\(\int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx\) [119]

Optimal result

Integrand size = 25, antiderivative size = 126 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {(3 A+5 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(3 A+5 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}} \]

[Out]

1/4*(A-B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)+1/16*(3*A+5*B)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)+1/32*(3*A+5
*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 = 0.17 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2829, 2729, 2728, 212} \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {(3 A+5 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(3 A+5 B) \sin (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}+\frac {(A-B) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]

[In]

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

[Out]

((3*A + 5*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)) + ((3*A + 5*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 2729

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d
*(2*n + 1))), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

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)]

Rubi steps \begin{align*} \text {integral}& = \frac {(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(3 A+5 B) \int \frac {1}{(a+a \cos (c+d x))^{3/2}} \, dx}{8 a} \\ & = \frac {(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(3 A+5 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(3 A+5 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 {(3 A+5 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {(3 A+5 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 {(3 A+5 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 {(3 A+5 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.63 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {4 (3 A+5 B) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right )+(7 A+B+(3 A+5 B) \cos (c+d x)) \sin (c+d x)}{16 d (a (1+\cos (c+d x)))^{5/2}} \]

[In]

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

[Out]

(4*(3*A + 5*B)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^5 + (7*A + B + (3*A + 5*B)*Cos[c + d*x])*Sin[c + d*x
])/(16*d*(a*(1 + Cos[c + d*x]))^(5/2))

Maple [B] (verified)

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

Time = 4.24 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.32

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

[In]

int((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)*(3*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+5*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+3*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)+5*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*s
in(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*c
os(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 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {2} {\left ({\left (3 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + 5 \, B\right )} \cos \left (d x + c\right ) + 3 \, A + 5 \, B\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (3 \, A + 5 \, B\right )} \cos \left (d x + c\right ) + 7 \, A + B\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

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

[Out]

1/64*(sqrt(2)*((3*A + 5*B)*cos(d*x + c)^3 + 3*(3*A + 5*B)*cos(d*x + c)^2 + 3*(3*A + 5*B)*cos(d*x + c) + 3*A +
5*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*((3*A + 5*B)*cos(d*x + c) + 7*A + 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]

\[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {A + B \cos {\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

Integral((A + B*cos(c + d*x))/(a*(cos(c + d*x) + 1))**(5/2), x)

Maxima [F(-1)]

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

[In]

integrate((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 = 0.75 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.54 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\frac {\sqrt {2} {\left (3 \, A \sqrt {a} + 5 \, B \sqrt {a}\right )} \log \left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\sqrt {2} {\left (3 \, A \sqrt {a} + 5 \, B \sqrt {a}\right )} \log \left (-\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, {\left (3 \, \sqrt {2} A \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, \sqrt {2} B \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, \sqrt {2} A \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, \sqrt {2} B \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{64 \, d} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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