3.376 \(\int \frac{A+B \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=345 \[ \frac{\left (2 a^2 A+3 a b B-5 A b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d \left (a^2-b^2\right )}+\frac{\left (4 a^2 A b-2 a^3 B+3 a b^2 B-5 A b^3\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d \left (a^2-b^2\right )}+\frac{b \left (7 a^2 A b-5 a^3 B+3 a b^2 B-5 A b^3\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d (a-b) (a+b)^2}+\frac{b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{\left (2 a^2 A+3 a b B-5 A b^2\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x)}-\frac{\left (4 a^2 A b-2 a^3 B+3 a b^2 B-5 A b^3\right ) \sin (c+d x)}{a^3 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)}} \]

[Out]

((4*a^2*A*b - 5*A*b^3 - 2*a^3*B + 3*a*b^2*B)*EllipticE[(c + d*x)/2, 2])/(a^3*(a^2 - b^2)*d) + ((2*a^2*A - 5*A*
b^2 + 3*a*b*B)*EllipticF[(c + d*x)/2, 2])/(3*a^2*(a^2 - b^2)*d) + (b*(7*a^2*A*b - 5*A*b^3 - 5*a^3*B + 3*a*b^2*
B)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a^3*(a - b)*(a + b)^2*d) + ((2*a^2*A - 5*A*b^2 + 3*a*b*B)*Sin[c
+ d*x])/(3*a^2*(a^2 - b^2)*d*Cos[c + d*x]^(3/2)) - ((4*a^2*A*b - 5*A*b^3 - 2*a^3*B + 3*a*b^2*B)*Sin[c + d*x])
/(a^3*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]) + (b*(A*b - a*B)*Sin[c + d*x])/(a*(a^2 - b^2)*d*Cos[c + d*x]^(3/2)*(a
+ b*Cos[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.29213, antiderivative size = 345, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {3000, 3055, 3059, 2639, 3002, 2641, 2805} \[ \frac{\left (2 a^2 A+3 a b B-5 A b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d \left (a^2-b^2\right )}+\frac{\left (4 a^2 A b-2 a^3 B+3 a b^2 B-5 A b^3\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d \left (a^2-b^2\right )}+\frac{b \left (7 a^2 A b-5 a^3 B+3 a b^2 B-5 A b^3\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d (a-b) (a+b)^2}+\frac{b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{\left (2 a^2 A+3 a b B-5 A b^2\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x)}-\frac{\left (4 a^2 A b-2 a^3 B+3 a b^2 B-5 A b^3\right ) \sin (c+d x)}{a^3 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[c + d*x])/(Cos[c + d*x]^(5/2)*(a + b*Cos[c + d*x])^2),x]

[Out]

((4*a^2*A*b - 5*A*b^3 - 2*a^3*B + 3*a*b^2*B)*EllipticE[(c + d*x)/2, 2])/(a^3*(a^2 - b^2)*d) + ((2*a^2*A - 5*A*
b^2 + 3*a*b*B)*EllipticF[(c + d*x)/2, 2])/(3*a^2*(a^2 - b^2)*d) + (b*(7*a^2*A*b - 5*A*b^3 - 5*a^3*B + 3*a*b^2*
B)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a^3*(a - b)*(a + b)^2*d) + ((2*a^2*A - 5*A*b^2 + 3*a*b*B)*Sin[c
+ d*x])/(3*a^2*(a^2 - b^2)*d*Cos[c + d*x]^(3/2)) - ((4*a^2*A*b - 5*A*b^3 - 2*a^3*B + 3*a*b^2*B)*Sin[c + d*x])
/(a^3*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]) + (b*(A*b - a*B)*Sin[c + d*x])/(a*(a^2 - b^2)*d*Cos[c + d*x]^(3/2)*(a
+ b*Cos[c + d*x]))

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 3059

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]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 3002

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 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[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]

Rubi steps

\begin{align*} \int \frac{A+B \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx &=\frac{b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{\int \frac{\frac{1}{2} \left (2 a^2 A-5 A b^2+3 a b B\right )-a (A b-a B) \cos (c+d x)+\frac{3}{2} b (A b-a B) \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{\left (2 a^2 A-5 A b^2+3 a b B\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x)}+\frac{b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{2 \int \frac{-\frac{3}{4} \left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right )+\frac{1}{2} a \left (a^2 A+2 A b^2-3 a b B\right ) \cos (c+d x)+\frac{1}{4} b \left (2 a^2 A-5 A b^2+3 a b B\right ) \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{3 a^2 \left (a^2-b^2\right )}\\ &=\frac{\left (2 a^2 A-5 A b^2+3 a b B\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x)}-\frac{\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)}}+\frac{b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{4 \int \frac{\frac{1}{8} \left (2 a^4 A+16 a^2 A b^2-15 A b^4-12 a^3 b B+9 a b^3 B\right )+\frac{1}{4} a \left (7 a^2 A b-10 A b^3-3 a^3 B+6 a b^2 B\right ) \cos (c+d x)+\frac{3}{8} b \left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 a^3 \left (a^2-b^2\right )}\\ &=\frac{\left (2 a^2 A-5 A b^2+3 a b B\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x)}-\frac{\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)}}+\frac{b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}-\frac{4 \int \frac{-\frac{1}{8} b \left (2 a^4 A+16 a^2 A b^2-15 A b^4-12 a^3 b B+9 a b^3 B\right )-\frac{1}{8} a b^2 \left (2 a^2 A-5 A b^2+3 a b B\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 a^3 b \left (a^2-b^2\right )}+\frac{\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac{\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (2 a^2 A-5 A b^2+3 a b B\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x)}-\frac{\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)}}+\frac{b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{\left (2 a^2 A-5 A b^2+3 a b B\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a^2 \left (a^2-b^2\right )}+\frac{\left (b \left (7 a^2 A b-5 A b^3-5 a^3 B+3 a b^2 B\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac{\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (2 a^2 A-5 A b^2+3 a b B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 \left (a^2-b^2\right ) d}+\frac{b \left (7 a^2 A b-5 A b^3-5 a^3 B+3 a b^2 B\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 (a-b) (a+b)^2 d}+\frac{\left (2 a^2 A-5 A b^2+3 a b B\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x)}-\frac{\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)}}+\frac{b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 6.31555, size = 367, normalized size = 1.06 \[ \frac{4 \sqrt{\cos (c+d x)} \left (\frac{3 b^3 (A b-a B) \sin (c+d x)}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}+2 \tan (c+d x) (a A \sec (c+d x)+3 a B-6 A b)\right )+\frac{\frac{2 \left (44 a^2 A b^2+4 a^4 A-30 a^3 b B+27 a b^3 B-45 A b^4\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}-\frac{8 a \left (-7 a^2 A b+3 a^3 B-6 a b^2 B+10 A b^3\right ) \left ((a+b) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-a \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{b (a+b)}-\frac{6 \left (-4 a^2 A b+2 a^3 B-3 a b^2 B+5 A b^3\right ) \sin (c+d x) \left (\left (2 a^2-b^2\right ) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a b \sqrt{\sin ^2(c+d x)}}}{(a-b) (a+b)}}{12 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cos[c + d*x])/(Cos[c + d*x]^(5/2)*(a + b*Cos[c + d*x])^2),x]

[Out]

(((2*(4*a^4*A + 44*a^2*A*b^2 - 45*A*b^4 - 30*a^3*b*B + 27*a*b^3*B)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/
(a + b) - (8*a*(-7*a^2*A*b + 10*A*b^3 + 3*a^3*B - 6*a*b^2*B)*((a + b)*EllipticF[(c + d*x)/2, 2] - a*EllipticPi
[(2*b)/(a + b), (c + d*x)/2, 2]))/(b*(a + b)) - (6*(-4*a^2*A*b + 5*A*b^3 + 2*a^3*B - 3*a*b^2*B)*(-2*a*b*Ellipt
icE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (2*a^2 - b^2)*El
lipticPi[-(b/a), -ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b*Sqrt[Sin[c + d*x]^2]))/((a - b)*(a + b))
+ 4*Sqrt[Cos[c + d*x]]*((3*b^3*(A*b - a*B)*Sin[c + d*x])/((a^2 - b^2)*(a + b*Cos[c + d*x])) + 2*(-6*A*b + 3*a
*B + a*A*Sec[c + d*x])*Tan[c + d*x]))/(12*a^3*d)

________________________________________________________________________________________

Maple [B]  time = 16.085, size = 1031, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^2,x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-4*b^2*(2*A*b-B*a)/a^3/(-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)*Ellipt
icPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+2*(-2*A*b+B*a)/a^3*(-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(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))+2*
(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*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/(2*sin(1/2*d*x+1/2*c)^2-1)+2*(A*b-B*a)*b/a^2*(-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*co
s(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*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))+1/2*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)*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*(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))+
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)))+2*A/a^2*(-1/6*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)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(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)*Ellip
ticF(cos(1/2*d*x+1/2*c),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

________________________________________________________________________________________

Maxima [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+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

Timed out

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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((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

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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+B*cos(d*x+c))/cos(d*x+c)**(5/2)/(a+b*cos(d*x+c))**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)/((b*cos(d*x + c) + a)^2*cos(d*x + c)^(5/2)), x)