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

[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)

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Rubi [A]  time = 0.255076, antiderivative size = 124, normalized size of antiderivative = 1., 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 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]

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 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]

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*}

Mathematica [A]  time = 1.45006, 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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Maple [A]  time = 0.75, size = 77, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ){a}^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2} \left ( 35\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+120\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+159\,\sin \left ( dx+c \right ) +106 \right ) }{315\,d\cos \left ( dx+c \right ) }{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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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Fricas [A]  time = 1.5655, size = 396, normalized size = 3.19 \begin{align*} \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 )}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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]

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