3.502 \(\int \cos (c+d x) (a+b \cos (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=249 \[ \frac{2 \left (3 a^2+5 b^2\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{21 d}-\frac{2 \left (2 a^2 b^2+3 a^4-5 b^4\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{21 b d \sqrt{a+b \cos (c+d x)}}+\frac{2 a \left (3 a^2+29 b^2\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{21 b d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 d}+\frac{2 a \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{7 d} \]

[Out]

(2*a*(3*a^2 + 29*b^2)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(21*b*d*Sqrt[(a + b*Cos[
c + d*x])/(a + b)]) - (2*(3*a^4 + 2*a^2*b^2 - 5*b^4)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2,
(2*b)/(a + b)])/(21*b*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(3*a^2 + 5*b^2)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])
/(21*d) + (2*a*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(7*d) + (2*(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7
*d)

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Rubi [A]  time = 0.359508, antiderivative size = 249, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2753, 2752, 2663, 2661, 2655, 2653} \[ \frac{2 \left (3 a^2+5 b^2\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{21 d}-\frac{2 \left (2 a^2 b^2+3 a^4-5 b^4\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{21 b d \sqrt{a+b \cos (c+d x)}}+\frac{2 a \left (3 a^2+29 b^2\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{21 b d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 d}+\frac{2 a \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*(a + b*Cos[c + d*x])^(5/2),x]

[Out]

(2*a*(3*a^2 + 29*b^2)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(21*b*d*Sqrt[(a + b*Cos[
c + d*x])/(a + b)]) - (2*(3*a^4 + 2*a^2*b^2 - 5*b^4)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2,
(2*b)/(a + b)])/(21*b*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(3*a^2 + 5*b^2)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])
/(21*d) + (2*a*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(7*d) + (2*(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7
*d)

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 \cos (c+d x) (a+b \cos (c+d x))^{5/2} \, dx &=\frac{2 (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac{2}{7} \int \left (\frac{5 b}{2}+\frac{5}{2} a \cos (c+d x)\right ) (a+b \cos (c+d x))^{3/2} \, dx\\ &=\frac{2 a (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 d}+\frac{2 (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 (10 a b+\frac{5}{4} \left (3 a^2+5 b^2\right ) \cos (c+d x)\right ) \, dx\\ &=\frac{2 \left (3 a^2+5 b^2\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac{2 a (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 d}+\frac{2 (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac{8}{105} \int \frac{\frac{5}{8} b \left (27 a^2+5 b^2\right )+\frac{5}{8} a \left (3 a^2+29 b^2\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx\\ &=\frac{2 \left (3 a^2+5 b^2\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac{2 a (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 d}+\frac{2 (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac{\left (a \left (3 a^2+29 b^2\right )\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{21 b}-\frac{\left (3 a^4+2 a^2 b^2-5 b^4\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{21 b}\\ &=\frac{2 \left (3 a^2+5 b^2\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac{2 a (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 d}+\frac{2 (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac{\left (a \left (3 a^2+29 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}{21 b \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{\left (\left (3 a^4+2 a^2 b^2-5 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}{21 b \sqrt{a+b \cos (c+d x)}}\\ &=\frac{2 a \left (3 a^2+29 b^2\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{21 b d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{2 \left (3 a^4+2 a^2 b^2-5 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 )}{21 b d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (3 a^2+5 b^2\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac{2 a (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{7 d}+\frac{2 (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}\\ \end{align*}

Mathematica [A]  time = 0.886519, size = 214, normalized size = 0.86 \[ \frac{b \sin (c+d x) \left (b \left (72 a^2+29 b^2\right ) \cos (c+d x)+36 a^3+24 a b^2 \cos (2 (c+d x))+44 a b^2+3 b^3 \cos (3 (c+d x))\right )-4 \left (2 a^2 b^2+3 a^4-5 b^4\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+4 a \left (3 a^2 b+3 a^3+29 a b^2+29 b^3\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{42 b d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*(a + b*Cos[c + d*x])^(5/2),x]

[Out]

(4*a*(3*a^3 + 3*a^2*b + 29*a*b^2 + 29*b^3)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2, (2*b)/(a
+ b)] - 4*(3*a^4 + 2*a^2*b^2 - 5*b^4)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)]
+ b*(36*a^3 + 44*a*b^2 + b*(72*a^2 + 29*b^2)*Cos[c + d*x] + 24*a*b^2*Cos[2*(c + d*x)] + 3*b^3*Cos[3*(c + d*x)
])*Sin[c + d*x])/(42*b*d*Sqrt[a + b*Cos[c + d*x]])

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Maple [B]  time = 2.899, size = 827, normalized size = 3.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(a+b*cos(d*x+c))^(5/2),x)

[Out]

-2/21*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(48*cos(1/2*d*x+1/2*c)^9*b^4+96*cos(1/2*d*x+
1/2*c)^7*a*b^3-120*cos(1/2*d*x+1/2*c)^7*b^4+72*cos(1/2*d*x+1/2*c)^5*a^2*b^2-192*cos(1/2*d*x+1/2*c)^5*a*b^3+128
*cos(1/2*d*x+1/2*c)^5*b^4+18*cos(1/2*d*x+1/2*c)^3*a^3*b-108*cos(1/2*d*x+1/2*c)^3*a^2*b^2+130*cos(1/2*d*x+1/2*c
)^3*a*b^3-72*cos(1/2*d*x+1/2*c)^3*b^4+3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(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^4-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(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^3*b+29*(sin(1/2*d*x+1/2*c)^2)^(1/2)*
((2*b*cos(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^2*b^2-29*(sin(
1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(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*b^3-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*
x+1/2*c),(-2*b/(a-b))^(1/2))*a^4-2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)*E
llipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b^2+5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c
)^2+a-b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^4-18*cos(1/2*d*x+1/2*c)*a^3*b+36*cos(
1/2*d*x+1/2*c)*a^2*b^2-34*cos(1/2*d*x+1/2*c)*a*b^3+16*cos(1/2*d*x+1/2*c)*b^4)/b/(-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 \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c) + a)^(5/2)*cos(d*x + c), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{2} \cos \left (d x + c\right )^{3} + 2 \, a b \cos \left (d x + c\right )^{2} + a^{2} \cos \left (d x + c\right )\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(cos(d*x+c)*(a+b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral((b^2*cos(d*x + c)^3 + 2*a*b*cos(d*x + c)^2 + a^2*cos(d*x + c))*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(cos(d*x+c)*(a+b*cos(d*x+c))**(5/2),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(cos(d*x+c)*(a+b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out