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

Optimal. Leaf size=119 \[ -\frac{b (A+2 C) \sin ^3(c+d x) \sqrt{b \cos (c+d x)}}{3 d \sqrt{\cos (c+d x)}}+\frac{b (A+C) \sin (c+d x) \sqrt{b \cos (c+d x)}}{d \sqrt{\cos (c+d x)}}+\frac{b C \sin ^5(c+d x) \sqrt{b \cos (c+d x)}}{5 d \sqrt{\cos (c+d x)}} \]

[Out]

(b*(A + C)*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) - (b*(A + 2*C)*Sqrt[b*Cos[c + d*x]]*Sin[c
 + d*x]^3)/(3*d*Sqrt[Cos[c + d*x]]) + (b*C*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x]^5)/(5*d*Sqrt[Cos[c + d*x]])

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Rubi [A]  time = 0.0575295, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086, Rules used = {17, 3013, 373} \[ -\frac{b (A+2 C) \sin ^3(c+d x) \sqrt{b \cos (c+d x)}}{3 d \sqrt{\cos (c+d x)}}+\frac{b (A+C) \sin (c+d x) \sqrt{b \cos (c+d x)}}{d \sqrt{\cos (c+d x)}}+\frac{b C \sin ^5(c+d x) \sqrt{b \cos (c+d x)}}{5 d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(A + C)*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) - (b*(A + 2*C)*Sqrt[b*Cos[c + d*x]]*Sin[c
 + d*x]^3)/(3*d*Sqrt[Cos[c + d*x]]) + (b*C*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x]^5)/(5*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]

Rule 373

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

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

Mathematica [A]  time = 0.243569, size = 70, normalized size = 0.59 \[ \frac{\sin (c+d x) (b \cos (c+d x))^{3/2} (4 (5 A+7 C) \cos (2 (c+d x))+100 A+3 C \cos (4 (c+d x))+89 C)}{120 d \cos ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Cos[c + d*x])^(3/2)*(100*A + 89*C + 4*(5*A + 7*C)*Cos[2*(c + d*x)] + 3*C*Cos[4*(c + d*x)])*Sin[c + d*x])/(
120*d*Cos[c + d*x]^(3/2))

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Maple [A]  time = 0.303, size = 70, normalized size = 0.6 \begin{align*}{\frac{ \left ( 3\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}+5\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+4\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+10\,A+8\,C \right ) \sin \left ( dx+c \right ) }{15\,d} \left ( b\cos \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}} \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15/d*(3*C*cos(d*x+c)^4+5*A*cos(d*x+c)^2+4*C*cos(d*x+c)^2+10*A+8*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 = 2.11192, size = 158, normalized size = 1.33 \begin{align*} \frac{20 \,{\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 )} A \sqrt{b} +{\left (3 \, b \sin \left (5 \, d x + 5 \, c\right ) + 25 \, b \sin \left (\frac{3}{5} \, \arctan \left (\sin \left (5 \, d x + 5 \, c\right ), \cos \left (5 \, d x + 5 \, c\right )\right )\right ) + 150 \, b \sin \left (\frac{1}{5} \, \arctan \left (\sin \left (5 \, d x + 5 \, c\right ), \cos \left (5 \, d x + 5 \, c\right )\right )\right )\right )} C \sqrt{b}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(20*(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))))*A*sqrt(b) + (3*b*sin
(5*d*x + 5*c) + 25*b*sin(3/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 150*b*sin(1/5*arctan2(sin(5*d*x +
5*c), cos(5*d*x + 5*c))))*C*sqrt(b))/d

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Fricas [A]  time = 1.52038, size = 182, normalized size = 1.53 \begin{align*} \frac{{\left (3 \, C b \cos \left (d x + c\right )^{4} +{\left (5 \, A + 4 \, C\right )} b \cos \left (d x + c\right )^{2} + 2 \,{\left (5 \, A + 4 \, C\right )} b\right )} \sqrt{b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d \sqrt{\cos \left (d x + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*C*b*cos(d*x + c)^4 + (5*A + 4*C)*b*cos(d*x + c)^2 + 2*(5*A + 4*C)*b)*sqrt(b*cos(d*x + c))*sin(d*x + c)
/(d*sqrt(cos(d*x + c)))

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

[Out]

Timed out