∫ ∘ _______ cos (c + dx) ∘ ________ b cos(c + dx) (A + B cos(c + dx) + C cos2(c + dx)) dx

3.290 \(\int \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=143 \[ \frac {(3 A+2 C) \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d \sqrt {\cos (c+d x)}}+\frac {B x \sqrt {b \cos (c+d x)}}{2 \sqrt {\cos (c+d x)}}+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}{2 d}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}{3 d} \]

[Out]

1/3*C*cos(d*x+c)^(3/2)*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d+1/2*B*x*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2)+1/3*(3*

A+2*C)*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)+1/2*B*sin(d*x+c)*cos(d*x+c)^(1/2)*(b*cos(d*x+c))^(1/

2)/d

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Rubi [A] time = 0.06, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {17, 3023, 2734} \[ \frac {(3 A+2 C) \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d \sqrt {\cos (c+d x)}}+\frac {B x \sqrt {b \cos (c+d x)}}{2 \sqrt {\cos (c+d x)}}+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}{2 d}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}{3 d} \]

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(B*x*Sqrt[b*Cos[c + d*x]])/(2*Sqrt[Cos[c + d*x]]) + ((3*A + 2*C)*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[

Cos[c + d*x]]) + (B*Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(2*d) + (C*Cos[c + d*x]^(3/2)*Sqrt[b

*Cos[c + d*x]]*Sin[c + d*x])/(3*d)

Rule 17

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(a^(m + 1/2)*b^(n - 1/2)*Sqrt[b*v])/Sqrt[a*v]

, Int[u*v^(m + n), x], x] /; FreeQ[{a, b, m}, x] && !IntegerQ[m] && IGtQ[n + 1/2, 0] && IntegerQ[m + n]

Rule 2734

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((2*a*c

+ b*d)*x)/2, x] + (-Simp[((b*c + a*d)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;

FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

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]

Rubi steps

\begin {align*} \int \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac {\sqrt {b \cos (c+d x)} \int \cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx}{\sqrt {\cos (c+d x)}}\\ &=\frac {C \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {\sqrt {b \cos (c+d x)} \int \cos (c+d x) (3 A+2 C+3 B \cos (c+d x)) \, dx}{3 \sqrt {\cos (c+d x)}}\\ &=\frac {B x \sqrt {b \cos (c+d x)}}{2 \sqrt {\cos (c+d x)}}+\frac {(3 A+2 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {B \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d}+\frac {C \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A] time = 0.19, size = 75, normalized size = 0.52 \[ \frac {\sqrt {b \cos (c+d x)} (3 (4 A+3 C) \sin (c+d x)+3 B \sin (2 (c+d x))+6 B c+6 B d x+C \sin (3 (c+d x)))}{12 d \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(Sqrt[b*Cos[c + d*x]]*(6*B*c + 6*B*d*x + 3*(4*A + 3*C)*Sin[c + d*x] + 3*B*Sin[2*(c + d*x)] + C*Sin[3*(c + d*x)

]))/(12*d*Sqrt[Cos[c + d*x]])

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fricas [A] time = 0.77, size = 236, normalized size = 1.65 \[ \left [\frac {3 \, B \sqrt {-b} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, {\left (2 \, C \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 6 \, A + 4 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )}, \frac {3 \, B \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (2 \, C \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 6 \, A + 4 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )}\right ] \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2)*(b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/12*(3*B*sqrt(-b)*cos(d*x + c)*log(2*b*cos(d*x + c)^2 - 2*sqrt(b*cos(d*x + c))*sqrt(-b)*sqrt(cos(d*x + c))*s

in(d*x + c) - b) + 2*(2*C*cos(d*x + c)^2 + 3*B*cos(d*x + c) + 6*A + 4*C)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c

))*sin(d*x + c))/(d*cos(d*x + c)), 1/6*(3*B*sqrt(b)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x

+ c)^(3/2)))*cos(d*x + c) + (2*C*cos(d*x + c)^2 + 3*B*cos(d*x + c) + 6*A + 4*C)*sqrt(b*cos(d*x + c))*sqrt(cos(

d*x + c))*sin(d*x + c))/(d*cos(d*x + c))]

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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2)*(b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/

x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check si