3.129 \(\int \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=49 \[ \frac {4 (a \sin (c+d x)+a)^{9/2}}{9 a^2 d}-\frac {2 (a \sin (c+d x)+a)^{11/2}}{11 a^3 d} \]

[Out]

4/9*(a+a*sin(d*x+c))^(9/2)/a^2/d-2/11*(a+a*sin(d*x+c))^(11/2)/a^3/d

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Rubi [A]  time = 0.07, 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} \[ \frac {4 (a \sin (c+d x)+a)^{9/2}}{9 a^2 d}-\frac {2 (a \sin (c+d x)+a)^{11/2}}{11 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(4*(a + a*Sin[c + d*x])^(9/2))/(9*a^2*d) - (2*(a + a*Sin[c + d*x])^(11/2))/(11*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) (a+a \sin (c+d x))^{5/2} \, dx &=\frac {\operatorname {Subst}\left (\int (a-x) (a+x)^{7/2} \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (2 a (a+x)^{7/2}-(a+x)^{9/2}\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac {4 (a+a \sin (c+d x))^{9/2}}{9 a^2 d}-\frac {2 (a+a \sin (c+d x))^{11/2}}{11 a^3 d}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 41, normalized size = 0.84 \[ -\frac {2 (\sin (c+d x)+1)^2 (9 \sin (c+d x)-13) (a (\sin (c+d x)+1))^{5/2}}{99 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-2*(1 + Sin[c + d*x])^2*(a*(1 + Sin[c + d*x]))^(5/2)*(-13 + 9*Sin[c + d*x]))/(99*d)

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fricas [B]  time = 0.78, size = 88, normalized size = 1.80 \[ -\frac {2 \, {\left (23 \, a^{2} \cos \left (d x + c\right )^{4} - 4 \, a^{2} \cos \left (d x + c\right )^{2} - 32 \, a^{2} + {\left (9 \, a^{2} \cos \left (d x + c\right )^{4} - 20 \, a^{2} \cos \left (d x + c\right )^{2} - 32 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{99 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-2/99*(23*a^2*cos(d*x + c)^4 - 4*a^2*cos(d*x + c)^2 - 32*a^2 + (9*a^2*cos(d*x + c)^4 - 20*a^2*cos(d*x + c)^2 -
32*a^2)*sin(d*x + c))*sqrt(a*sin(d*x + c) + a)/d

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giac [B]  time = 4.54, size = 339, normalized size = 6.92 \[ -\frac {1}{55440} \, \sqrt {2} {\left (\frac {1980 \, a^{2} \cos \left (\frac {1}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {9240 \, a^{2} \cos \left (\frac {1}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {1540 \, a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {9}{2} \, d x + \frac {9}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {5544 \, a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {385 \, a^{2} \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 {9}{2} \, d x + \frac {9}{2} \, c\right )}{d} - \frac {2079 \, a^{2} \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 {48510 \, a^{2} \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 {315 \, a^{2} \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 {11}{2} \, d x + \frac {11}{2} \, c\right )}{d} - \frac {1485 \, a^{2} \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 {16170 \, a^{2} \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 )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-1/55440*sqrt(2)*(1980*a^2*cos(1/4*pi + 7/2*d*x + 7/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d + 9240*a^2*cos(
1/4*pi + 3/2*d*x + 3/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d + 1540*a^2*cos(-1/4*pi + 9/2*d*x + 9/2*c)*sgn(
cos(-1/4*pi + 1/2*d*x + 1/2*c))/d + 5544*a^2*cos(-1/4*pi + 5/2*d*x + 5/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)
)/d + 385*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(1/4*pi + 9/2*d*x + 9/2*c)/d - 2079*a^2*sgn(cos(-1/4*pi +
1/2*d*x + 1/2*c))*sin(1/4*pi + 5/2*d*x + 5/2*c)/d - 48510*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(1/4*pi
+ 1/2*d*x + 1/2*c)/d + 315*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 11/2*d*x + 11/2*c)/d - 1485*a
^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 7/2*d*x + 7/2*c)/d - 16170*a^2*sgn(cos(-1/4*pi + 1/2*d*x
+ 1/2*c))*sin(-1/4*pi + 3/2*d*x + 3/2*c)/d)*sqrt(a)

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maple [A]  time = 0.16, size = 31, normalized size = 0.63 \[ -\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {9}{2}} \left (9 \sin \left (d x +c \right )-13\right )}{99 a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/99/a^2*(a+a*sin(d*x+c))^(9/2)*(9*sin(d*x+c)-13)/d

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maxima [A]  time = 0.47, size = 38, normalized size = 0.78 \[ -\frac {2 \, {\left (9 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {11}{2}} - 22 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a\right )}}{99 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-2/99*(9*(a*sin(d*x + c) + a)^(11/2) - 22*(a*sin(d*x + c) + a)^(9/2)*a)/(a^3*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\cos \left (c+d\,x\right )}^3\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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