3.2.29 \(\int \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\) [129]

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

[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 = 0.04, 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)^{9/2}}{9 a^2 d}-\frac {2 (a \sin (c+d x)+a)^{11/2}}{11 a^3 d} \end {gather*}

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 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) (a+a \sin (c+d x))^{5/2} \, dx &=\frac {\text {Subst}\left (\int (a-x) (a+x)^{7/2} \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac {\text {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 = 0.08, size = 41, normalized size = 0.84 \begin {gather*} -\frac {2 (1+\sin (c+d x))^2 (a (1+\sin (c+d x)))^{5/2} (-13+9 \sin (c+d x))}{99 d} \end {gather*}

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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Maple [A]
time = 0.34, size = 31, normalized size = 0.63

method	result	size
default	\(-\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {9}{2}} \left (9 \sin \left (d x +c \right )-13\right )}{99 a^{2} d}\)	\(31\)

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,method=_RETURNVERBOSE)

[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 = 0.30, size = 38, normalized size = 0.78 \begin {gather*} -\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} \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))^(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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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (41) = 82\).
time = 0.36, size = 88, normalized size = 1.80 \begin {gather*} -\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} \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))^(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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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (42) = 84\).
time = 48.11, size = 335, normalized size = 6.84 \begin {gather*} \begin {cases} \frac {8 a^{2} \sqrt {a \sin {\left (c + d x \right )} + a} \sin ^{5}{\left (c + d x \right )}}{77 d} + \frac {272 a^{2} \sqrt {a \sin {\left (c + d x \right )} + a} \sin ^{4}{\left (c + d x \right )}}{693 d} + \frac {2 a^{2} \sqrt {a \sin {\left (c + d x \right )} + a} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{7 d} + \frac {368 a^{2} \sqrt {a \sin {\left (c + d x \right )} + a} \sin ^{3}{\left (c + d x \right )}}{693 d} + \frac {6 a^{2} \sqrt {a \sin {\left (c + d x \right )} + a} \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{7 d} + \frac {64 a^{2} \sqrt {a \sin {\left (c + d x \right )} + a} \sin ^{2}{\left (c + d x \right )}}{231 d} + \frac {6 a^{2} \sqrt {a \sin {\left (c + d x \right )} + a} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{7 d} + \frac {8 a^{2} \sqrt {a \sin {\left (c + d x \right )} + a} \sin {\left (c + d x \right )}}{693 d} + \frac {2 a^{2} \sqrt {a \sin {\left (c + d x \right )} + a} \cos ^{2}{\left (c + d x \right )}}{7 d} - \frac {16 a^{2} \sqrt {a \sin {\left (c + d x \right )} + a}}{693 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{\frac {5}{2}} \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \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))**(5/2),x)

[Out]

Piecewise((8*a**2*sqrt(a*sin(c + d*x) + a)*sin(c + d*x)**5/(77*d) + 272*a**2*sqrt(a*sin(c + d*x) + a)*sin(c +
d*x)**4/(693*d) + 2*a**2*sqrt(a*sin(c + d*x) + a)*sin(c + d*x)**3*cos(c + d*x)**2/(7*d) + 368*a**2*sqrt(a*sin(
c + d*x) + a)*sin(c + d*x)**3/(693*d) + 6*a**2*sqrt(a*sin(c + d*x) + a)*sin(c + d*x)**2*cos(c + d*x)**2/(7*d)
+ 64*a**2*sqrt(a*sin(c + d*x) + a)*sin(c + d*x)**2/(231*d) + 6*a**2*sqrt(a*sin(c + d*x) + a)*sin(c + d*x)*cos(
c + d*x)**2/(7*d) + 8*a**2*sqrt(a*sin(c + d*x) + a)*sin(c + d*x)/(693*d) + 2*a**2*sqrt(a*sin(c + d*x) + a)*cos
(c + d*x)**2/(7*d) - 16*a**2*sqrt(a*sin(c + d*x) + a)/(693*d), Ne(d, 0)), (x*(a*sin(c) + a)**(5/2)*cos(c)**3,
True))

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Giac [A]
time = 4.75, size = 76, normalized size = 1.55 \begin {gather*} -\frac {64 \, \sqrt {2} {\left (9 \, a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 11 \, a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sqrt {a}}{99 \, 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))^(5/2),x, algorithm="giac")

[Out]

-64/99*sqrt(2)*(9*a^2*cos(-1/4*pi + 1/2*d*x + 1/2*c)^11*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) - 11*a^2*cos(-1/4*
pi + 1/2*d*x + 1/2*c)^9*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\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2} \,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))^(5/2),x)

[Out]

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

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