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

Optimal. Leaf size=49 \[ \frac {4 (a \sin (c+d x)+a)^{7/2}}{7 a^2 d}-\frac {2 (a \sin (c+d x)+a)^{9/2}}{9 a^3 d} \]

[Out]

4/7*(a+a*sin(d*x+c))^(7/2)/a^2/d-2/9*(a+a*sin(d*x+c))^(9/2)/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} \[ \frac {4 (a \sin (c+d x)+a)^{7/2}}{7 a^2 d}-\frac {2 (a \sin (c+d x)+a)^{9/2}}{9 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.09, size = 41, normalized size = 0.84 \[ -\frac {2 (\sin (c+d x)+1)^2 (7 \sin (c+d x)-11) (a (\sin (c+d x)+1))^{3/2}}{63 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(1 + Sin[c + d*x])^2*(a*(1 + Sin[c + d*x]))^(3/2)*(-11 + 7*Sin[c + d*x]))/(63*d)

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fricas [A]  time = 0.70, size = 66, normalized size = 1.35 \[ -\frac {2 \, {\left (7 \, a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{2} - 2 \, {\left (5 \, a \cos \left (d x + c\right )^{2} + 8 \, a\right )} \sin \left (d x + c\right ) - 16 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{63 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/63*(7*a*cos(d*x + c)^4 - 2*a*cos(d*x + c)^2 - 2*(5*a*cos(d*x + c)^2 + 8*a)*sin(d*x + c) - 16*a)*sqrt(a*sin(
d*x + c) + a)/d

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giac [B]  time = 0.70, size = 257, normalized size = 5.24 \[ -\frac {1}{2520} \, \sqrt {2} {\left (\frac {45 \, a \cos \left (\frac {1}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {210 \, a \cos \left (\frac {1}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {35 \, a \cos \left (-\frac {1}{4} \, \pi + \frac {9}{2} \, d x + \frac {9}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {126 \, a \cos \left (-\frac {1}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} - \frac {126 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} - \frac {1890 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d} - \frac {90 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right )}{d} - \frac {630 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2520*sqrt(2)*(45*a*cos(1/4*pi + 7/2*d*x + 7/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d + 210*a*cos(1/4*pi +
3/2*d*x + 3/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d + 35*a*cos(-1/4*pi + 9/2*d*x + 9/2*c)*sgn(cos(-1/4*pi
+ 1/2*d*x + 1/2*c))/d + 126*a*cos(-1/4*pi + 5/2*d*x + 5/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d - 126*a*sgn
(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(1/4*pi + 5/2*d*x + 5/2*c)/d - 1890*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*
sin(1/4*pi + 1/2*d*x + 1/2*c)/d - 90*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 7/2*d*x + 7/2*c)/d -
630*a*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 = 0.15, size = 31, normalized size = 0.63 \[ -\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {7}{2}} \left (7 \sin \left (d x +c \right )-11\right )}{63 a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/63/a^2*(a+a*sin(d*x+c))^(7/2)*(7*sin(d*x+c)-11)/d

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maxima [A]  time = 0.32, size = 38, normalized size = 0.78 \[ -\frac {2 \, {\left (7 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 18 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a\right )}}{63 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/63*(7*(a*sin(d*x + c) + a)^(9/2) - 18*(a*sin(d*x + c) + a)^(7/2)*a)/(a^3*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 127.62, size = 252, normalized size = 5.14 \[ \begin {cases} \frac {8 a \sqrt {a \sin {\left (c + d x \right )} + a} \sin ^{4}{\left (c + d x \right )}}{45 d} + \frac {152 a \sqrt {a \sin {\left (c + d x \right )} + a} \sin ^{3}{\left (c + d x \right )}}{315 d} + \frac {2 a \sqrt {a \sin {\left (c + d x \right )} + a} \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {8 a \sqrt {a \sin {\left (c + d x \right )} + a} \sin ^{2}{\left (c + d x \right )}}{21 d} + \frac {4 a \sqrt {a \sin {\left (c + d x \right )} + a} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {8 a \sqrt {a \sin {\left (c + d x \right )} + a} \sin {\left (c + d x \right )}}{315 d} + \frac {2 a \sqrt {a \sin {\left (c + d x \right )} + a} \cos ^{2}{\left (c + d x \right )}}{5 d} - \frac {16 a \sqrt {a \sin {\left (c + d x \right )} + a}}{315 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{\frac {3}{2}} \cos ^{3}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((8*a*sqrt(a*sin(c + d*x) + a)*sin(c + d*x)**4/(45*d) + 152*a*sqrt(a*sin(c + d*x) + a)*sin(c + d*x)**
3/(315*d) + 2*a*sqrt(a*sin(c + d*x) + a)*sin(c + d*x)**2*cos(c + d*x)**2/(5*d) + 8*a*sqrt(a*sin(c + d*x) + a)*
sin(c + d*x)**2/(21*d) + 4*a*sqrt(a*sin(c + d*x) + a)*sin(c + d*x)*cos(c + d*x)**2/(5*d) + 8*a*sqrt(a*sin(c +
d*x) + a)*sin(c + d*x)/(315*d) + 2*a*sqrt(a*sin(c + d*x) + a)*cos(c + d*x)**2/(5*d) - 16*a*sqrt(a*sin(c + d*x)
+ a)/(315*d), Ne(d, 0)), (x*(a*sin(c) + a)**(3/2)*cos(c)**3, True))

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