3.2.45 \(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \cos (c+d x))}{a+a \cos (c+d x)} \, dx\) [145]

Optimal. Leaf size=156 \[ -\frac {3 (5 A-7 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d}+\frac {5 (A-B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a d}+\frac {5 (A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d}-\frac {(5 A-7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 a d}+\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))} \]

[Out]

-3/5*(5*A-7*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/d+5/3*(
A-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/d-1/5*(5*A-7*B)*c
os(d*x+c)^(3/2)*sin(d*x+c)/a/d+(A-B)*cos(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))+5/3*(A-B)*sin(d*x+c)*cos(d
*x+c)^(1/2)/a/d

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Rubi [A]
time = 0.13, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3056, 2827, 2715, 2720, 2719} \begin {gather*} \frac {5 (A-B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a d}-\frac {3 (5 A-7 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}-\frac {(5 A-7 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d}+\frac {5 (A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d} \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]),x]

[Out]

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

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

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)} \, dx &=\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}+\frac {\int \cos ^{\frac {3}{2}}(c+d x) \left (\frac {5}{2} a (A-B)-\frac {1}{2} a (5 A-7 B) \cos (c+d x)\right ) \, dx}{a^2}\\ &=\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}-\frac {(5 A-7 B) \int \cos ^{\frac {5}{2}}(c+d x) \, dx}{2 a}+\frac {(5 (A-B)) \int \cos ^{\frac {3}{2}}(c+d x) \, dx}{2 a}\\ &=\frac {5 (A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d}-\frac {(5 A-7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 a d}+\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}-\frac {(3 (5 A-7 B)) \int \sqrt {\cos (c+d x)} \, dx}{10 a}+\frac {(5 (A-B)) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a}\\ &=-\frac {3 (5 A-7 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d}+\frac {5 (A-B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a d}+\frac {5 (A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d}-\frac {(5 A-7 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 a d}+\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.67, size = 1182, normalized size = 7.58 \begin {gather*} -\frac {3 i A \cos ^2\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 )}{4 (a+a \cos (c+d x))}+\frac {21 i B \cos ^2\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 )}{20 (a+a \cos (c+d x))}+\frac {\cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \left (\frac {2 (5 A-5 B+10 A \cos (c)-16 B \cos (c)) \csc (c)}{5 d}+\frac {4 (A-B) \cos (d x) \sin (c)}{3 d}+\frac {2 B \cos (2 d x) \sin (2 c)}{5 d}+\frac {2 \sec \left (\frac {c}{2}\right ) \sec \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 )}{d}+\frac {4 (A-B) \cos (c) \sin (d x)}{3 d}+\frac {2 B \cos (2 c) \sin (2 d x)}{5 d}\right )}{a+a \cos (c+d x)}-\frac {5 A \cos ^2\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)) \sqrt {1+\cot ^2(c)}}+\frac {5 B \cos ^2\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)) \sqrt {1+\cot ^2(c)}} \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]),x]

[Out]

(((-3*I)/4)*A*Cos[c/2 + (d*x)/2]^2*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)]*Sq
rt[(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*C
os[c + d*x]) + (((21*I)/20)*B*Cos[c/2 + (d*x)/2]^2*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))*Si
n[c])))/(a + a*Cos[c + d*x]) + (Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*((2*(5*A - 5*B + 10*A*Cos[c] - 16*B*Co
s[c])*Csc[c])/(5*d) + (4*(A - B)*Cos[d*x]*Sin[c])/(3*d) + (2*B*Cos[2*d*x]*Sin[2*c])/(5*d) + (2*Sec[c/2]*Sec[c/
2 + (d*x)/2]*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2]))/d + (4*(A - B)*Cos[c]*Sin[d*x])/(3*d) + (2*B*Cos[2*c]*Sin[2*d*
x])/(5*d)))/(a + a*Cos[c + d*x]) - (5*A*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin
[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]]]])/(3*d*(a + a*Cos[c + d*x])
*Sqrt[1 + Cot[c]^2]) + (5*B*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTa
n[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])*Sqrt[1 + Co
t[c]^2])

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Maple [A]
time = 0.31, size = 281, normalized size = 1.80

method	result	size
default	\(-\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (25 A \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+45 A \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-25 B \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-63 B \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )+48 B \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-40 A -56 B \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (90 A -30 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-35 A +23 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{15 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(281\)

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)),x,method=_RETURNVERBOSE)

[Out]

-1/15*((2*cos(1/2*d*x+1/2*c)^2-1)*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)^(1/2)
*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(25*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+45*A*EllipticE(cos(1/2*d*x+1/2*c
),2^(1/2))-25*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-63*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+48*B*sin(1/2
*d*x+1/2*c)^8+(-40*A-56*B)*sin(1/2*d*x+1/2*c)^6+(90*A-30*B)*sin(1/2*d*x+1/2*c)^4+(-35*A+23*B)*sin(1/2*d*x+1/2*
c)^2)/a/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)/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)),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.13, size = 269, normalized size = 1.72 \begin {gather*} \frac {2 \, {\left (6 \, B \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right ) + 25 \, A - 25 \, B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 25 \, {\left (\sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 25 \, {\left (\sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 9 \, {\left (\sqrt {2} {\left (5 i \, A - 7 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (5 i \, A - 7 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 ) - 9 \, {\left (\sqrt {2} {\left (-5 i \, A + 7 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-5 i \, A + 7 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 )}{30 \, {\left (a d \cos \left (d x + c\right ) + a 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)),x, algorithm="fricas")

[Out]

1/30*(2*(6*B*cos(d*x + c)^2 + 2*(5*A - 2*B)*cos(d*x + c) + 25*A - 25*B)*sqrt(cos(d*x + c))*sin(d*x + c) - 25*(
sqrt(2)*(I*A - I*B)*cos(d*x + c) + sqrt(2)*(I*A - I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x +
c)) - 25*(sqrt(2)*(-I*A + I*B)*cos(d*x + c) + sqrt(2)*(-I*A + I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) -
I*sin(d*x + c)) - 9*(sqrt(2)*(5*I*A - 7*I*B)*cos(d*x + c) + sqrt(2)*(5*I*A - 7*I*B))*weierstrassZeta(-4, 0, we
ierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 9*(sqrt(2)*(-5*I*A + 7*I*B)*cos(d*x + c) + sqrt(2)*
(-5*I*A + 7*I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a*d*cos(
d*x + c) + a*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)),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)),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*cos(d*x + c)^(5/2)/(a*cos(d*x + c) + a), 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 )}{a+a\,\cos \left (c+d\,x\right )} \,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)),x)

[Out]

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

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