3.337 \(\int \frac{\cos ^2(c+d x) \sin ^2(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx\)

Optimal. Leaf size=92 \[ -\frac{2 \cos ^3(c+d x) \sqrt{a \sin (c+d x)+a}}{7 a d}+\frac{12 \cos ^3(c+d x)}{35 d \sqrt{a \sin (c+d x)+a}}-\frac{22 a \cos ^3(c+d x)}{105 d (a \sin (c+d x)+a)^{3/2}} \]

[Out]

(-22*a*Cos[c + d*x]^3)/(105*d*(a + a*Sin[c + d*x])^(3/2)) + (12*Cos[c + d*x]^3)/(35*d*Sqrt[a + a*Sin[c + d*x]]
) - (2*Cos[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(7*a*d)

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Rubi [A]  time = 0.342793, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {2877, 2856, 2674, 2673} \[ -\frac{2 \cos ^3(c+d x) \sqrt{a \sin (c+d x)+a}}{7 a d}+\frac{12 \cos ^3(c+d x)}{35 d \sqrt{a \sin (c+d x)+a}}-\frac{22 a \cos ^3(c+d x)}{105 d (a \sin (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-22*a*Cos[c + d*x]^3)/(105*d*(a + a*Sin[c + d*x])^(3/2)) + (12*Cos[c + d*x]^3)/(35*d*Sqrt[a + a*Sin[c + d*x]]
) - (2*Cos[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(7*a*d)

Rule 2877

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^2*((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)/(a*f*g*(2*m + p + 1)), x] - Dist[1/(a^
2*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1)*(a*m - b*(2*m + p + 1)*Sin[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && LeQ[m, -2^(-1)] && NeQ[2*m + p + 1, 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 \frac{\cos ^2(c+d x) \sin ^2(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx &=\frac{\cos ^3(c+d x)}{2 d \sqrt{a+a \sin (c+d x)}}-\frac{\int \cos ^2(c+d x) \left (-\frac{a}{2}-2 a \sin (c+d x)\right ) \sqrt{a+a \sin (c+d x)} \, dx}{2 a^2}\\ &=\frac{\cos ^3(c+d x)}{2 d \sqrt{a+a \sin (c+d x)}}-\frac{2 \cos ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{7 a d}+\frac{11 \int \cos ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{28 a}\\ &=\frac{12 \cos ^3(c+d x)}{35 d \sqrt{a+a \sin (c+d x)}}-\frac{2 \cos ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{7 a d}+\frac{11}{35} \int \frac{\cos ^2(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=-\frac{22 a \cos ^3(c+d x)}{105 d (a+a \sin (c+d x))^{3/2}}+\frac{12 \cos ^3(c+d x)}{35 d \sqrt{a+a \sin (c+d x)}}-\frac{2 \cos ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{7 a d}\\ \end{align*}

Mathematica [A]  time = 0.338654, size = 87, normalized size = 0.95 \[ -\frac{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) (24 \sin (c+d x)-15 \cos (2 (c+d x))+31)}{105 d \sqrt{a (\sin (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(31 - 15*Cos[2*(c + d*x)] + 24
*Sin[c + d*x]))/(105*d*Sqrt[a*(1 + Sin[c + d*x])])

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Maple [A]  time = 1.152, size = 64, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ) \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2} \left ( 15\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+12\,\sin \left ( dx+c \right ) +8 \right ) }{105\,d\cos \left ( dx+c \right ) }{\frac{1}{\sqrt{a \left ( 1+\sin \left ( dx+c \right ) \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))^(1/2),x)

[Out]

-2/105/d*(1+sin(d*x+c))*(sin(d*x+c)-1)^2*(15*sin(d*x+c)^2+12*sin(d*x+c)+8)/cos(d*x+c)/(a*(1+sin(d*x+c)))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{2} \sin \left (d x + c\right )^{2}}{\sqrt{a \sin \left (d x + c\right ) + a}}\,{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))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.63284, size = 320, normalized size = 3.48 \begin{align*} -\frac{2 \,{\left (15 \, \cos \left (d x + c\right )^{4} - 3 \, \cos \left (d x + c\right )^{3} - 29 \, \cos \left (d x + c\right )^{2} +{\left (15 \, \cos \left (d x + c\right )^{3} + 18 \, \cos \left (d x + c\right )^{2} - 11 \, \cos \left (d x + c\right ) - 22\right )} \sin \left (d x + c\right ) + 11 \, \cos \left (d x + c\right ) + 22\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{105 \,{\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a 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))^(1/2),x, algorithm="fricas")

[Out]

-2/105*(15*cos(d*x + c)^4 - 3*cos(d*x + c)^3 - 29*cos(d*x + c)^2 + (15*cos(d*x + c)^3 + 18*cos(d*x + c)^2 - 11
*cos(d*x + c) - 22)*sin(d*x + c) + 11*cos(d*x + c) + 22)*sqrt(a*sin(d*x + c) + a)/(a*d*cos(d*x + c) + a*d*sin(
d*x + c) + a*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{\sqrt{a \left (\sin{\left (c + d x \right )} + 1\right )}}\, dx \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))**(1/2),x)

[Out]

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

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Giac [B]  time = 2.34651, size = 277, normalized size = 3.01 \begin{align*} -\frac{\frac{2 \,{\left ({\left ({\left ({\left ({\left (\frac{2 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right ) \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{9}} + \frac{7 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{9}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{35 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{9}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{35 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{9}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{7 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{9}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \frac{2 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{9}}\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{7}{2}}} + \frac{11 \, \sqrt{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{\frac{25}{2}}}}{13440 \, d} \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))^(1/2),x, algorithm="giac")

[Out]

-1/13440*(2*(((((2*sgn(tan(1/2*d*x + 1/2*c) + 1)*tan(1/2*d*x + 1/2*c)^2/a^9 + 7*sgn(tan(1/2*d*x + 1/2*c) + 1)/
a^9)*tan(1/2*d*x + 1/2*c) - 35*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^9)*tan(1/2*d*x + 1/2*c) + 35*sgn(tan(1/2*d*x +
1/2*c) + 1)/a^9)*tan(1/2*d*x + 1/2*c) - 7*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^9)*tan(1/2*d*x + 1/2*c)^2 - 2*sgn(ta
n(1/2*d*x + 1/2*c) + 1)/a^9)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(7/2) + 11*sqrt(2)*sgn(tan(1/2*d*x + 1/2*c) + 1)/a
^(25/2))/d