3.107 \(\int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=261 \[ -\frac {(15 A-19 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(273 A-397 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{210 a^2 d}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {(7 A-11 B) \sin (c+d x) \cos ^3(c+d x)}{14 a d \sqrt {a \cos (c+d x)+a}}+\frac {(63 A-67 B) \sin (c+d x) \cos ^2(c+d x)}{70 a d \sqrt {a \cos (c+d x)+a}}+\frac {(651 A-799 B) \sin (c+d x)}{105 a d \sqrt {a \cos (c+d x)+a}} \]

[Out]

1/2*(A-B)*cos(d*x+c)^4*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)-1/4*(15*A-19*B)*arctanh(1/2*sin(d*x+c)*a^(1/2)*2^(1
/2)/(a+a*cos(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)+1/105*(651*A-799*B)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)+1/70*(
63*A-67*B)*cos(d*x+c)^2*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)-1/14*(7*A-11*B)*cos(d*x+c)^3*sin(d*x+c)/a/d/(a+a
*cos(d*x+c))^(1/2)-1/210*(273*A-397*B)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/a^2/d

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Rubi [A]  time = 0.79, 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} \[ -\frac {(273 A-397 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{210 a^2 d}-\frac {(15 A-19 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {(7 A-11 B) \sin (c+d x) \cos ^3(c+d x)}{14 a d \sqrt {a \cos (c+d x)+a}}+\frac {(63 A-67 B) \sin (c+d x) \cos ^2(c+d x)}{70 a d \sqrt {a \cos (c+d x)+a}}+\frac {(651 A-799 B) \sin (c+d x)}{105 a d \sqrt {a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^(3/2),x]

[Out]

-((15*A - 19*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) + ((
A - B)*Cos[c + d*x]^4*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((651*A - 799*B)*Sin[c + d*x])/(105*a*d
*Sqrt[a + a*Cos[c + d*x]]) + ((63*A - 67*B)*Cos[c + d*x]^2*Sin[c + d*x])/(70*a*d*Sqrt[a + a*Cos[c + d*x]]) - (
(7*A - 11*B)*Cos[c + d*x]^3*Sin[c + d*x])/(14*a*d*Sqrt[a + a*Cos[c + d*x]]) - ((273*A - 397*B)*Sqrt[a + a*Cos[
c + d*x]]*Sin[c + d*x])/(210*a^2*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))^{3/2}} \, dx &=\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\cos ^3(c+d x) \left (4 a (A-B)-\frac {1}{2} a (7 A-11 B) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {(7 A-11 B) \cos ^3(c+d x) \sin (c+d x)}{14 a d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\cos ^2(c+d x) \left (-\frac {3}{2} a^2 (7 A-11 B)+\frac {1}{4} a^2 (63 A-67 B) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{7 a^3}\\ &=\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(63 A-67 B) \cos ^2(c+d x) \sin (c+d x)}{70 a d \sqrt {a+a \cos (c+d x)}}-\frac {(7 A-11 B) \cos ^3(c+d x) \sin (c+d x)}{14 a d \sqrt {a+a \cos (c+d x)}}+\frac {2 \int \frac {\cos (c+d x) \left (\frac {1}{2} a^3 (63 A-67 B)-\frac {1}{8} a^3 (273 A-397 B) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{35 a^4}\\ &=\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(63 A-67 B) \cos ^2(c+d x) \sin (c+d x)}{70 a d \sqrt {a+a \cos (c+d x)}}-\frac {(7 A-11 B) \cos ^3(c+d x) \sin (c+d x)}{14 a d \sqrt {a+a \cos (c+d x)}}+\frac {2 \int \frac {\frac {1}{2} a^3 (63 A-67 B) \cos (c+d x)-\frac {1}{8} a^3 (273 A-397 B) \cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{35 a^4}\\ &=\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(63 A-67 B) \cos ^2(c+d x) \sin (c+d x)}{70 a d \sqrt {a+a \cos (c+d x)}}-\frac {(7 A-11 B) \cos ^3(c+d x) \sin (c+d x)}{14 a d \sqrt {a+a \cos (c+d x)}}-\frac {(273 A-397 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{210 a^2 d}+\frac {4 \int \frac {-\frac {1}{16} a^4 (273 A-397 B)+\frac {1}{8} a^4 (651 A-799 B) \cos (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{105 a^5}\\ &=\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(651 A-799 B) \sin (c+d x)}{105 a d \sqrt {a+a \cos (c+d x)}}+\frac {(63 A-67 B) \cos ^2(c+d x) \sin (c+d x)}{70 a d \sqrt {a+a \cos (c+d x)}}-\frac {(7 A-11 B) \cos ^3(c+d x) \sin (c+d x)}{14 a d \sqrt {a+a \cos (c+d x)}}-\frac {(273 A-397 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{210 a^2 d}-\frac {(15 A-19 B) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{4 a}\\ &=\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(651 A-799 B) \sin (c+d x)}{105 a d \sqrt {a+a \cos (c+d x)}}+\frac {(63 A-67 B) \cos ^2(c+d x) \sin (c+d x)}{70 a d \sqrt {a+a \cos (c+d x)}}-\frac {(7 A-11 B) \cos ^3(c+d x) \sin (c+d x)}{14 a d \sqrt {a+a \cos (c+d x)}}-\frac {(273 A-397 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{210 a^2 d}+\frac {(15 A-19 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 )}{2 a d}\\ &=-\frac {(15 A-19 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(651 A-799 B) \sin (c+d x)}{105 a d \sqrt {a+a \cos (c+d x)}}+\frac {(63 A-67 B) \cos ^2(c+d x) \sin (c+d x)}{70 a d \sqrt {a+a \cos (c+d x)}}-\frac {(7 A-11 B) \cos ^3(c+d x) \sin (c+d x)}{14 a d \sqrt {a+a \cos (c+d x)}}-\frac {(273 A-397 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{210 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 1.17, size = 167, normalized size = 0.64 \[ \frac {105 (15 A-19 B) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-\frac {1}{2} \sin \left (\frac {1}{2} (c+d x)\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right ) (6 (273 A-277 B) \cos (c+d x)+(256 B-84 A) \cos (2 (c+d x))+42 A \cos (3 (c+d x))+1974 A-18 B \cos (3 (c+d x))+15 B \cos (4 (c+d x))-2161 B)}{105 d \left (\sin ^2\left (\frac {1}{2} (c+d x)\right )-1\right ) (a (\cos (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^(3/2),x]

[Out]

(105*(15*A - 19*B)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^5 - (Cos[(c + d*x)/2]^3*(1974*A - 2161*B + 6*(27
3*A - 277*B)*Cos[c + d*x] + (-84*A + 256*B)*Cos[2*(c + d*x)] + 42*A*Cos[3*(c + d*x)] - 18*B*Cos[3*(c + d*x)] +
15*B*Cos[4*(c + d*x)])*Sin[(c + d*x)/2])/2)/(105*d*(a*(1 + Cos[c + d*x]))^(3/2)*(-1 + Sin[(c + d*x)/2]^2))

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fricas [A]  time = 0.96, size = 241, normalized size = 0.92 \[ -\frac {105 \, \sqrt {2} {\left ({\left (15 \, A - 19 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (15 \, A - 19 \, B\right )} \cos \left (d x + c\right ) + 15 \, A - 19 \, B\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left (60 \, B \cos \left (d x + c\right )^{4} + 12 \, {\left (7 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{3} - 28 \, {\left (3 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2} + 12 \, {\left (63 \, A - 67 \, B\right )} \cos \left (d x + c\right ) + 1029 \, A - 1201 \, B\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{840 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

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))^(3/2),x, algorithm="fricas")

[Out]

-1/840*(105*sqrt(2)*((15*A - 19*B)*cos(d*x + c)^2 + 2*(15*A - 19*B)*cos(d*x + c) + 15*A - 19*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*(60*B*cos(d*x + c)^4 + 12*(7*A - 3*B)*cos(d*x + c)^3 - 28*(3*A - 7*B)*cos(d*
x + c)^2 + 12*(63*A - 67*B)*cos(d*x + c) + 1029*A - 1201*B)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(a^2*d*cos(
d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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giac [A]  time = 2.40, size = 254, normalized size = 0.97 \[ \frac {\frac {105 \, {\left (15 \, \sqrt {2} A - 19 \, \sqrt {2} B\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {3}{2}}} + \frac {{\left ({\left ({\left ({\left (\frac {105 \, {\left (\sqrt {2} A a^{5} - \sqrt {2} B a^{5}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{3}} + \frac {4 \, {\left (693 \, \sqrt {2} A a^{5} - 877 \, \sqrt {2} B a^{5}\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {14 \, {\left (453 \, \sqrt {2} A a^{5} - 517 \, \sqrt {2} B a^{5}\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {140 \, {\left (39 \, \sqrt {2} A a^{5} - 47 \, \sqrt {2} B a^{5}\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1785 \, {\left (\sqrt {2} A a^{5} - \sqrt {2} B a^{5}\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {7}{2}}}}{420 \, d} \]

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))^(3/2),x, algorithm="giac")

[Out]

1/420*(105*(15*sqrt(2)*A - 19*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^(3/2) + ((((105*(sqrt(2)*A*a^5 - sqrt(2)*B*a^5)*tan(1/2*d*x + 1/2*c)^2/a^3 + 4*(693*sqrt(2)*A*a^5 -
877*sqrt(2)*B*a^5)/a^3)*tan(1/2*d*x + 1/2*c)^2 + 14*(453*sqrt(2)*A*a^5 - 517*sqrt(2)*B*a^5)/a^3)*tan(1/2*d*x +
1/2*c)^2 + 140*(39*sqrt(2)*A*a^5 - 47*sqrt(2)*B*a^5)/a^3)*tan(1/2*d*x + 1/2*c)^2 + 1785*(sqrt(2)*A*a^5 - sqrt
(2)*B*a^5)/a^3)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(7/2))/d

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maple [A]  time = 0.80, size = 448, normalized size = 1.72 \[ \frac {\sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (960 B \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-96 \sqrt {2}\, \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (7 A +17 B \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+224 \sqrt {2}\, \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (3 A +8 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+35 \sqrt {2}\, \left (45 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 ) a -48 A \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-57 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 ) a +16 B \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1575 \sqrt {2}\, \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 ) a A +1995 \sqrt {2}\, \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 ) a B +1785 A \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-1785 B \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{420 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{\frac {5}{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))^(3/2),x)

[Out]

1/420/cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(960*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*
sin(1/2*d*x+1/2*c)^8-96*2^(1/2)*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(7*A+17*B)*sin(1/2*d*x+1/2*c)^6+224*2^(
1/2)*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(3*A+8*B)*sin(1/2*d*x+1/2*c)^4+35*2^(1/2)*(45*A*ln(4/cos(1/2*d*x+1
/2*c)*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a))*a-48*A*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-57*B*ln(4/cos(
1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a))*a+16*B*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))*sin(
1/2*d*x+1/2*c)^2-1575*2^(1/2)*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a))*a*A+1995*2^(
1/2)*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a))*a*B+1785*A*2^(1/2)*(a*sin(1/2*d*x+1/2
*c)^2)^(1/2)*a^(1/2)-1785*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/a^(5/2)/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))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^4\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

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))^(3/2),x)

[Out]

int((cos(c + d*x)^4*(A + B*cos(c + d*x)))/(a + a*cos(c + d*x))^(3/2), 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))**(3/2),x)

[Out]

Timed out

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