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

Optimal. Leaf size=76 \[ \frac {b (A+C) \sin (c+d x) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {b C \sin ^3(c+d x) \sqrt {b \cos (c+d x)}}{3 d \sqrt {\cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 76, 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} \[ \frac {b (A+C) \sin (c+d x) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {b C \sin ^3(c+d x) \sqrt {b \cos (c+d x)}}{3 d \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.03, size = 53, normalized size = 0.70 \[ \frac {b \sin (c+d x) \sqrt {b \cos (c+d x)} (6 A+C \cos (2 (c+d x))+5 C)}{6 d \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 50, normalized size = 0.66 \[ \frac {{\left (C b \cos \left (d x + c\right )^{2} + {\left (3 \, A + 2 \, C\right )} b\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, d \sqrt {\cos \left (d x + c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.22, size = 47, normalized size = 0.62 \[ \frac {\left (C \left (\cos ^{2}\left (d x +c \right )\right )+3 A +2 C \right ) \left (b \cos \left (d x +c \right )\right )^{\frac {3}{2}} \sin \left (d x +c \right )}{3 d \cos \left (d x +c \right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.08, size = 60, normalized size = 0.79 \[ \frac {12 \, A b^{\frac {3}{2}} \sin \left (d x + c\right ) + {\left (b \sin \left (3 \, d x + 3 \, c\right ) + 9 \, b \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )} C \sqrt {b}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*A*b^(3/2)*sin(d*x + c) + (b*sin(3*d*x + 3*c) + 9*b*sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)
)))*C*sqrt(b))/d

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mupad [B]  time = 0.60, size = 54, normalized size = 0.71 \[ \frac {b\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (12\,A\,\sin \left (c+d\,x\right )+9\,C\,\sin \left (c+d\,x\right )+C\,\sin \left (3\,c+3\,d\,x\right )\right )}{12\,d\,\sqrt {\cos \left (c+d\,x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*(b*cos(c + d*x))^(1/2)*(12*A*sin(c + d*x) + 9*C*sin(c + d*x) + C*sin(3*c + 3*d*x)))/(12*d*cos(c + d*x)^(1/2
))

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

[Out]

Timed out

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