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

Optimal. Leaf size=74 \[ \frac {(A+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {b \cos (c+d x)}}-\frac {C \sin ^3(c+d x) \sqrt {\cos (c+d x)}}{3 d \sqrt {b \cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {17, 3013} \[ \frac {(A+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {b \cos (c+d x)}}-\frac {C \sin ^3(c+d x) \sqrt {\cos (c+d x)}}{3 d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A + C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[b*Cos[c + d*x]]) - (C*Sqrt[Cos[c + d*x]]*Sin[c + d*x]^3)/(3*
d*Sqrt[b*Cos[c + d*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 3013

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Dist[f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 52, normalized size = 0.70 \[ \frac {\sin (c+d x) \sqrt {\cos (c+d x)} (6 A+C \cos (2 (c+d x))+5 C)}{6 d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Cos[c + d*x]]*(6*A + 5*C + C*Cos[2*(c + d*x)])*Sin[c + d*x])/(6*d*Sqrt[b*Cos[c + d*x]])

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fricas [A]  time = 0.43, size = 49, normalized size = 0.66 \[ \frac {{\left (C \cos \left (d x + c\right )^{2} + 3 \, A + 2 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, b d \sqrt {\cos \left (d x + c\right )}} \]

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]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{\sqrt {b \cos \left (d x + c\right )}}\,{d x} \]

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]

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

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maple [A]  time = 0.26, size = 47, normalized size = 0.64 \[ \frac {\left (C \left (\cos ^{2}\left (d x +c \right )\right )+3 A +2 C \right ) \left (\sqrt {\cos }\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d \sqrt {b \cos \left (d x +c \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(3/2)*(A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(1/2),x)

[Out]

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

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maxima [A]  time = 1.08, size = 57, normalized size = 0.77 \[ \frac {\frac {C {\left (\sin \left (3 \, d x + 3 \, c\right ) + 9 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )}}{\sqrt {b}} + \frac {12 \, A \sin \left (d x + c\right )}{\sqrt {b}}}{12 \, d} \]

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="maxima")

[Out]

1/12*(C*(sin(3*d*x + 3*c) + 9*sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))))/sqrt(b) + 12*A*sin(d*x + c
)/sqrt(b))/d

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mupad [B]  time = 0.95, size = 75, normalized size = 1.01 \[ \frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (12\,A\,\sin \left (2\,c+2\,d\,x\right )+10\,C\,\sin \left (2\,c+2\,d\,x\right )+C\,\sin \left (4\,c+4\,d\,x\right )\right )}{12\,b\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^(3/2)*(A + C*cos(c + d*x)^2))/(b*cos(c + d*x))^(1/2),x)

[Out]

(cos(c + d*x)^(1/2)*(b*cos(c + d*x))^(1/2)*(12*A*sin(2*c + 2*d*x) + 10*C*sin(2*c + 2*d*x) + C*sin(4*c + 4*d*x)
))/(12*b*d*(cos(2*c + 2*d*x) + 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)

[Out]

Timed out

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