∫ cos3 _ 2 (c + dx)(b cos(c + dx))5∕2 dx

3.161 \(\int \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=107 \[ \frac {3 b^2 x \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {b^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}{4 d}+\frac {3 b^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}{8 d} \]

[Out]

1/4*b^2*cos(d*x+c)^(5/2)*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d+3/8*b^2*x*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2)+3/8

*b^2*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 = 0.03, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {17, 2635, 8} \[ \frac {3 b^2 x \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {b^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}{4 d}+\frac {3 b^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}{8 d} \]

Antiderivative was successfully veriﬁed.

[In]

Int[Cos[c + d*x]^(3/2)*(b*Cos[c + d*x])^(5/2),x]

[Out]

(3*b^2*x*Sqrt[b*Cos[c + d*x]])/(8*Sqrt[Cos[c + d*x]]) + (3*b^2*Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*x]]*Sin[c +

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

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Mathematica [A] time = 0.08, size = 55, normalized size = 0.51 \[ \frac {(12 (c+d x)+8 \sin (2 (c+d x))+\sin (4 (c+d x))) (b \cos (c+d x))^{5/2}}{32 d \cos ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[Cos[c + d*x]^(3/2)*(b*Cos[c + d*x])^(5/2),x]

[Out]

((b*Cos[c + d*x])^(5/2)*(12*(c + d*x) + 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)]))/(32*d*Cos[c + d*x]^(5/2))

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fricas [A] time = 0.68, size = 193, normalized size = 1.80 \[ \left [\frac {3 \, \sqrt {-b} b^{2} \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, {\left (2 \, b^{2} \cos \left (d x + c\right )^{2} + 3 \, b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{16 \, d}, \frac {3 \, b^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) + {\left (2 \, b^{2} \cos \left (d x + c\right )^{2} + 3 \, b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{8 \, d}\right ] \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)*(b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[1/16*(3*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) + 2*(2*b^2*cos(d*x + c)^2 + 3*b^2)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/d, 1/8*(3*b^(5/

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

t(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/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)^(3/2)*(b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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