3.1064 \(\int \frac{a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=271 \[ -\frac{2 b \left (-5 a^2 C+4 a b B-3 b^2 C\right ) \sin (c+d x)}{3 d \left (a^2-b^2\right )^2 \sqrt{a+b \cos (c+d x)}}-\frac{2 b (b B-2 a C) \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac{2 (b B-2 a C) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (-5 a^2 C+4 a b B-3 b^2 C\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 d \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}} \]

[Out]

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

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Rubi [A]  time = 0.451462, antiderivative size = 271, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 50, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.14, Rules used = {24, 2754, 2752, 2663, 2661, 2655, 2653} \[ -\frac{2 b \left (-5 a^2 C+4 a b B-3 b^2 C\right ) \sin (c+d x)}{3 d \left (a^2-b^2\right )^2 \sqrt{a+b \cos (c+d x)}}-\frac{2 b (b B-2 a C) \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac{2 (b B-2 a C) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (-5 a^2 C+4 a b B-3 b^2 C\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 d \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 24

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((A_.) + (B_.)*(v_) + (C_.)*(v_)^2), x_Symbol] :> Dist[1/b^2, Int[u*(a + b*
v)^(m + 1)*Simp[b*B - a*C + b*C*v, x], x], x] /; FreeQ[{a, b, A, B, C}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0] &&
LeQ[m, -1]

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 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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{7/2}} \, dx &=\frac{\int \frac{b^2 (b B-a C)+b^3 C \cos (c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx}{b^2}\\ &=-\frac{2 b (b B-2 a C) \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} b^2 \left (a b B-a^2 C-b^2 C\right )+\frac{1}{2} b^3 (b B-2 a C) \cos (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=-\frac{2 b (b B-2 a C) \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac{2 b \left (4 a b B-5 a^2 C-3 b^2 C\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} b^2 \left (3 a^2 b B+b^3 B-3 a^3 C-5 a b^2 C\right )+\frac{1}{4} b^3 \left (4 a b B-5 a^2 C-3 b^2 C\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{3 b^2 \left (a^2-b^2\right )^2}\\ &=-\frac{2 b (b B-2 a C) \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac{2 b \left (4 a b B-5 a^2 C-3 b^2 C\right ) \sin (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}-\frac{(b B-2 a C) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{3 \left (a^2-b^2\right )}+\frac{\left (4 a b B-5 a^2 C-3 b^2 C\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{3 \left (a^2-b^2\right )^2}\\ &=-\frac{2 b (b B-2 a C) \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac{2 b \left (4 a b B-5 a^2 C-3 b^2 C\right ) \sin (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}+\frac{\left (\left (4 a b B-5 a^2 C-3 b^2 C\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 ((b B-2 a C) \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 (4 a b B-5 a^2 C-3 b^2 C\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{2 (b B-2 a C) \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 (b B-2 a C) \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac{2 b \left (4 a b B-5 a^2 C-3 b^2 C\right ) \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 = 1.54474, size = 193, normalized size = 0.71 \[ \frac{2 \left (\frac{b \sin (c+d x) \left (b \left (5 a^2 C-4 a b B+3 b^2 C\right ) \cos (c+d x)-5 a^2 b B+7 a^3 C+a b^2 C+b^3 B\right )}{\left (a^2-b^2\right )^2}-\frac{\left (\frac{a+b \cos (c+d x)}{a+b}\right )^{3/2} \left (\left (5 a^2 C-4 a b B+3 b^2 C\right ) E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )-(a-b) (2 a C-b B) F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )\right )}{(a-b)^2}\right )}{3 d (a+b \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*b*B - a^2*C + b^2*B*Cos[c + d*x] + b^2*C*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)*((-4*a*b*B + 5*a^2*C + 3*b^2*C)*EllipticE[(c + d*x)/2, (2*b)/(a +
b)] - (a - b)*(-(b*B) + 2*a*C)*EllipticF[(c + d*x)/2, (2*b)/(a + b)]))/(a - b)^2) + (b*(-5*a^2*b*B + b^3*B + 7
*a^3*C + a*b^2*C + b*(-4*a*b*B + 5*a^2*C + 3*b^2*C)*Cos[c + d*x])*Sin[c + d*x])/(a^2 - b^2)^2))/(3*d*(a + b*Co
s[c + d*x])^(3/2))

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Maple [B]  time = 3.672, size = 744, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*b*B-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^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)*(2*C/sin(1/2*d*x+1/2*c)^2/(-2*sin(1/2*d*x+1/2*c
)^2*b+a+b)/(a^2-b^2)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*((sin(1/2*d*x+1/2*c)^2)^(1/2
)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a-(sin(
1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b
/(a-b))^(1/2))*b+2*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)+2*(B*b-2*C*a)*(1/6/b/(a-b)/(a+b)*cos(1/2*d*x+1/2
*c)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2+1/2*(a-b)/b)^2+8/3*b*si
n(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*b*sin(1/2*d*x+1/2*c)^4+(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)^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*b*
sin(1/2*d*x+1/2*c)^4+(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))-Ellip
ticE(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., size = 0, normalized size = 0. \begin{align*} \int \frac{C b^{2} \cos \left (d x + c\right )^{2} + B b^{2} \cos \left (d x + c\right ) - C a^{2} + B a b}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C b \cos \left (d x + c\right ) - C a + B b\right )} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*b*cos(d*x + c) - C*a + B*b)*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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C b^{2} \cos \left (d x + c\right )^{2} + B b^{2} \cos \left (d x + c\right ) - C a^{2} + B a b}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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