3.1060 \(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=449 \[ -\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{5 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}+\frac {2 \sin (c+d x) \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right )}{15 b d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (2 a^3 C+3 a^2 b B-2 a b^2 (4 A+5 C)+5 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^2 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}+\frac {2 \sin (c+d x) \left (2 a^4 C+3 a^3 b B-a^2 b^2 (23 A+19 C)+29 a b^3 B-3 b^4 (3 A+5 C)\right )}{15 b d \left (a^2-b^2\right )^3 \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (2 a^4 C+3 a^3 b B-a^2 b^2 (23 A+19 C)+29 a b^3 B-3 b^4 (3 A+5 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^2 d \left (a^2-b^2\right )^3 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \]

[Out]

-2/5*(A*b^2-a*(B*b-C*a))*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^(5/2)+2/15*(3*a^2*b*B+5*b^3*B+2*a^3*C-2*a*b
^2*(4*A+5*C))*sin(d*x+c)/b/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^(3/2)+2/15*(3*a^3*b*B+29*a*b^3*B+2*a^4*C-3*b^4*(3*A+
5*C)-a^2*b^2*(23*A+19*C))*sin(d*x+c)/b/(a^2-b^2)^3/d/(a+b*cos(d*x+c))^(1/2)-2/15*(3*a^3*b*B+29*a*b^3*B+2*a^4*C
-3*b^4*(3*A+5*C)-a^2*b^2*(23*A+19*C))*(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^2/(a^2-b^2)^3/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+2/15*(3*
a^2*b*B+5*b^3*B+2*a^3*C-2*a*b^2*(4*A+5*C))*(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^2/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.74, antiderivative size = 449, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3021, 2754, 2752, 2663, 2661, 2655, 2653} \[ \frac {2 \sin (c+d x) \left (-a^2 b^2 (23 A+19 C)+3 a^3 b B+2 a^4 C+29 a b^3 B-3 b^4 (3 A+5 C)\right )}{15 b d \left (a^2-b^2\right )^3 \sqrt {a+b \cos (c+d x)}}+\frac {2 \sin (c+d x) \left (3 a^2 b B+2 a^3 C-2 a b^2 (4 A+5 C)+5 b^3 B\right )}{15 b d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^{3/2}}-\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{5 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}+\frac {2 \left (3 a^2 b B+2 a^3 C-2 a b^2 (4 A+5 C)+5 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^2 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (-a^2 b^2 (23 A+19 C)+3 a^3 b B+2 a^4 C+29 a b^3 B-3 b^4 (3 A+5 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^2 d \left (a^2-b^2\right )^3 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(a + b*Cos[c + d*x])^(7/2),x]

[Out]

(-2*(3*a^3*b*B + 29*a*b^3*B + 2*a^4*C - 3*b^4*(3*A + 5*C) - a^2*b^2*(23*A + 19*C))*Sqrt[a + b*Cos[c + d*x]]*El
lipticE[(c + d*x)/2, (2*b)/(a + b)])/(15*b^2*(a^2 - b^2)^3*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + (2*(3*a^2*b
*B + 5*b^3*B + 2*a^3*C - 2*a*b^2*(4*A + 5*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/
(a + b)])/(15*b^2*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x]]) - (2*(A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(5*b*(a
^2 - b^2)*d*(a + b*Cos[c + d*x])^(5/2)) + (2*(3*a^2*b*B + 5*b^3*B + 2*a^3*C - 2*a*b^2*(4*A + 5*C))*Sin[c + d*x
])/(15*b*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x])^(3/2)) + (2*(3*a^3*b*B + 29*a*b^3*B + 2*a^4*C - 3*b^4*(3*A + 5*C
) - a^2*b^2*(23*A + 19*C))*Sin[c + d*x])/(15*b*(a^2 - b^2)^3*d*Sqrt[a + b*Cos[c + d*x]])

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 2754

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

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(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))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{7/2}} \, dx &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{5 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{5/2}}-\frac {2 \int \frac {\frac {5}{2} b (b B-a (A+C))+\frac {1}{2} \left (3 A b^2-3 a b B-2 a^2 C+5 b^2 C\right ) \cos (c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx}{5 b \left (a^2-b^2\right )}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{5 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{5/2}}+\frac {2 \left (3 a^2 b B+5 b^3 B+2 a^3 C-2 a b^2 (4 A+5 C)\right ) \sin (c+d x)}{15 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^{3/2}}+\frac {4 \int \frac {-\frac {3}{4} b \left (8 a b B-a^2 (5 A+3 C)-b^2 (3 A+5 C)\right )+\frac {1}{4} \left (3 a^2 b B+5 b^3 B+2 a^3 C-2 a b^2 (4 A+5 C)\right ) \cos (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx}{15 b \left (a^2-b^2\right )^2}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{5 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{5/2}}+\frac {2 \left (3 a^2 b B+5 b^3 B+2 a^3 C-2 a b^2 (4 A+5 C)\right ) \sin (c+d x)}{15 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (3 a^3 b B+29 a b^3 B+2 a^4 C-3 b^4 (3 A+5 C)-a^2 b^2 (23 A+19 C)\right ) \sin (c+d x)}{15 b \left (a^2-b^2\right )^3 d \sqrt {a+b \cos (c+d x)}}-\frac {8 \int \frac {\frac {1}{8} b \left (27 a^2 b B+5 b^3 B-a^3 (15 A+7 C)-a b^2 (17 A+25 C)\right )+\frac {1}{8} \left (3 a^3 b B+29 a b^3 B+2 a^4 C-3 b^4 (3 A+5 C)-a^2 b^2 (23 A+19 C)\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{15 b \left (a^2-b^2\right )^3}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{5 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{5/2}}+\frac {2 \left (3 a^2 b B+5 b^3 B+2 a^3 C-2 a b^2 (4 A+5 C)\right ) \sin (c+d x)}{15 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (3 a^3 b B+29 a b^3 B+2 a^4 C-3 b^4 (3 A+5 C)-a^2 b^2 (23 A+19 C)\right ) \sin (c+d x)}{15 b \left (a^2-b^2\right )^3 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (3 a^2 b B+5 b^3 B+2 a^3 C-2 a b^2 (4 A+5 C)\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{15 b^2 \left (a^2-b^2\right )^2}-\frac {\left (3 a^3 b B+29 a b^3 B+2 a^4 C-3 b^4 (3 A+5 C)-a^2 b^2 (23 A+19 C)\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{15 b^2 \left (a^2-b^2\right )^3}\\ &=-\frac {2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{5 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{5/2}}+\frac {2 \left (3 a^2 b B+5 b^3 B+2 a^3 C-2 a b^2 (4 A+5 C)\right ) \sin (c+d x)}{15 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (3 a^3 b B+29 a b^3 B+2 a^4 C-3 b^4 (3 A+5 C)-a^2 b^2 (23 A+19 C)\right ) \sin (c+d x)}{15 b \left (a^2-b^2\right )^3 d \sqrt {a+b \cos (c+d x)}}-\frac {\left (\left (3 a^3 b B+29 a b^3 B+2 a^4 C-3 b^4 (3 A+5 C)-a^2 b^2 (23 A+19 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^2 \left (a^2-b^2\right )^3 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (\left (3 a^2 b B+5 b^3 B+2 a^3 C-2 a b^2 (4 A+5 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^2 \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}\\ &=-\frac {2 \left (3 a^3 b B+29 a b^3 B+2 a^4 C-3 b^4 (3 A+5 C)-a^2 b^2 (23 A+19 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^2 \left (a^2-b^2\right )^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \left (3 a^2 b B+5 b^3 B+2 a^3 C-2 a b^2 (4 A+5 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^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{5 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{5/2}}+\frac {2 \left (3 a^2 b B+5 b^3 B+2 a^3 C-2 a b^2 (4 A+5 C)\right ) \sin (c+d x)}{15 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (3 a^3 b B+29 a b^3 B+2 a^4 C-3 b^4 (3 A+5 C)-a^2 b^2 (23 A+19 C)\right ) \sin (c+d x)}{15 b \left (a^2-b^2\right )^3 d \sqrt {a+b \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 3.89, size = 433, normalized size = 0.96 \[ \frac {2 \left (\frac {\left (\frac {a+b \cos (c+d x)}{a+b}\right )^{5/2} \left (b^2 \left (a^3 (15 A+7 C)-27 a^2 b B+a b^2 (17 A+25 C)-5 b^3 B\right ) F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )+\left (-2 a^4 C-3 a^3 b B+a^2 b^2 (23 A+19 C)-29 a b^3 B+3 b^4 (3 A+5 C)\right ) \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )\right )}{(a-b)^3 (a+b)}+\frac {b \sin (c+d x) \left (-2 a^6 C-18 a^5 b B+68 a^4 A b^2+48 a^4 b^2 C-53 a^3 b^3 B+13 a^2 A b^4+35 a^2 b^4 C+b^2 \cos (2 (c+d x)) \left (-2 a^4 C-3 a^3 b B+a^2 b^2 (23 A+19 C)-29 a b^3 B+3 b^4 (3 A+5 C)\right )+2 b \cos (c+d x) \left (-6 a^5 C-9 a^4 b B+2 a^3 b^2 (27 A+25 C)-60 a^2 b^3 B+10 a b^4 (A+2 C)+5 b^5 B\right )-25 a b^5 B+15 A b^6+15 b^6 C\right )}{2 \left (b^2-a^2\right )^3}\right )}{15 b^2 d (a+b \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(a + b*Cos[c + d*x])^(7/2),x]

[Out]

(2*((((a + b*Cos[c + d*x])/(a + b))^(5/2)*(b^2*(-27*a^2*b*B - 5*b^3*B + a^3*(15*A + 7*C) + a*b^2*(17*A + 25*C)
)*EllipticF[(c + d*x)/2, (2*b)/(a + b)] + (-3*a^3*b*B - 29*a*b^3*B - 2*a^4*C + 3*b^4*(3*A + 5*C) + a^2*b^2*(23
*A + 19*C))*((a + b)*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - a*EllipticF[(c + d*x)/2, (2*b)/(a + b)])))/((a -
b)^3*(a + b)) + (b*(68*a^4*A*b^2 + 13*a^2*A*b^4 + 15*A*b^6 - 18*a^5*b*B - 53*a^3*b^3*B - 25*a*b^5*B - 2*a^6*C
+ 48*a^4*b^2*C + 35*a^2*b^4*C + 15*b^6*C + 2*b*(-9*a^4*b*B - 60*a^2*b^3*B + 5*b^5*B - 6*a^5*C + 10*a*b^4*(A +
2*C) + 2*a^3*b^2*(27*A + 25*C))*Cos[c + d*x] + b^2*(-3*a^3*b*B - 29*a*b^3*B - 2*a^4*C + 3*b^4*(3*A + 5*C) + a^
2*b^2*(23*A + 19*C))*Cos[2*(c + d*x)])*Sin[c + d*x])/(2*(-a^2 + b^2)^3)))/(15*b^2*d*(a + b*Cos[c + d*x])^(5/2)
)

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fricas [F]  time = 1.24, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right ) + a}}{b^{4} \cos \left (d x + c\right )^{4} + 4 \, a b^{3} \cos \left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \cos \left (d x + c\right )^{2} + 4 \, a^{3} b \cos \left (d x + c\right ) + a^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c) + a)/(b^4*cos(d*x + c)^4 + 4*a*b^3*cos(d*
x + c)^3 + 6*a^2*b^2*cos(d*x + c)^2 + 4*a^3*b*cos(d*x + c) + a^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/(b*cos(d*x + c) + a)^(7/2), x)

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maple [B]  time = 18.54, size = 1316, normalized size = 2.93 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(7/2),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)*(2*C/b^2/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))*(sin(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)+2*(B*b-2*C*a)/b^2*(1/6/b/(a-b)/(a+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)/(cos(1/2*d*x+1/2*c)^2+1/2/b*(a-b))^2+
8/3*b*sin(1/2*d*x+1/2*c)^2/(a-b)^2/(a+b)^2*cos(1/2*d*x+1/2*c)*a/(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+
1/2*c)^2)^(1/2)+(3*a-b)/(3*a^3+3*a^2*b-3*a*b^2-3*b^3)*(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))-4/3*a/(a-b)/(a+b)^2*(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*(A*b^2-B*a*b+C*a^2)/b^2*(1/20/b^2/(a-b)/(a+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)/(cos(1/2*d*x+1/2*c)^2+1/2/b*(a-b))^3+4
/15*a/b/(a-b)^2/(a+b)^2*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)/(cos(1
/2*d*x+1/2*c)^2+1/2/b*(a-b))^2+2/15*b*sin(1/2*d*x+1/2*c)^2/(a-b)^3/(a+b)^3*cos(1/2*d*x+1/2*c)*(23*a^2+9*b^2)/(
-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)+(15*a^2-8*a*b+9*b^2)/(15*a^5+15*a^4*b-30*a^3*b^2-
30*a^2*b^3+15*a*b^4+15*b^5)*(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))-1/15*(23
*a^2+9*b^2)/(a-b)^2/(a+b)^3*(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))-Ellipti
cE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2)))))/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 \[ \int \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/(b*cos(d*x + c) + a)^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(a + b*cos(c + d*x))^(7/2),x)

[Out]

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(a + b*cos(c + d*x))^(7/2), 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((A+B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**(7/2),x)

[Out]

Timed out

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