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

Optimal. Leaf size=95 \[ -\frac {64 a^3 \cos ^3(c+d x)}{105 d (a \sin (c+d x)+a)^{3/2}}-\frac {16 a^2 \cos ^3(c+d x)}{35 d \sqrt {a \sin (c+d x)+a}}-\frac {2 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d} \]

[Out]

-64/105*a^3*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(3/2)-16/35*a^2*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(1/2)-2/7*a*cos(d*
x+c)^3*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.17, antiderivative size = 95, 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 = {2674, 2673} \[ -\frac {16 a^2 \cos ^3(c+d x)}{35 d \sqrt {a \sin (c+d x)+a}}-\frac {64 a^3 \cos ^3(c+d x)}{105 d (a \sin (c+d x)+a)^{3/2}}-\frac {2 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-64*a^3*Cos[c + d*x]^3)/(105*d*(a + a*Sin[c + d*x])^(3/2)) - (16*a^2*Cos[c + d*x]^3)/(35*d*Sqrt[a + a*Sin[c +
d*x]]) - (2*a*Cos[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(7*d)

Rule 2673

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m - 1)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rule 2674

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[(a*(2*m + p - 1))/(m + p), Int[(
g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx &=-\frac {2 a \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{7 d}+\frac {1}{7} (8 a) \int \cos ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {16 a^2 \cos ^3(c+d x)}{35 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{7 d}+\frac {1}{35} \left (32 a^2\right ) \int \frac {\cos ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=-\frac {64 a^3 \cos ^3(c+d x)}{105 d (a+a \sin (c+d x))^{3/2}}-\frac {16 a^2 \cos ^3(c+d x)}{35 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{7 d}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 59, normalized size = 0.62 \[ -\frac {2 \left (15 \sin ^2(c+d x)+54 \sin (c+d x)+71\right ) \cos ^3(c+d x) (a (\sin (c+d x)+1))^{3/2}}{105 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[c + d*x]^3*(a*(1 + Sin[c + d*x]))^(3/2)*(71 + 54*Sin[c + d*x] + 15*Sin[c + d*x]^2))/(105*d*(1 + Sin[c
+ d*x])^3)

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fricas [A]  time = 0.65, size = 122, normalized size = 1.28 \[ -\frac {2 \, {\left (15 \, a \cos \left (d x + c\right )^{4} + 39 \, a \cos \left (d x + c\right )^{3} - 8 \, a \cos \left (d x + c\right )^{2} + 32 \, a \cos \left (d x + c\right ) + {\left (15 \, a \cos \left (d x + c\right )^{3} - 24 \, a \cos \left (d x + c\right )^{2} - 32 \, a \cos \left (d x + c\right ) - 64 \, a\right )} \sin \left (d x + c\right ) + 64 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{105 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(15*a*cos(d*x + c)^4 + 39*a*cos(d*x + c)^3 - 8*a*cos(d*x + c)^2 + 32*a*cos(d*x + c) + (15*a*cos(d*x + c
)^3 - 24*a*cos(d*x + c)^2 - 32*a*cos(d*x + c) - 64*a)*sin(d*x + c) + 64*a)*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x
+ c) + d*sin(d*x + c) + d)

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giac [B]  time = 0.64, size = 226, normalized size = 2.38 \[ -\frac {1}{420} \, \sqrt {2} {\left (\frac {21 \, a \cos \left (\frac {1}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {105 \, a \cos \left (\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {15 \, a \cos \left (-\frac {1}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {35 \, a \cos \left (-\frac {1}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} - \frac {70 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d} - \frac {42 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} - \frac {420 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/420*sqrt(2)*(21*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 + 105*a*cos(1/4*pi +
1/2*d*x + 1/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d + 15*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 + 35*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 - 70*a*sgn(co
s(-1/4*pi + 1/2*d*x + 1/2*c))*sin(1/4*pi + 3/2*d*x + 3/2*c)/d - 42*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 - 420*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)*sqrt
(a)

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maple [A]  time = 0.27, size = 67, normalized size = 0.71 \[ -\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{2} \left (\sin \left (d x +c \right )-1\right )^{2} \left (15 \left (\sin ^{2}\left (d x +c \right )\right )+54 \sin \left (d x +c \right )+71\right )}{105 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)^2*(15*sin(d*x+c)^2+54*sin(d*x+c)+71)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2
)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \cos ^{2}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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