\(\int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx\) [109]

Optimal result

Integrand size = 33, antiderivative size = 171 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {(7 A-11 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(9 A-13 B) \sin (c+d x)}{3 a d \sqrt {a+a \cos (c+d x)}}-\frac {(3 A-7 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{6 a^2 d} \]

[Out]

1/2*(A-B)*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)-1/4*(7*A-11*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)+1/3*(9*A-13*B)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)-1/6*(3*A-7*B
)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/a^2/d

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3056, 3047, 3102, 2830, 2728, 212} \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {(7 A-11 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(3 A-7 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{6 a^2 d}+\frac {(A-B) \sin (c+d x) \cos ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}+\frac {(9 A-13 B) \sin (c+d x)}{3 a d \sqrt {a \cos (c+d x)+a}} \]

[In]

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

[Out]

-1/2*((7*A - 11*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]^2*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((9*A - 13*B)*Sin[c + d*x])/(3*a*d*Sqr
t[a + a*Cos[c + d*x]]) - ((3*A - 7*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(6*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 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*} \text {integral}& = \frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\cos (c+d x) \left (2 a (A-B)-\frac {1}{2} a (3 A-7 B) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2} \\ & = \frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {2 a (A-B) \cos (c+d x)-\frac {1}{2} a (3 A-7 B) \cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2} \\ & = \frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {(3 A-7 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{6 a^2 d}+\frac {\int \frac {-\frac {1}{4} a^2 (3 A-7 B)+\frac {1}{2} a^2 (9 A-13 B) \cos (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{3 a^3} \\ & = \frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(9 A-13 B) \sin (c+d x)}{3 a d \sqrt {a+a \cos (c+d x)}}-\frac {(3 A-7 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{6 a^2 d}-\frac {(7 A-11 B) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{4 a} \\ & = \frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(9 A-13 B) \sin (c+d x)}{3 a d \sqrt {a+a \cos (c+d x)}}-\frac {(3 A-7 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{6 a^2 d}+\frac {(7 A-11 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 {(7 A-11 B) \text {arctanh}\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 ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(9 A-13 B) \sin (c+d x)}{3 a d \sqrt {a+a \cos (c+d x)}}-\frac {(3 A-7 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{6 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.57 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {-3 (7 A-11 B) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos \left (\frac {1}{2} (c+d x)\right )+(15 A-17 B+12 (A-B) \cos (c+d x)+2 B \cos (2 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{6 a d \sqrt {a (1+\cos (c+d x))}} \]

[In]

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

[Out]

(-3*(7*A - 11*B)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2] + (15*A - 17*B + 12*(A - B)*Cos[c + d*x] + 2*B*Cos
[2*(c + d*x)])*Tan[(c + d*x)/2])/(6*a*d*Sqrt[a*(1 + Cos[c + d*x])])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(326\) vs. \(2(148)=296\).

Time = 4.22 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.91

method	result	size
default	\(\frac {\sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (16 B \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {2}\, \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-21 A \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \sqrt {2}\, \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +33 B \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \sqrt {2}\, \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +24 A \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {2}\, \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-40 B \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {2}\, \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+3 A \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-3 B \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{12 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{\frac {5}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(327\)
parts	\(\frac {A \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-7 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {2}\, \sqrt {a}+\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{\frac {5}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {B \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (16 \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-33 \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -27 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+33 \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \right )}{12 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{\frac {5}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(409\)

[In]

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

[Out]

1/12/cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(16*B*cos(1/2*d*x+1/2*c)^4*2^(1/2)*a^(1/2)*(a*sin(1/2*d
*x+1/2*c)^2)^(1/2)-21*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)^2*a+33*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)^2*a+24*A*cos(1/2*d*x+1/2*c)^2*2^(1/2)*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-40*B*cos(1/2*d*x+1/2*c
)^2*2^(1/2)*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+3*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-3*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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.20 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {3 \, \sqrt {2} {\left ({\left (7 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (7 \, A - 11 \, B\right )} \cos \left (d x + c\right ) + 7 \, A - 11 \, B\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left (4 \, B \cos \left (d x + c\right )^{2} + 12 \, {\left (A - B\right )} \cos \left (d x + c\right ) + 15 \, A - 19 \, B\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{24 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-1/24*(3*sqrt(2)*((7*A - 11*B)*cos(d*x + c)^2 + 2*(7*A - 11*B)*cos(d*x + c) + 7*A - 11*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*(4*B*cos(d*x + c)^2 + 12*(A - B)*cos(d*x + c) + 15*A - 19*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)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F(-2)]

Exception generated. \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(cos(d*x+c)^2*(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:Unable to divide, perhaps due to rounding error%%%{%%{[%%%{%%{[-54043195528445952,0]:[1,0,-2]%%},[6]%%%},0]
:[1,0,%%%{-

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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