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

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

[Out]

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

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Rubi [A]  time = 0.0262921, 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} \[ \frac{A b^2 x \sqrt{b \cos (c+d x)}}{\sqrt{\cos (c+d x)}}+\frac{b^2 C x \sqrt{b \cos (c+d x)}}{2 \sqrt{\cos (c+d x)}}+\frac{b^2 C \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.108027, size = 52, normalized size = 0.53 \[ \frac{(b \cos (c+d x))^{5/2} (2 (2 A+C) (c+d x)+C \sin (2 (c+d x)))}{4 d \cos ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.234, 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 ( 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)^(5/2),x)

[Out]

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

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Maxima [A]  time = 1.97368, size = 80, normalized size = 0.81 \begin{align*} \frac{8 \, A b^{\frac{5}{2}} \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) +{\left (2 \,{\left (d x + c\right )} b^{2} + b^{2} \sin \left (2 \, d x + 2 \, c\right )\right )} C \sqrt{b}}{4 \, 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)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.67028, size = 475, normalized size = 4.8 \begin{align*} \left [\frac{2 \, \sqrt{b \cos \left (d x + c\right )} C b^{2} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) +{\left (2 \, A + 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 )}{4 \, d}, \frac{\sqrt{b \cos \left (d x + c\right )} C b^{2} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) +{\left (2 \, A + 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 )}{2 \, 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)^(5/2),x, algorithm="fricas")

[Out]

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

[Out]

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