3.2.61 \(\int \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{5/2} \, dx\) [161]

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

[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 = 0.02, 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, 2715, 8} \begin {gather*} \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} \end {gather*}

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 2715

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] && Integ
erQ[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 = 0.05, size = 55, normalized size = 0.51 \begin {gather*} \frac {(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 ^{\frac {5}{2}}(c+d x)} \end {gather*}

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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Maple [A]
time = 0.16, size = 62, normalized size = 0.58

method	result	size
default	\(\frac {\left (b \cos \left (d x +c \right )\right )^{\frac {5}{2}} \left (2 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+3 \sin \left (d x +c \right ) \cos \left (d x +c \right )+3 d x +3 c \right )}{8 d \cos \left (d x +c \right )^{\frac {5}{2}}}\)	\(62\)
risch	\(\frac {3 b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{i \left (d x +c \right )} x}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {i b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{5 i \left (d x +c \right )}}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{-i \left (d x +c \right )}}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {7 i b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \cos \left (3 d x +3 c \right )}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {9 b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \sin \left (3 d x +3 c \right )}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}\)	\(243\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(3/2)*(b*cos(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.57, size = 59, normalized size = 0.55 \begin {gather*} \frac {{\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 )} \sqrt {b}}{32 \, d} \end {gather*}

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="maxima")

[Out]

1/32*(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))))*sq
rt(b)/d

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Fricas [A]
time = 0.40, size = 193, normalized size = 1.80 \begin {gather*} \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 ] \end {gather*}

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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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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)

[Out]

Timed out

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

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: Simplification assuming sageVARc near 0S
implification assuming sageVARc near 0Simplification assuming sageVARc near 0Simplification assuming sageVARc
near 0(3*sageVARb^2*s

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Mupad [B]
time = 1.03, size = 78, normalized size = 0.73 \begin {gather*} \frac {b^2\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (8\,\sin \left (c+d\,x\right )+9\,\sin \left (3\,c+3\,d\,x\right )+\sin \left (5\,c+5\,d\,x\right )+24\,d\,x\,\cos \left (c+d\,x\right )\right )}{32\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^(3/2)*(b*cos(c + d*x))^(5/2),x)

[Out]

(b^2*cos(c + d*x)^(1/2)*(b*cos(c + d*x))^(1/2)*(8*sin(c + d*x) + 9*sin(3*c + 3*d*x) + sin(5*c + 5*d*x) + 24*d*
x*cos(c + d*x)))/(32*d*(cos(2*c + 2*d*x) + 1))

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