∫ cos3 _ 2 (c + dx)∘ ________ b cos(c + dx) (A + C cos2(c + dx)) dx

3.90 \(\int \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)} (A+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=113 \[ \frac {x (4 A+3 C) \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {(4 A+3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}{8 d}+\frac {C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}{4 d} \]

[Out]

1/4*C*cos(d*x+c)^(5/2)*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d+1/8*(4*A+3*C)*x*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2)

+1/8*(4*A+3*C)*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 = 0.07, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {17, 3014, 2635, 8} \[ \frac {x (4 A+3 C) \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {(4 A+3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}{8 d}+\frac {C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}{4 d} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((4*A + 3*C)*x*Sqrt[b*Cos[c + d*x]])/(8*Sqrt[Cos[c + d*x]]) + ((4*A + 3*C)*Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d

*x]]*Sin[c + d*x])/(8*d) + (C*Cos[c + d*x]^(5/2)*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(4*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),

x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer

Q[2*n]

Rule 3014

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[

e + f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e + f*

x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && !LtQ[m, -1]

Rubi steps

\begin {align*} \int \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac {\sqrt {b \cos (c+d x)} \int \cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx}{\sqrt {\cos (c+d x)}}\\ &=\frac {C \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d}+\frac {\left ((4 A+3 C) \sqrt {b \cos (c+d x)}\right ) \int \cos ^2(c+d x) \, dx}{4 \sqrt {\cos (c+d x)}}\\ &=\frac {(4 A+3 C) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d}+\frac {C \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d}+\frac {\left ((4 A+3 C) \sqrt {b \cos (c+d x)}\right ) \int 1 \, dx}{8 \sqrt {\cos (c+d x)}}\\ &=\frac {(4 A+3 C) x \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {(4 A+3 C) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d}+\frac {C \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A] time = 0.18, size = 67, normalized size = 0.59 \[ \frac {\sqrt {b \cos (c+d x)} (4 (4 A+3 C) (c+d x)+8 (A+C) \sin (2 (c+d x))+C \sin (4 (c+d x)))}{32 d \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[b*Cos[c + d*x]]*(4*(4*A + 3*C)*(c + d*x) + 8*(A + C)*Sin[2*(c + d*x)] + C*Sin[4*(c + d*x)]))/(32*d*Sqrt[

Cos[c + d*x]])

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fricas [A] time = 0.53, size = 200, normalized size = 1.77 \[ \left [\frac {2 \, {\left (2 \, C \cos \left (d x + c\right )^{2} + 4 \, A + 3 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (4 \, A + 3 \, C\right )} \sqrt {-b} \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right )}{16 \, d}, \frac {{\left (2 \, C \cos \left (d x + c\right )^{2} + 4 \, A + 3 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (4 \, A + 3 \, C\right )} \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right )}{8 \, d}\right ] \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)*(A+C*cos(d*x+c)^2)*(b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/16*(2*(2*C*cos(d*x + c)^2 + 4*A + 3*C)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + (4*A + 3*C)*s

qrt(-b)*log(2*b*cos(d*x + c)^2 - 2*sqrt(b*cos(d*x + c))*sqrt(-b)*sqrt(cos(d*x + c))*sin(d*x + c) - b))/d, 1/8*

((2*C*cos(d*x + c)^2 + 4*A + 3*C)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + (4*A + 3*C)*sqrt(b)*a

rctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2))))/d]

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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(cos(d*x+c)^(3/2)*(A+C*cos(d*x+c)^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