3.3.2 \(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx\) [202]

Optimal. Leaf size=246 \[ \frac {(2 A-5 B) \text {ArcSin}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{5/2} d}-\frac {(43 A-115 B) \text {ArcTan}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(7 A-15 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {(11 A-35 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}} \]

[Out]

(2*A-5*B)*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d+1/4*(A-B)*cos(d*x+c)^(5/2)*sin(d*x+c)/d/
(a+a*cos(d*x+c))^(5/2)+1/16*(7*A-15*B)*cos(d*x+c)^(3/2)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)-1/32*(43*A-115*B
)*arctan(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)-1/16*(11*A-
35*B)*sin(d*x+c)*cos(d*x+c)^(1/2)/a^2/d/(a+a*cos(d*x+c))^(1/2)

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Rubi [A]
time = 0.53, antiderivative size = 246, 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 = {3056, 3062, 3061, 2861, 211, 2853, 222} \begin {gather*} \frac {(2 A-5 B) \text {ArcSin}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{5/2} d}-\frac {(43 A-115 B) \text {ArcTan}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(11 A-35 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{16 a^2 d \sqrt {a \cos (c+d x)+a}}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}+\frac {(7 A-15 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \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])^(5/2),x]

[Out]

((2*A - 5*B)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(a^(5/2)*d) - ((43*A - 115*B)*ArcTan[(Sq
rt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + ((A - B)*
Cos[c + d*x]^(5/2)*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) + ((7*A - 15*B)*Cos[c + d*x]^(3/2)*Sin[c + d
*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)) - ((11*A - 35*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(16*a^2*d*Sqrt[a +
a*Cos[c + d*x]])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 2853

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, a/b]

Rule 2861

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[-2*(a/f), Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c +
d*Sin[e + f*x]]))], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 -
d^2, 0]

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])

Rule 3061

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin
[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A*b - a*B)/b, Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*
x]]), x], x] + Dist[B/b, Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*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]

Rule 3062

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)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*
(m + n + 1))), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*b*c
*(m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{
a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] &&
(IntegerQ[n] || EqQ[m + 1/2, 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))^{5/2}} \, dx &=\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (\frac {5}{2} a (A-B)-a (A-5 B) \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(7 A-15 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\sqrt {\cos (c+d x)} \left (\frac {3}{4} a^2 (7 A-15 B)-\frac {1}{2} a^2 (11 A-35 B) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(7 A-15 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {(11 A-35 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {-\frac {1}{4} a^3 (11 A-35 B)+4 a^3 (2 A-5 B) \cos (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{8 a^5}\\ &=\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(7 A-15 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {(11 A-35 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(43 A-115 B) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{32 a^2}+\frac {(2 A-5 B) \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx}{2 a^3}\\ &=\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(7 A-15 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {(11 A-35 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {(43 A-115 B) \text {Subst}\left (\int \frac {1}{2 a^2+a x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 a d}-\frac {(2 A-5 B) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{a}}} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^3 d}\\ &=\frac {(2 A-5 B) \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{5/2} d}-\frac {(43 A-115 B) \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(A-B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(7 A-15 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {(11 A-35 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 3.49, size = 376, normalized size = 1.53 \begin {gather*} \frac {\cos ^5\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\sqrt {2} e^{\frac {1}{2} i (c+d x)} \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (32 A d x-80 B d x-16 i (2 A-5 B) \sinh ^{-1}\left (e^{i (c+d x)}\right )+i \sqrt {2} (43 A-115 B) \log \left (1+e^{i (c+d x)}\right )+32 i A \log \left (1+\sqrt {1+e^{2 i (c+d x)}}\right )-80 i B \log \left (1+\sqrt {1+e^{2 i (c+d x)}}\right )-43 i \sqrt {2} A \log \left (1-e^{i (c+d x)}+\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}\right )+115 i \sqrt {2} B \log \left (1-e^{i (c+d x)}+\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}\right )\right )}{\sqrt {1+e^{2 i (c+d x)}}}+\sqrt {\cos (c+d x)} (-11 A+43 B+(-15 A+55 B) \cos (c+d x)+8 B \cos (2 (c+d x))) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{8 d (a (1+\cos (c+d x)))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^5*((Sqrt[2]*E^((I/2)*(c + d*x))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]*(32*A*d*x -
80*B*d*x - (16*I)*(2*A - 5*B)*ArcSinh[E^(I*(c + d*x))] + I*Sqrt[2]*(43*A - 115*B)*Log[1 + E^(I*(c + d*x))] + (
32*I)*A*Log[1 + Sqrt[1 + E^((2*I)*(c + d*x))]] - (80*I)*B*Log[1 + Sqrt[1 + E^((2*I)*(c + d*x))]] - (43*I)*Sqrt
[2]*A*Log[1 - E^(I*(c + d*x)) + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]] + (115*I)*Sqrt[2]*B*Log[1 - E^(I*(c + d
*x)) + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]]))/Sqrt[1 + E^((2*I)*(c + d*x))] + Sqrt[Cos[c + d*x]]*(-11*A + 43
*B + (-15*A + 55*B)*Cos[c + d*x] + 8*B*Cos[2*(c + d*x)])*Sec[(c + d*x)/2]^3*Tan[(c + d*x)/2]))/(8*d*(a*(1 + Co
s[c + d*x]))^(5/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(646\) vs. \(2(209)=418\).
time = 0.38, size = 647, normalized size = 2.63

method	result	size
default	\(-\frac {\sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (-1+\cos \left (d x +c \right )\right )^{5} \left (\cos ^{\frac {5}{2}}\left (d x +c \right )\right ) \left (30 A \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (\cos ^{3}\left (d x +c \right )\right )+22 A \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (\cos ^{2}\left (d x +c \right )\right )+43 A \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-32 B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{4}\left (d x +c \right )\right )-115 B \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-30 A \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \cos \left (d x +c \right )+64 A \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )+43 A \sqrt {2}\, \sin \left (d x +c \right ) \cos \left (d x +c \right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-78 B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{3}\left (d x +c \right )\right )-160 B \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )-115 B \sqrt {2}\, \sin \left (d x +c \right ) \cos \left (d x +c \right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-22 A \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}+64 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right ) \sin \left (d x +c \right ) \cos \left (d x +c \right )+40 B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right )-160 B \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right ) \sin \left (d x +c \right ) \cos \left (d x +c \right )+70 B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )\right )}{32 d \sin \left (d x +c \right )^{11} \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} a^{3}}\)	\(647\)

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))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/32/d*(a*(1+cos(d*x+c)))^(1/2)*(-1+cos(d*x+c))^5*cos(d*x+c)^(5/2)*(30*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*co
s(d*x+c)^3+22*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)^2+43*A*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)*arcsin((-1
+cos(d*x+c))/sin(d*x+c))-32*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^4-115*B*2^(1/2)*cos(d*x+c)^2*sin(d*
x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-30*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)+64*A*cos(d*x+c)^2*si
n(d*x+c)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))+43*A*2^(1/2)*sin(d*x+c)*cos(d*x+c)*ar
csin((-1+cos(d*x+c))/sin(d*x+c))-78*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^3-160*B*cos(d*x+c)^2*sin(d*
x+c)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))-115*B*2^(1/2)*sin(d*x+c)*cos(d*x+c)*arcsi
n((-1+cos(d*x+c))/sin(d*x+c))-22*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+64*A*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos
(d*x+c)))^(1/2)/cos(d*x+c))*sin(d*x+c)*cos(d*x+c)+40*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2-160*B*ar
ctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*sin(d*x+c)*cos(d*x+c)+70*B*(cos(d*x+c)/(1+cos(d*
x+c)))^(1/2)*cos(d*x+c))/sin(d*x+c)^11/(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)/a^3

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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))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 8.39, size = 302, normalized size = 1.23 \begin {gather*} \frac {\sqrt {2} {\left ({\left (43 \, A - 115 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (43 \, A - 115 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (43 \, A - 115 \, B\right )} \cos \left (d x + c\right ) + 43 \, A - 115 \, B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 2 \, {\left (16 \, B \cos \left (d x + c\right )^{2} - 5 \, {\left (3 \, A - 11 \, B\right )} \cos \left (d x + c\right ) - 11 \, A + 35 \, B\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 32 \, {\left ({\left (2 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, A - 5 \, B\right )} \cos \left (d x + c\right ) + 2 \, A - 5 \, B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{32 \, {\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))^(5/2),x, algorithm="fricas")

[Out]

1/32*(sqrt(2)*((43*A - 115*B)*cos(d*x + c)^3 + 3*(43*A - 115*B)*cos(d*x + c)^2 + 3*(43*A - 115*B)*cos(d*x + c)
+ 43*A - 115*B)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) +
2*(16*B*cos(d*x + c)^2 - 5*(3*A - 11*B)*cos(d*x + c) - 11*A + 35*B)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c)
)*sin(d*x + c) - 32*((2*A - 5*B)*cos(d*x + c)^3 + 3*(2*A - 5*B)*cos(d*x + c)^2 + 3*(2*A - 5*B)*cos(d*x + c) +
2*A - 5*B)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*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*cos(d*x + c) + a^3*d)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \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))**(5/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4369 deep

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Giac [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))^(5/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \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 )}^{5/2}} \,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))^(5/2),x)

[Out]

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

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