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

Optimal. Leaf size=99 \[ \frac{A x \sqrt{\cos (c+d x)}}{b^2 \sqrt{b \cos (c+d x)}}+\frac{C x \sqrt{\cos (c+d x)}}{2 b^2 \sqrt{b \cos (c+d x)}}+\frac{C \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 b^2 d \sqrt{b \cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0268343, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086, Rules used = {17, 2635, 8} \[ \frac{A x \sqrt{\cos (c+d x)}}{b^2 \sqrt{b \cos (c+d x)}}+\frac{C x \sqrt{\cos (c+d x)}}{2 b^2 \sqrt{b \cos (c+d x)}}+\frac{C \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 b^2 d \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0583793, size = 55, normalized size = 0.56 \[ \frac{\sqrt{\cos (c+d x)} (2 (2 A+C) (c+d x)+C \sin (2 (c+d x)))}{4 b^2 d \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.279, size = 54, normalized size = 0.6 \begin{align*}{\frac{C\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +2\,A \left ( dx+c \right ) +C \left ( dx+c \right ) }{2\,d} \left ( \cos \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}} \left ( b\cos \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}} \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))^(5/2),x)

[Out]

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

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Maxima [A]  time = 2.45914, size = 70, normalized size = 0.71 \begin{align*} \frac{\frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C}{b^{\frac{5}{2}}} + \frac{8 \, A \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{b^{\frac{5}{2}}}}{4 \, 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))^(5/2),x, algorithm="maxima")

[Out]

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*C/b^(5/2) + 8*A*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/b^(5/2))/d

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Fricas [A]  time = 1.98551, size = 475, normalized size = 4.8 \begin{align*} \left [\frac{2 \, \sqrt{b \cos \left (d x + c\right )} C \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) -{\left (2 \, A + 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 )}{4 \, b^{3} d}, \frac{\sqrt{b \cos \left (d x + c\right )} C \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) +{\left (2 \, A + 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 )}{2 \, b^{3} 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))^(5/2),x, algorithm="fricas")

[Out]

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

[Out]

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