3.841 \(\int \frac {\cos ^2(c+d x) (B \cos (c+d x)+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=387 \[ \frac {2 a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (-6 a^2 C+5 a b B+b^2 C\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b^2 d \left (a^2-b^2\right )}+\frac {2 \left (-24 a^3 C+20 a^2 b B+9 a b^2 C-5 b^3 B\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{15 b^3 d \left (a^2-b^2\right )}+\frac {2 \left (-48 a^3 C+40 a^2 b B-12 a b^2 C+5 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^4 d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (-48 a^4 C+40 a^3 b B+24 a^2 b^2 C-25 a b^3 B+9 b^4 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^4 d \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \]

[Out]

2*a*(B*b-C*a)*cos(d*x+c)^2*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)+2/15*(20*B*a^2*b-5*B*b^3-24*C*a^3+9
*C*a*b^2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b^3/(a^2-b^2)/d-2/5*(5*B*a*b-6*C*a^2+C*b^2)*cos(d*x+c)*sin(d*x+c)*
(a+b*cos(d*x+c))^(1/2)/b^2/(a^2-b^2)/d-2/15*(40*B*a^3*b-25*B*a*b^3-48*C*a^4+24*C*a^2*b^2+9*C*b^4)*(cos(1/2*d*x
+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/
2)/b^4/(a^2-b^2)/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+2/15*(40*B*a^2*b+5*B*b^3-48*C*a^3-12*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)*(b/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b
))^(1/2)/b^4/d/(a+b*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.87, antiderivative size = 387, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3029, 2989, 3049, 3023, 2752, 2663, 2661, 2655, 2653} \[ \frac {2 a (b B-a C) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (-6 a^2 C+5 a b B+b^2 C\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b^2 d \left (a^2-b^2\right )}+\frac {2 \left (20 a^2 b B-24 a^3 C+9 a b^2 C-5 b^3 B\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{15 b^3 d \left (a^2-b^2\right )}+\frac {2 \left (40 a^2 b B-48 a^3 C-12 a b^2 C+5 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^4 d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (24 a^2 b^2 C+40 a^3 b B-48 a^4 C-25 a b^3 B+9 b^4 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^4 d \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^(3/2),x]

[Out]

(-2*(40*a^3*b*B - 25*a*b^3*B - 48*a^4*C + 24*a^2*b^2*C + 9*b^4*C)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)
/2, (2*b)/(a + b)])/(15*b^4*(a^2 - b^2)*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + (2*(40*a^2*b*B + 5*b^3*B - 48*
a^3*C - 12*a*b^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(15*b^4*d*Sqrt[a
+ b*Cos[c + d*x]]) + (2*a*(b*B - a*C)*Cos[c + d*x]^2*Sin[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]])
+ (2*(20*a^2*b*B - 5*b^3*B - 24*a^3*C + 9*a*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(15*b^3*(a^2 - b^2)
*d) - (2*(5*a*b*B - 6*a^2*C + b^2*C)*Cos[c + d*x]*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(5*b^2*(a^2 - b^2)*d)

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2752

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 2989

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((b*c - a*d)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)
*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[
e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (
A*b + a*B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) - a*(b*c - a*d)*(B*c - A*d)
*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^2,
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] && GtQ[m, 1] && LtQ[n, -1]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

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 3049

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] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*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] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx &=\int \frac {\cos ^3(c+d x) (B+C \cos (c+d x))}{(a+b \cos (c+d x))^{3/2}} \, dx\\ &=\frac {2 a (b B-a C) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {2 \int \frac {\cos (c+d x) \left (-2 a (b B-a C)+\frac {1}{2} b (b B-a C) \cos (c+d x)+\frac {1}{2} \left (5 a b B-6 a^2 C+b^2 C\right ) \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac {2 a (b B-a C) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (5 a b B-6 a^2 C+b^2 C\right ) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {4 \int \frac {\frac {1}{2} a \left (5 a b B-6 a^2 C+b^2 C\right )-\frac {1}{4} b \left (5 a b B-2 a^2 C-3 b^2 C\right ) \cos (c+d x)-\frac {1}{4} \left (20 a^2 b B-5 b^3 B-24 a^3 C+9 a b^2 C\right ) \cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{5 b^2 \left (a^2-b^2\right )}\\ &=\frac {2 a (b B-a C) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (20 a^2 b B-5 b^3 B-24 a^3 C+9 a b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^3 \left (a^2-b^2\right ) d}-\frac {2 \left (5 a b B-6 a^2 C+b^2 C\right ) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {8 \int \frac {\frac {1}{8} b \left (10 a^2 b B+5 b^3 B-12 a^3 C-3 a b^2 C\right )+\frac {1}{8} \left (40 a^3 b B-25 a b^3 B-48 a^4 C+24 a^2 b^2 C+9 b^4 C\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{15 b^3 \left (a^2-b^2\right )}\\ &=\frac {2 a (b B-a C) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (20 a^2 b B-5 b^3 B-24 a^3 C+9 a b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^3 \left (a^2-b^2\right ) d}-\frac {2 \left (5 a b B-6 a^2 C+b^2 C\right ) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}+\frac {\left (40 a^2 b B+5 b^3 B-48 a^3 C-12 a b^2 C\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{15 b^4}-\frac {\left (40 a^3 b B-25 a b^3 B-48 a^4 C+24 a^2 b^2 C+9 b^4 C\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{15 b^4 \left (a^2-b^2\right )}\\ &=\frac {2 a (b B-a C) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (20 a^2 b B-5 b^3 B-24 a^3 C+9 a b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^3 \left (a^2-b^2\right ) d}-\frac {2 \left (5 a b B-6 a^2 C+b^2 C\right ) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {\left (\left (40 a^3 b B-25 a b^3 B-48 a^4 C+24 a^2 b^2 C+9 b^4 C\right ) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{15 b^4 \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (\left (40 a^2 b B+5 b^3 B-48 a^3 C-12 a b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}} \, dx}{15 b^4 \sqrt {a+b \cos (c+d x)}}\\ &=-\frac {2 \left (40 a^3 b B-25 a b^3 B-48 a^4 C+24 a^2 b^2 C+9 b^4 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^4 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \left (40 a^2 b B+5 b^3 B-48 a^3 C-12 a b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^4 d \sqrt {a+b \cos (c+d x)}}+\frac {2 a (b B-a C) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (20 a^2 b B-5 b^3 B-24 a^3 C+9 a b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^3 \left (a^2-b^2\right ) d}-\frac {2 \left (5 a b B-6 a^2 C+b^2 C\right ) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}\\ \end {align*}

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Mathematica [A]  time = 1.80, size = 304, normalized size = 0.79 \[ \frac {\frac {30 a^3 b (a C-b B) \sin (c+d x)}{b^2-a^2}+\frac {2 b^2 \left (12 a^3 C-10 a^2 b B+3 a b^2 C-5 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{(a-b) (a+b)}+\frac {2 \left (48 a^4 C-40 a^3 b B-24 a^2 b^2 C+25 a b^3 B-9 b^4 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )}{(a-b) (a+b)}+3 b^2 C \sin (2 (c+d x)) (a+b \cos (c+d x))+2 b (5 b B-9 a C) \sin (c+d x) (a+b \cos (c+d x))}{15 b^4 d \sqrt {a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^2*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^(3/2),x]

[Out]

((2*b^2*(-10*a^2*b*B - 5*b^3*B + 12*a^3*C + 3*a*b^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/
2, (2*b)/(a + b)])/((a - b)*(a + b)) + (2*(-40*a^3*b*B + 25*a*b^3*B + 48*a^4*C - 24*a^2*b^2*C - 9*b^4*C)*Sqrt[
(a + b*Cos[c + d*x])/(a + b)]*((a + b)*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - a*EllipticF[(c + d*x)/2, (2*b)/
(a + b)]))/((a - b)*(a + b)) + (30*a^3*b*(-(b*B) + a*C)*Sin[c + d*x])/(-a^2 + b^2) + 2*b*(5*b*B - 9*a*C)*(a +
b*Cos[c + d*x])*Sin[c + d*x] + 3*b^2*C*(a + b*Cos[c + d*x])*Sin[2*(c + d*x)])/(15*b^4*d*Sqrt[a + b*Cos[c + d*x
]])

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fricas [F]  time = 0.88, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{4} + B \cos \left (d x + c\right )^{3}\right )} \sqrt {b \cos \left (d x + c\right ) + a}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^4 + B*cos(d*x + c)^3)*sqrt(b*cos(d*x + c) + a)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x +
c) + a^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \cos \left (d x + c\right )^{2}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*cos(d*x + c)^2/(b*cos(d*x + c) + a)^(3/2), x)

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maple [B]  time = 8.94, size = 1308, normalized size = 3.38 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/2),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(16/b*C*(-1/10/b*cos(1/2*d*x+1/2*c)^3*(-2*sin(1
/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)-1/60/b^2*(-4*a+12*b)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/
2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)+1/60/b^2*(-4*a+12*b)*(a-b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2
*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1
/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-1/60*(4*a^2-15*a*b+27*b^2)/b^3*(a-b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/
2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos
(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))))+8/b^2*(B*b-C*a-3*C*b)*(
-1/6/b*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)+1/6/b*(a-b)*(sin(1/2*d*
x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/
2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-1/12/b^2*(-2*a+6*b)*(a-b)*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1
/2)*(EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))))+2/b^4
*(B*a*b+2*B*b^2-C*a^2-2*C*a*b-3*C*b^2)*(a-b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b
))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b
))^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2)))+2*(B*a^2*b+B*a*b^2+B*b^3-C*a^3-C*a^2*b-C*a*b^2-C*b
^3)/b^4*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(
a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-2*a^3*(B*b-C*a)/b^4/sin(1/2*
d*x+1/2*c)^2/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)/(a^2-b^2)*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^
(1/2)*((-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(s
in(1/2*d*x+1/2*c)^2)^(1/2)*a-(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),
(-2*b/(a-b))^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*b+2*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2))/sin(1/2*d*x+1
/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \cos \left (d x + c\right )^{2}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*cos(d*x + c)^2/(b*cos(d*x + c) + a)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^2\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^2*(B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + b*cos(c + d*x))^(3/2),x)

[Out]

int((cos(c + d*x)^2*(B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + b*cos(c + d*x))^(3/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**(3/2),x)

[Out]

Timed out

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