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

Optimal. Leaf size=98 \[ \frac {16 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (c+d x)+a}}{d^3}+\frac {8 x \sqrt {a \sin (c+d x)+a}}{d^2}-\frac {2 x^2 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (c+d x)+a}}{d} \]

[Out]

8*x*(a+a*sin(d*x+c))^(1/2)/d^2+16*cot(1/2*c+1/4*Pi+1/2*d*x)*(a+a*sin(d*x+c))^(1/2)/d^3-2*x^2*cot(1/2*c+1/4*Pi+
1/2*d*x)*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.10, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3319, 3296, 2638} \[ \frac {8 x \sqrt {a \sin (c+d x)+a}}{d^2}+\frac {16 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (c+d x)+a}}{d^3}-\frac {2 x^2 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (c+d x)+a}}{d} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(8*x*Sqrt[a + a*Sin[c + d*x]])/d^2 + (16*Cot[c/2 + Pi/4 + (d*x)/2]*Sqrt[a + a*Sin[c + d*x]])/d^3 - (2*x^2*Cot[
c/2 + Pi/4 + (d*x)/2]*Sqrt[a + a*Sin[c + d*x]])/d

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3319

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[((2*a)^IntPart[n
]*(a + b*Sin[e + f*x])^FracPart[n])/Sin[e/2 + (a*Pi)/(4*b) + (f*x)/2]^(2*FracPart[n]), Int[(c + d*x)^m*Sin[e/2
+ (a*Pi)/(4*b) + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[n
+ 1/2] && (GtQ[n, 0] || IGtQ[m, 0])

Rubi steps

\begin {align*} \int x^2 \sqrt {a+a \sin (c+d x)} \, dx &=\left (\csc \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sqrt {a+a \sin (c+d x)}\right ) \int x^2 \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \, dx\\ &=-\frac {2 x^2 \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sqrt {a+a \sin (c+d x)}}{d}+\frac {\left (4 \csc \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sqrt {a+a \sin (c+d x)}\right ) \int x \cos \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \, dx}{d}\\ &=\frac {8 x \sqrt {a+a \sin (c+d x)}}{d^2}-\frac {2 x^2 \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sqrt {a+a \sin (c+d x)}}{d}-\frac {\left (8 \csc \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sqrt {a+a \sin (c+d x)}\right ) \int \cos \left (\frac {c}{2}-\frac {\pi }{4}+\frac {d x}{2}\right ) \, dx}{d^2}\\ &=\frac {8 x \sqrt {a+a \sin (c+d x)}}{d^2}+\frac {16 \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sqrt {a+a \sin (c+d x)}}{d^3}-\frac {2 x^2 \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sqrt {a+a \sin (c+d x)}}{d}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 92, normalized size = 0.94 \[ -\frac {2 \sqrt {a (\sin (c+d x)+1)} \left (\left (d^2 x^2-4 d x-8\right ) \cos \left (\frac {1}{2} (c+d x)\right )-\left (d^2 x^2+4 d x-8\right ) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{d^3 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*((-8 - 4*d*x + d^2*x^2)*Cos[(c + d*x)/2] - (-8 + 4*d*x + d^2*x^2)*Sin[(c + d*x)/2])*Sqrt[a*(1 + Sin[c + d*
x])])/(d^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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giac [A]  time = 0.60, size = 92, normalized size = 0.94 \[ 2 \, \sqrt {2} \sqrt {a} {\left (\frac {4 \, x \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d^{2}} + \frac {{\left (d^{2} x^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 8 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(2)*sqrt(a)*(4*x*cos(-1/4*pi + 1/2*d*x + 1/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d^2 + (d^2*x^2*sgn(c
os(-1/4*pi + 1/2*d*x + 1/2*c)) - 8*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))*sin(-1/4*pi + 1/2*d*x + 1/2*c)/d^3)

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maple [C]  time = 0.07, size = 119, normalized size = 1.21 \[ -\frac {i \sqrt {2}\, \sqrt {-a \left (-2-2 \sin \left (d x +c \right )\right )}\, \left (-i d^{2} x^{2}+d^{2} x^{2} {\mathrm e}^{i \left (d x +c \right )}+4 i d x \,{\mathrm e}^{i \left (d x +c \right )}-4 d x +8 i-8 \,{\mathrm e}^{i \left (d x +c \right )}\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1+2 i {\mathrm e}^{i \left (d x +c \right )}\right ) d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*2^(1/2)*(-a*(-2-2*sin(d*x+c)))^(1/2)/(exp(2*I*(d*x+c))-1+2*I*exp(I*(d*x+c)))*(-I*d^2*x^2+d^2*x^2*exp(I*(d*x
+c))+4*I*d*x*exp(I*(d*x+c))-4*d*x+8*I-8*exp(I*(d*x+c)))*(exp(I*(d*x+c))+I)/d^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.84, size = 64, normalized size = 0.65 \[ \frac {2\,\sqrt {a\,\left (\sin \left (c+d\,x\right )+1\right )}\,\left (8\,\cos \left (c+d\,x\right )+4\,d\,x-d^2\,x^2\,\cos \left (c+d\,x\right )+4\,d\,x\,\sin \left (c+d\,x\right )\right )}{d^3\,\left (\sin \left (c+d\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + a*sin(c + d*x))^(1/2),x)

[Out]

(2*(a*(sin(c + d*x) + 1))^(1/2)*(8*cos(c + d*x) + 4*d*x - d^2*x^2*cos(c + d*x) + 4*d*x*sin(c + d*x)))/(d^3*(si
n(c + d*x) + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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