3.116 \(\int \frac{\cos ^{\frac{5}{2}}(c+d x) (A+C \cos ^2(c+d x))}{\sqrt{b \cos (c+d x)}} \, dx\)

Optimal. Leaf size=113 \[ \frac{x (4 A+3 C) \sqrt{\cos (c+d x)}}{8 \sqrt{b \cos (c+d x)}}+\frac{(4 A+3 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{8 d \sqrt{b \cos (c+d x)}}+\frac{C \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x)}{4 d \sqrt{b \cos (c+d x)}} \]

[Out]

((4*A + 3*C)*x*Sqrt[Cos[c + d*x]])/(8*Sqrt[b*Cos[c + d*x]]) + ((4*A + 3*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(8
*d*Sqrt[b*Cos[c + d*x]]) + (C*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(4*d*Sqrt[b*Cos[c + d*x]])

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Rubi [A]  time = 0.0593178, 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} \[ \frac{x (4 A+3 C) \sqrt{\cos (c+d x)}}{8 \sqrt{b \cos (c+d x)}}+\frac{(4 A+3 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{8 d \sqrt{b \cos (c+d x)}}+\frac{C \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x)}{4 d \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^(5/2)*(A + C*Cos[c + d*x]^2))/Sqrt[b*Cos[c + d*x]],x]

[Out]

((4*A + 3*C)*x*Sqrt[Cos[c + d*x]])/(8*Sqrt[b*Cos[c + d*x]]) + ((4*A + 3*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(8
*d*Sqrt[b*Cos[c + d*x]]) + (C*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(4*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 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 \frac{\cos ^{\frac{5}{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 ^2(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx}{\sqrt{b \cos (c+d x)}}\\ &=\frac{C \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{4 d \sqrt{b \cos (c+d x)}}+\frac{\left ((4 A+3 C) \sqrt{\cos (c+d x)}\right ) \int \cos ^2(c+d x) \, dx}{4 \sqrt{b \cos (c+d x)}}\\ &=\frac{(4 A+3 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{8 d \sqrt{b \cos (c+d x)}}+\frac{C \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{4 d \sqrt{b \cos (c+d x)}}+\frac{\left ((4 A+3 C) \sqrt{\cos (c+d x)}\right ) \int 1 \, dx}{8 \sqrt{b \cos (c+d x)}}\\ &=\frac{(4 A+3 C) x \sqrt{\cos (c+d x)}}{8 \sqrt{b \cos (c+d x)}}+\frac{(4 A+3 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{8 d \sqrt{b \cos (c+d x)}}+\frac{C \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{4 d \sqrt{b \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.133435, size = 67, normalized size = 0.59 \[ \frac{\sqrt{\cos (c+d x)} (4 (4 A+3 C) (c+d x)+8 (A+C) \sin (2 (c+d x))+C \sin (4 (c+d x)))}{32 d \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^(5/2)*(A + C*Cos[c + d*x]^2))/Sqrt[b*Cos[c + d*x]],x]

[Out]

(Sqrt[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[b*
Cos[c + d*x]])

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Maple [A]  time = 0.506, 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{\cos \left ( dx+c \right ) }{\frac{1}{\sqrt{b\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(5/2)*(A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(1/2),x)

[Out]

1/8/d*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))/(b*cos(d*x+c))^(1/2)

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Maxima [A]  time = 2.1218, size = 101, normalized size = 0.89 \begin{align*} \frac{\frac{8 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A}{\sqrt{b}} + \frac{{\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)^(5/2)*(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 = 1.80576, size = 560, normalized size = 4.96 \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 \, b 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 \, b d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(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))/(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)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2))))/(b*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)**(5/2)*(A+C*cos(d*x+c)**2)/(b*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac{5}{2}}}{\sqrt{b \cos \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(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)^(5/2)/sqrt(b*cos(d*x + c)), x)