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

Optimal. Leaf size=239 \[ -\frac {2 a \left (5 a^2 C+5 A b^2+3 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 d}+\frac {2 \left (7 a^2 C+b^2 (7 A+5 C)\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{21 b^3 d}+\frac {2 \left (21 a^4 C+7 a^2 b^2 (3 A+C)+b^4 (7 A+5 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 b^5 d}-\frac {2 a^3 \left (a^2 C+A b^2\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 d (a+b)}-\frac {2 a C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b^2 d}+\frac {2 C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 b d} \]

[Out]

-2/5*a*(5*A*b^2+5*C*a^2+3*C*b^2)*(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^4/d+2/21*(21*a^4*C+7*a^2*b^2*(3*A+C)+b^4*(7*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^5/d-2*a^3*(A*b^2+C*a^2)*(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))/b^5/(a+b)/d-2/5*a*C*cos(d*x+c)^(3/2)*sin(d*x+c)/b^2/d+2/7*
C*cos(d*x+c)^(5/2)*sin(d*x+c)/b/d+2/21*(7*a^2*C+b^2*(7*A+5*C))*sin(d*x+c)*cos(d*x+c)^(1/2)/b^3/d

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Rubi [A]  time = 1.12, antiderivative size = 239, 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 = {3050, 3049, 3059, 2639, 3002, 2641, 2805} \[ \frac {2 \left (7 a^2 b^2 (3 A+C)+21 a^4 C+b^4 (7 A+5 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 b^5 d}-\frac {2 a \left (5 a^2 C+5 A b^2+3 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 d}-\frac {2 a^3 \left (a^2 C+A b^2\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 d (a+b)}+\frac {2 \left (7 a^2 C+b^2 (7 A+5 C)\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{21 b^3 d}-\frac {2 a C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b^2 d}+\frac {2 C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*(5*A*b^2 + 5*a^2*C + 3*b^2*C)*EllipticE[(c + d*x)/2, 2])/(5*b^4*d) + (2*(21*a^4*C + 7*a^2*b^2*(3*A + C)
+ b^4*(7*A + 5*C))*EllipticF[(c + d*x)/2, 2])/(21*b^5*d) - (2*a^3*(A*b^2 + a^2*C)*EllipticPi[(2*b)/(a + b), (c
+ d*x)/2, 2])/(b^5*(a + b)*d) + (2*(7*a^2*C + b^2*(7*A + 5*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(21*b^3*d) -
(2*a*C*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*b^2*d) + (2*C*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(7*b*d)

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 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]

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 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])))

Rule 3050

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (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)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*
x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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])))

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]

Rubi steps

\begin {align*} \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx &=\frac {2 C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 b d}+\frac {2 \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (\frac {5 a C}{2}+\frac {1}{2} b (7 A+5 C) \cos (c+d x)-\frac {7}{2} a C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{7 b}\\ &=-\frac {2 a C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b^2 d}+\frac {2 C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 b d}+\frac {4 \int \frac {\sqrt {\cos (c+d x)} \left (-\frac {21 a^2 C}{4}+a b C \cos (c+d x)+\frac {5}{4} \left (7 a^2 C+b^2 (7 A+5 C)\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{35 b^2}\\ &=\frac {2 \left (7 a^2 C+b^2 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 b^3 d}-\frac {2 a C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b^2 d}+\frac {2 C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 b d}+\frac {8 \int \frac {\frac {5}{8} a \left (7 a^2 C+b^2 (7 A+5 C)\right )+\frac {1}{8} b \left (35 A b^2-28 a^2 C+25 b^2 C\right ) \cos (c+d x)-\frac {21}{8} a \left (5 A b^2+5 a^2 C+3 b^2 C\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{105 b^3}\\ &=\frac {2 \left (7 a^2 C+b^2 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 b^3 d}-\frac {2 a C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b^2 d}+\frac {2 C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 b d}-\frac {8 \int \frac {-\frac {5}{8} a b \left (7 a^2 C+b^2 (7 A+5 C)\right )-\frac {5}{8} \left (21 a^4 C+7 a^2 b^2 (3 A+C)+b^4 (7 A+5 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{105 b^4}-\frac {\left (a \left (5 A b^2+5 a^2 C+3 b^2 C\right )\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 b^4}\\ &=-\frac {2 a \left (5 A b^2+5 a^2 C+3 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 d}+\frac {2 \left (7 a^2 C+b^2 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 b^3 d}-\frac {2 a C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b^2 d}+\frac {2 C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 b d}-\frac {\left (a^3 \left (A b^2+a^2 C\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{b^5}+\frac {\left (21 a^4 C+7 a^2 b^2 (3 A+C)+b^4 (7 A+5 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 b^5}\\ &=-\frac {2 a \left (5 A b^2+5 a^2 C+3 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 d}+\frac {2 \left (21 a^4 C+7 a^2 b^2 (3 A+C)+b^4 (7 A+5 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 b^5 d}-\frac {2 a^3 \left (A b^2+a^2 C\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 (a+b) d}+\frac {2 \left (7 a^2 C+b^2 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 b^3 d}-\frac {2 a C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b^2 d}+\frac {2 C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 b d}\\ \end {align*}

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Mathematica [A]  time = 2.26, size = 291, normalized size = 1.22 \[ \frac {-\frac {2 a \left (35 a^2 C+35 A b^2+13 b^2 C\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (70 a^2 C-42 a b C \cos (c+d x)+70 A b^2+15 b^2 C \cos (2 (c+d x))+65 b^2 C\right )+\frac {4 \left (-28 a^2 C+35 A b^2+25 b^2 C\right ) \left ((a+b) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-a \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{a+b}-\frac {42 \left (5 a^2 C+5 A b^2+3 b^2 C\right ) \sin (c+d x) \left (\left (b^2-2 a^2\right ) \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{b^2 \sqrt {\sin ^2(c+d x)}}}{210 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*a*(35*A*b^2 + 35*a^2*C + 13*b^2*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + (4*(35*A*b^2 - 28
*a^2*C + 25*b^2*C)*((a + b)*EllipticF[(c + d*x)/2, 2] - a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]))/(a + b)
+ 2*Sqrt[Cos[c + d*x]]*(70*A*b^2 + 70*a^2*C + 65*b^2*C - 42*a*b*C*Cos[c + d*x] + 15*b^2*C*Cos[2*(c + d*x)])*Si
n[c + d*x] - (42*(5*A*b^2 + 5*a^2*C + 3*b^2*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])/(b^2*Sqrt[Sin[c + d*x]^2]))/(210*b^3*d)

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{b \cos \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 2.92, size = 1244, normalized size = 5.21 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*((240*C*a*b^4-240*C*b^5)*cos(1/2*d*x+1/2*c)*sin
(1/2*d*x+1/2*c)^8+(168*C*a^2*b^3-528*C*a*b^4+360*C*b^5)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(140*A*a*b^4-1
40*A*b^5+140*C*a^3*b^2-308*C*a^2*b^3+448*C*a*b^4-280*C*b^5)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-70*A*a*b
^4+70*A*b^5-70*C*a^3*b^2+112*C*a^2*b^3-122*C*a*b^4+80*C*b^5)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+105*A*(si
n(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))*a^2*b^3-105*A
*(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))*a*b^4+105
*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^3*b^2
-105*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2
*b^3+35*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*
a*b^4-35*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))
*b^5-105*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a
-b),2^(1/2))*a^3*b^2+105*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))*a^4*b-105*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))*a^3*b^2+63*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))*a^2*b^3-63*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))*a*b^4+105*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)*EllipticF(c
os(1/2*d*x+1/2*c),2^(1/2))*a^5-105*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)*EllipticF(c
os(1/2*d*x+1/2*c),2^(1/2))*a^4*b+35*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)*EllipticF(
cos(1/2*d*x+1/2*c),2^(1/2))*a^3*b^2-35*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)*Ellipti
cF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b^3+25*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)*Elli
pticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^4-25*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)*Ell
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^5-105*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)*Ell
ipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))*a^5)/b^5/(a-b)/(-2*sin(1/2*d*x+1/2*c)^4+sin(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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{b \cos \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )}{a+b\,\cos \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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