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

Optimal. Leaf size=199 \[ -\frac{2 a \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{5 b d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (a^2+3 b^2\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{5 b d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}+\frac{2 a \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.247547, antiderivative size = 199, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{2 a \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{5 b d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (a^2+3 b^2\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{5 b d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}+\frac{2 a \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a^2 + 3*b^2)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(5*b*d*Sqrt[(a + b*Cos[c + d*
x])/(a + b)]) - (2*a*(a^2 - b^2)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(5*
b*d*Sqrt[a + b*Cos[c + d*x]]) + (2*a*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(5*d) + (2*(a + b*Cos[c + d*x])^(3
/2)*Sin[c + d*x])/(5*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))^{3/2} \, dx &=\frac{2 (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac{2}{5} \int \left (\frac{3 b}{2}+\frac{3}{2} a \cos (c+d x)\right ) \sqrt{a+b \cos (c+d x)} \, dx\\ &=\frac{2 a \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 d}+\frac{2 (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac{4}{15} \int \frac{3 a b+\frac{3}{4} \left (a^2+3 b^2\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx\\ &=\frac{2 a \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 d}+\frac{2 (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}-\frac{\left (a \left (a^2-b^2\right )\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{5 b}+\frac{\left (a^2+3 b^2\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{5 b}\\ &=\frac{2 a \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 d}+\frac{2 (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac{\left (\left (a^2+3 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}{5 b \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{\left (a \left (a^2-b^2\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}{5 b \sqrt{a+b \cos (c+d x)}}\\ &=\frac{2 \left (a^2+3 b^2\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{5 b d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{2 a \left (a^2-b^2\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 )}{5 b d \sqrt{a+b \cos (c+d x)}}+\frac{2 a \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 d}+\frac{2 (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.777986, size = 174, normalized size = 0.87 \[ \frac{b \sin (c+d x) \left (4 a^2+6 a b \cos (c+d x)+b^2 \cos (2 (c+d x))+b^2\right )-2 a \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+2 \left (a^2 b+a^3+3 a b^2+3 b^3\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{5 b d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a^3 + a^2*b + 3*a*b^2 + 3*b^3)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2, (2*b)/(a + b)] -
2*a*(a^2 - b^2)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)] + b*(4*a^2 + b^2 + 6*
a*b*Cos[c + d*x] + b^2*Cos[2*(c + d*x)])*Sin[c + d*x])/(5*b*d*Sqrt[a + b*Cos[c + d*x]])

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Maple [B]  time = 2.925, size = 663, 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))^(3/2),x)

[Out]

-2/5*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(8*cos(1/2*d*x+1/2*c)^7*b^3+12*cos(1/2*d*x+1/
2*c)^5*a*b^2-16*cos(1/2*d*x+1/2*c)^5*b^3+4*cos(1/2*d*x+1/2*c)^3*a^2*b-18*cos(1/2*d*x+1/2*c)^3*a*b^2+10*cos(1/2
*d*x+1/2*c)^3*b^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^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)*E
llipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^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^2*b+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*b^2
-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))*b^3-4*cos(1/2*d*x+1/2*c)*a^2*b+6*cos(1/2*d*x+1/2*c)*a*b^2-2*cos(1/2*d*x+1/2*c)*b^3)/b/(-2*b*si
n(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{3}{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))^(3/2),x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

Timed out