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

Optimal. Leaf size=285 \[ -\frac{2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{105 b d}-\frac{2 \left (a^2-b^2\right ) \left (-6 a^2 C+35 A b^2+25 b^2 C\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b^2 d \sqrt{a+b \cos (c+d x)}}+\frac{4 a \left (-3 a^2 C+70 A b^2+41 b^2 C\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 C \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d}-\frac{4 a C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 b d} \]

[Out]

(4*a*(70*A*b^2 - 3*a^2*C + 41*b^2*C)*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*(a^2 - b^2)*(35*A*b^2 - 6*a^2*C + 25*b^2*C)*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*C - 5*b^2*
(7*A + 5*C))*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(105*b*d) - (4*a*C*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x]
)/(35*b*d) + (2*C*(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*b*d)

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Rubi [A]  time = 0.469181, antiderivative size = 285, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {3024, 2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac{2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{105 b d}-\frac{2 \left (a^2-b^2\right ) \left (-6 a^2 C+35 A b^2+25 b^2 C\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b^2 d \sqrt{a+b \cos (c+d x)}}+\frac{4 a \left (-3 a^2 C+70 A b^2+41 b^2 C\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 C \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d}-\frac{4 a C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*(70*A*b^2 - 3*a^2*C + 41*b^2*C)*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*(a^2 - b^2)*(35*A*b^2 - 6*a^2*C + 25*b^2*C)*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*C - 5*b^2*
(7*A + 5*C))*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(105*b*d) - (4*a*C*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x]
)/(35*b*d) + (2*C*(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*b*d)

Rule 3024

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

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} \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac{2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}+\frac{2 \int (a+b \cos (c+d x))^{3/2} \left (\frac{1}{2} b (7 A+5 C)-a C \cos (c+d x)\right ) \, dx}{7 b}\\ &=-\frac{4 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b d}+\frac{2 C (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{1}{4} a b (35 A+19 C)-\frac{1}{4} \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \cos (c+d x)\right ) \, dx}{35 b}\\ &=-\frac{2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 b d}-\frac{4 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b d}+\frac{2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}+\frac{8 \int \frac{\frac{1}{8} b \left (5 b^2 (7 A+5 C)+3 a^2 (35 A+17 C)\right )+\frac{1}{4} a \left (70 A b^2-3 a^2 C+41 b^2 C\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{105 b}\\ &=-\frac{2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 b d}-\frac{4 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b d}+\frac{2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}-\frac{\left (\left (a^2-b^2\right ) \left (35 A b^2-6 a^2 C+25 b^2 C\right )\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{105 b^2}+\frac{\left (2 a \left (70 A b^2-3 a^2 C+41 b^2 C\right )\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{105 b^2}\\ &=-\frac{2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 b d}-\frac{4 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b d}+\frac{2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}+\frac{\left (2 a \left (70 A b^2-3 a^2 C+41 b^2 C\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^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{\left (\left (a^2-b^2\right ) \left (35 A b^2-6 a^2 C+25 b^2 C\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 (70 A b^2-3 a^2 C+41 b^2 C\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 (a^2-b^2\right ) \left (35 A b^2-6 a^2 C+25 b^2 C\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 C-5 b^2 (7 A+5 C)\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 b d}-\frac{4 a C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b d}+\frac{2 C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}\\ \end{align*}

Mathematica [A]  time = 0.836575, size = 224, normalized size = 0.79 \[ \frac{2 b \sin (c+d x) (a+b \cos (c+d x)) \left (6 a^2 C+48 a b C \cos (c+d x)+70 A b^2+15 b^2 C \cos (2 (c+d x))+65 b^2 C\right )+4 \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \left (b^2 \left (3 a^2 (35 A+17 C)+5 b^2 (7 A+5 C)\right ) F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )-2 a \left (3 a^2 C-70 A b^2-41 b^2 C\right ) \left ((a+b) E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )-a F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )\right )\right )}{210 b^2 d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(b^2*(5*b^2*(7*A + 5*C) + 3*a^2*(35*A + 17*C))*EllipticF[(c + d*x)/2, (2
*b)/(a + b)] - 2*a*(-70*A*b^2 + 3*a^2*C - 41*b^2*C)*((a + b)*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - a*Ellipti
cF[(c + d*x)/2, (2*b)/(a + b)])) + 2*b*(a + b*Cos[c + d*x])*(70*A*b^2 + 6*a^2*C + 65*b^2*C + 48*a*b*C*Cos[c +
d*x] + 15*b^2*C*Cos[2*(c + d*x)])*Sin[c + d*x])/(210*b^2*d*Sqrt[a + b*Cos[c + d*x]])

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Maple [B]  time = 0.403, size = 1131, normalized size = 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+C*cos(d*x+c)^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*C*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1
/2*c)^8+(-312*C*a*b^3-360*C*b^4)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(140*A*b^4+108*C*a^2*b^2+312*C*a*b^3+
280*C*b^4)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-70*A*a*b^3-70*A*b^4-6*C*a^3*b-54*C*a^2*b^2-128*C*a*b^3-80
*C*b^4)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+140*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^2-140*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^3-35*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^2+35*A*b^4*(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))-6*C*(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^4+6*C*(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^3*b+82*C*(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^2-82*C*(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^3+6*C*(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^4-31*C
*(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^2+25*C*b^4*(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/(-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 (C \cos \left (d x + c\right )^{2} + 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+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

Timed out