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

Optimal. Leaf size=223 \[ -\frac{(5 A+4 C) \sin ^3(c+d x) \sqrt{b \cos (c+d x)}}{15 d \sqrt{\cos (c+d x)}}+\frac{(5 A+4 C) \sin (c+d x) \sqrt{b \cos (c+d x)}}{5 d \sqrt{\cos (c+d x)}}+\frac{3 B x \sqrt{b \cos (c+d x)}}{8 \sqrt{\cos (c+d x)}}+\frac{B \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) \sqrt{b \cos (c+d x)}}{4 d}+\frac{3 B \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}}{8 d}+\frac{C \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x) \sqrt{b \cos (c+d x)}}{5 d} \]

[Out]

(3*B*x*Sqrt[b*Cos[c + d*x]])/(8*Sqrt[Cos[c + d*x]]) + ((5*A + 4*C)*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(5*d*Sqr
t[Cos[c + d*x]]) + (3*B*Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(8*d) + (B*Cos[c + d*x]^(5/2)*Sq
rt[b*Cos[c + d*x]]*Sin[c + d*x])/(4*d) + (C*Cos[c + d*x]^(7/2)*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(5*d) - ((5*
A + 4*C)*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x]^3)/(15*d*Sqrt[Cos[c + d*x]])

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Rubi [A]  time = 0.12537, antiderivative size = 223, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.14, Rules used = {17, 3023, 2748, 2633, 2635, 8} \[ -\frac{(5 A+4 C) \sin ^3(c+d x) \sqrt{b \cos (c+d x)}}{15 d \sqrt{\cos (c+d x)}}+\frac{(5 A+4 C) \sin (c+d x) \sqrt{b \cos (c+d x)}}{5 d \sqrt{\cos (c+d x)}}+\frac{3 B x \sqrt{b \cos (c+d x)}}{8 \sqrt{\cos (c+d x)}}+\frac{B \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) \sqrt{b \cos (c+d x)}}{4 d}+\frac{3 B \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}}{8 d}+\frac{C \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x) \sqrt{b \cos (c+d x)}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*B*x*Sqrt[b*Cos[c + d*x]])/(8*Sqrt[Cos[c + d*x]]) + ((5*A + 4*C)*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(5*d*Sqr
t[Cos[c + d*x]]) + (3*B*Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(8*d) + (B*Cos[c + d*x]^(5/2)*Sq
rt[b*Cos[c + d*x]]*Sin[c + d*x])/(4*d) + (C*Cos[c + d*x]^(7/2)*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(5*d) - ((5*
A + 4*C)*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x]^3)/(15*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 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

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 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]

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{5}{2}}(c+d x) \sqrt{b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac{\sqrt{b \cos (c+d x)} \int \cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx}{\sqrt{\cos (c+d x)}}\\ &=\frac{C \cos ^{\frac{7}{2}}(c+d x) \sqrt{b \cos (c+d x)} \sin (c+d x)}{5 d}+\frac{\sqrt{b \cos (c+d x)} \int \cos ^3(c+d x) (5 A+4 C+5 B \cos (c+d x)) \, dx}{5 \sqrt{\cos (c+d x)}}\\ &=\frac{C \cos ^{\frac{7}{2}}(c+d x) \sqrt{b \cos (c+d x)} \sin (c+d x)}{5 d}+\frac{\left (B \sqrt{b \cos (c+d x)}\right ) \int \cos ^4(c+d x) \, dx}{\sqrt{\cos (c+d x)}}+\frac{\left ((5 A+4 C) \sqrt{b \cos (c+d x)}\right ) \int \cos ^3(c+d x) \, dx}{5 \sqrt{\cos (c+d x)}}\\ &=\frac{B \cos ^{\frac{5}{2}}(c+d x) \sqrt{b \cos (c+d x)} \sin (c+d x)}{4 d}+\frac{C \cos ^{\frac{7}{2}}(c+d x) \sqrt{b \cos (c+d x)} \sin (c+d x)}{5 d}+\frac{\left (3 B \sqrt{b \cos (c+d x)}\right ) \int \cos ^2(c+d x) \, dx}{4 \sqrt{\cos (c+d x)}}-\frac{\left ((5 A+4 C) \sqrt{b \cos (c+d x)}\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d \sqrt{\cos (c+d x)}}\\ &=\frac{(5 A+4 C) \sqrt{b \cos (c+d x)} \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{3 B \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)} \sin (c+d x)}{8 d}+\frac{B \cos ^{\frac{5}{2}}(c+d x) \sqrt{b \cos (c+d x)} \sin (c+d x)}{4 d}+\frac{C \cos ^{\frac{7}{2}}(c+d x) \sqrt{b \cos (c+d x)} \sin (c+d x)}{5 d}-\frac{(5 A+4 C) \sqrt{b \cos (c+d x)} \sin ^3(c+d x)}{15 d \sqrt{\cos (c+d x)}}+\frac{\left (3 B \sqrt{b \cos (c+d x)}\right ) \int 1 \, dx}{8 \sqrt{\cos (c+d x)}}\\ &=\frac{3 B x \sqrt{b \cos (c+d x)}}{8 \sqrt{\cos (c+d x)}}+\frac{(5 A+4 C) \sqrt{b \cos (c+d x)} \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{3 B \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)} \sin (c+d x)}{8 d}+\frac{B \cos ^{\frac{5}{2}}(c+d x) \sqrt{b \cos (c+d x)} \sin (c+d x)}{4 d}+\frac{C \cos ^{\frac{7}{2}}(c+d x) \sqrt{b \cos (c+d x)} \sin (c+d x)}{5 d}-\frac{(5 A+4 C) \sqrt{b \cos (c+d x)} \sin ^3(c+d x)}{15 d \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.285483, size = 109, normalized size = 0.49 \[ \frac{\sqrt{b \cos (c+d x)} (60 (6 A+5 C) \sin (c+d x)+40 A \sin (3 (c+d x))+120 B \sin (2 (c+d x))+15 B \sin (4 (c+d x))+180 B c+180 B d x+50 C \sin (3 (c+d x))+6 C \sin (5 (c+d x)))}{480 d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*Cos[c + d*x]]*(180*B*c + 180*B*d*x + 60*(6*A + 5*C)*Sin[c + d*x] + 120*B*Sin[2*(c + d*x)] + 40*A*Sin[3
*(c + d*x)] + 50*C*Sin[3*(c + d*x)] + 15*B*Sin[4*(c + d*x)] + 6*C*Sin[5*(c + d*x)]))/(480*d*Sqrt[Cos[c + d*x]]
)

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Maple [A]  time = 0.354, size = 134, normalized size = 0.6 \begin{align*}{\frac{24\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}\sin \left ( dx+c \right ) +30\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +40\,A\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+32\,C\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+45\,B\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) +80\,A\sin \left ( dx+c \right ) +45\,B \left ( dx+c \right ) +64\,\sin \left ( dx+c \right ) C}{120\,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)^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*(b*cos(d*x+c))^(1/2),x)

[Out]

1/120/d*(b*cos(d*x+c))^(1/2)*(24*C*cos(d*x+c)^4*sin(d*x+c)+30*B*cos(d*x+c)^3*sin(d*x+c)+40*A*sin(d*x+c)*cos(d*
x+c)^2+32*C*sin(d*x+c)*cos(d*x+c)^2+45*B*sin(d*x+c)*cos(d*x+c)+80*A*sin(d*x+c)+45*B*(d*x+c)+64*sin(d*x+c)*C)/c
os(d*x+c)^(1/2)

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Maxima [A]  time = 2.33069, size = 215, normalized size = 0.96 \begin{align*} \frac{15 \,{\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 )} B \sqrt{b} + 2 \, C \sqrt{b}{\left (3 \, \sin \left (5 \, d x + 5 \, c\right ) + 25 \, \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 \, \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 )} + 40 \, A \sqrt{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 )}}{480 \, 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)+C*cos(d*x+c)^2)*(b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/480*(15*(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))))*B*sqrt(b
) + 2*C*sqrt(b)*(3*sin(5*d*x + 5*c) + 25*sin(3/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 150*sin(1/5*ar
ctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c)))) + 40*A*sqrt(b)*(sin(3*d*x + 3*c) + 9*sin(1/3*arctan2(sin(3*d*x + 3
*c), cos(3*d*x + 3*c)))))/d

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Fricas [A]  time = 2.03138, size = 815, normalized size = 3.65 \begin{align*} \left [\frac{45 \, B \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 (24 \, C \cos \left (d x + c\right )^{4} + 30 \, B \cos \left (d x + c\right )^{3} + 8 \,{\left (5 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 45 \, B \cos \left (d x + c\right ) + 80 \, A + 64 \, C\right )} \sqrt{b \cos \left (d x + c\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )}, \frac{45 \, B \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 (24 \, C \cos \left (d x + c\right )^{4} + 30 \, B \cos \left (d x + c\right )^{3} + 8 \,{\left (5 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 45 \, B \cos \left (d x + c\right ) + 80 \, A + 64 \, C\right )} \sqrt{b \cos \left (d x + c\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{120 \, 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)+C*cos(d*x+c)^2)*(b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/240*(45*B*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*(24*C*cos(d*x + c)^4 + 30*B*cos(d*x + c)^3 + 8*(5*A + 4*C)*cos(d*x + c)^2 + 45*B*cos(d*
x + c) + 80*A + 64*C)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)), 1/120*(45*B*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) + (24*C*cos(d*x + c)^4
+ 30*B*cos(d*x + c)^3 + 8*(5*A + 4*C)*cos(d*x + c)^2 + 45*B*cos(d*x + c) + 80*A + 64*C)*sqrt(b*cos(d*x + c))*
sqrt(cos(d*x + c))*sin(d*x + c))/(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)+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)^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*(b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError