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

Optimal. Leaf size=124 \[ -\frac {64 a^3 \cos ^3(c+d x)}{315 d (a \sin (c+d x)+a)^{3/2}}-\frac {16 a^2 \cos ^3(c+d x)}{105 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 d}-\frac {2 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{21 d} \]

[Out]

-64/315*a^3*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(3/2)-2/9*cos(d*x+c)^3*(a+a*sin(d*x+c))^(3/2)/d-16/105*a^2*cos(d*x
+c)^3/d/(a+a*sin(d*x+c))^(1/2)-2/21*a*cos(d*x+c)^3*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.26, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2856, 2674, 2673} \[ -\frac {16 a^2 \cos ^3(c+d x)}{105 d \sqrt {a \sin (c+d x)+a}}-\frac {64 a^3 \cos ^3(c+d x)}{315 d (a \sin (c+d x)+a)^{3/2}}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 d}-\frac {2 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{21 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-64*a^3*Cos[c + d*x]^3)/(315*d*(a + a*Sin[c + d*x])^(3/2)) - (16*a^2*Cos[c + d*x]^3)/(105*d*Sqrt[a + a*Sin[c
+ d*x]]) - (2*a*Cos[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(21*d) - (2*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^(3/2)
)/(9*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]

Rule 2856

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

Rubi steps

\begin {align*} \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx &=-\frac {2 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{9 d}+\frac {1}{3} \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)}}{21 d}-\frac {2 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{9 d}+\frac {1}{21} (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)}{105 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{21 d}-\frac {2 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{9 d}+\frac {1}{105} \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)}{315 d (a+a \sin (c+d x))^{3/2}}-\frac {16 a^2 \cos ^3(c+d x)}{105 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{21 d}-\frac {2 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{9 d}\\ \end {align*}

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Mathematica [A]  time = 1.43, size = 100, normalized size = 0.81 \[ \frac {a \sqrt {a (\sin (c+d x)+1)} \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 (-741 \sin (c+d x)+35 \sin (3 (c+d x))+240 \cos (2 (c+d x))-664)}{630 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*Sqrt[a*(1 + Sin[c + d*x])]*(-664 + 240*Cos[2*(c + d*x)] - 741*Sin[c
+ d*x] + 35*Sin[3*(c + d*x)]))/(630*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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fricas [A]  time = 0.49, size = 145, normalized size = 1.17 \[ \frac {2 \, {\left (35 \, a \cos \left (d x + c\right )^{5} - 50 \, a \cos \left (d x + c\right )^{4} - 109 \, a \cos \left (d x + c\right )^{3} + 8 \, a \cos \left (d x + c\right )^{2} - 32 \, a \cos \left (d x + c\right ) - {\left (35 \, a \cos \left (d x + c\right )^{4} + 85 \, a \cos \left (d x + c\right )^{3} - 24 \, a \cos \left (d x + c\right )^{2} - 32 \, a \cos \left (d x + c\right ) - 64 \, a\right )} \sin \left (d x + c\right ) - 64 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{315 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*a*cos(d*x + c)^5 - 50*a*cos(d*x + c)^4 - 109*a*cos(d*x + c)^3 + 8*a*cos(d*x + c)^2 - 32*a*cos(d*x +
c) - (35*a*cos(d*x + c)^4 + 85*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 = 0.38, size = 226, normalized size = 1.82 \[ -\frac {1}{2520} \, \sqrt {2} {\left (\frac {126 \, a \cos \left (\frac {1}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {630 \, a \cos \left (\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {90 \, a \cos \left (-\frac {1}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {210 \, a \cos \left (-\frac {1}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {45 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right )}{d} + \frac {35 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {9}{2} \, d x + \frac {9}{2} \, c\right )}{d} - \frac {630 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2520*sqrt(2)*(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 + 630*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 + 90*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 + 45*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 + 35*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*si
n(-1/4*pi + 9/2*d*x + 9/2*c)/d - 630*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)*s
qrt(a)

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maple [A]  time = 0.89, size = 77, normalized size = 0.62 \[ -\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{2} \left (\sin \left (d x +c \right )-1\right )^{2} \left (35 \left (\sin ^{3}\left (d x +c \right )\right )+120 \left (\sin ^{2}\left (d x +c \right )\right )+159 \sin \left (d x +c \right )+106\right )}{315 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)^2*(35*sin(d*x+c)^3+120*sin(d*x+c)^2+159*sin(d*x+c)+106)/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 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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