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

Optimal. Leaf size=156 \[ -\frac{48 a^2 \cos ^3(c+d x)}{385 d \sqrt{a \sin (c+d x)+a}}-\frac{64 a^3 \cos ^3(c+d x)}{385 d (a \sin (c+d x)+a)^{3/2}}-\frac{2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 a d}+\frac{4 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{33 d}-\frac{6 a \cos ^3(c+d x) \sqrt{a \sin (c+d x)+a}}{77 d} \]

[Out]

(-64*a^3*Cos[c + d*x]^3)/(385*d*(a + a*Sin[c + d*x])^(3/2)) - (48*a^2*Cos[c + d*x]^3)/(385*d*Sqrt[a + a*Sin[c
+ d*x]]) - (6*a*Cos[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(77*d) + (4*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^(3/2)
)/(33*d) - (2*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^(5/2))/(11*a*d)

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Rubi [A]  time = 0.431542, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {2878, 2856, 2674, 2673} \[ -\frac{48 a^2 \cos ^3(c+d x)}{385 d \sqrt{a \sin (c+d x)+a}}-\frac{64 a^3 \cos ^3(c+d x)}{385 d (a \sin (c+d x)+a)^{3/2}}-\frac{2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 a d}+\frac{4 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{33 d}-\frac{6 a \cos ^3(c+d x) \sqrt{a \sin (c+d x)+a}}{77 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-64*a^3*Cos[c + d*x]^3)/(385*d*(a + a*Sin[c + d*x])^(3/2)) - (48*a^2*Cos[c + d*x]^3)/(385*d*Sqrt[a + a*Sin[c
+ d*x]]) - (6*a*Cos[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(77*d) + (4*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^(3/2)
)/(33*d) - (2*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^(5/2))/(11*a*d)

Rule 2878

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

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 ^2(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))^{5/2}}{11 a d}+\frac{2 \int \cos ^2(c+d x) \left (\frac{5 a}{2}-3 a \sin (c+d x)\right ) (a+a \sin (c+d x))^{3/2} \, dx}{11 a}\\ &=\frac{4 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{33 d}-\frac{2 \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 a d}+\frac{3}{11} \int \cos ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\\ &=-\frac{6 a \cos ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{77 d}+\frac{4 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{33 d}-\frac{2 \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 a d}+\frac{1}{77} (24 a) \int \cos ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{48 a^2 \cos ^3(c+d x)}{385 d \sqrt{a+a \sin (c+d x)}}-\frac{6 a \cos ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{77 d}+\frac{4 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{33 d}-\frac{2 \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 a d}+\frac{1}{385} \left (96 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)}{385 d (a+a \sin (c+d x))^{3/2}}-\frac{48 a^2 \cos ^3(c+d x)}{385 d \sqrt{a+a \sin (c+d x)}}-\frac{6 a \cos ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{77 d}+\frac{4 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{33 d}-\frac{2 \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 a d}\\ \end{align*}

Mathematica [A]  time = 1.91679, size = 110, normalized size = 0.71 \[ -\frac{a \sqrt{a (\sin (c+d x)+1)} \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3 (5076 \sin (c+d x)-700 \sin (3 (c+d x))-2280 \cos (2 (c+d x))+105 \cos (4 (c+d x))+4159)}{4620 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*Sin[c + d*x]^2*(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])]*(4159 - 2280*Cos[2*(c + d*x)] + 105*Cos
[4*(c + d*x)] + 5076*Sin[c + d*x] - 700*Sin[3*(c + d*x)]))/(4620*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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Maple [A]  time = 0.868, size = 87, 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 ( 105\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}+350\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+465\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+372\,\sin \left ( dx+c \right ) +248 \right ) }{1155\,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)^2*(a+a*sin(d*x+c))^(3/2),x)

[Out]

-2/1155*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)^2*(105*sin(d*x+c)^4+350*sin(d*x+c)^3+465*sin(d*x+c)^2+372*sin(d*x+c)
+248)/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 )^{2}\,{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)^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*sin(d*x + c)^2, x)

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Fricas [A]  time = 1.63129, size = 468, normalized size = 3. \begin{align*} \frac{2 \,{\left (105 \, a \cos \left (d x + c\right )^{6} + 245 \, a \cos \left (d x + c\right )^{5} - 185 \, a \cos \left (d x + c\right )^{4} - 397 \, a \cos \left (d x + c\right )^{3} + 24 \, a \cos \left (d x + c\right )^{2} - 96 \, a \cos \left (d x + c\right ) +{\left (105 \, a \cos \left (d x + c\right )^{5} - 140 \, a \cos \left (d x + c\right )^{4} - 325 \, a \cos \left (d x + c\right )^{3} + 72 \, a \cos \left (d x + c\right )^{2} + 96 \, a \cos \left (d x + c\right ) + 192 \, a\right )} \sin \left (d x + c\right ) - 192 \, a\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{1155 \,{\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)^2*(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

2/1155*(105*a*cos(d*x + c)^6 + 245*a*cos(d*x + c)^5 - 185*a*cos(d*x + c)^4 - 397*a*cos(d*x + c)^3 + 24*a*cos(d
*x + c)^2 - 96*a*cos(d*x + c) + (105*a*cos(d*x + c)^5 - 140*a*cos(d*x + c)^4 - 325*a*cos(d*x + c)^3 + 72*a*cos
(d*x + c)^2 + 96*a*cos(d*x + c) + 192*a)*sin(d*x + c) - 192*a)*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*si
n(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)**2*(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 )^{2}\,{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)^2*(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)^2, x)