3.876 \(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (B \cos (c+d x)+C \cos ^2(c+d x))}{a+b \cos (c+d x)} \, dx\)
Optimal. Leaf size=246 \[ \frac {2 a^4 (b B-a C) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 d (a+b)}+\frac {2 \left (5 a^2+3 b^2\right ) (b B-a C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 d}-\frac {2 \left (-7 a^2 C+7 a b B-5 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{21 b^3 d}-\frac {2 \left (-21 a^4 C+21 a^3 b B-7 a^2 b^2 C+7 a b^3 B-5 b^4 C\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 b^5 d}+\frac {2 (b B-a C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b^2 d}+\frac {2 C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 b d} \]
[Out]
2/5*(5*a^2+3*b^2)*(B*b-C*a)*(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*B*a^3*b+7*B*a*b^3-21*C*a^4-7*C*a^2*b^2-5*C*b^4)*(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^4*(B*b-C*a)*(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*(B*b-C*a)*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*B*a*b-7*C*a^2-5*C*b^2)*sin(d*x+c)*cos(d*x+c)^(1/2)/b^3/d
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Rubi [A] time = 1.20, antiderivative size = 246, normalized size of antiderivative = 1.00,
number of steps used = 9, number of rules used = 8, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used =
{3029, 2990, 3049, 3059, 2639, 3002, 2641, 2805} \[ -\frac {2 \left (-7 a^2 b^2 C+21 a^3 b B-21 a^4 C+7 a b^3 B-5 b^4 C\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 b^5 d}+\frac {2 \left (5 a^2+3 b^2\right ) (b B-a C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 d}+\frac {2 a^4 (b B-a C) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 d (a+b)}-\frac {2 \left (-7 a^2 C+7 a b B-5 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{21 b^3 d}+\frac {2 (b B-a C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b^2 d}+\frac {2 C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 b d} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^(5/2)*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x]),x]
[Out]
(2*(5*a^2 + 3*b^2)*(b*B - a*C)*EllipticE[(c + d*x)/2, 2])/(5*b^4*d) - (2*(21*a^3*b*B + 7*a*b^3*B - 21*a^4*C -
7*a^2*b^2*C - 5*b^4*C)*EllipticF[(c + d*x)/2, 2])/(21*b^5*d) + (2*a^4*(b*B - a*C)*EllipticPi[(2*b)/(a + b), (c
+ d*x)/2, 2])/(b^5*(a + b)*d) - (2*(7*a*b*B - 7*a^2*C - 5*b^2*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(21*b^3*d)
+ (2*(b*B - 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 2990
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*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x
])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*(m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c -
b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x]^2, x], x], x] /; F
reeQ[{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] && GtQ[m,
1] && !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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 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])))
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 (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx &=\int \frac {\cos ^{\frac {7}{2}}(c+d x) (B+C \cos (c+d x))}{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 {5}{2} b C \cos (c+d x)+\frac {7}{2} (b B-a C) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{7 b}\\ &=\frac {2 (b B-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}{4} a (b B-a C)+\frac {1}{4} b (21 b B+4 a C) \cos (c+d x)-\frac {5}{4} \left (7 a b B-7 a^2 C-5 b^2 C\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{35 b^2}\\ &=-\frac {2 \left (7 a b B-7 a^2 C-5 b^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 b^3 d}+\frac {2 (b B-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 b B-7 a^2 C-5 b^2 C\right )+\frac {1}{8} b \left (28 a b B-28 a^2 C+25 b^2 C\right ) \cos (c+d x)+\frac {21}{8} \left (5 a^2+3 b^2\right ) (b B-a C) \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 b B-7 a^2 C-5 b^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 b^3 d}+\frac {2 (b B-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 b B-7 a^2 C-5 b^2 C\right )+\frac {5}{8} \left (21 a^3 b B+7 a b^3 B-21 a^4 C-7 a^2 b^2 C-5 b^4 C\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{105 b^4}+\frac {\left (\left (5 a^2+3 b^2\right ) (b B-a C)\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 b^4}\\ &=\frac {2 \left (5 a^2+3 b^2\right ) (b B-a C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 d}-\frac {2 \left (7 a b B-7 a^2 C-5 b^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 b^3 d}+\frac {2 (b B-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^4 (b B-a C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{b^5}-\frac {\left (21 a^3 b B+7 a b^3 B-21 a^4 C-7 a^2 b^2 C-5 b^4 C\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 b^5}\\ &=\frac {2 \left (5 a^2+3 b^2\right ) (b B-a C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 d}-\frac {2 \left (21 a^3 b B+7 a b^3 B-21 a^4 C-7 a^2 b^2 C-5 b^4 C\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 b^5 d}+\frac {2 a^4 (b B-a C) \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 b B-7 a^2 C-5 b^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 b^3 d}+\frac {2 (b B-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 = 2.84, size = 305, normalized size = 1.24 \[ \frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (70 a^2 C+42 b (b B-a C) \cos (c+d x)-70 a b B+15 b^2 C \cos (2 (c+d x))+65 b^2 C\right )+\frac {4 \left (-28 a^2 C+28 a b B+25 b^2 C\right ) \left ((a+b) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-a \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{a+b}-\frac {42 \left (5 a^2+3 b^2\right ) (a C-b B) \sin (c+d x) \left (\left (b^2-2 a^2\right ) \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{a b^2 \sqrt {\sin ^2(c+d x)}}+\frac {2 \left (-35 a^3 C+35 a^2 b B-13 a b^2 C+63 b^3 B\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}}{210 b^3 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cos[c + d*x]^(5/2)*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x]),x]
[Out]
((2*(35*a^2*b*B + 63*b^3*B - 35*a^3*C - 13*a*b^2*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + (4*(2
8*a*b*B - 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*B + 70*a^2*C + 65*b^2*C + 42*b*(b*B - a*C)*Cos[c + d*x] + 15*b^2*C
*Cos[2*(c + d*x)])*Sin[c + d*x] - (42*(5*a^2 + 3*b^2)*(-(b*B) + a*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[Sq
rt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*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)*(B*cos(d*x+c)+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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{b \cos \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(5/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*cos(d*x + c)^(5/2)/(b*cos(d*x + c) + a), x)
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maple [B] time = 2.80, size = 1375, normalized size = 5.59 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^(5/2)*(B*cos(d*x+c)+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*B*a*b^4+168*B*b^5+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*B*a^2*b^3+308*B*a*b^4-168*B*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*B*a^2*b^3-112*B*a*b^4+42*B*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*B*(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^4*b-105*B*(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+105*B*(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*B*(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+63*B*(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))*b^5-105*B*(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^4*b+105*B*(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*B*(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*B*(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^4-105*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)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),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)*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(cos(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(cos(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)*EllipticF(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)*EllipticF(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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{b \cos \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(5/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="maxima")
[Out]
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*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 \[ \int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )\right )}{a+b\,\cos \left (c+d\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((cos(c + d*x)^(5/2)*(B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + b*cos(c + d*x)),x)
[Out]
int((cos(c + d*x)^(5/2)*(B*cos(c + d*x) + 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)*(B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+b*cos(d*x+c)),x)
[Out]
Timed out
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