3.140 \(\int \cos ^{\frac{7}{2}}(c+d x) \sqrt{b \cos (c+d x)} \, dx\)

Optimal. Leaf size=98 \[ \frac{3 x \sqrt{b \cos (c+d x)}}{8 \sqrt{\cos (c+d x)}}+\frac{\sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) \sqrt{b \cos (c+d x)}}{4 d}+\frac{3 \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}}{8 d} \]

[Out]

(3*x*Sqrt[b*Cos[c + d*x]])/(8*Sqrt[Cos[c + d*x]]) + (3*Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(
8*d) + (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 = 0.0271258, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {17, 2635, 8} \[ \frac{3 x \sqrt{b \cos (c+d x)}}{8 \sqrt{\cos (c+d x)}}+\frac{\sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) \sqrt{b \cos (c+d x)}}{4 d}+\frac{3 \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(7/2)*Sqrt[b*Cos[c + d*x]],x]

[Out]

(3*x*Sqrt[b*Cos[c + d*x]])/(8*Sqrt[Cos[c + d*x]]) + (3*Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(
8*d) + (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 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{7}{2}}(c+d x) \sqrt{b \cos (c+d x)} \, dx &=\frac{\sqrt{b \cos (c+d x)} \int \cos ^4(c+d x) \, dx}{\sqrt{\cos (c+d x)}}\\ &=\frac{\cos ^{\frac{5}{2}}(c+d x) \sqrt{b \cos (c+d x)} \sin (c+d x)}{4 d}+\frac{\left (3 \sqrt{b \cos (c+d x)}\right ) \int \cos ^2(c+d x) \, dx}{4 \sqrt{\cos (c+d x)}}\\ &=\frac{3 \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)} \sin (c+d x)}{8 d}+\frac{\cos ^{\frac{5}{2}}(c+d x) \sqrt{b \cos (c+d x)} \sin (c+d x)}{4 d}+\frac{\left (3 \sqrt{b \cos (c+d x)}\right ) \int 1 \, dx}{8 \sqrt{\cos (c+d x)}}\\ &=\frac{3 x \sqrt{b \cos (c+d x)}}{8 \sqrt{\cos (c+d x)}}+\frac{3 \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)} \sin (c+d x)}{8 d}+\frac{\cos ^{\frac{5}{2}}(c+d x) \sqrt{b \cos (c+d x)} \sin (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.0950277, size = 55, normalized size = 0.56 \[ \frac{(12 (c+d x)+8 \sin (2 (c+d x))+\sin (4 (c+d x))) \sqrt{b \cos (c+d x)}}{32 d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^(7/2)*Sqrt[b*Cos[c + d*x]],x]

[Out]

(Sqrt[b*Cos[c + d*x]]*(12*(c + d*x) + 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)]))/(32*d*Sqrt[Cos[c + d*x]])

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Maple [A]  time = 0.316, size = 62, normalized size = 0.6 \begin{align*}{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +3\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +3\,dx+3\,c}{8\,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)^(7/2)*(b*cos(d*x+c))^(1/2),x)

[Out]

1/8/d*(b*cos(d*x+c))^(1/2)*(2*cos(d*x+c)^3*sin(d*x+c)+3*cos(d*x+c)*sin(d*x+c)+3*d*x+3*c)/cos(d*x+c)^(1/2)

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Maxima [A]  time = 1.81787, size = 66, normalized size = 0.67 \begin{align*} \frac{{\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 )} \sqrt{b}}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(7/2)*(b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/32*(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))))*sqrt(b)/d

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Fricas [A]  time = 1.93805, size = 495, normalized size = 5.05 \begin{align*} \left [\frac{2 \, \sqrt{b \cos \left (d x + c\right )}{\left (2 \, \cos \left (d x + c\right )^{2} + 3\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, \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 )}{16 \, d}, \frac{\sqrt{b \cos \left (d x + c\right )}{\left (2 \, \cos \left (d x + c\right )^{2} + 3\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, \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 )}{8 \, d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(7/2)*(b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/16*(2*sqrt(b*cos(d*x + c))*(2*cos(d*x + c)^2 + 3)*sqrt(cos(d*x + c))*sin(d*x + c) + 3*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))/d, 1/8*(sqrt(b*cos(d*x + c)
)*(2*cos(d*x + c)^2 + 3)*sqrt(cos(d*x + c))*sin(d*x + c) + 3*sqrt(b)*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(cos(d*x+c)**(7/2)*(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)^(7/2)*(b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError