3.1.74 \(\int (a+b \sin ^2(x)) \, dx\) [74]

Optimal. Leaf size=19 \[ a x+\frac {b x}{2}-\frac {1}{2} b \cos (x) \sin (x) \]

[Out]

a*x+1/2*b*x-1/2*b*cos(x)*sin(x)

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Rubi [A]
time = 0.01, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2715, 8} \begin {gather*} a x+\frac {b x}{2}-\frac {1}{2} b \sin (x) \cos (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[a + b*Sin[x]^2,x]

[Out]

a*x + (b*x)/2 - (b*Cos[x]*Sin[x])/2

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2715

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] && Integ
erQ[2*n]

Rubi steps

\begin {align*} \int \left (a+b \sin ^2(x)\right ) \, dx &=a x+b \int \sin ^2(x) \, dx\\ &=a x-\frac {1}{2} b \cos (x) \sin (x)+\frac {1}{2} b \int 1 \, dx\\ &=a x+\frac {b x}{2}-\frac {1}{2} b \cos (x) \sin (x)\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 19, normalized size = 1.00 \begin {gather*} a x+\frac {b x}{2}-\frac {1}{4} b \sin (2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[a + b*Sin[x]^2,x]

[Out]

a*x + (b*x)/2 - (b*Sin[2*x])/4

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Maple [A]
time = 0.08, size = 17, normalized size = 0.89

method	result	size
risch	\(a x +\frac {b x}{2}-\frac {b \sin \left (2 x \right )}{4}\)	\(16\)
default	\(a x +b \left (-\frac {\sin \left (x \right ) \cos \left (x \right )}{2}+\frac {x}{2}\right )\)	\(17\)
norman	\(\frac {b \left (\tan ^{3}\left (\frac {x}{2}\right )\right )+\left (a +\frac {b}{2}\right ) x +\left (a +\frac {b}{2}\right ) x \left (\tan ^{4}\left (\frac {x}{2}\right )\right )+\left (2 a +b \right ) x \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-b \tan \left (\frac {x}{2}\right )}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{2}}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a+b*sin(x)^2,x,method=_RETURNVERBOSE)

[Out]

a*x+b*(-1/2*sin(x)*cos(x)+1/2*x)

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Maxima [A]
time = 0.29, size = 17, normalized size = 0.89 \begin {gather*} \frac {1}{4} \, b {\left (2 \, x - \sin \left (2 \, x\right )\right )} + a x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*(2*x - sin(2*x)) + a*x

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Fricas [A]
time = 0.39, size = 16, normalized size = 0.84 \begin {gather*} -\frac {1}{2} \, b \cos \left (x\right ) \sin \left (x\right ) + \frac {1}{2} \, {\left (2 \, a + b\right )} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a+b*sin(x)^2,x, algorithm="fricas")

[Out]

-1/2*b*cos(x)*sin(x) + 1/2*(2*a + b)*x

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Sympy [A]
time = 0.01, size = 15, normalized size = 0.79 \begin {gather*} a x + b \left (\frac {x}{2} - \frac {\sin {\left (x \right )} \cos {\left (x \right )}}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a+b*sin(x)**2,x)

[Out]

a*x + b*(x/2 - sin(x)*cos(x)/2)

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Giac [A]
time = 0.53, size = 17, normalized size = 0.89 \begin {gather*} \frac {1}{4} \, b {\left (2 \, x - \sin \left (2 \, x\right )\right )} + a x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*(2*x - sin(2*x)) + a*x

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Mupad [B]
time = 13.31, size = 15, normalized size = 0.79 \begin {gather*} x\,\left (a+\frac {b}{2}\right )-\frac {b\,\sin \left (2\,x\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a + b*sin(x)^2,x)

[Out]

x*(a + b/2) - (b*sin(2*x))/4

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