3.1.34 \(\int \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\) [34]

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

[Out]

-2/5*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/a/d-14/15*a*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)+4/15*cos(d*x+c)*(a+a*si
n(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.07, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2838, 2830, 2725} \begin {gather*} -\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}} \end {gather*}

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 2725

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]

Rule 2830

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*S
in[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 2838

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)]

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*}

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Mathematica [A]
time = 0.13, size = 117, normalized size = 1.36 \begin {gather*} -\frac {\sqrt {a (1+\sin (c+d x))} \left (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 )-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 )\right )}{30 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 1.70, size = 63, normalized size = 0.73

method	result	size
default	\(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a \left (\sin \left (d x +c \right )-1\right ) \left (3 \left (\sin ^{2}\left (d x +c \right )\right )+4 \sin \left (d x +c \right )+8\right )}{15 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(63\)

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,method=_RETURNVERBOSE)

[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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.43, size = 92, normalized size = 1.07 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \sin ^{2}{\left (c + d x \right )}\, dx \end {gather*}

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 [A]
time = 0.57, size = 93, normalized size = 1.08 \begin {gather*} \frac {\sqrt {2} {\left (30 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )} \sqrt {a}}{30 \, d} \end {gather*}

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]

1/30*sqrt(2)*(30*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c) + 5*sgn(cos(-1/4*pi + 1/2*
d*x + 1/2*c))*sin(-3/4*pi + 3/2*d*x + 3/2*c) + 3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-5/4*pi + 5/2*d*x + 5
/2*c))*sqrt(a)/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\sin \left (c+d\,x\right )}^2\,\sqrt {a+a\,\sin \left (c+d\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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