3.130 \(\int x (a+a \sin (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=165 \[ \frac {8 a \sin ^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (e+f x)+a}}{9 f^2}+\frac {16 a \sqrt {a \sin (e+f x)+a}}{3 f^2}-\frac {4 a x \sin \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \cos \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (e+f x)+a}}{3 f}-\frac {8 a x \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (e+f x)+a}}{3 f} \]

[Out]

16/3*a*(a+a*sin(f*x+e))^(1/2)/f^2-8/3*a*x*cot(1/2*e+1/4*Pi+1/2*f*x)*(a+a*sin(f*x+e))^(1/2)/f-4/3*a*x*cos(1/2*e
+1/4*Pi+1/2*f*x)*sin(1/2*e+1/4*Pi+1/2*f*x)*(a+a*sin(f*x+e))^(1/2)/f+8/9*a*sin(1/2*e+1/4*Pi+1/2*f*x)^2*(a+a*sin
(f*x+e))^(1/2)/f^2

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Rubi [A]  time = 0.09, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3319, 3310, 3296, 2638} \[ \frac {8 a \sin ^2\left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (e+f x)+a}}{9 f^2}+\frac {16 a \sqrt {a \sin (e+f x)+a}}{3 f^2}-\frac {4 a x \sin \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \cos \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (e+f x)+a}}{3 f}-\frac {8 a x \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (e+f x)+a}}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + a*Sin[e + f*x])^(3/2),x]

[Out]

(16*a*Sqrt[a + a*Sin[e + f*x]])/(3*f^2) - (8*a*x*Cot[e/2 + Pi/4 + (f*x)/2]*Sqrt[a + a*Sin[e + f*x]])/(3*f) - (
4*a*x*Cos[e/2 + Pi/4 + (f*x)/2]*Sin[e/2 + Pi/4 + (f*x)/2]*Sqrt[a + a*Sin[e + f*x]])/(3*f) + (8*a*Sin[e/2 + Pi/
4 + (f*x)/2]^2*Sqrt[a + a*Sin[e + f*x]])/(9*f^2)

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 3310

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*(b*Sin[e + f*x])^n)/(f^2*n
^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[(b*(c + d*x)*Cos[e + f*
x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

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 (a+a \sin (e+f x))^{3/2} \, dx &=\left (2 a \csc \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \sqrt {a+a \sin (e+f x)}\right ) \int x \sin ^3\left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \, dx\\ &=-\frac {4 a x \cos \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \sin \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \sqrt {a+a \sin (e+f x)}}{3 f}+\frac {8 a \sin ^2\left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \sqrt {a+a \sin (e+f x)}}{9 f^2}+\frac {1}{3} \left (4 a \csc \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \sqrt {a+a \sin (e+f x)}\right ) \int x \sin \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \, dx\\ &=-\frac {8 a x \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \sqrt {a+a \sin (e+f x)}}{3 f}-\frac {4 a x \cos \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \sin \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \sqrt {a+a \sin (e+f x)}}{3 f}+\frac {8 a \sin ^2\left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \sqrt {a+a \sin (e+f x)}}{9 f^2}+\frac {\left (8 a \csc \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \sqrt {a+a \sin (e+f x)}\right ) \int \cos \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \, dx}{3 f}\\ &=\frac {16 a \sqrt {a+a \sin (e+f x)}}{3 f^2}-\frac {8 a x \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \sqrt {a+a \sin (e+f x)}}{3 f}-\frac {4 a x \cos \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \sin \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \sqrt {a+a \sin (e+f x)}}{3 f}+\frac {8 a \sin ^2\left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \sqrt {a+a \sin (e+f x)}}{9 f^2}\\ \end {align*}

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Mathematica [A]  time = 0.71, size = 113, normalized size = 0.68 \[ -\frac {(a (\sin (e+f x)+1))^{3/2} \left (27 (f x-2) \cos \left (\frac {1}{2} (e+f x)\right )+(3 f x+2) \cos \left (\frac {3}{2} (e+f x)\right )+2 \sin \left (\frac {1}{2} (e+f x)\right ) ((3 f x-2) \cos (e+f x)-4 (3 f x+7))\right )}{9 f^2 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + a*Sin[e + f*x])^(3/2),x]

[Out]

-1/9*((27*(-2 + f*x)*Cos[(e + f*x)/2] + (2 + 3*f*x)*Cos[(3*(e + f*x))/2] + 2*(-4*(7 + 3*f*x) + (-2 + 3*f*x)*Co
s[e + f*x])*Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(3/2))/(f^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)

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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*(a+a*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+a*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)sqrt(2*a)*(2*a*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*sin(1/4*(2*f*x+2*exp(1)+pi))/f^2+2/9*a*sign(cos(1/2*(f*x
+exp(1))-1/4*pi))*sin(1/4*(6*f*x+6*exp(1)-pi))/f^2+4*a*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*cos(1/4*(2*f*x-pi)+1
/2*exp(1))/f^2+2*a*x*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*sin(1/4*(2*f*x-pi)+1/2*exp(1))/f-a*x*sign(cos(1/2*(f*x
+exp(1))-1/4*pi))*cos(1/4*(2*f*x+2*exp(1)+pi))/f-1/3*a*x*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*cos(1/4*(6*f*x+6*e
xp(1)-pi))/f)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int x \left (a +a \sin \left (f x +e \right )\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+a*sin(f*x+e))^(3/2),x)

[Out]

int(x*(a+a*sin(f*x+e))^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} x\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+a*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(3/2)*x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + a*sin(e + f*x))^(3/2),x)

[Out]

int(x*(a + a*sin(e + f*x))^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+a*sin(f*x+e))**(3/2),x)

[Out]

Integral(x*(a*(sin(e + f*x) + 1))**(3/2), x)

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