3.543 \(\int \frac {1}{(a+b \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=221 \[ -\frac {8 a b \sin (c+d x)}{3 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}-\frac {2 b \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}+\frac {8 a \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \]

[Out]

-2/3*b*sin(d*x+c)/(a^2-b^2)/d/(a+b*cos(d*x+c))^(3/2)-8/3*a*b*sin(d*x+c)/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^(1/2)+8
/3*a*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))*(a+
b*cos(d*x+c))^(1/2)/(a^2-b^2)^2/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-2/3*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+
1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b))^(1/2)/(a^2-b^2)/d/(a+b*c
os(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.23, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2664, 2754, 2752, 2663, 2661, 2655, 2653} \[ -\frac {8 a b \sin (c+d x)}{3 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}-\frac {2 b \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}+\frac {8 a \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^(-5/2),x]

[Out]

(8*a*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(3*(a^2 - b^2)^2*d*Sqrt[(a + b*Cos[c + d*
x])/(a + b)]) - (2*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(3*(a^2 - b^2)*d*
Sqrt[a + b*Cos[c + d*x]]) - (2*b*Sin[c + d*x])/(3*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2)) - (8*a*b*Sin[c + d
*x])/(3*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x]])

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2664

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

Rule 2752

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 2754

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 + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \frac {1}{(a+b \cos (c+d x))^{5/2}} \, dx &=-\frac {2 b \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {2 \int \frac {-\frac {3 a}{2}+\frac {1}{2} b \cos (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx}{3 \left (a^2-b^2\right )}\\ &=-\frac {2 b \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {8 a b \sin (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {4 \int \frac {\frac {1}{4} \left (3 a^2+b^2\right )+a b \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{3 \left (a^2-b^2\right )^2}\\ &=-\frac {2 b \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {8 a b \sin (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {(4 a) \int \sqrt {a+b \cos (c+d x)} \, dx}{3 \left (a^2-b^2\right )^2}-\frac {\int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{3 \left (a^2-b^2\right )}\\ &=-\frac {2 b \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {8 a b \sin (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (4 a \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{3 \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}} \, dx}{3 \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\\ &=\frac {8 a \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 \left (a^2-b^2\right )^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {2 b \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {8 a b \sin (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.93, size = 158, normalized size = 0.71 \[ \frac {2 b \sin (c+d x) \left (-5 a^2-4 a b \cos (c+d x)+b^2\right )-2 (a-b) (a+b)^2 \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{3/2} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )+8 a (a+b)^2 \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{3/2} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d (a-b)^2 (a+b)^2 (a+b \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a*(a + b)^2*((a + b*Cos[c + d*x])/(a + b))^(3/2)*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - 2*(a - b)*(a + b)^
2*((a + b*Cos[c + d*x])/(a + b))^(3/2)*EllipticF[(c + d*x)/2, (2*b)/(a + b)] + 2*b*(-5*a^2 + b^2 - 4*a*b*Cos[c
+ d*x])*Sin[c + d*x])/(3*(a - b)^2*(a + b)^2*d*(a + b*Cos[c + d*x])^(3/2))

________________________________________________________________________________________

fricas [F]  time = 1.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \cos \left (d x + c\right ) + a}}{b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right )^{2} + 3 \, a^{2} b \cos \left (d x + c\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*cos(d*x + c) + a)/(b^3*cos(d*x + c)^3 + 3*a*b^2*cos(d*x + c)^2 + 3*a^2*b*cos(d*x + c) + a^3),
x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)^(-5/2), x)

________________________________________________________________________________________

maple [A]  time = 1.66, size = 489, normalized size = 2.21 \[ -\frac {\sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +\left (a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{3 b \left (a -b \right ) \left (a +b \right ) \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {a -b}{2 b}\right )^{2}}+\frac {16 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a}{3 \left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}+\frac {2 \left (3 a -b \right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b}{a -b}}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )}{\left (3 a^{3}+3 a^{2} b -3 b^{2} a -3 b^{3}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +\left (a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}-\frac {8 a \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b}{a -b}}\, \left (\EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )-\EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )\right )}{3 \left (a -b \right ) \left (a +b \right )^{2} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +\left (a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cos(d*x+c))^(5/2),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(1/3/b/(a-b)/(a+b)*cos(1/2*d*x+1/2*c)*(-2*sin(1
/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2+1/2/b*(a-b))^2+16/3*b*sin(1/2*d*x+1/
2*c)^2/(a-b)^2/(a+b)^2*cos(1/2*d*x+1/2*c)*a/(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*(3
*a-b)/(3*a^3+3*a^2*b-3*a*b^2-3*b^3)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/
(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-
8/3*a/(a-b)/(a+b)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+
1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-EllipticE(cos(1
/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))))/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)^(1/2)/d

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c) + a)^(-5/2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*cos(c + d*x))^(5/2),x)

[Out]

int(1/(a + b*cos(c + d*x))^(5/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \cos {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c))**(5/2),x)

[Out]

Integral((a + b*cos(c + d*x))**(-5/2), x)

________________________________________________________________________________________