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

Optimal. Leaf size=258 \[ -\frac {2 \left (6 a^2-25 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{105 b d}-\frac {4 a \left (3 a^2-41 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \left (6 a^4-31 a^2 b^2+25 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d}-\frac {4 a \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 b d} \]

[Out]

-4/35*a*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/b/d+2/7*(a+b*cos(d*x+c))^(5/2)*sin(d*x+c)/b/d-2/105*(6*a^2-25*b^2)*s
in(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b/d-4/105*a*(3*a^2-41*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*El
lipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/b^2/d/((a+b*cos(d*x+c))/(a+b))^(1/2
)+2/105*(6*a^4-31*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)/b^2/d/(a+b*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.39, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2791, 2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac {2 \left (6 a^2-25 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{105 b d}+\frac {2 \left (-31 a^2 b^2+6 a^4+25 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^2 d \sqrt {a+b \cos (c+d x)}}-\frac {4 a \left (3 a^2-41 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d}-\frac {4 a \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a*(3*a^2 - 41*b^2)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(105*b^2*d*Sqrt[(a + b*
Cos[c + d*x])/(a + b)]) + (2*(6*a^4 - 31*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*b^2*d*Sqrt[a + b*Cos[c + d*x]]) - (2*(6*a^2 - 25*b^2)*Sqrt[a + b*Cos[c + d*x]]*Sin
[c + d*x])/(105*b*d) - (4*a*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(35*b*d) + (2*(a + b*Cos[c + d*x])^(5/2)*
Sin[c + d*x])/(7*b*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 2791

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> -Simp[
(d^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[(a + b*Sin[e + f*x
])^m*Simp[b*(d^2*(m + 1) + c^2*(m + 2)) - d*(a*d - 2*b*c*(m + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+b \cos (c+d x))^{3/2} \, dx &=\frac {2 (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}+\frac {2 \int \left (\frac {5 b}{2}-a \cos (c+d x)\right ) (a+b \cos (c+d x))^{3/2} \, dx}{7 b}\\ &=-\frac {4 a (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b d}+\frac {2 (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}+\frac {4 \int \sqrt {a+b \cos (c+d x)} \left (\frac {19 a b}{4}-\frac {1}{4} \left (6 a^2-25 b^2\right ) \cos (c+d x)\right ) \, dx}{35 b}\\ &=-\frac {2 \left (6 a^2-25 b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b d}-\frac {4 a (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b d}+\frac {2 (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}+\frac {8 \int \frac {\frac {1}{8} b \left (51 a^2+25 b^2\right )-\frac {1}{4} a \left (3 a^2-41 b^2\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{105 b}\\ &=-\frac {2 \left (6 a^2-25 b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b d}-\frac {4 a (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b d}+\frac {2 (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}+\frac {1}{105} \left (2 a \left (41-\frac {3 a^2}{b^2}\right )\right ) \int \sqrt {a+b \cos (c+d x)} \, dx+\frac {\left (6 a^4-31 a^2 b^2+25 b^4\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{105 b^2}\\ &=-\frac {2 \left (6 a^2-25 b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b d}-\frac {4 a (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b d}+\frac {2 (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}+\frac {\left (2 a \left (41-\frac {3 a^2}{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 (6 a^4-31 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 b^2 \sqrt {a+b \cos (c+d x)}}\\ &=\frac {4 a \left (41-\frac {3 a^2}{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 (6 a^4-31 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 b^2 d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (6 a^2-25 b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b d}-\frac {4 a (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b d}+\frac {2 (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}\\ \end {align*}

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Mathematica [A]  time = 1.15, size = 214, normalized size = 0.83 \[ \frac {4 \left (6 a^4-31 a^2 b^2+25 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )+b \sin (c+d x) \left (12 a^3+b \left (108 a^2+145 b^2\right ) \cos (c+d x)+78 a b^2 \cos (2 (c+d x))+178 a b^2+15 b^3 \cos (3 (c+d x))\right )-8 a \left (3 a^3+3 a^2 b-41 a b^2-41 b^3\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{210 b^2 d \sqrt {a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a*(3*a^3 + 3*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*(6*a^4 - 31*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*(12*a^3 + 178*a*b^2 + b*(108*a^2 + 145*b^2)*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*b^2*d*Sqrt[a + b*Cos[c + d*x]])

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fricas [F]  time = 1.33, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )^{2}\right )} \sqrt {b \cos \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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maple [B]  time = 0.70, size = 827, normalized size = 3.21 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+b*cos(d*x+c))^(3/2),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+108*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
+640*cos(1/2*d*x+1/2*c)^5*b^4+6*cos(1/2*d*x+1/2*c)^3*a^3*b-162*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+6*(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^4-31*(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^4-6*(si
n(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+6*(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^3-6*cos(1/2*d*x+1/2*c)*a^3*b+54
*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)/b^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)/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 \[ \int {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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