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

Optimal. Leaf size=105 \[ -\frac{2 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{3 a d}+\frac{4 \cos (c+d x)}{3 d \sqrt{a \sin (c+d x)+a}}-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{\sqrt{a} d} \]

[Out]

-((Sqrt[2]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(Sqrt[a]*d)) + (4*Cos[c + d*x])
/(3*d*Sqrt[a + a*Sin[c + d*x]]) - (2*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*a*d)

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Rubi [A]  time = 0.119386, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2759, 2751, 2649, 206} \[ -\frac{2 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{3 a d}+\frac{4 \cos (c+d x)}{3 d \sqrt{a \sin (c+d x)+a}}-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(Sqrt[a]*d)) + (4*Cos[c + d*x])
/(3*d*Sqrt[a + a*Sin[c + d*x]]) - (2*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*a*d)

Rule 2759

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

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 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 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])

Rubi steps

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

Mathematica [C]  time = 0.217577, size = 105, normalized size = 1. \[ -\frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (-2 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3-(6+6 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (c+d x)\right )-1\right )\right )\right )}{3 d \sqrt{a (\sin (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((-6 - 6*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(c + d*x)/4])] - 2*(Cos[(c + d*x)/2] - Sin[(
c + d*x)/2])^3)*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))/(3*d*Sqrt[a*(1 + Sin[c + d*x])])

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Maple [A]  time = 0.566, size = 96, normalized size = 0.9 \begin{align*} -{\frac{1+\sin \left ( dx+c \right ) }{3\,\cos \left ( dx+c \right ){a}^{2}d}\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) } \left ( 3\,{a}^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) -2\, \left ( a-a\sin \left ( dx+c \right ) \right ) ^{3/2} \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(sin(d*x+c)^2/(a+a*sin(d*x+c))^(1/2),x)

[Out]

-1/3*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(3*a^(3/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^
(1/2))-2*(a-a*sin(d*x+c))^(3/2))/a^2/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 \frac{\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(sin(d*x+c)^2/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.73476, size = 585, normalized size = 5.57 \begin{align*} \frac{\frac{3 \, \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}} - 4 \,{\left (\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 \sin \left (d x + c\right ) + a}}{6 \,{\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(sin(d*x+c)^2/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/6*(3*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) - 4*(cos(d*x + c)^2 + (cos(d*x + c) + 2)*s
in(d*x + c) - cos(d*x + c) - 2)*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 )}}{\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(sin(d*x+c)**2/(a+a*sin(d*x+c))**(1/2),x)

[Out]

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

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Giac [B]  time = 3.32096, size = 332, normalized size = 3.16 \begin{align*} \frac{\frac{6 \, \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} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )} - \frac{{\left ({\left (\frac{\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 )}{a^{5}} - \frac{3 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{5}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{3 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{5}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{5}}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{3}{2}}} - \frac{2 \,{\left (3 \, \sqrt{2} a^{\frac{15}{2}} \arctan \left (\frac{\sqrt{a}}{\sqrt{-a}}\right ) + \sqrt{2} \sqrt{-a} a\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{\sqrt{-a} a^{\frac{15}{2}}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^2/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/3*(6*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)*sgn(tan(1/2*d*x + 1/2*c) + 1)) - (((sgn(tan(1/2*d*x + 1/2*c) + 1)*tan(1/2*d*x + 1/2*c)/
a^5 - 3*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^5)*tan(1/2*d*x + 1/2*c) + 3*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^5)*tan(1/2
*d*x + 1/2*c) - sgn(tan(1/2*d*x + 1/2*c) + 1)/a^5)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2) - 2*(3*sqrt(2)*a^(15/2
)*arctan(sqrt(a)/sqrt(-a)) + sqrt(2)*sqrt(-a)*a)*sgn(tan(1/2*d*x + 1/2*c) + 1)/(sqrt(-a)*a^(15/2)))/d