∫ ∘ _______ cos (c + dx) (b cos(c + dx))3∕2 dx

3.152 \(\int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=65 \[ \frac {b x \sqrt {b \cos (c+d x)}}{2 \sqrt {\cos (c+d x)}}+\frac {b \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}{2 d} \]

[Out]

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

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Rubi [A] time = 0.01, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {17, 2635, 8} \[ \frac {b x \sqrt {b \cos (c+d x)}}{2 \sqrt {\cos (c+d x)}}+\frac {b \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}{2 d} \]

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[Cos[c + d*x]]*(b*Cos[c + d*x])^(3/2),x]

[Out]

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

2*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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]

Rubi steps

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

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Mathematica [A] time = 0.05, size = 45, normalized size = 0.69 \[ \frac {(2 (c+d x)+\sin (2 (c+d x))) (b \cos (c+d x))^{3/2}}{4 d \cos ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sqrt[Cos[c + d*x]]*(b*Cos[c + d*x])^(3/2),x]

[Out]

((b*Cos[c + d*x])^(3/2)*(2*(c + d*x) + Sin[2*(c + d*x)]))/(4*d*Cos[c + d*x]^(3/2))

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fricas [A] time = 0.70, size = 153, normalized size = 2.35 \[ \left [\frac {2 \, \sqrt {b \cos \left (d x + c\right )} b \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + \sqrt {-b} b \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right )}{4 \, d}, \frac {\sqrt {b \cos \left (d x + c\right )} b \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right )}{2 \, d}\right ] \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)*(b*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/4*(2*sqrt(b*cos(d*x + c))*b*sqrt(cos(d*x + c))*sin(d*x + c) + sqrt(-b)*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, 1/2*(sqrt(b*cos(d*x + c))*b*sqrt(cos(d*x + c))

*sin(d*x + c) + b^(3/2)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2))))/d]

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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)*(b*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/

x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check si