3.546 \(\int \frac {1}{(a+b \cos (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=282 \[ -\frac {2 b \left (23 a^2+9 b^2\right ) \sin (c+d x)}{15 d \left (a^2-b^2\right )^3 \sqrt {a+b \cos (c+d x)}}-\frac {16 a b \sin (c+d x)}{15 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^{3/2}}-\frac {2 b \sin (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}-\frac {16 a \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (23 a^2+9 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 d \left (a^2-b^2\right )^3 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \]

[Out]

-2/5*b*sin(d*x+c)/(a^2-b^2)/d/(a+b*cos(d*x+c))^(5/2)-16/15*a*b*sin(d*x+c)/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^(3/2)
-2/15*b*(23*a^2+9*b^2)*sin(d*x+c)/(a^2-b^2)^3/d/(a+b*cos(d*x+c))^(1/2)+2/15*(23*a^2+9*b^2)*(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)^3/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-16/15*a*(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)^2/d/(a+b*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.36, antiderivative size = 282, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ -\frac {2 b \left (23 a^2+9 b^2\right ) \sin (c+d x)}{15 d \left (a^2-b^2\right )^3 \sqrt {a+b \cos (c+d x)}}-\frac {16 a b \sin (c+d x)}{15 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^{3/2}}-\frac {2 b \sin (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{5/2}}-\frac {16 a \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (23 a^2+9 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 d \left (a^2-b^2\right )^3 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^(-7/2),x]

[Out]

(2*(23*a^2 + 9*b^2)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(15*(a^2 - b^2)^3*d*Sqrt[(
a + b*Cos[c + d*x])/(a + b)]) - (16*a*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)]
)/(15*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x]]) - (2*b*Sin[c + d*x])/(5*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(5/
2)) - (16*a*b*Sin[c + d*x])/(15*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x])^(3/2)) - (2*b*(23*a^2 + 9*b^2)*Sin[c + d*
x])/(15*(a^2 - b^2)^3*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))^{7/2}} \, dx &=-\frac {2 b \sin (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{5/2}}-\frac {2 \int \frac {-\frac {5 a}{2}+\frac {3}{2} b \cos (c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx}{5 \left (a^2-b^2\right )}\\ &=-\frac {2 b \sin (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{5/2}}-\frac {16 a b \sin (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^{3/2}}+\frac {4 \int \frac {\frac {3}{4} \left (5 a^2+3 b^2\right )-2 a b \cos (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx}{15 \left (a^2-b^2\right )^2}\\ &=-\frac {2 b \sin (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{5/2}}-\frac {16 a b \sin (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^{3/2}}-\frac {2 b \left (23 a^2+9 b^2\right ) \sin (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \cos (c+d x)}}-\frac {8 \int \frac {-\frac {1}{8} a \left (15 a^2+17 b^2\right )-\frac {1}{8} b \left (23 a^2+9 b^2\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{15 \left (a^2-b^2\right )^3}\\ &=-\frac {2 b \sin (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{5/2}}-\frac {16 a b \sin (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^{3/2}}-\frac {2 b \left (23 a^2+9 b^2\right ) \sin (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \cos (c+d x)}}-\frac {(8 a) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{15 \left (a^2-b^2\right )^2}+\frac {\left (23 a^2+9 b^2\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{15 \left (a^2-b^2\right )^3}\\ &=-\frac {2 b \sin (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{5/2}}-\frac {16 a b \sin (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^{3/2}}-\frac {2 b \left (23 a^2+9 b^2\right ) \sin (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (\left (23 a^2+9 b^2\right ) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{15 \left (a^2-b^2\right )^3 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (8 a \sqrt {\frac {a+b \cos (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}} \, dx}{15 \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}\\ &=\frac {2 \left (23 a^2+9 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 \left (a^2-b^2\right )^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {16 a \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}-\frac {2 b \sin (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{5/2}}-\frac {16 a b \sin (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^{3/2}}-\frac {2 b \left (23 a^2+9 b^2\right ) \sin (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 1.42, size = 189, normalized size = 0.67 \[ \frac {2 \left (\frac {\left (\frac {a+b \cos (c+d x)}{a+b}\right )^{5/2} \left (\left (23 a^2+9 b^2\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )+8 a (b-a) F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )}{(a-b)^3}+\frac {b \sin (c+d x) \left (34 a^4+b^2 \left (23 a^2+9 b^2\right ) \cos ^2(c+d x)+2 a b \left (27 a^2+5 b^2\right ) \cos (c+d x)-5 a^2 b^2+3 b^4\right )}{\left (b^2-a^2\right )^3}\right )}{15 d (a+b \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^(-7/2),x]

[Out]

(2*((((a + b*Cos[c + d*x])/(a + b))^(5/2)*((23*a^2 + 9*b^2)*EllipticE[(c + d*x)/2, (2*b)/(a + b)] + 8*a*(-a +
b)*EllipticF[(c + d*x)/2, (2*b)/(a + b)]))/(a - b)^3 + (b*(34*a^4 - 5*a^2*b^2 + 3*b^4 + 2*a*b*(27*a^2 + 5*b^2)
*Cos[c + d*x] + b^2*(23*a^2 + 9*b^2)*Cos[c + d*x]^2)*Sin[c + d*x])/(-a^2 + b^2)^3))/(15*d*(a + b*Cos[c + d*x])
^(5/2))

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fricas [F]  time = 2.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \cos \left (d x + c\right ) + a}}{b^{4} \cos \left (d x + c\right )^{4} + 4 \, a b^{3} \cos \left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \cos \left (d x + c\right )^{2} + 4 \, a^{3} b \cos \left (d x + c\right ) + a^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*cos(d*x + c) + a)/(b^4*cos(d*x + c)^4 + 4*a*b^3*cos(d*x + c)^3 + 6*a^2*b^2*cos(d*x + c)^2 + 4*
a^3*b*cos(d*x + c) + a^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)^(-7/2), x)

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maple [A]  time = 2.36, size = 616, normalized size = 2.18 \[ -\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 )}}{10 b^{2} \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 )^{3}}+\frac {8 a \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 )}}{15 b \left (a -b \right )^{2} \left (a +b \right )^{2} \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {a -b}{2 b}\right )^{2}}+\frac {4 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (23 a^{2}+9 b^{2}\right )}{15 \left (a -b \right )^{3} \left (a +b \right )^{3} \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 (15 a^{2}-8 a b +9 b^{2}\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 (15 a^{5}+15 a^{4} b -30 a^{3} b^{2}-30 a^{2} b^{3}+15 a \,b^{4}+15 b^{5}\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 {2 \left (23 a^{2}+9 b^{2}\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}}\, \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 )}{15 \left (a -b \right )^{2} \left (a +b \right )^{3} \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))^(7/2),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)*(1/10/b^2/(a-b)/(a+b)*cos(1/2*d*x+1/2*c)*(-2*si
n(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))^3+8/15*a/b/(a-b)^2/(
a+b)^2*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+4/15*b*sin(1/2*d*x+1/2*c)^2/(a-b)^3/(a+b)^3*cos(1/2*d*x+1/2*c)*(23*a^2+9*b^2)/(-(-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*(15*a^2-8*a*b+9*b^2)/(15*a^5+15*a^4*b-30*a^3*b^2-30*a^2*b^3+15*a
*b^4+15*b^5)*(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))-2/15*(23*a^2+9*b^2)/(a-
b)^2/(a+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))-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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c) + a)^(-7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*cos(c + d*x))^(7/2),x)

[Out]

int(1/(a + b*cos(c + d*x))^(7/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c))**(7/2),x)

[Out]

Timed out

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