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

Optimal. Leaf size=225 \[ -\frac{2 \left (a^2-b^2\right ) (3 a B+5 A b) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (3 a^2 B+20 a A b+9 b^2 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 (3 a B+5 A b) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{15 d}+\frac{2 B \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d} \]

[Out]

(2*(20*a*A*b + 3*a^2*B + 9*b^2*B)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(15*b*d*Sqrt
[(a + b*Cos[c + d*x])/(a + b)]) - (2*(a^2 - b^2)*(5*A*b + 3*a*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[
(c + d*x)/2, (2*b)/(a + b)])/(15*b*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(5*A*b + 3*a*B)*Sqrt[a + b*Cos[c + d*x]]*S
in[c + d*x])/(15*d) + (2*B*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d)

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Rubi [A]  time = 0.352248, antiderivative size = 225, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac{2 \left (a^2-b^2\right ) (3 a B+5 A b) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (3 a^2 B+20 a A b+9 b^2 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 (3 a B+5 A b) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{15 d}+\frac{2 B \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(20*a*A*b + 3*a^2*B + 9*b^2*B)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(15*b*d*Sqrt
[(a + b*Cos[c + d*x])/(a + b)]) - (2*(a^2 - b^2)*(5*A*b + 3*a*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[
(c + d*x)/2, (2*b)/(a + b)])/(15*b*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(5*A*b + 3*a*B)*Sqrt[a + b*Cos[c + d*x]]*S
in[c + d*x])/(15*d) + (2*B*(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 (a+b \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx &=\frac{2 B (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac{2}{5} \int \sqrt{a+b \cos (c+d x)} \left (\frac{1}{2} (5 a A+3 b B)+\frac{1}{2} (5 A b+3 a B) \cos (c+d x)\right ) \, dx\\ &=\frac{2 (5 A b+3 a B) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 B (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac{4}{15} \int \frac{\frac{1}{4} \left (15 a^2 A+5 A b^2+12 a b B\right )+\frac{1}{4} \left (20 a A b+3 a^2 B+9 b^2 B\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx\\ &=\frac{2 (5 A b+3 a B) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 B (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}-\frac{\left (\left (a^2-b^2\right ) (5 A b+3 a B)\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{15 b}+\frac{\left (20 a A b+3 a^2 B+9 b^2 B\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{15 b}\\ &=\frac{2 (5 A b+3 a B) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 B (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac{\left (\left (20 a A b+3 a^2 B+9 b^2 B\right ) \sqrt{a+b \cos (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}} \, dx}{15 b \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{\left (\left (a^2-b^2\right ) (5 A b+3 a B) \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}{15 b \sqrt{a+b \cos (c+d x)}}\\ &=\frac{2 \left (20 a A b+3 a^2 B+9 b^2 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{2 \left (a^2-b^2\right ) (5 A b+3 a B) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b d \sqrt{a+b \cos (c+d x)}}+\frac{2 (5 A b+3 a B) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 B (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.733757, size = 203, normalized size = 0.9 \[ \frac{2 \left (b \left (15 a^2 A+12 a b B+5 A b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+\left (3 a^2 B+20 a A b+9 b^2 B\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \left ((a+b) E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )-a F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )\right )+b \sin (c+d x) (a+b \cos (c+d x)) (6 a B+5 A b+3 b B \cos (c+d x))\right )}{15 b d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*(15*a^2*A + 5*A*b^2 + 12*a*b*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)]
+ (20*a*A*b + 3*a^2*B + 9*b^2*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*((a + b)*EllipticE[(c + d*x)/2, (2*b)/(a
+ b)] - a*EllipticF[(c + d*x)/2, (2*b)/(a + b)]) + b*(a + b*Cos[c + d*x])*(5*A*b + 6*a*B + 3*b*B*Cos[c + d*x])
*Sin[c + d*x]))/(15*b*d*Sqrt[a + b*Cos[c + d*x]])

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Maple [B]  time = 4.003, size = 993, normalized size = 4.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*B*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^6+(20*A*b^3+36*B*a*b^2+24*B*b^3)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-10*A*a*b^2-10*A*b^3-12*B*a^2*b
-18*B*a*b^2-6*B*b^3)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-5*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(
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^2*b+5*A*b^3*(sin(1/2*d*
x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(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))+20*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(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-20*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(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*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/
(a-b)*sin(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+3*a*B*(sin(
1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(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^2+3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*Ellipt
icE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3-3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(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+9*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*
(-2*b/(a-b)*sin(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^2-9*B
*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(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)/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 )}{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

Timed out