*d*x+1/2*c)^2
+a-b)-1/2/a/(a+b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(
1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/(a^2-b^2)*b/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(
-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/
2*c),2^(1/2))+1/2/(a^2-b^2)*b/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x
+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3*a/(a^2-b^2)/(-2*a*b+2*b^2)*b*(si
n(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/
2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+1/a/(a^2-b^2)/(-2*a*b+2*b^2)*b^3*(sin(1/2*d*x+1/2*c)^2)^(
1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2
*d*x+1/2*c),-2*b/(a-b),2^(1/2)))+4*B/a^3*b^2/(-2*a*b+2*b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c
)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(
1/2))+2/a*(-B*b+C*a)*(-1/2/a*b^2/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(
1/2)/(2*b*cos(1/2*d*x+1/2*c)^2+a-b)^2-3/4*b^2*(3*a^2-b^2)/a^2/(a^2-b^2)^2*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1
/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*b*cos(1/2*d*x+1/2*c)^2+a-b)-7/8/(a+b)/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^
(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2
*d*x+1/2*c),2^(1/2))+1/4/(a+b)/(a^2-b^2)/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*
sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b+3/8/(a+b)/(a^2-b^2)/a
^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^
2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^2-9/8*b/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*
d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2
))+3/8*b^3/a^2/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*
c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+9/8*b/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1
/2*d*x+1/2*c),2^(1/2))-3/8*b^3/a^2/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/
(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-15/4*a^2/(a^2-b^2)^
2/(-2*a*b+2*b^2)*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin
(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+3/2/(a^2-b^2)^2/(-2*a*b+2*b^2)*b^3*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^
(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))-3/4/a^2/(a^2-b^2)^2/(-2*a*b+2*b^2)*b^5*(sin(1/2*d*x+1/
2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticP
i(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
Fricas [F(-1)]
Timed out. \[
\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^3} \, dx=\text {Timed out}
\]
[In]
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^3,x, algorithm="fricas")
[Out]
Timed out
Sympy [F(-1)]
Timed out. \[
\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^3} \, dx=\text {Timed out}
\]
[In]
integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(5/2)/(a+b*cos(d*x+c))**3,x)
[Out]
Timed out
Maxima [F(-1)]
Timed out. \[
\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^3} \, dx=\text {Timed out}
\]
[In]
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^3,x, algorithm="maxima")
[Out]
Timed out
Giac [F]
\[
\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^3} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x }
\]
[In]
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^3,x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))/((b*cos(d*x + c) + a)^3*cos(d*x + c)^(5/2)), x)
Mupad [F(-1)]
Timed out. \[
\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^3} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x
\]
[In]
int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + b*cos(c + d*x))^3),x)
[Out]
int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + b*cos(c + d*x))^3), x)