3.2.60 \(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx\) [160]

Optimal. Leaf size=188 \[ -\frac {(9 A-49 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(3 A-13 B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(3 A-8 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(3 A-13 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right )} \]

[Out]

-1/10*(9*A-49*B)*(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))/a^3/d+1
/6*(3*A-13*B)*(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))/a^3/d+1/5*
(A-B)*cos(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^3+1/15*(3*A-8*B)*cos(d*x+c)^(3/2)*sin(d*x+c)/a/d/(a+a*cos
(d*x+c))^2+1/6*(3*A-13*B)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a^3+a^3*cos(d*x+c))

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Rubi [A]
time = 0.31, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {3056, 2827, 2720, 2719} \begin {gather*} \frac {(3 A-13 B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac {(9 A-49 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(3 A-13 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{6 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac {(3 A-8 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^(5/2)*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^3,x]

[Out]

-1/10*((9*A - 49*B)*EllipticE[(c + d*x)/2, 2])/(a^3*d) + ((3*A - 13*B)*EllipticF[(c + d*x)/2, 2])/(6*a^3*d) +
((A - B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + ((3*A - 8*B)*Cos[c + d*x]^(3/2)*Sin[c
+ d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) + ((3*A - 13*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(6*d*(a^3 + a^3*Cos[
c + d*x]))

Rule 2719

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

Rule 2720

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

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3056

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 - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x]
)^n/(a*f*(2*m + 1))), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rubi steps

\begin {align*} \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx &=\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (\frac {5}{2} a (A-B)-\frac {1}{2} a (A-11 B) \cos (c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(3 A-8 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\sqrt {\cos (c+d x)} \left (\frac {3}{2} a^2 (3 A-8 B)-\frac {1}{2} a^2 (6 A-41 B) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(3 A-8 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(3 A-13 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \frac {\frac {5}{4} a^3 (3 A-13 B)-\frac {3}{4} a^3 (9 A-49 B) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{15 a^6}\\ &=\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(3 A-8 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(3 A-13 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right )}-\frac {(9 A-49 B) \int \sqrt {\cos (c+d x)} \, dx}{20 a^3}+\frac {(3 A-13 B) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{12 a^3}\\ &=-\frac {(9 A-49 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(3 A-13 B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(3 A-8 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(3 A-13 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.95, size = 1273, normalized size = 6.77 \begin {gather*} -\frac {9 i A \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (\frac {2 e^{2 i d x} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right ) \sqrt {e^{-i d x} \left (2 \left (1+e^{2 i d x}\right ) \cos (c)+2 i \left (-1+e^{2 i d x}\right ) \sin (c)\right )} \sqrt {1+e^{2 i d x} \cos (2 c)+i e^{2 i d x} \sin (2 c)}}{3 i d \left (1+e^{2 i d x}\right ) \cos (c)-3 d \left (-1+e^{2 i d x}\right ) \sin (c)}-\frac {2 \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right ) \sqrt {e^{-i d x} \left (2 \left (1+e^{2 i d x}\right ) \cos (c)+2 i \left (-1+e^{2 i d x}\right ) \sin (c)\right )} \sqrt {1+e^{2 i d x} \cos (2 c)+i e^{2 i d x} \sin (2 c)}}{-i d \left (1+e^{2 i d x}\right ) \cos (c)+d \left (-1+e^{2 i d x}\right ) \sin (c)}\right )}{10 (a+a \cos (c+d x))^3}+\frac {49 i B \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (\frac {2 e^{2 i d x} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right ) \sqrt {e^{-i d x} \left (2 \left (1+e^{2 i d x}\right ) \cos (c)+2 i \left (-1+e^{2 i d x}\right ) \sin (c)\right )} \sqrt {1+e^{2 i d x} \cos (2 c)+i e^{2 i d x} \sin (2 c)}}{3 i d \left (1+e^{2 i d x}\right ) \cos (c)-3 d \left (-1+e^{2 i d x}\right ) \sin (c)}-\frac {2 \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right ) \sqrt {e^{-i d x} \left (2 \left (1+e^{2 i d x}\right ) \cos (c)+2 i \left (-1+e^{2 i d x}\right ) \sin (c)\right )} \sqrt {1+e^{2 i d x} \cos (2 c)+i e^{2 i d x} \sin (2 c)}}{-i d \left (1+e^{2 i d x}\right ) \cos (c)+d \left (-1+e^{2 i d x}\right ) \sin (c)}\right )}{10 (a+a \cos (c+d x))^3}-\frac {2 A \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\text {ArcTan}(\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \sec (d x-\text {ArcTan}(\cot (c))) \sqrt {1-\sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {1+\sin (d x-\text {ArcTan}(\cot (c)))}}{d (a+a \cos (c+d x))^3 \sqrt {1+\cot ^2(c)}}+\frac {26 B \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\text {ArcTan}(\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \sec (d x-\text {ArcTan}(\cot (c))) \sqrt {1-\sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {1+\sin (d x-\text {ArcTan}(\cot (c)))}}{3 d (a+a \cos (c+d x))^3 \sqrt {1+\cot ^2(c)}}+\frac {\cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \left (-\frac {4 (-9 A+29 B+20 B \cos (c)) \csc (c)}{5 d}+\frac {4 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (9 A \sin \left (\frac {d x}{2}\right )-29 B \sin \left (\frac {d x}{2}\right )\right )}{5 d}-\frac {4 \sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (9 A \sin \left (\frac {d x}{2}\right )-14 B \sin \left (\frac {d x}{2}\right )\right )}{15 d}+\frac {2 \sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sin \left (\frac {d x}{2}\right )-B \sin \left (\frac {d x}{2}\right )\right )}{5 d}-\frac {4 (9 A-14 B) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{15 d}+\frac {2 (A-B) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{5 d}\right )}{(a+a \cos (c+d x))^3} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(Cos[c + d*x]^(5/2)*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^3,x]

[Out]

(((-9*I)/10)*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4, 7/4, -(E^(
(2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(
I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))*Cos[c] - 3*d
*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*S
qrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2
*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[c])))/(a + a*
Cos[c + d*x])^3 + (((49*I)/10)*B*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/
2, 3/4, 7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)
*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*
d*x))*Cos[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c]
+ I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E
^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))
*Sin[c])))/(a + a*Cos[c + d*x])^3 - (2*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Si
n[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[
1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(a + a*Cos[c + d*x])^
3*Sqrt[1 + Cot[c]^2]) + (26*B*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - Arc
Tan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^
2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(a + a*Cos[c + d*x])^3*Sqrt[1
+ Cot[c]^2]) + (Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*((-4*(-9*A + 29*B + 20*B*Cos[c])*Csc[c])/(5*d) + (4*Se
c[c/2]*Sec[c/2 + (d*x)/2]*(9*A*Sin[(d*x)/2] - 29*B*Sin[(d*x)/2]))/(5*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(9*
A*Sin[(d*x)/2] - 14*B*Sin[(d*x)/2]))/(15*d) + (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2
]))/(5*d) - (4*(9*A - 14*B)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(15*d) + (2*(A - B)*Sec[c/2 + (d*x)/2]^4*Tan[c/2])/
(5*d)))/(a + a*Cos[c + d*x])^3

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(450\) vs. \(2(224)=448\).
time = 0.35, size = 451, normalized size = 2.40

method	result	size
default	\(-\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (108 A \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+30 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+54 A \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-348 B \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-130 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-294 B \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-198 A \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+578 B \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+114 A \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-264 B \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-27 A \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+37 B \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 A -3 B \right )}{60 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(451\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

-1/60*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(108*A*cos(1/2*d*x+1/2*c)^8+30*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5+54*
A*cos(1/2*d*x+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/
2*c),2^(1/2))-348*B*cos(1/2*d*x+1/2*c)^8-130*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)*
EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5-294*B*cos(1/2*d*x+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(
1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-198*A*cos(1/2*d*x+1/2*c)^6+578*B*
cos(1/2*d*x+1/2*c)^6+114*A*cos(1/2*d*x+1/2*c)^4-264*B*cos(1/2*d*x+1/2*c)^4-27*A*cos(1/2*d*x+1/2*c)^2+37*B*cos(
1/2*d*x+1/2*c)^2+3*A-3*B)/a^3/cos(1/2*d*x+1/2*c)^5/(-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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*cos(d*x + c)^(5/2)/(a*cos(d*x + c) + a)^3, x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.13, size = 467, normalized size = 2.48 \begin {gather*} \frac {2 \, {\left (3 \, {\left (9 \, A - 29 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (18 \, A - 73 \, B\right )} \cos \left (d x + c\right ) + 15 \, A - 65 \, B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 5 \, {\left (\sqrt {2} {\left (3 i \, A - 13 i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (3 i \, A - 13 i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (3 i \, A - 13 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (3 i \, A - 13 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 \, {\left (\sqrt {2} {\left (-3 i \, A + 13 i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-3 i \, A + 13 i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-3 i \, A + 13 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A + 13 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, {\left (\sqrt {2} {\left (9 i \, A - 49 i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (9 i \, A - 49 i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (9 i \, A - 49 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (9 i \, A - 49 i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, {\left (\sqrt {2} {\left (-9 i \, A + 49 i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-9 i \, A + 49 i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-9 i \, A + 49 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-9 i \, A + 49 i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

1/60*(2*(3*(9*A - 29*B)*cos(d*x + c)^2 + 2*(18*A - 73*B)*cos(d*x + c) + 15*A - 65*B)*sqrt(cos(d*x + c))*sin(d*
x + c) - 5*(sqrt(2)*(3*I*A - 13*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(3*I*A - 13*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(3
*I*A - 13*I*B)*cos(d*x + c) + sqrt(2)*(3*I*A - 13*I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x +
c)) - 5*(sqrt(2)*(-3*I*A + 13*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(-3*I*A + 13*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(-3
*I*A + 13*I*B)*cos(d*x + c) + sqrt(2)*(-3*I*A + 13*I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x +
c)) - 3*(sqrt(2)*(9*I*A - 49*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(9*I*A - 49*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(9*I
*A - 49*I*B)*cos(d*x + c) + sqrt(2)*(9*I*A - 49*I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*
x + c) + I*sin(d*x + c))) - 3*(sqrt(2)*(-9*I*A + 49*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(-9*I*A + 49*I*B)*cos(d*x
+ c)^2 + 3*sqrt(2)*(-9*I*A + 49*I*B)*cos(d*x + c) + sqrt(2)*(-9*I*A + 49*I*B))*weierstrassZeta(-4, 0, weierstr
assPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*c
os(d*x + c) + a^3*d)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**(5/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))**3,x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*cos(d*x + c)^(5/2)/(a*cos(d*x + c) + a)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^(5/2)*(A + B*cos(c + d*x)))/(a + a*cos(c + d*x))^3,x)

[Out]

int((cos(c + d*x)^(5/2)*(A + B*cos(c + d*x)))/(a + a*cos(c + d*x))^3, x)

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