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]

-22/105*a*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(3/2)+12/35*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(1/2)-2/7*cos(d*x+c)^3*(
a+a*sin(d*x+c))^(1/2)/a/d

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Rubi [A]  time = 0.34, antiderivative size = 92, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.35, 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]

-1/105*((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]))/(d*Sqrt[a*(1 + Sin[c + d*x])])

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fricas [A]  time = 0.49, size = 115, normalized size = 1.25 \[ -\frac {2 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 3 \, \cos \left (d x + c\right )^{3} - 29 \, \cos \left (d x + c\right )^{2} + {\left (15 \, \cos \left (d x + c\right )^{3} + 18 \, \cos \left (d x + c\right )^{2} - 11 \, \cos \left (d x + c\right ) - 22\right )} \sin \left (d x + c\right ) + 11 \, \cos \left (d x + c\right ) + 22\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{105 \, {\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a d\right )}} \]

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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giac [B]  time = 0.56, size = 220, normalized size = 2.39 \[ \frac {4 \, {\left (\frac {11 \, \sqrt {2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {a}} - \frac {2 \, {\left ({\left (\frac {7 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {35 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {35 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + \frac {7 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {2 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {7}{2}}}\right )}}{105 \, d} \]

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]

4/105*(11*sqrt(2)*sgn(tan(1/2*d*x + 1/2*c) + 1)/sqrt(a) - 2*((7*a^3/sgn(tan(1/2*d*x + 1/2*c) + 1) - (35*a^3/sg
n(tan(1/2*d*x + 1/2*c) + 1) - (35*a^3/sgn(tan(1/2*d*x + 1/2*c) + 1) - (2*a^3*tan(1/2*d*x + 1/2*c)^2/sgn(tan(1/
2*d*x + 1/2*c) + 1) + 7*a^3/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c))*tan(1/2
*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c)^2 + 2*a^3/sgn(tan(1/2*d*x + 1/2*c) + 1))/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(7
/2))/d

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maple [A]  time = 1.51, size = 64, normalized size = 0.70 \[ -\frac {2 \left (1+\sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right )^{2} \left (15 \left (\sin ^{2}\left (d x +c \right )\right )+12 \sin \left (d x +c \right )+8\right )}{105 d \cos \left (d x +c \right ) \sqrt {a \left (1+\sin \left (d x +c \right )\right )}} \]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^2*sin(c + d*x)^2)/(a + a*sin(c + d*x))^(1/2),x)

[Out]

int((cos(c + d*x)^2*sin(c + d*x)^2)/(a + a*sin(c + d*x))^(1/2), x)

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

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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