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

Optimal. Leaf size=389 \[ \frac{2 \left (-a^2 b^2 (23 A+19 C)+2 a^4 C-3 b^4 (3 A+5 C)\right ) \sin (c+d x)}{15 b d \left (a^2-b^2\right )^3 \sqrt{a+b \cos (c+d x)}}-\frac{4 a \left (a^2 (-C)+4 A b^2+5 b^2 C\right ) \sin (c+d x)}{15 b d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^{3/2}}-\frac{2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{5 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}-\frac{4 a \left (4 A b^2-C \left (a^2-5 b^2\right )\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (-a^2 b^2 (23 A+19 C)+2 a^4 C-3 b^4 (3 A+5 C)\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^2 d \left (a^2-b^2\right )^3 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}} \]

[Out]

(-2*(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]]*EllipticE[(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)]) - (4*a*(4*A*b^2 - (a^2 - 5*b^2)*C)*Sqr
t[(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^2*C)*Sin[c + d*x])/(5*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(5/2)) - (4*a*(4*A*b^2 -
a^2*C + 5*b^2*C)*Sin[c + d*x])/(15*b*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x])^(3/2)) + (2*(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]])

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Rubi [A]  time = 0.605387, antiderivative size = 389, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {3022, 2754, 2752, 2663, 2661, 2655, 2653} \[ \frac{2 \left (-a^2 b^2 (23 A+19 C)+2 a^4 C-3 b^4 (3 A+5 C)\right ) \sin (c+d x)}{15 b d \left (a^2-b^2\right )^3 \sqrt{a+b \cos (c+d x)}}-\frac{4 a \left (a^2 (-C)+4 A b^2+5 b^2 C\right ) \sin (c+d x)}{15 b d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^{3/2}}-\frac{2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{5 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}-\frac{4 a \left (4 A b^2-C \left (a^2-5 b^2\right )\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (-a^2 b^2 (23 A+19 C)+2 a^4 C-3 b^4 (3 A+5 C)\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^2 d \left (a^2-b^2\right )^3 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(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]]*EllipticE[(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)]) - (4*a*(4*A*b^2 - (a^2 - 5*b^2)*C)*Sqr
t[(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^2*C)*Sin[c + d*x])/(5*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(5/2)) - (4*a*(4*A*b^2 -
a^2*C + 5*b^2*C)*Sin[c + d*x])/(15*b*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x])^(3/2)) + (2*(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 3022

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

Rubi steps

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

Mathematica [A]  time = 2.56816, size = 314, normalized size = 0.81 \[ \frac{2 \left (\frac{b \sin (c+d x) \left (-4 a b \left (-a^2 b^2 (27 A+25 C)+3 a^4 C-5 b^4 (A+2 C)\right ) \cos (c+d x)+\left (a^2 b^4 (23 A+19 C)-2 a^4 b^2 C+3 b^6 (3 A+5 C)\right ) \cos (2 (c+d x))+68 a^4 A b^2+13 a^2 A b^4+48 a^4 b^2 C+35 a^2 b^4 C-2 a^6 C+15 A b^6+15 b^6 C\right )}{2 \left (b^2-a^2\right )^3}+\frac{\left (\frac{a+b \cos (c+d x)}{a+b}\right )^{5/2} \left (2 a (a-b) \left (C \left (a^2-5 b^2\right )-4 A b^2\right ) F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+\left (a^2 b^2 (23 A+19 C)-2 a^4 C+3 b^4 (3 A+5 C)\right ) E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )\right )}{(a-b)^3}\right )}{15 b^2 d (a+b \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + 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)*((-2*a^4*C + 3*b^4*(3*A + 5*C) + a^2*b^2*(23*A + 19*C))*EllipticE[(c
+ d*x)/2, (2*b)/(a + b)] + 2*a*(a - b)*(-4*A*b^2 + (a^2 - 5*b^2)*C)*EllipticF[(c + d*x)/2, (2*b)/(a + b)]))/(
a - b)^3 + (b*(68*a^4*A*b^2 + 13*a^2*A*b^4 + 15*A*b^6 - 2*a^6*C + 48*a^4*b^2*C + 35*a^2*b^4*C + 15*b^6*C - 4*a
*b*(3*a^4*C - 5*b^4*(A + 2*C) - a^2*b^2*(27*A + 25*C))*Cos[c + d*x] + (-2*a^4*b^2*C + 3*b^6*(3*A + 5*C) + a^2*
b^4*(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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Maple [B]  time = 2.747, size = 1305, normalized size = 3.4 \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)/(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*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*((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))*a-(
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))*b+2*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)+2*(A*b^2+C*a^2)/b^2*(1/20/b^2/(a-b)/(a+b)*co
s(1/2*d*x+1/2*c)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2+1/2*(a-b)/
b)^3+4/15*a/b/(a+b)^2/(a-b)^2*cos(1/2*d*x+1/2*c)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/
(cos(1/2*d*x+1/2*c)^2+1/2*(a-b)/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*b*sin(1/2*d*x+1/2*c)^4+(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)^3/(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*b*sin(1/2*d*x+1/2*c)^4+(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))-E
llipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))))-4*a*C/b^2*(1/6/b/(a-b)/(a+b)*cos(1/2*d*x+1/2*c)*(-2*b*sin(1/
2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2+1/2*(a-b)/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*b*
sin(1/2*d*x+1/2*c)^4+(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)^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*b*sin(1/2*d*x+1/2*
c)^4+(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)))))/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., size = 0, normalized size = 0. \begin{align*} \int \frac{C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \cos \left (d x + c\right )^{2} + 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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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