\(\int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx\) [118]

Optimal result

Integrand size = 31, antiderivative size = 126 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {(5 A+19 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(5 A-13 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}} \]

[Out]

-1/4*(A-B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)+1/16*(5*A-13*B)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)+1/32*(5*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^(5/2)/d*2^(1/2)

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3047, 3098, 2829, 2728, 212} \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {(5 A+19 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(5 A-13 B) \sin (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac {(A-B) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]

[In]

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

[Out]

((5*A + 19*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)*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) + ((5*A - 13*B)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x]
)^(3/2))

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 2829

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*
c - a*d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, 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 3098

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_
.)*(x_)]^2), x_Symbol] :> Simp[(A*b - a*B + b*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {A \cos (c+d x)+B \cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx \\ & = -\frac {(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {\int \frac {-\frac {5}{2} a (A-B)-4 a B \cos (c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2} \\ & = -\frac {(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(5 A-13 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(5 A+19 B) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{32 a^2} \\ & = -\frac {(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(5 A-13 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {(5 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 )}{16 a^2 d} \\ & = \frac {(5 A+19 B) \text {arctanh}\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) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(5 A-13 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.69 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {2 (5 A+19 B) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right )+(A-9 B+(5 A-13 B) \cos (c+d x)) \tan \left (\frac {1}{2} (c+d x)\right )}{16 a d (a (1+\cos (c+d x)))^{3/2}} \]

[In]

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

[Out]

(2*(5*A + 19*B)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^3 + (A - 9*B + (5*A - 13*B)*Cos[c + d*x])*Tan[(c +
d*x)/2])/(16*a*d*(a*(1 + Cos[c + d*x]))^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(291\) vs. \(2(107)=214\).

Time = 4.32 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.32

method	result	size
default	\(\frac {\sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (5 A \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +19 B \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +5 A \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {2}\, \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-13 B \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {2}\, \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-2 A \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+2 B \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{32 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{\frac {7}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(292\)
parts	\(\frac {A \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (5 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {2}\, \sqrt {a}-2 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{32 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{\frac {7}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {B \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (19 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-13 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {2}\, \sqrt {a}+2 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{32 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{\frac {7}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(350\)

[In]

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

[Out]

1/32/cos(1/2*d*x+1/2*c)^3*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(5*A*2^(1/2)*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))*cos(1/2*d*x+1/2*c)^4*a+19*B*2^(1/2)*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))*cos(1/2*d*x+1/2*c)^4*a+5*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)-13*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)-2*A*a^(1/2)*2^(1/2)*(a
*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (107) = 214\).

Time = 0.31 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.77 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {2} {\left ({\left (5 \, A + 19 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A + 19 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (5 \, A + 19 \, B\right )} \cos \left (d x + c\right ) + 5 \, A + 19 \, B\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (5 \, A - 13 \, B\right )} \cos \left (d x + c\right ) + A - 9 \, B\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

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

[Out]

1/64*(sqrt(2)*((5*A + 19*B)*cos(d*x + c)^3 + 3*(5*A + 19*B)*cos(d*x + c)^2 + 3*(5*A + 19*B)*cos(d*x + c) + 5*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*((5*A - 13*B)*cos(d*x + c) + A - 9*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)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Giac [A] (verification not implemented)

none

Time = 1.38 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.63 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\frac {\sqrt {2} {\left (5 \, A \sqrt {a} + 19 \, B \sqrt {a}\right )} \log \left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\sqrt {2} {\left (5 \, A \sqrt {a} + 19 \, B \sqrt {a}\right )} \log \left (-\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, {\left (5 \, \sqrt {2} A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 13 \, \sqrt {2} B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, \sqrt {2} A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 11 \, \sqrt {2} B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{64 \, d} \]

[In]

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

[Out]

1/64*(sqrt(2)*(5*A*sqrt(a) + 19*B*sqrt(a))*log(sin(1/2*d*x + 1/2*c) + 1)/(a^3*sgn(cos(1/2*d*x + 1/2*c))) - sqr
t(2)*(5*A*sqrt(a) + 19*B*sqrt(a))*log(-sin(1/2*d*x + 1/2*c) + 1)/(a^3*sgn(cos(1/2*d*x + 1/2*c))) - 2*(5*sqrt(2
)*A*sqrt(a)*sin(1/2*d*x + 1/2*c)^3 - 13*sqrt(2)*B*sqrt(a)*sin(1/2*d*x + 1/2*c)^3 - 3*sqrt(2)*A*sqrt(a)*sin(1/2
*d*x + 1/2*c) + 11*sqrt(2)*B*sqrt(a)*sin(1/2*d*x + 1/2*c))/((sin(1/2*d*x + 1/2*c)^2 - 1)^2*a^3*sgn(cos(1/2*d*x
+ 1/2*c))))/d

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {\cos \left (c+d\,x\right )\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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