3.2.5 \(\int \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\) [105]

Optimal. Leaf size=49 \[ \frac {4 (a+a \sin (c+d x))^{5/2}}{5 a^2 d}-\frac {2 (a+a \sin (c+d x))^{7/2}}{7 a^3 d} \]

[Out]

4/5*(a+a*sin(d*x+c))^(5/2)/a^2/d-2/7*(a+a*sin(d*x+c))^(7/2)/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)^{5/2}}{5 a^2 d}-\frac {2 (a \sin (c+d x)+a)^{7/2}}{7 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])^(5/2))/(5*a^2*d) - (2*(a + a*Sin[c + d*x])^(7/2))/(7*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 \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx &=\frac {\text {Subst}\left (\int (a-x) (a+x)^{3/2} \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac {\text {Subst}\left (\int \left (2 a (a+x)^{3/2}-(a+x)^{5/2}\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac {4 (a+a \sin (c+d x))^{5/2}}{5 a^2 d}-\frac {2 (a+a \sin (c+d x))^{7/2}}{7 a^3 d}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 54, normalized size = 1.10 \begin {gather*} -\frac {2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4 \sqrt {a (1+\sin (c+d x))} (-9+5 \sin (c+d x))}{35 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*Sqrt[a*(1 + Sin[c + d*x])]*(-9 + 5*Sin[c + d*x]))/(35*d)

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Maple [A]
time = 0.27, size = 31, normalized size = 0.63

method	result	size
default	\(-\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}} \left (5 \sin \left (d x +c \right )-9\right )}{35 a^{2} d}\)	\(31\)

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]

-2/35/a^2*(a+a*sin(d*x+c))^(5/2)*(5*sin(d*x+c)-9)/d

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Maxima [A]
time = 0.31, size = 38, normalized size = 0.78 \begin {gather*} -\frac {2 \, {\left (5 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} - 14 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a\right )}}{35 \, a^{3} 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]

-2/35*(5*(a*sin(d*x + c) + a)^(7/2) - 14*(a*sin(d*x + c) + a)^(5/2)*a)/(a^3*d)

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Fricas [A]
time = 0.36, size = 46, normalized size = 0.94 \begin {gather*} \frac {2 \, {\left (\cos \left (d x + c\right )^{2} + {\left (5 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) + 8\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{35 \, 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]

2/35*(cos(d*x + c)^2 + (5*cos(d*x + c)^2 + 8)*sin(d*x + c) + 8)*sqrt(a*sin(d*x + c) + 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 = 5.47, size = 70, normalized size = 1.43 \begin {gather*} -\frac {16 \, \sqrt {2} {\left (5 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 7 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sqrt {a}}{35 \, 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="giac")

[Out]

-16/35*sqrt(2)*(5*cos(-1/4*pi + 1/2*d*x + 1/2*c)^7*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) - 7*cos(-1/4*pi + 1/2*d
*x + 1/2*c)^5*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))*sqrt(a)/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int {\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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