3.646 \(\int (a+b \cos (c+d x))^{3/2} (a^2-b^2 \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=246 \[ \frac{2 b \left (41 a^2-25 b^2\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{105 d}-\frac{2 \left (-66 a^2 b^2+41 a^4+25 b^4\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 d \sqrt{a+b \cos (c+d x)}}+\frac{4 a \left (73 a^2-41 b^2\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{2 b \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 d}+\frac{4 a b \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 d} \]

[Out]

(4*a*(73*a^2 - 41*b^2)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(105*d*Sqrt[(a + b*Cos[
c + d*x])/(a + b)]) - (2*(41*a^4 - 66*a^2*b^2 + 25*b^4)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)
/2, (2*b)/(a + b)])/(105*d*Sqrt[a + b*Cos[c + d*x]]) + (2*b*(41*a^2 - 25*b^2)*Sqrt[a + b*Cos[c + d*x]]*Sin[c +
d*x])/(105*d) + (4*a*b*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(35*d) - (2*b*(a + b*Cos[c + d*x])^(5/2)*Sin[
c + d*x])/(7*d)

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Rubi [A]  time = 0.45753, antiderivative size = 246, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.219, Rules used = {3016, 2753, 2752, 2663, 2661, 2655, 2653} \[ \frac{2 b \left (41 a^2-25 b^2\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{105 d}-\frac{2 \left (-66 a^2 b^2+41 a^4+25 b^4\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 d \sqrt{a+b \cos (c+d x)}}+\frac{4 a \left (73 a^2-41 b^2\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{2 b \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 d}+\frac{4 a b \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^(3/2)*(a^2 - b^2*Cos[c + d*x]^2),x]

[Out]

(4*a*(73*a^2 - 41*b^2)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(105*d*Sqrt[(a + b*Cos[
c + d*x])/(a + b)]) - (2*(41*a^4 - 66*a^2*b^2 + 25*b^4)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)
/2, (2*b)/(a + b)])/(105*d*Sqrt[a + b*Cos[c + d*x]]) + (2*b*(41*a^2 - 25*b^2)*Sqrt[a + b*Cos[c + d*x]]*Sin[c +
d*x])/(105*d) + (4*a*b*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(35*d) - (2*b*(a + b*Cos[c + d*x])^(5/2)*Sin[
c + d*x])/(7*d)

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]

Rule 2753

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[
b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*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] && GtQ[m, 0] && 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 (a+b \cos (c+d x))^{3/2} \left (a^2-b^2 \cos ^2(c+d x)\right ) \, dx &=-\int (-a+b \cos (c+d x)) (a+b \cos (c+d x))^{5/2} \, dx\\ &=-\frac{2 b (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}-\frac{2}{7} \int (a+b \cos (c+d x))^{3/2} \left (\frac{1}{2} \left (-7 a^2+5 b^2\right )-a b \cos (c+d x)\right ) \, dx\\ &=\frac{4 a b (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}-\frac{2 b (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}-\frac{4}{35} \int \sqrt{a+b \cos (c+d x)} \left (-\frac{1}{4} a \left (35 a^2-19 b^2\right )-\frac{1}{4} b \left (41 a^2-25 b^2\right ) \cos (c+d x)\right ) \, dx\\ &=\frac{2 b \left (41 a^2-25 b^2\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 d}+\frac{4 a b (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}-\frac{2 b (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}-\frac{8}{105} \int \frac{\frac{1}{8} \left (-105 a^4+16 a^2 b^2+25 b^4\right )-\frac{1}{4} a b \left (73 a^2-41 b^2\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx\\ &=\frac{2 b \left (41 a^2-25 b^2\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 d}+\frac{4 a b (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}-\frac{2 b (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac{1}{105} \left (2 a \left (73 a^2-41 b^2\right )\right ) \int \sqrt{a+b \cos (c+d x)} \, dx-\frac{1}{105} \left (41 a^4-66 a^2 b^2+25 b^4\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx\\ &=\frac{2 b \left (41 a^2-25 b^2\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 d}+\frac{4 a b (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}-\frac{2 b (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac{\left (2 a \left (73 a^2-41 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}{105 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{\left (\left (41 a^4-66 a^2 b^2+25 b^4\right ) \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}{105 \sqrt{a+b \cos (c+d x)}}\\ &=\frac{4 a \left (73 a^2-41 b^2\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{2 \left (41 a^4-66 a^2 b^2+25 b^4\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 d \sqrt{a+b \cos (c+d x)}}+\frac{2 b \left (41 a^2-25 b^2\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 d}+\frac{4 a b (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}-\frac{2 b (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}\\ \end{align*}

Mathematica [A]  time = 1.16437, size = 212, normalized size = 0.86 \[ \frac{-b \sin (c+d x) \left (\left (145 b^3-32 a^2 b\right ) \cos (c+d x)-128 a^3+78 a b^2 \cos (2 (c+d x))+178 a b^2+15 b^3 \cos (3 (c+d x))\right )-4 \left (-66 a^2 b^2+41 a^4+25 b^4\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+8 a \left (73 a^2 b+73 a^3-41 a b^2-41 b^3\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{210 d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^(3/2)*(a^2 - b^2*Cos[c + d*x]^2),x]

[Out]

(8*a*(73*a^3 + 73*a^2*b - 41*a*b^2 - 41*b^3)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2, (2*b)/(
a + b)] - 4*(41*a^4 - 66*a^2*b^2 + 25*b^4)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a
+ b)] - b*(-128*a^3 + 178*a*b^2 + (-32*a^2*b + 145*b^3)*Cos[c + d*x] + 78*a*b^2*Cos[2*(c + d*x)] + 15*b^3*Cos[
3*(c + d*x)])*Sin[c + d*x])/(210*d*Sqrt[a + b*Cos[c + d*x]])

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Maple [B]  time = 0.437, size = 824, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^(3/2)*(a^2-b^2*cos(d*x+c)^2),x)

[Out]

2/105*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*cos(1/2*d*x+1/2*c)^9*b^4+312*cos(1/2*d*
x+1/2*c)^7*a*b^3-600*cos(1/2*d*x+1/2*c)^7*b^4-32*cos(1/2*d*x+1/2*c)^5*a^2*b^2-624*cos(1/2*d*x+1/2*c)^5*a*b^3+6
40*cos(1/2*d*x+1/2*c)^5*b^4-64*cos(1/2*d*x+1/2*c)^3*a^3*b+48*cos(1/2*d*x+1/2*c)^3*a^2*b^2+440*cos(1/2*d*x+1/2*
c)^3*a*b^3-360*cos(1/2*d*x+1/2*c)^3*b^4+41*(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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^4-66*(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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b^2+25*(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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^4-146*(s
in(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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-
b))^(1/2))*a^4+146*(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)*EllipticE(cos(1/2
*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3*b+82*(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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b^2-82*(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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^3+64*cos(1/2*d*x+1/2*c)*a^3*
b-16*cos(1/2*d*x+1/2*c)*a^2*b^2-128*cos(1/2*d*x+1/2*c)*a*b^3+80*cos(1/2*d*x+1/2*c)*b^4)/(-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*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{\left (b^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )}{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^(3/2)*(a^2-b^2*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

-integrate((b^2*cos(d*x + c)^2 - a^2)*(b*cos(d*x + c) + a)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (b^{3} \cos \left (d x + c\right )^{3} + a b^{2} \cos \left (d x + c\right )^{2} - a^{2} b \cos \left (d x + c\right ) - a^{3}\right )} \sqrt{b \cos \left (d x + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^(3/2)*(a^2-b^2*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [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*cos(d*x+c))^(3/2)*(a^2-b^2*cos(d*x+c)^2),x, algorithm="giac")

[Out]

Timed out