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

Optimal. Leaf size=122 \[ \frac{b^2 x (4 A+3 C) \sqrt{b \cos (c+d x)}}{8 \sqrt{\cos (c+d x)}}+\frac{b^2 (4 A+3 C) \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}}{8 d}+\frac{b^2 C \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) \sqrt{b \cos (c+d x)}}{4 d} \]

[Out]

(b^2*(4*A + 3*C)*x*Sqrt[b*Cos[c + d*x]])/(8*Sqrt[Cos[c + d*x]]) + (b^2*(4*A + 3*C)*Sqrt[Cos[c + d*x]]*Sqrt[b*C
os[c + d*x]]*Sin[c + d*x])/(8*d) + (b^2*C*Cos[c + d*x]^(5/2)*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(4*d)

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Rubi [A]  time = 0.0537611, antiderivative size = 122, 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} \[ \frac{b^2 x (4 A+3 C) \sqrt{b \cos (c+d x)}}{8 \sqrt{\cos (c+d x)}}+\frac{b^2 (4 A+3 C) \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}}{8 d}+\frac{b^2 C \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) \sqrt{b \cos (c+d x)}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*(4*A + 3*C)*x*Sqrt[b*Cos[c + d*x]])/(8*Sqrt[Cos[c + d*x]]) + (b^2*(4*A + 3*C)*Sqrt[Cos[c + d*x]]*Sqrt[b*C
os[c + d*x]]*Sin[c + d*x])/(8*d) + (b^2*C*Cos[c + d*x]^(5/2)*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(4*d)

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

Mathematica [A]  time = 0.179927, size = 67, normalized size = 0.55 \[ \frac{(b \cos (c+d x))^{5/2} (4 (4 A+3 C) (c+d x)+8 (A+C) \sin (2 (c+d x))+C \sin (4 (c+d x)))}{32 d \cos ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Cos[c + d*x])^(5/2)*(4*(4*A + 3*C)*(c + d*x) + 8*(A + C)*Sin[2*(c + d*x)] + C*Sin[4*(c + d*x)]))/(32*d*Cos
[c + d*x]^(5/2))

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Maple [A]  time = 0.431, size = 88, normalized size = 0.7 \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} \left ( b\cos \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}} \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/d*(b*cos(d*x+c))^(5/2)*(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))/cos(d*x+c)^(5/2)

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Maxima [A]  time = 2.13532, size = 124, normalized size = 1.02 \begin{align*} \frac{8 \,{\left (2 \,{\left (d x + c\right )} b^{2} + b^{2} \sin \left (2 \, d x + 2 \, c\right )\right )} A \sqrt{b} +{\left (12 \,{\left (d x + c\right )} b^{2} + b^{2} \sin \left (4 \, d x + 4 \, c\right ) + 8 \, b^{2} \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((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

1/32*(8*(2*(d*x + c)*b^2 + b^2*sin(2*d*x + 2*c))*A*sqrt(b) + (12*(d*x + c)*b^2 + b^2*sin(4*d*x + 4*c) + 8*b^2*
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 = 1.72187, size = 582, normalized size = 4.77 \begin{align*} \left [\frac{{\left (4 \, A + 3 \, C\right )} \sqrt{-b} b^{2} \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 b^{2} \cos \left (d x + c\right )^{2} +{\left (4 \, A + 3 \, C\right )} b^{2}\right )} \sqrt{b \cos \left (d x + c\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{16 \, d}, \frac{{\left (4 \, A + 3 \, C\right )} b^{\frac{5}{2}} \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 ) +{\left (2 \, C b^{2} \cos \left (d x + c\right )^{2} +{\left (4 \, A + 3 \, C\right )} b^{2}\right )} \sqrt{b \cos \left (d x + c\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{8 \, d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*((4*A + 3*C)*sqrt(-b)*b^2*log(2*b*cos(d*x + c)^2 - 2*sqrt(b*cos(d*x + c))*sqrt(-b)*sqrt(cos(d*x + c))*si
n(d*x + c) - b) + 2*(2*C*b^2*cos(d*x + c)^2 + (4*A + 3*C)*b^2)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x
+ c))/d, 1/8*((4*A + 3*C)*b^(5/2)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2))) + (2
*C*b^2*cos(d*x + c)^2 + (4*A + 3*C)*b^2)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/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((b*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2)/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 )} \left (b \cos \left (d x + c\right )\right )^{\frac{5}{2}}}{\sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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