3.324 \(\int \frac{\cos ^{\frac{5}{2}}(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{(b \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=155 \[ \frac{(3 A+2 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 b d \sqrt{b \cos (c+d x)}}+\frac{B x \sqrt{\cos (c+d x)}}{2 b \sqrt{b \cos (c+d x)}}+\frac{B \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 b d \sqrt{b \cos (c+d x)}}+\frac{C \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{3 b d \sqrt{b \cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0600347, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.07, Rules used = {17, 3023, 2734} \[ \frac{(3 A+2 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 b d \sqrt{b \cos (c+d x)}}+\frac{B x \sqrt{\cos (c+d x)}}{2 b \sqrt{b \cos (c+d x)}}+\frac{B \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 b d \sqrt{b \cos (c+d x)}}+\frac{C \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{3 b d \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(b*Cos[c + d*x])^(3/2),x]

[Out]

(B*x*Sqrt[Cos[c + d*x]])/(2*b*Sqrt[b*Cos[c + d*x]]) + ((3*A + 2*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*b*d*Sqr
t[b*Cos[c + d*x]]) + (B*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*b*d*Sqrt[b*Cos[c + d*x]]) + (C*Cos[c + d*x]^(5/2)*
Sin[c + d*x])/(3*b*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 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 2734

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((2*a*c
+ b*d)*x)/2, x] + (-Simp[((b*c + a*d)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rubi steps

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

Mathematica [A]  time = 0.169384, size = 75, normalized size = 0.48 \[ \frac{\cos ^{\frac{3}{2}}(c+d x) (3 (4 A+3 C) \sin (c+d x)+3 B \sin (2 (c+d x))+6 B c+6 B d x+C \sin (3 (c+d x)))}{12 d (b \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(b*Cos[c + d*x])^(3/2),x]

[Out]

(Cos[c + d*x]^(3/2)*(6*B*c + 6*B*d*x + 3*(4*A + 3*C)*Sin[c + d*x] + 3*B*Sin[2*(c + d*x)] + C*Sin[3*(c + d*x)])
)/(12*d*(b*Cos[c + d*x])^(3/2))

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Maple [A]  time = 0.313, size = 83, normalized size = 0.5 \begin{align*}{\frac{2\,C\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+3\,B\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) +6\,A\sin \left ( dx+c \right ) +3\,B \left ( dx+c \right ) +4\,\sin \left ( dx+c \right ) C}{6\,d} \left ( \cos \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}} \left ( b\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)^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(3/2),x)

[Out]

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

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Maxima [A]  time = 2.33386, size = 108, normalized size = 0.7 \begin{align*} \frac{\frac{3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B}{b^{\frac{3}{2}}} + \frac{C{\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 )}}{b^{\frac{3}{2}}} + \frac{12 \, A \sin \left (d x + c\right )}{b^{\frac{3}{2}}}}{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)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

1/12*(3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B/b^(3/2) + C*(sin(3*d*x + 3*c) + 9*sin(1/3*arctan2(sin(3*d*x + 3*c),
cos(3*d*x + 3*c))))/b^(3/2) + 12*A*sin(d*x + c)/b^(3/2))/d

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Fricas [A]  time = 2.05172, size = 667, normalized size = 4.3 \begin{align*} \left [-\frac{3 \, 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 (2 \, C \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 6 \, A + 4 \, C\right )} \sqrt{b \cos \left (d x + c\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{12 \, b^{2} d \cos \left (d x + c\right )}, \frac{3 \, 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 (2 \, C \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 6 \, A + 4 \, C\right )} \sqrt{b \cos \left (d x + c\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{6 \, b^{2} 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))^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)^(5/2)/(b*cos(d*x + c))^(3/2), x)