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

Optimal. Leaf size=136 \[ \frac{A x \sqrt{\cos (c+d x)}}{2 \sqrt{b \cos (c+d x)}}+\frac{A \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 d \sqrt{b \cos (c+d x)}}-\frac{B \sin ^3(c+d x) \sqrt{\cos (c+d x)}}{3 d \sqrt{b \cos (c+d x)}}+\frac{B \sin (c+d x) \sqrt{\cos (c+d x)}}{d \sqrt{b \cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0556648, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {17, 2748, 2635, 8, 2633} \[ \frac{A x \sqrt{\cos (c+d x)}}{2 \sqrt{b \cos (c+d x)}}+\frac{A \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 d \sqrt{b \cos (c+d x)}}-\frac{B \sin ^3(c+d x) \sqrt{\cos (c+d x)}}{3 d \sqrt{b \cos (c+d x)}}+\frac{B \sin (c+d x) \sqrt{\cos (c+d x)}}{d \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*x*Sqrt[Cos[c + d*x]])/(2*Sqrt[b*Cos[c + d*x]]) + (B*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[b*Cos[c + d*x]
]) + (A*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[b*Cos[c + d*x]]) - (B*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 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

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

Mathematica [A]  time = 0.0983586, size = 69, normalized size = 0.51 \[ \frac{\sqrt{\cos (c+d x)} (3 A \sin (2 (c+d x))+6 A c+6 A d x+9 B \sin (c+d x)+B \sin (3 (c+d x)))}{12 d \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Cos[c + d*x]]*(6*A*c + 6*A*d*x + 9*B*Sin[c + d*x] + 3*A*Sin[2*(c + d*x)] + B*Sin[3*(c + d*x)]))/(12*d*Sq
rt[b*Cos[c + d*x]])

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Maple [A]  time = 0.325, size = 74, normalized size = 0.5 \begin{align*}{\frac{2\,B\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+3\,A\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +3\,A \left ( dx+c \right ) +4\,B\sin \left ( dx+c \right ) }{6\,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+B*cos(d*x+c))/(b*cos(d*x+c))^(1/2),x)

[Out]

1/6/d*cos(d*x+c)^(1/2)*(2*B*sin(d*x+c)*cos(d*x+c)^2+3*A*cos(d*x+c)*sin(d*x+c)+3*A*(d*x+c)+4*B*sin(d*x+c))/(b*c
os(d*x+c))^(1/2)

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Maxima [A]  time = 2.04084, size = 92, normalized size = 0.68 \begin{align*} \frac{\frac{3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A}{\sqrt{b}} + \frac{B{\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}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/12*(3*(2*d*x + 2*c + sin(2*d*x + 2*c))*A/sqrt(b) + B*(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))/d

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Fricas [A]  time = 1.68327, size = 645, normalized size = 4.74 \begin{align*} \left [-\frac{3 \, A \sqrt{-b} \cos \left (d x + c\right ) \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 ) - 2 \,{\left (2 \, B \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) + 4 \, B\right )} \sqrt{b \cos \left (d x + c\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{12 \, b d \cos \left (d x + c\right )}, \frac{3 \, A \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 ) \cos \left (d x + c\right ) +{\left (2 \, B \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) + 4 \, B\right )} \sqrt{b \cos \left (d x + c\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{6 \, b d \cos \left (d x + c\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[-1/12*(3*A*sqrt(-b)*cos(d*x + c)*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) - 2*(2*B*cos(d*x + c)^2 + 3*A*cos(d*x + c) + 4*B)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*si
n(d*x + c))/(b*d*cos(d*x + c)), 1/6*(3*A*sqrt(b)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c
)^(3/2)))*cos(d*x + c) + (2*B*cos(d*x + c)^2 + 3*A*cos(d*x + c) + 4*B)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))
*sin(d*x + c))/(b*d*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)**(5/2)*(A+B*cos(d*x+c))/(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 (B \cos \left (d x + c\right ) + 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+B*cos(d*x+c))/(b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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