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

Optimal. Leaf size=98 \[ \frac {10 b^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {2 \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b d}+\frac {10 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{21 d} \]

[Out]

2/7*(b*cos(d*x+c))^(5/2)*sin(d*x+c)/b/d+10/21*b^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(si
n(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/d/(b*cos(d*x+c))^(1/2)+10/21*b*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.06, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {16, 2635, 2642, 2641} \[ \frac {10 b^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {2 \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b d}+\frac {10 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{21 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(10*b^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*d*Sqrt[b*Cos[c + d*x]]) + (10*b*Sqrt[b*Cos[c + d*x]]
*Sin[c + d*x])/(21*d) + (2*(b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*b*d)

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (b \cos (c+d x))^{3/2} \, dx &=\frac {\int (b \cos (c+d x))^{7/2} \, dx}{b^2}\\ &=\frac {2 (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}+\frac {5}{7} \int (b \cos (c+d x))^{3/2} \, dx\\ &=\frac {10 b \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}+\frac {1}{21} \left (5 b^2\right ) \int \frac {1}{\sqrt {b \cos (c+d x)}} \, dx\\ &=\frac {10 b \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}+\frac {\left (5 b^2 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 \sqrt {b \cos (c+d x)}}\\ &=\frac {10 b^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {10 b \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 73, normalized size = 0.74 \[ \frac {(b \cos (c+d x))^{3/2} \left (20 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+(23 \sin (c+d x)+3 \sin (3 (c+d x))) \sqrt {\cos (c+d x)}\right )}{42 d \cos ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Cos[c + d*x])^(3/2)*(20*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(23*Sin[c + d*x] + 3*Sin[3*(c + d*x
)])))/(42*d*Cos[c + d*x]^(3/2))

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \cos \left (d x + c\right )} b \cos \left (d x + c\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*cos(d*x + c))*b*cos(d*x + c)^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}} \cos \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.12, size = 210, normalized size = 2.14 \[ -\frac {2 \sqrt {b \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b^{2} \left (48 \left (\cos ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-120 \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+128 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-72 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+16 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{21 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/21*(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b^2*(48*cos(1/2*d*x+1/2*c)^9-120*cos(1/2*d*x+1
/2*c)^7+128*cos(1/2*d*x+1/2*c)^5-72*cos(1/2*d*x+1/2*c)^3+5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)
^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+16*cos(1/2*d*x+1/2*c))/(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d
*x+1/2*c)^2))^(1/2)/sin(1/2*d*x+1/2*c)/(b*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}} \cos \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^2\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(b*cos(d*x+c))**(3/2),x)

[Out]

Timed out

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