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]

((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)

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Rubi [A]  time = 0.072712, antiderivative size = 113, normalized size of antiderivative = 1., 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 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 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]

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 8

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

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*}

Mathematica [A]  time = 0.168615, 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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Maple [A]  time = 0.489, size = 88, normalized size = 0.8 \begin{align*}{\frac{2\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +4\,A\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +3\,C\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +4\,A \left ( dx+c \right ) +3\,C \left ( dx+c \right ) }{8\,d}\sqrt{b\cos \left ( dx+c \right ) }{\frac{1}{\sqrt{\cos \left ( dx+c \right ) }}}} \end{align*}

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]

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

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Maxima [A]  time = 2.05079, size = 101, normalized size = 0.89 \begin{align*} \frac{8 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A \sqrt{b} +{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (\frac{1}{2} \, \arctan \left (\sin \left (4 \, d x + 4 \, c\right ), \cos \left (4 \, d x + 4 \, c\right )\right )\right )\right )} C \sqrt{b}}{32 \, d} \end{align*}

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]

1/32*(8*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*sqrt(b) + (12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(1/2*arctan2(sin
(4*d*x + 4*c), cos(4*d*x + 4*c))))*C*sqrt(b))/d

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Fricas [A]  time = 1.74705, size = 549, normalized size = 4.86 \begin{align*} \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 ] \end{align*}

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

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