3.115 \(\int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=261 \[ \frac {(163 A-283 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(475 A-787 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{240 a^3 d}-\frac {(85 A-157 B) \sin (c+d x) \cos ^2(c+d x)}{80 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {(985 A-1729 B) \sin (c+d x)}{120 a^2 d \sqrt {a \cos (c+d x)+a}}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}+\frac {(13 A-21 B) \sin (c+d x) \cos ^3(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

[Out]

1/4*(A-B)*cos(d*x+c)^4*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)+1/16*(13*A-21*B)*cos(d*x+c)^3*sin(d*x+c)/a/d/(a+a*c
os(d*x+c))^(3/2)+1/32*(163*A-283*B)*arctanh(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d*2
^(1/2)-1/120*(985*A-1729*B)*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(1/2)-1/80*(85*A-157*B)*cos(d*x+c)^2*sin(d*x+c)/
a^2/d/(a+a*cos(d*x+c))^(1/2)+1/240*(475*A-787*B)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/a^3/d

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Rubi [A]  time = 0.80, antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {2977, 2983, 2968, 3023, 2751, 2649, 206} \[ -\frac {(85 A-157 B) \sin (c+d x) \cos ^2(c+d x)}{80 a^2 d \sqrt {a \cos (c+d x)+a}}+\frac {(475 A-787 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{240 a^3 d}-\frac {(985 A-1729 B) \sin (c+d x)}{120 a^2 d \sqrt {a \cos (c+d x)+a}}+\frac {(163 A-283 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}+\frac {(13 A-21 B) \sin (c+d x) \cos ^3(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

((163*A - 283*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) +
((A - B)*Cos[c + d*x]^4*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) + ((13*A - 21*B)*Cos[c + d*x]^3*Sin[c +
d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)) - ((985*A - 1729*B)*Sin[c + d*x])/(120*a^2*d*Sqrt[a + a*Cos[c + d*x
]]) - ((85*A - 157*B)*Cos[c + d*x]^2*Sin[c + d*x])/(80*a^2*d*Sqrt[a + a*Cos[c + d*x]]) + ((475*A - 787*B)*Sqrt
[a + a*Cos[c + d*x]]*Sin[c + d*x])/(240*a^3*d)

Rule 206

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

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2751

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

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 2977

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 2983

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] && (I
ntegerQ[n] || EqQ[m + 1/2, 0])

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((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 + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx &=\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {\int \frac {\cos ^3(c+d x) \left (4 a (A-B)-\frac {1}{2} a (5 A-13 B) \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(13 A-21 B) \cos ^3(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\cos ^2(c+d x) \left (\frac {3}{2} a^2 (13 A-21 B)-\frac {1}{4} a^2 (85 A-157 B) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(13 A-21 B) \cos ^3(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {(85 A-157 B) \cos ^2(c+d x) \sin (c+d x)}{80 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\cos (c+d x) \left (-\frac {1}{2} a^3 (85 A-157 B)+\frac {1}{8} a^3 (475 A-787 B) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{20 a^5}\\ &=\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(13 A-21 B) \cos ^3(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {(85 A-157 B) \cos ^2(c+d x) \sin (c+d x)}{80 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {-\frac {1}{2} a^3 (85 A-157 B) \cos (c+d x)+\frac {1}{8} a^3 (475 A-787 B) \cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{20 a^5}\\ &=\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(13 A-21 B) \cos ^3(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {(85 A-157 B) \cos ^2(c+d x) \sin (c+d x)}{80 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {(475 A-787 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{240 a^3 d}+\frac {\int \frac {\frac {1}{16} a^4 (475 A-787 B)-\frac {1}{8} a^4 (985 A-1729 B) \cos (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{30 a^6}\\ &=\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(13 A-21 B) \cos ^3(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {(985 A-1729 B) \sin (c+d x)}{120 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(85 A-157 B) \cos ^2(c+d x) \sin (c+d x)}{80 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {(475 A-787 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{240 a^3 d}+\frac {(163 A-283 B) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(13 A-21 B) \cos ^3(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {(985 A-1729 B) \sin (c+d x)}{120 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(85 A-157 B) \cos ^2(c+d x) \sin (c+d x)}{80 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {(475 A-787 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{240 a^3 d}-\frac {(163 A-283 B) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{16 a^2 d}\\ &=\frac {(163 A-283 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(13 A-21 B) \cos ^3(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {(985 A-1729 B) \sin (c+d x)}{120 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(85 A-157 B) \cos ^2(c+d x) \sin (c+d x)}{80 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {(475 A-787 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{240 a^3 d}\\ \end {align*}

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Mathematica [A]  time = 1.64, size = 139, normalized size = 0.53 \[ \frac {\tan \left (\frac {1}{2} (c+d x)\right ) (-5 (479 A-887 B) \cos (c+d x)+(832 B-400 A) \cos (2 (c+d x))+40 A \cos (3 (c+d x))-1895 A-40 B \cos (3 (c+d x))+12 B \cos (4 (c+d x))+3491 B)+30 (163 A-283 B) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{240 a d (a (\cos (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

(30*(163*A - 283*B)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^3 + (-1895*A + 3491*B - 5*(479*A - 887*B)*Cos[c
+ d*x] + (-400*A + 832*B)*Cos[2*(c + d*x)] + 40*A*Cos[3*(c + d*x)] - 40*B*Cos[3*(c + d*x)] + 12*B*Cos[4*(c +
d*x)])*Tan[(c + d*x)/2])/(240*a*d*(a*(1 + Cos[c + d*x]))^(3/2))

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fricas [A]  time = 0.80, size = 270, normalized size = 1.03 \[ -\frac {15 \, \sqrt {2} {\left ({\left (163 \, A - 283 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (163 \, A - 283 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (163 \, A - 283 \, B\right )} \cos \left (d x + c\right ) + 163 \, A - 283 \, B\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left (96 \, B \cos \left (d x + c\right )^{4} + 160 \, {\left (A - B\right )} \cos \left (d x + c\right )^{3} - 32 \, {\left (25 \, A - 49 \, B\right )} \cos \left (d x + c\right )^{2} - 5 \, {\left (503 \, A - 911 \, B\right )} \cos \left (d x + c\right ) - 1495 \, A + 2671 \, B\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{960 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/960*(15*sqrt(2)*((163*A - 283*B)*cos(d*x + c)^3 + 3*(163*A - 283*B)*cos(d*x + c)^2 + 3*(163*A - 283*B)*cos(
d*x + c) + 163*A - 283*B)*sqrt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sin(d*x
+ c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*(96*B*cos(d*x + c)^4 + 160*(A - B)*c
os(d*x + c)^3 - 32*(25*A - 49*B)*cos(d*x + c)^2 - 5*(503*A - 911*B)*cos(d*x + c) - 1495*A + 2671*B)*sqrt(a*cos
(d*x + c) + 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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giac [A]  time = 2.69, size = 257, normalized size = 0.98 \[ -\frac {\frac {15 \, {\left (163 \, \sqrt {2} A - 283 \, \sqrt {2} B\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {5}{2}}} - \frac {{\left ({\left ({\left (15 \, {\left (\frac {2 \, {\left (\sqrt {2} A a^{2} - \sqrt {2} B a^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{2}} - \frac {21 \, \sqrt {2} A a^{2} - 29 \, \sqrt {2} B a^{2}}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {3685 \, \sqrt {2} A a^{2} - 6733 \, \sqrt {2} B a^{2}}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {5 \, {\left (1133 \, \sqrt {2} A a^{2} - 1973 \, \sqrt {2} B a^{2}\right )}}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {15 \, {\left (155 \, \sqrt {2} A a^{2} - 291 \, \sqrt {2} B a^{2}\right )}}{a^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {5}{2}}}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-1/480*(15*(163*sqrt(2)*A - 283*sqrt(2)*B)*log(abs(-sqrt(a)*tan(1/2*d*x + 1/2*c) + sqrt(a*tan(1/2*d*x + 1/2*c)
^2 + a)))/a^(5/2) - (((15*(2*(sqrt(2)*A*a^2 - sqrt(2)*B*a^2)*tan(1/2*d*x + 1/2*c)^2/a^2 - (21*sqrt(2)*A*a^2 -
29*sqrt(2)*B*a^2)/a^2)*tan(1/2*d*x + 1/2*c)^2 - (3685*sqrt(2)*A*a^2 - 6733*sqrt(2)*B*a^2)/a^2)*tan(1/2*d*x + 1
/2*c)^2 - 5*(1133*sqrt(2)*A*a^2 - 1973*sqrt(2)*B*a^2)/a^2)*tan(1/2*d*x + 1/2*c)^2 - 15*(155*sqrt(2)*A*a^2 - 29
1*sqrt(2)*B*a^2)/a^2)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(5/2))/d

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maple [B]  time = 0.82, size = 467, normalized size = 1.79 \[ \frac {\sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (768 B \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+640 A \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2176 B \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2445 A \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \sqrt {2}\, \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -4245 B \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \sqrt {2}\, \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2560 A \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5248 B \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-435 A \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+555 B \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+30 A \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-30 B \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{480 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{\frac {7}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(5/2),x)

[Out]

1/480/cos(1/2*d*x+1/2*c)^3*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(768*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2
)*cos(1/2*d*x+1/2*c)^8+640*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^6-2176*B*2^(1/2
)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^6+2445*A*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^
(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^4*a-4245*B*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^
(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^4*a-2560*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^
(1/2)*cos(1/2*d*x+1/2*c)^4+5248*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^4-435*A*a^
(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)^2+555*B*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^
2)^(1/2)*cos(1/2*d*x+1/2*c)^2+30*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-30*B*2^(1/2)*(a*sin(1/2*d*x+
1/2*c)^2)^(1/2)*a^(1/2))/a^(7/2)/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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maxima [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)^4*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^4\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^4*(A + B*cos(c + d*x)))/(a + a*cos(c + d*x))^(5/2),x)

[Out]

int((cos(c + d*x)^4*(A + B*cos(c + d*x)))/(a + a*cos(c + d*x))^(5/2), 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)**4*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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