3.345 \(\int x \cos (a+b x) \sin ^{\frac {3}{2}}(a+b x) \, dx\)

Optimal. Leaf size=65 \[ -\frac {12 E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{25 b^2}+\frac {4 \sin ^{\frac {3}{2}}(a+b x) \cos (a+b x)}{25 b^2}+\frac {2 x \sin ^{\frac {5}{2}}(a+b x)}{5 b} \]

[Out]

12/25*(sin(1/2*a+1/4*Pi+1/2*b*x)^2)^(1/2)/sin(1/2*a+1/4*Pi+1/2*b*x)*EllipticE(cos(1/2*a+1/4*Pi+1/2*b*x),2^(1/2
))/b^2+4/25*cos(b*x+a)*sin(b*x+a)^(3/2)/b^2+2/5*x*sin(b*x+a)^(5/2)/b

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Rubi [A]  time = 0.03, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3443, 2635, 2639} \[ -\frac {12 E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{25 b^2}+\frac {4 \sin ^{\frac {3}{2}}(a+b x) \cos (a+b x)}{25 b^2}+\frac {2 x \sin ^{\frac {5}{2}}(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[x*Cos[a + b*x]*Sin[a + b*x]^(3/2),x]

[Out]

(-12*EllipticE[(a - Pi/2 + b*x)/2, 2])/(25*b^2) + (4*Cos[a + b*x]*Sin[a + b*x]^(3/2))/(25*b^2) + (2*x*Sin[a +
b*x]^(5/2))/(5*b)

Rule 2635

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

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 3443

Int[Cos[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sin[(a_.) + (b_.)*(x_)^(n_.)]^(p_.), x_Symbol] :> Simp[(x^(m - n
+ 1)*Sin[a + b*x^n]^(p + 1))/(b*n*(p + 1)), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Sin[a + b*x^n]^
(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]

Rubi steps

\begin {align*} \int x \cos (a+b x) \sin ^{\frac {3}{2}}(a+b x) \, dx &=\frac {2 x \sin ^{\frac {5}{2}}(a+b x)}{5 b}-\frac {2 \int \sin ^{\frac {5}{2}}(a+b x) \, dx}{5 b}\\ &=\frac {4 \cos (a+b x) \sin ^{\frac {3}{2}}(a+b x)}{25 b^2}+\frac {2 x \sin ^{\frac {5}{2}}(a+b x)}{5 b}-\frac {6 \int \sqrt {\sin (a+b x)} \, dx}{25 b}\\ &=-\frac {12 E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right )}{25 b^2}+\frac {4 \cos (a+b x) \sin ^{\frac {3}{2}}(a+b x)}{25 b^2}+\frac {2 x \sin ^{\frac {5}{2}}(a+b x)}{5 b}\\ \end {align*}

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Mathematica [C]  time = 0.90, size = 108, normalized size = 1.66 \[ \frac {\sqrt {\sin (a+b x)} \left (4 \tan \left (\frac {1}{2} (a+b x)\right ) \sqrt {\sec ^2\left (\frac {1}{2} (a+b x)\right )} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )+2 \sin (2 (a+b x))-5 b x \cos (2 (a+b x))-12 \tan \left (\frac {1}{2} (a+b x)\right )+5 b x\right )}{25 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Cos[a + b*x]*Sin[a + b*x]^(3/2),x]

[Out]

(Sqrt[Sin[a + b*x]]*(5*b*x - 5*b*x*Cos[2*(a + b*x)] + 2*Sin[2*(a + b*x)] - 12*Tan[(a + b*x)/2] + 4*Hypergeomet
ric2F1[1/2, 3/4, 7/4, -Tan[(a + b*x)/2]^2]*Sqrt[Sec[(a + b*x)/2]^2]*Tan[(a + b*x)/2]))/(25*b^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*cos(b*x+a)*sin(b*x+a)^(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]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \cos \left (b x + a\right ) \sin \left (b x + a\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*cos(b*x + a)*sin(b*x + a)^(3/2), x)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int x \cos \left (b x +a \right ) \left (\sin ^{\frac {3}{2}}\left (b x +a \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cos(b*x+a)*sin(b*x+a)^(3/2),x)

[Out]

int(x*cos(b*x+a)*sin(b*x+a)^(3/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*cos(b*x + a)*sin(b*x + a)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x\,\cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cos(a + b*x)*sin(a + b*x)^(3/2),x)

[Out]

int(x*cos(a + b*x)*sin(a + b*x)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cos(b*x+a)*sin(b*x+a)**(3/2),x)

[Out]

Integral(x*sin(a + b*x)**(3/2)*cos(a + b*x), x)

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