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

Optimal. Leaf size=297 \[ \frac{2 \left (-6 a^2 C+21 a b B+25 b^2 C\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{105 b d}-\frac{2 \left (a^2-b^2\right ) \left (-6 a^2 C+21 a b B+25 b^2 C\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b^2 d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (21 a^2 b B-6 a^3 C+82 a b^2 C+63 b^3 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 (7 b B-2 a C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 b d}+\frac{2 C \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d} \]

[Out]

(2*(21*a^2*b*B + 63*b^3*B - 6*a^3*C + 82*a*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)*(21*a*b*B - 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*(21*
a*b*B - 6*a^2*C + 25*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(105*b*d) + (2*(7*b*B - 2*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 = 0.454838, antiderivative size = 297, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.206, Rules used = {3023, 2753, 2752, 2663, 2661, 2655, 2653} \[ \frac{2 \left (-6 a^2 C+21 a b B+25 b^2 C\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{105 b d}-\frac{2 \left (a^2-b^2\right ) \left (-6 a^2 C+21 a b B+25 b^2 C\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b^2 d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (21 a^2 b B-6 a^3 C+82 a b^2 C+63 b^3 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{105 b^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 (7 b B-2 a C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 b d}+\frac{2 C \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(21*a^2*b*B + 63*b^3*B - 6*a^3*C + 82*a*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)*(21*a*b*B - 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*(21*
a*b*B - 6*a^2*C + 25*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(105*b*d) + (2*(7*b*B - 2*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 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]

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 (B \cos (c+d x)+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{5 b C}{2}+\frac{1}{2} (7 b B-2 a C) \cos (c+d x)\right ) \, dx}{7 b}\\ &=\frac{2 (7 b B-2 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} b (21 b B+19 a C)+\frac{1}{4} \left (21 a b B-6 a^2 C+25 b^2 C\right ) \cos (c+d x)\right ) \, dx}{35 b}\\ &=\frac{2 \left (21 a b B-6 a^2 C+25 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 b d}+\frac{2 (7 b B-2 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 (84 a b B+51 a^2 C+25 b^2 C\right )+\frac{1}{8} \left (21 a^2 b B+63 b^3 B-6 a^3 C+82 a b^2 C\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{105 b}\\ &=\frac{2 \left (21 a b B-6 a^2 C+25 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 b d}+\frac{2 (7 b B-2 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 (21 a b B-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 (21 a^2 b B+63 b^3 B-6 a^3 C+82 a b^2 C\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{105 b^2}\\ &=\frac{2 \left (21 a b B-6 a^2 C+25 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 b d}+\frac{2 (7 b B-2 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 (21 a^2 b B+63 b^3 B-6 a^3 C+82 a 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 (21 a b B-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{2 \left (21 a^2 b B+63 b^3 B-6 a^3 C+82 a 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 (21 a b B-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 (21 a b B-6 a^2 C+25 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{105 b d}+\frac{2 (7 b B-2 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 = 1.09261, size = 233, normalized size = 0.78 \[ \frac{b (a+b \cos (c+d x)) \left (\left (12 a^2 C+168 a b B+115 b^2 C\right ) \sin (c+d x)+3 b (2 (8 a C+7 b B) \sin (2 (c+d x))+5 b C \sin (3 (c+d x)))\right )+4 \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \left (b^2 \left (51 a^2 C+84 a b B+25 b^2 C\right ) F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+\left (21 a^2 b B-6 a^3 C+82 a b^2 C+63 b^3 B\right ) \left ((a+b) E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )-a F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )\right )\right )}{210 b^2 d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(b^2*(84*a*b*B + 51*a^2*C + 25*b^2*C)*EllipticF[(c + d*x)/2, (2*b)/(a +
b)] + (21*a^2*b*B + 63*b^3*B - 6*a^3*C + 82*a*b^2*C)*((a + b)*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - a*Ellipt
icF[(c + d*x)/2, (2*b)/(a + b)])) + b*(a + b*Cos[c + d*x])*((168*a*b*B + 12*a^2*C + 115*b^2*C)*Sin[c + d*x] +
3*b*(2*(7*b*B + 8*a*C)*Sin[2*(c + d*x)] + 5*b*C*Sin[3*(c + d*x)])))/(210*b^2*d*Sqrt[a + b*Cos[c + d*x]])

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Maple [B]  time = 0.85, size = 1305, 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)*(B*cos(d*x+c)+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+(-168*B*b^4-312*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)+(252*B*a*b^3+168*B*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)+(-84*B*a^2*b^2-126*B*a*b^3-42*B*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)+21*B*EllipticE(cos(1/2*d*x
+1/2*c),(-2*b/(a-b))^(1/2))*(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)*a
^3*b-21*B*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(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)*a^2*b^2+63*B*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(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)*a*b^3-63*B*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(
a-b))^(1/2))*(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)*b^4-21*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*b+21*B*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)*Ell
ipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*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)*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)*Ellipti
cF(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)/si
n(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} + B \cos \left (d x + c\right )\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)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

Timed out