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

Optimal. Leaf size=140 \[ -\frac {2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{a^{3/2} d}-\frac {4 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{5 a^2 d}-\frac {2 \cos ^3(c+d x)}{5 a d \sqrt {a \sin (c+d x)+a}}+\frac {18 \cos (c+d x)}{5 a d \sqrt {a \sin (c+d x)+a}} \]

[Out]

-2*arctanh(1/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)+18/5*cos(d*x+c)/a/d/(a+a*s
in(d*x+c))^(1/2)-2/5*cos(d*x+c)^3/a/d/(a+a*sin(d*x+c))^(1/2)-4/5*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/a^2/d

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Rubi [A]  time = 0.35, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2878, 2858, 2751, 2649, 206} \[ -\frac {4 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{5 a^2 d}-\frac {2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{a^{3/2} d}-\frac {2 \cos ^3(c+d x)}{5 a d \sqrt {a \sin (c+d x)+a}}+\frac {18 \cos (c+d x)}{5 a d \sqrt {a \sin (c+d x)+a}} \]

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]

(-2*Sqrt[2]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(a^(3/2)*d) + (18*Cos[c + d*x]
)/(5*a*d*Sqrt[a + a*Sin[c + d*x]]) - (2*Cos[c + d*x]^3)/(5*a*d*Sqrt[a + a*Sin[c + d*x]]) - (4*Cos[c + d*x]*Sqr
t[a + a*Sin[c + d*x]])/(5*a^2*d)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2751

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

Rule 2858

Int[cos[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_
)]), x_Symbol] :> Simp[(d*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 2))/(b^2*f*(m + 3)), x] - Dist[1/(b^2*(m + 3)
), Int[(a + b*Sin[e + f*x])^(m + 1)*(b*d*(m + 2) - a*c*(m + 3) + (b*c*(m + 3) - a*d*(m + 4))*Sin[e + f*x]), x]
, x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GeQ[m, -3/2] && LtQ[m, 0]

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]

Rubi steps

\begin {align*} \int \frac {\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)}{5 a d \sqrt {a+a \sin (c+d x)}}+\frac {2 \int \frac {\cos ^2(c+d x) \left (-\frac {a}{2}-3 a \sin (c+d x)\right )}{(a+a \sin (c+d x))^{3/2}} \, dx}{5 a}\\ &=-\frac {2 \cos ^3(c+d x)}{5 a d \sqrt {a+a \sin (c+d x)}}-\frac {4 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{5 a^2 d}-\frac {4 \int \frac {-\frac {3 a^2}{4}+\frac {27}{4} a^2 \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{15 a^3}\\ &=\frac {18 \cos (c+d x)}{5 a d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos ^3(c+d x)}{5 a d \sqrt {a+a \sin (c+d x)}}-\frac {4 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{5 a^2 d}+\frac {2 \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx}{a}\\ &=\frac {18 \cos (c+d x)}{5 a d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos ^3(c+d x)}{5 a d \sqrt {a+a \sin (c+d x)}}-\frac {4 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{5 a^2 d}-\frac {4 \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{a d}\\ &=-\frac {2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{a^{3/2} d}+\frac {18 \cos (c+d x)}{5 a d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos ^3(c+d x)}{5 a d \sqrt {a+a \sin (c+d x)}}-\frac {4 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{5 a^2 d}\\ \end {align*}

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Mathematica [C]  time = 0.29, size = 150, normalized size = 1.07 \[ \frac {\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (-30 \sin \left (\frac {1}{2} (c+d x)\right )-5 \sin \left (\frac {3}{2} (c+d x)\right )+\sin \left (\frac {5}{2} (c+d x)\right )+30 \cos \left (\frac {1}{2} (c+d x)\right )-5 \cos \left (\frac {3}{2} (c+d x)\right )-\cos \left (\frac {5}{2} (c+d x)\right )+(40+40 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (c+d x)\right )-1\right )\right )\right )}{10 d (a (\sin (c+d x)+1))^{3/2}} \]

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]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*((40 + 40*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(c +
d*x)/4])] + 30*Cos[(c + d*x)/2] - 5*Cos[(3*(c + d*x))/2] - Cos[(5*(c + d*x))/2] - 30*Sin[(c + d*x)/2] - 5*Sin
[(3*(c + d*x))/2] + Sin[(5*(c + d*x))/2]))/(10*d*(a*(1 + Sin[c + d*x]))^(3/2))

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fricas [A]  time = 0.50, size = 236, normalized size = 1.69 \[ \frac {\frac {5 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) - \frac {2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (d x + c\right ) + 2}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right )}{\sqrt {a}} - 2 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 9\right )} \sin \left (d x + c\right ) - 7 \, \cos \left (d x + c\right ) - 9\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{5 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \]

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))^(3/2),x, algorithm="fricas")

[Out]

1/5*(5*sqrt(2)*(a*cos(d*x + c) + a*sin(d*x + c) + a)*log(-(cos(d*x + c)^2 - (cos(d*x + c) - 2)*sin(d*x + c) -
2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*(cos(d*x + c) - sin(d*x + c) + 1)/sqrt(a) + 3*cos(d*x + c) + 2)/(cos(d*x +
c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2))/sqrt(a) - 2*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 - (
cos(d*x + c)^2 - 2*cos(d*x + c) - 9)*sin(d*x + c) - 7*cos(d*x + c) - 9)*sqrt(a*sin(d*x + c) + a))/(a^2*d*cos(d
*x + c) + a^2*d*sin(d*x + c) + a^2*d)

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giac [B]  time = 0.67, size = 303, normalized size = 2.16 \[ -\frac {2 \, {\left (\frac {\sqrt {2} {\left (10 \, a \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) + 9 \, \sqrt {-a} \sqrt {a}\right )} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {-a} a^{2}} - \frac {10 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} + \sqrt {a}\right )}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + \frac {2 \, {\left ({\left ({\left ({\left ({\left (\frac {3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - \frac {5 \, a}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {10 \, a}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {10 \, a}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {5 \, a}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {3 \, a}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {5}{2}}}\right )}}{5 \, d} \]

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))^(3/2),x, algorithm="giac")

[Out]

-2/5*(sqrt(2)*(10*a*arctan(sqrt(a)/sqrt(-a)) + 9*sqrt(-a)*sqrt(a))*sgn(tan(1/2*d*x + 1/2*c) + 1)/(sqrt(-a)*a^2
) - 10*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a) + sqrt(a
))/sqrt(-a))/(sqrt(-a)*a*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 2*(((((3*a*tan(1/2*d*x + 1/2*c)/sgn(tan(1/2*d*x + 1/
2*c) + 1) - 5*a/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) + 10*a/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(
1/2*d*x + 1/2*c) - 10*a/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) + 5*a/sgn(tan(1/2*d*x + 1/2*c) + 1
))*tan(1/2*d*x + 1/2*c) - 3*a/sgn(tan(1/2*d*x + 1/2*c) + 1))/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(5/2))/d

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maple [A]  time = 1.68, size = 112, normalized size = 0.80 \[ \frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (-5 a^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )+\left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}}+5 a^{2} \sqrt {a -a \sin \left (d x +c \right )}\right )}{5 d \,a^{4} \cos \left (d x +c \right ) \sqrt {a \left (1+\sin \left (d x +c \right )\right )}} \]

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))^(3/2),x)

[Out]

2/5/d/a^4*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(-5*a^(5/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1
/2)/a^(1/2))+(a-a*sin(d*x+c))^(5/2)+5*a^2*(a-a*sin(d*x+c))^(1/2))/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 \[ \int \frac {\cos \left (d x + c\right )^{2} \sin \left (d x + c\right )^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

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))^(3/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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))**(3/2),x)

[Out]

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

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