3.2.7 \(\int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx\) [107]

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

[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 = 0.52, 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 = {3056, 3062, 3047, 3102, 2830, 2728, 212} \begin {gather*} -\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}} \end {gather*}

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]

-1/2*((15*A - 19*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(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*Co
s[c + d*x]]*Sin[c + d*x])/(210*a^2*d)

Rule 212

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

Rule 2728

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 2830

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*S
in[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 3047

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 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 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])

Rule 3102

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) \text {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 = 1.13, size = 167, normalized size = 0.64 \begin {gather*} \frac {105 (15 A-19 B) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right )-\frac {1}{2} \cos ^3\left (\frac {1}{2} (c+d x)\right ) (1974 A-2161 B+6 (273 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 \left (\frac {1}{2} (c+d x)\right )}{105 d (a (1+\cos (c+d x)))^{3/2} \left (-1+\sin ^2\left (\frac {1}{2} (c+d x)\right )\right )} \end {gather*}

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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Maple [A]
time = 0.40, size = 448, normalized size = 1.72

method	result	size
default	\(\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 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (7 A +17 B \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+224 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \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}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-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}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\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 {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-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}\)	\(448\)

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,method=_RETURNVERBOSE)

[Out]

1/420/cos(1/2*d*x+1/2*c)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(960*B*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)*
sin(1/2*d*x+1/2*c)^8-96*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)*(7*A+17*B)*sin(1/2*d*x+1/2*c)^6+224*2^(
1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(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)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+a))*a-48*A*a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-57*B*ln(4/cos(
1/2*d*x+1/2*c)*(a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+a))*a+16*B*a^(1/2)*(sin(1/2*d*x+1/2*c)^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)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+a))*a*A+1995*2^(
1/2)*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+a))*a*B+1785*A*a^(1/2)*2^(1/2)*(sin(1/2*d
*x+1/2*c)^2*a)^(1/2)-1785*B*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(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)] 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)^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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Fricas [A]
time = 0.37, size = 241, normalized size = 0.92 \begin {gather*} -\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 )}} \end {gather*}

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4848 deep

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(co

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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 )}^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 \end {gather*}

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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