3.105 \(\int \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\)

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

[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 = 0.06, 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 = {2667, 43} \[ \frac {4 (a \sin (c+d x)+a)^{5/2}}{5 a^2 d}-\frac {2 (a \sin (c+d x)+a)^{7/2}}{7 a^3 d} \]

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 43

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 2667

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 {\operatorname {Subst}\left (\int (a-x) (a+x)^{3/2} \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac {\operatorname {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 = 0.23, size = 54, normalized size = 1.10 \[ -\frac {2 (5 \sin (c+d x)-9) \sqrt {a (\sin (c+d x)+1)} \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4}{35 d} \]

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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fricas [A]  time = 0.78, size = 46, normalized size = 0.94 \[ \frac {2 \, {\left (\cos \left (d x + c\right )^{2} + {\left (5 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) + 8\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{35 \, d} \]

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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giac [B]  time = 0.69, size = 129, normalized size = 2.63 \[ \frac {1}{140} \, \sqrt {2} \sqrt {a} {\left (\frac {7 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} + \frac {105 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d} + \frac {5 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right )}{d} + \frac {35 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d}\right )} \]

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]

1/140*sqrt(2)*sqrt(a)*(7*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(1/4*pi + 5/2*d*x + 5/2*c)/d + 105*sgn(cos(-1/
4*pi + 1/2*d*x + 1/2*c))*sin(1/4*pi + 1/2*d*x + 1/2*c)/d + 5*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi +
7/2*d*x + 7/2*c)/d + 35*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 3/2*d*x + 3/2*c)/d)

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maple [A]  time = 0.15, size = 31, normalized size = 0.63 \[ -\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}} \left (5 \sin \left (d x +c \right )-9\right )}{35 a^{2} d} \]

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)

[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 = 0.44, size = 38, normalized size = 0.78 \[ -\frac {2 \, {\left (5 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} - 14 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a\right )}}{35 \, a^{3} d} \]

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

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

[Out]

Timed out

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