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

Optimal. Leaf size=243 \[ -\frac {2 b \left (5 a^2+3 b^2\right ) \sin (c+d x)}{3 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}-\frac {4 a b \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {4 a \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (5 a^2+3 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \]

[Out]

-4/3*a*b*sin(d*x+c)/(a^2-b^2)/d/(a+b*cos(d*x+c))^(3/2)-2/3*b*(5*a^2+3*b^2)*sin(d*x+c)/(a^2-b^2)^2/d/(a+b*cos(d
*x+c))^(1/2)+2/3*(5*a^2+3*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)^2/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-4/3*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)/d/(a+b*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.34, antiderivative size = 243, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {3016, 2754, 2752, 2663, 2661, 2655, 2653} \[ -\frac {2 b \left (5 a^2+3 b^2\right ) \sin (c+d x)}{3 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}-\frac {4 a b \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {4 a \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (5 a^2+3 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3016

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

Rubi steps

\begin {align*} \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^{7/2}} \, dx &=-\int \frac {-a+b \cos (c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx\\ &=-\frac {4 a 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}{2} \left (a^2+b^2\right )-a b \cos (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx}{3 \left (a^2-b^2\right )}\\ &=-\frac {4 a b \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {2 b \left (5 a^2+3 b^2\right ) \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} a \left (3 a^2+5 b^2\right )-\frac {1}{4} b \left (5 a^2+3 b^2\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{3 \left (a^2-b^2\right )^2}\\ &=-\frac {4 a b \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {2 b \left (5 a^2+3 b^2\right ) \sin (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}-\frac {(2 a) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{3 \left (a^2-b^2\right )}+\frac {\left (5 a^2+3 b^2\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{3 \left (a^2-b^2\right )^2}\\ &=-\frac {4 a b \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {2 b \left (5 a^2+3 b^2\right ) \sin (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (\left (5 a^2+3 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}{3 \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (2 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}{3 \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\\ &=\frac {2 \left (5 a^2+3 b^2\right ) \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 {4 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 )}{3 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {4 a b \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {2 b \left (5 a^2+3 b^2\right ) \sin (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 1.01, size = 158, normalized size = 0.65 \[ \frac {2 \left (\frac {\left (\frac {a+b \cos (c+d x)}{a+b}\right )^{3/2} \left (\left (5 a^2+3 b^2\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )+2 a (b-a) F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )}{(a-b)^2}-\frac {b \sin (c+d x) \left (b \left (5 a^2+3 b^2\right ) \cos (c+d x)+a \left (7 a^2+b^2\right )\right )}{\left (a^2-b^2\right )^2}\right )}{3 d (a+b \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((((a + b*Cos[c + d*x])/(a + b))^(3/2)*((5*a^2 + 3*b^2)*EllipticE[(c + d*x)/2, (2*b)/(a + b)] + 2*a*(-a + b
)*EllipticF[(c + d*x)/2, (2*b)/(a + b)]))/(a - b)^2 - (b*(a*(7*a^2 + b^2) + b*(5*a^2 + 3*b^2)*Cos[c + d*x])*Si
n[c + d*x])/(a^2 - b^2)^2))/(3*d*(a + b*Cos[c + d*x])^(3/2))

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fricas [F]  time = 1.03, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {b \cos \left (d x + c\right ) + a} {\left (b \cos \left (d x + c\right ) - a\right )}}{b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right )^{2} + 3 \, a^{2} b \cos \left (d x + c\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(b*cos(d*x + c) + a)*(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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 7.17, size = 792, normalized size = 3.26 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-cos(d*x+c)^2*b^2+a^2)/(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)*(4*a*(1/6/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+8/3*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)+
(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)
)-4/3*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))))-4*b*sin(1/2*d*x+1/2*c)^2/(a-b)/(a+b)*cos(1/2*d*x+1/2*c)/(-(-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/(a+b)*(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/(a+b)*(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 \[ -\int \frac {b^{2} \cos \left (d x + c\right )^{2} - a^{2}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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