3.2.60 \(\int \frac {\cos ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\) [160]

Optimal. Leaf size=49 \[ \frac {4 (a+a \sin (c+d x))^{3/2}}{3 a^2 d}-\frac {2 (a+a \sin (c+d x))^{5/2}}{5 a^3 d} \]

[Out]

4/3*(a+a*sin(d*x+c))^(3/2)/a^2/d-2/5*(a+a*sin(d*x+c))^(5/2)/a^3/d

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Rubi [A]
time = 0.04, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2746, 45} \begin {gather*} \frac {4 (a \sin (c+d x)+a)^{3/2}}{3 a^2 d}-\frac {2 (a \sin (c+d x)+a)^{5/2}}{5 a^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(4*(a + a*Sin[c + d*x])^(3/2))/(3*a^2*d) - (2*(a + a*Sin[c + d*x])^(5/2))/(5*a^3*d)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx &=\frac {\text {Subst}\left (\int (a-x) \sqrt {a+x} \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac {\text {Subst}\left (\int \left (2 a \sqrt {a+x}-(a+x)^{3/2}\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac {4 (a+a \sin (c+d x))^{3/2}}{3 a^2 d}-\frac {2 (a+a \sin (c+d x))^{5/2}}{5 a^3 d}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 34, normalized size = 0.69 \begin {gather*} -\frac {2 (a (1+\sin (c+d x)))^{3/2} (-7+3 \sin (c+d x))}{15 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-2*(a*(1 + Sin[c + d*x]))^(3/2)*(-7 + 3*Sin[c + d*x]))/(15*a^2*d)

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Maple [A]
time = 0.26, size = 31, normalized size = 0.63

method	result	size
default	\(-\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \left (3 \sin \left (d x +c \right )-7\right )}{15 a^{2} d}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/15/a^2*(a+a*sin(d*x+c))^(3/2)*(3*sin(d*x+c)-7)/d

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Maxima [A]
time = 0.31, size = 75, normalized size = 1.53 \begin {gather*} \frac {2 \, {\left (15 \, \sqrt {a \sin \left (d x + c\right ) + a} - \frac {3 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} - 10 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{2}}{a^{2}}\right )}}{15 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

2/15*(15*sqrt(a*sin(d*x + c) + a) - (3*(a*sin(d*x + c) + a)^(5/2) - 10*(a*sin(d*x + c) + a)^(3/2)*a + 15*sqrt(
a*sin(d*x + c) + a)*a^2)/a^2)/(a*d)

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Fricas [A]
time = 0.33, size = 40, normalized size = 0.82 \begin {gather*} \frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 4 \, \sin \left (d x + c\right ) + 4\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{15 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*cos(d*x + c)^2 + 4*sin(d*x + c) + 4)*sqrt(a*sin(d*x + c) + a)/(a*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/(a+a*sin(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [A]
time = 5.26, size = 68, normalized size = 1.39 \begin {gather*} -\frac {8 \, {\left (3 \, \sqrt {2} \sqrt {a} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 5 \, \sqrt {2} \sqrt {a} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}\right )}}{15 \, a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/15*(3*sqrt(2)*sqrt(a)*cos(-1/4*pi + 1/2*d*x + 1/2*c)^5 - 5*sqrt(2)*sqrt(a)*cos(-1/4*pi + 1/2*d*x + 1/2*c)^3
)/(a*d*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^3}{\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3/(a + a*sin(c + d*x))^(1/2),x)

[Out]

int(cos(c + d*x)^3/(a + a*sin(c + d*x))^(1/2), x)

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