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

[Out]

4/35*a*b*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/d-2/7*b*(a+b*cos(d*x+c))^(5/2)*sin(d*x+c)/d+2/105*b*(41*a^2-25*b^2)
*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/d+4/105*a*(73*a^2-41*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*E
llipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-2
/105*(41*a^4-66*a^2*b^2+25*b^4)*(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)/d/(a+b*cos(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.46, antiderivative size = 246, normalized size of antiderivative = 1.00, 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 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 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 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 (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 = 1.15, size = 212, normalized size = 0.86 \[ \frac {-4 \left (41 a^4-66 a^2 b^2+25 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-b \sin (c+d x) \left (-128 a^3+\left (145 b^3-32 a^2 b\right ) \cos (c+d x)+78 a b^2 \cos (2 (c+d x))+178 a b^2+15 b^3 \cos (3 (c+d x))\right )+8 a \left (73 a^3+73 a^2 b-41 a b^2-41 b^3\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{210 d \sqrt {a+b \cos (c+d x)}} \]

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]])

________________________________________________________________________________________

fricas [F]  time = 1.25, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (b^{3} \cos \left (d x + c\right )^{3} + a b^{2} \cos \left (d x + c\right )^{2} - a^{2} b \cos \left (d x + c\right ) - a^{3}\right )} \sqrt {b \cos \left (d x + c\right ) + a}, x\right ) \]

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)

________________________________________________________________________________________

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

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]

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

________________________________________________________________________________________

maple [B]  time = 2.60, size = 824, normalized size = 3.35 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (a^2-b^2\,{\cos \left (c+d\,x\right )}^2\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2} \,d x \]

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))^(3/2),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________