3.486 \(\int \cos ^3(c+d x) \sqrt {a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=264 \[ \frac {2 \left (8 a^2+25 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{105 b^2 d}+\frac {2 a \left (8 a^2+19 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (8 a^4+17 a^2 b^2-25 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^3 d \sqrt {a+b \cos (c+d x)}}-\frac {8 a \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 b^2 d}+\frac {2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{3/2}}{7 b d} \]

[Out]

-8/35*a*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/b^2/d+2/7*cos(d*x+c)*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/b/d+2/105*(8*
a^2+25*b^2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b^2/d+2/105*a*(8*a^2+19*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)/b^3/d/((a+b*cos(d*x+
c))/(a+b))^(1/2)-2/105*(8*a^4+17*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^3/d/(a+b*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.41, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {2793, 3023, 2753, 2752, 2663, 2661, 2655, 2653} \[ \frac {2 \left (8 a^2+25 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{105 b^2 d}-\frac {2 \left (17 a^2 b^2+8 a^4-25 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^3 d \sqrt {a+b \cos (c+d x)}}+\frac {2 a \left (8 a^2+19 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {8 a \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 b^2 d}+\frac {2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{3/2}}{7 b d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*Sqrt[a + b*Cos[c + d*x]],x]

[Out]

(2*a*(8*a^2 + 19*b^2)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(105*b^3*d*Sqrt[(a + b*C
os[c + d*x])/(a + b)]) - (2*(8*a^4 + 17*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^3*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(8*a^2 + 25*b^2)*Sqrt[a + b*Cos[c + d*x]]*Sin[
c + d*x])/(105*b^2*d) - (8*a*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(35*b^2*d) + (2*Cos[c + d*x]*(a + b*Cos[
c + d*x])^(3/2)*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 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d
*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a*d
*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n -
2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] ||
(EqQ[a, 0] && NeQ[c, 0])))

Rule 3023

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

Rubi steps

\begin {align*} \int \cos ^3(c+d x) \sqrt {a+b \cos (c+d x)} \, dx &=\frac {2 \cos (c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 b d}+\frac {2 \int \sqrt {a+b \cos (c+d x)} \left (a+\frac {5}{2} b \cos (c+d x)-2 a \cos ^2(c+d x)\right ) \, dx}{7 b}\\ &=-\frac {8 a (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b^2 d}+\frac {2 \cos (c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 b d}+\frac {4 \int \sqrt {a+b \cos (c+d x)} \left (-\frac {a b}{2}+\frac {1}{4} \left (8 a^2+25 b^2\right ) \cos (c+d x)\right ) \, dx}{35 b^2}\\ &=\frac {2 \left (8 a^2+25 b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b^2 d}-\frac {8 a (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b^2 d}+\frac {2 \cos (c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 b d}+\frac {8 \int \frac {\frac {1}{8} b \left (2 a^2+25 b^2\right )+\frac {1}{8} a \left (8 a^2+19 b^2\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{105 b^2}\\ &=\frac {2 \left (8 a^2+25 b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b^2 d}-\frac {8 a (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b^2 d}+\frac {2 \cos (c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 b d}+\frac {\left (a \left (8 a^2+19 b^2\right )\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{105 b^3}-\frac {\left (8 a^4+17 a^2 b^2-25 b^4\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{105 b^3}\\ &=\frac {2 \left (8 a^2+25 b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b^2 d}-\frac {8 a (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b^2 d}+\frac {2 \cos (c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 b d}+\frac {\left (a \left (8 a^2+19 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 b^3 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (\left (8 a^4+17 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^3 \sqrt {a+b \cos (c+d x)}}\\ &=\frac {2 a \left (8 a^2+19 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^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (8 a^4+17 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^3 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (8 a^2+25 b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b^2 d}-\frac {8 a (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b^2 d}+\frac {2 \cos (c+d x) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 b d}\\ \end {align*}

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Mathematica [A]  time = 1.14, size = 214, normalized size = 0.81 \[ \frac {-4 \left (8 a^4+17 a^2 b^2-25 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )+b \sin (c+d x) \left (-16 a^3+\left (145 b^3-4 a^2 b\right ) \cos (c+d x)+36 a b^2 \cos (2 (c+d x))+136 a b^2+15 b^3 \cos (3 (c+d x))\right )+4 a \left (8 a^3+8 a^2 b+19 a b^2+19 b^3\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{210 b^3 d \sqrt {a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*Sqrt[a + b*Cos[c + d*x]],x]

[Out]

(4*a*(8*a^3 + 8*a^2*b + 19*a*b^2 + 19*b^3)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2, (2*b)/(a
+ b)] - 4*(8*a^4 + 17*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*(-16*a^3 + 136*a*b^2 + (-4*a^2*b + 145*b^3)*Cos[c + d*x] + 36*a*b^2*Cos[2*(c + d*x)] + 15*b^3*Cos[3*(c
+ d*x)])*Sin[c + d*x])/(210*b^3*d*Sqrt[a + b*Cos[c + d*x]])

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fricas [F]  time = 1.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \cos \left (d x + c\right ) + a} \cos \left (d x + c\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 1.09, size = 827, normalized size = 3.13 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+b*cos(d*x+c))^(1/2),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+144*cos(1/2*d
*x+1/2*c)^7*a*b^3-600*cos(1/2*d*x+1/2*c)^7*b^4-4*cos(1/2*d*x+1/2*c)^5*a^2*b^2-288*cos(1/2*d*x+1/2*c)^5*a*b^3+6
40*cos(1/2*d*x+1/2*c)^5*b^4-8*cos(1/2*d*x+1/2*c)^3*a^3*b+6*cos(1/2*d*x+1/2*c)^3*a^2*b^2+230*cos(1/2*d*x+1/2*c)
^3*a*b^3-360*cos(1/2*d*x+1/2*c)^3*b^4-8*(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^4-17*(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+8*(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^4-8*(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+19*(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)*El
lipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b^2-19*(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+8*cos(1/2*d*x+1/2*c)*a^3*b-2*cos(
1/2*d*x+1/2*c)*a^2*b^2-86*cos(1/2*d*x+1/2*c)*a*b^3+80*cos(1/2*d*x+1/2*c)*b^4)/b^3/(-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 \[ \int \sqrt {b \cos \left (d x + c\right ) + a} \cos \left (d x + c\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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