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

Optimal. Leaf size=86 \[ -\frac{2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}+\frac{4 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{15 d}-\frac{14 a \cos (c+d x)}{15 d \sqrt{a \sin (c+d x)+a}} \]

[Out]

(-14*a*Cos[c + d*x])/(15*d*Sqrt[a + a*Sin[c + d*x]]) + (4*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(15*d) - (2*C
os[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(5*a*d)

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Rubi [A]  time = 0.110794, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2759, 2751, 2646} \[ -\frac{2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}+\frac{4 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{15 d}-\frac{14 a \cos (c+d x)}{15 d \sqrt{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-14*a*Cos[c + d*x])/(15*d*Sqrt[a + a*Sin[c + d*x]]) + (4*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(15*d) - (2*C
os[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(5*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 2646

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

Rubi steps

\begin{align*} \int \sin ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx &=-\frac{2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 a d}+\frac{2 \int \left (\frac{3 a}{2}-a \sin (c+d x)\right ) \sqrt{a+a \sin (c+d x)} \, dx}{5 a}\\ &=\frac{4 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{15 d}-\frac{2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 a d}+\frac{7}{15} \int \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{14 a \cos (c+d x)}{15 d \sqrt{a+a \sin (c+d x)}}+\frac{4 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{15 d}-\frac{2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 a d}\\ \end{align*}

Mathematica [A]  time = 0.18153, size = 117, normalized size = 1.36 \[ -\frac{\sqrt{a (\sin (c+d x)+1)} \left (-30 \sin \left (\frac{1}{2} (c+d x)\right )+5 \sin \left (\frac{3}{2} (c+d x)\right )+3 \sin \left (\frac{5}{2} (c+d x)\right )+30 \cos \left (\frac{1}{2} (c+d x)\right )+5 \cos \left (\frac{3}{2} (c+d x)\right )-3 \cos \left (\frac{5}{2} (c+d x)\right )\right )}{30 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[a*(1 + Sin[c + d*x])]*(30*Cos[(c + d*x)/2] + 5*Cos[(3*(c + d*x))/2] - 3*Cos[(5*(c + d*x))/2] - 30*Sin[(
c + d*x)/2] + 5*Sin[(3*(c + d*x))/2] + 3*Sin[(5*(c + d*x))/2]))/(30*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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Maple [A]  time = 0.44, size = 63, normalized size = 0.7 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ) a \left ( \sin \left ( dx+c \right ) -1 \right ) \left ( 3\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+4\,\sin \left ( dx+c \right ) +8 \right ) }{15\,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(sin(d*x+c)^2*(a+a*sin(d*x+c))^(1/2),x)

[Out]

2/15*(1+sin(d*x+c))*a*(sin(d*x+c)-1)*(3*sin(d*x+c)^2+4*sin(d*x+c)+8)/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 \sqrt{a \sin \left (d x + c\right ) + a} \sin \left (d x + c\right )^{2}\,{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(sqrt(a*sin(d*x + c) + a)*sin(d*x + c)^2, x)

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Fricas [A]  time = 1.39477, size = 246, normalized size = 2.86 \begin{align*} \frac{2 \,{\left (3 \, \cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )^{2} -{\left (3 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) - 7\right )} \sin \left (d x + c\right ) - 11 \, \cos \left (d x + c\right ) - 7\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{15 \,{\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(sin(d*x+c)^2*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/15*(3*cos(d*x + c)^3 - cos(d*x + c)^2 - (3*cos(d*x + c)^2 + 4*cos(d*x + c) - 7)*sin(d*x + c) - 11*cos(d*x +
c) - 7)*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \sin \left (d x + c\right ) + a} \sin \left (d x + c\right )^{2}\,{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="giac")

[Out]

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