3.821 \(\int -\cos (1-x) \sin (1-x) \sqrt {1+\sin ^2(1-x)} \, dx\)

Optimal. Leaf size=18 \[ \frac {1}{3} \left (\sin ^2(1-x)+1\right )^{3/2} \]

[Out]

1/3*(1+sin(x-1)^2)^(3/2)

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Rubi [A]  time = 0.04, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {3198, 261} \[ \frac {1}{3} \left (\sin ^2(1-x)+1\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[-(Cos[1 - x]*Sin[1 - x]*Sqrt[1 + Sin[1 - x]^2]),x]

[Out]

(1 + Sin[1 - x]^2)^(3/2)/3

Rule 261

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

Rule 3198

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^
2)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(d*ff*x)^n*(1 - ff^2*x^2
)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[
(m - 1)/2]

Rubi steps

\begin {align*} \int -\cos (1-x) \sin (1-x) \sqrt {1+\sin ^2(1-x)} \, dx &=\operatorname {Subst}\left (\int x \sqrt {1+x^2} \, dx,x,\sin (1-x)\right )\\ &=\frac {1}{3} \left (1+\sin ^2(1-x)\right )^{3/2}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 18, normalized size = 1.00 \[ \frac {1}{3} \left (\sin ^2(1-x)+1\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Integrate[-(Cos[1 - x]*Sin[1 - x]*Sqrt[1 + Sin[1 - x]^2]),x]

[Out]

(1 + Sin[1 - x]^2)^(3/2)/3

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fricas [A]  time = 0.63, size = 14, normalized size = 0.78 \[ \frac {1}{3} \, {\left (-\cos \left (x - 1\right )^{2} + 2\right )}^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(-1+x)*sin(-1+x)*(1+sin(-1+x)^2)^(1/2),x, algorithm="fricas")

[Out]

1/3*(-cos(x - 1)^2 + 2)^(3/2)

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giac [A]  time = 0.13, size = 12, normalized size = 0.67 \[ \frac {1}{3} \, {\left (\sin \left (x - 1\right )^{2} + 1\right )}^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(-1+x)*sin(-1+x)*(1+sin(-1+x)^2)^(1/2),x, algorithm="giac")

[Out]

1/3*(sin(x - 1)^2 + 1)^(3/2)

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maple [A]  time = 0.07, size = 13, normalized size = 0.72 \[ \frac {\left (1+\sin ^{2}\left (-1+x \right )\right )^{\frac {3}{2}}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(-1+x)*sin(-1+x)*(1+sin(-1+x)^2)^(1/2),x)

[Out]

1/3*(1+sin(-1+x)^2)^(3/2)

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maxima [A]  time = 0.33, size = 12, normalized size = 0.67 \[ \frac {1}{3} \, {\left (\sin \left (x - 1\right )^{2} + 1\right )}^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(-1+x)*sin(-1+x)*(1+sin(-1+x)^2)^(1/2),x, algorithm="maxima")

[Out]

1/3*(sin(x - 1)^2 + 1)^(3/2)

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mupad [B]  time = 2.99, size = 12, normalized size = 0.67 \[ \frac {{\left ({\sin \left (x-1\right )}^2+1\right )}^{3/2}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x - 1)*sin(x - 1)*(sin(x - 1)^2 + 1)^(1/2),x)

[Out]

(sin(x - 1)^2 + 1)^(3/2)/3

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sympy [B]  time = 0.75, size = 32, normalized size = 1.78 \[ \frac {\sqrt {\sin ^{2}{\left (x - 1 \right )} + 1} \sin ^{2}{\left (x - 1 \right )}}{3} + \frac {\sqrt {\sin ^{2}{\left (x - 1 \right )} + 1}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(-1+x)*sin(-1+x)*(1+sin(-1+x)**2)**(1/2),x)

[Out]

sqrt(sin(x - 1)**2 + 1)*sin(x - 1)**2/3 + sqrt(sin(x - 1)**2 + 1)/3

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