3.2.89 \(\int \sin (a+b \sqrt {c+d x}) \, dx\) [189]

Optimal. Leaf size=54 \[ -\frac {2 \sqrt {c+d x} \cos \left (a+b \sqrt {c+d x}\right )}{b d}+\frac {2 \sin \left (a+b \sqrt {c+d x}\right )}{b^2 d} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3442, 3377, 2717} \begin {gather*} \frac {2 \sin \left (a+b \sqrt {c+d x}\right )}{b^2 d}-\frac {2 \sqrt {c+d x} \cos \left (a+b \sqrt {c+d x}\right )}{b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*Sqrt[c + d*x]],x]

[Out]

(-2*Sqrt[c + d*x]*Cos[a + b*Sqrt[c + d*x]])/(b*d) + (2*Sin[a + b*Sqrt[c + d*x]])/(b^2*d)

Rule 2717

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

Rule 3377

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 3442

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :> Dist[1/(n*f), Subst[Int[x
^(1/n - 1)*(a + b*Sin[c + d*x])^p, x], x, (e + f*x)^n], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && In
tegerQ[1/n]

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 50, normalized size = 0.93 \begin {gather*} \frac {-2 b \sqrt {c+d x} \cos \left (a+b \sqrt {c+d x}\right )+2 \sin \left (a+b \sqrt {c+d x}\right )}{b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*Sqrt[c + d*x]],x]

[Out]

(-2*b*Sqrt[c + d*x]*Cos[a + b*Sqrt[c + d*x]] + 2*Sin[a + b*Sqrt[c + d*x]])/(b^2*d)

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Maple [A]
time = 0.01, size = 61, normalized size = 1.13

method	result	size
derivativedivides	\(\frac {2 \sin \left (a +b \sqrt {d x +c}\right )-2 \left (a +b \sqrt {d x +c}\right ) \cos \left (a +b \sqrt {d x +c}\right )+2 a \cos \left (a +b \sqrt {d x +c}\right )}{b^{2} d}\)	\(61\)
default	\(\frac {2 \sin \left (a +b \sqrt {d x +c}\right )-2 \left (a +b \sqrt {d x +c}\right ) \cos \left (a +b \sqrt {d x +c}\right )+2 a \cos \left (a +b \sqrt {d x +c}\right )}{b^{2} d}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a+b*(d*x+c)^(1/2)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 62, normalized size = 1.15 \begin {gather*} -\frac {2 \, {\left ({\left (\sqrt {d x + c} b + a\right )} \cos \left (\sqrt {d x + c} b + a\right ) - a \cos \left (\sqrt {d x + c} b + a\right ) - \sin \left (\sqrt {d x + c} b + a\right )\right )}}{b^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*((sqrt(d*x + c)*b + a)*cos(sqrt(d*x + c)*b + a) - a*cos(sqrt(d*x + c)*b + a) - sin(sqrt(d*x + c)*b + a))/(b
^2*d)

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Fricas [A]
time = 0.37, size = 44, normalized size = 0.81 \begin {gather*} -\frac {2 \, {\left (\sqrt {d x + c} b \cos \left (\sqrt {d x + c} b + a\right ) - \sin \left (\sqrt {d x + c} b + a\right )\right )}}{b^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(sqrt(d*x + c)*b*cos(sqrt(d*x + c)*b + a) - sin(sqrt(d*x + c)*b + a))/(b^2*d)

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Sympy [A]
time = 0.15, size = 65, normalized size = 1.20 \begin {gather*} \begin {cases} x \sin {\left (a \right )} & \text {for}\: b = 0 \wedge \left (b = 0 \vee d = 0\right ) \\x \sin {\left (a + b \sqrt {c} \right )} & \text {for}\: d = 0 \\- \frac {2 \sqrt {c + d x} \cos {\left (a + b \sqrt {c + d x} \right )}}{b d} + \frac {2 \sin {\left (a + b \sqrt {c + d x} \right )}}{b^{2} d} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*sin(a), Eq(b, 0) & (Eq(b, 0) | Eq(d, 0))), (x*sin(a + b*sqrt(c)), Eq(d, 0)), (-2*sqrt(c + d*x)*co
s(a + b*sqrt(c + d*x))/(b*d) + 2*sin(a + b*sqrt(c + d*x))/(b**2*d), True))

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Giac [A]
time = 6.44, size = 44, normalized size = 0.81 \begin {gather*} -\frac {2 \, {\left (\sqrt {d x + c} b \cos \left (\sqrt {d x + c} b + a\right ) - \sin \left (\sqrt {d x + c} b + a\right )\right )}}{b^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(sqrt(d*x + c)*b*cos(sqrt(d*x + c)*b + a) - sin(sqrt(d*x + c)*b + a))/(b^2*d)

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Mupad [B]
time = 4.73, size = 43, normalized size = 0.80 \begin {gather*} \frac {2\,\left (\sin \left (a+b\,\sqrt {c+d\,x}\right )-b\,\cos \left (a+b\,\sqrt {c+d\,x}\right )\,\sqrt {c+d\,x}\right )}{b^2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*(c + d*x)^(1/2)),x)

[Out]

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

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