3.188 \(\int (e+f x) \sin (a+b \sqrt {c+d x}) \, dx\)

Optimal. Leaf size=185 \[ -\frac {12 f \sin \left (a+b \sqrt {c+d x}\right )}{b^4 d^2}+\frac {12 f \sqrt {c+d x} \cos \left (a+b \sqrt {c+d x}\right )}{b^3 d^2}+\frac {2 (d e-c f) \sin \left (a+b \sqrt {c+d x}\right )}{b^2 d^2}+\frac {6 f (c+d x) \sin \left (a+b \sqrt {c+d x}\right )}{b^2 d^2}-\frac {2 \sqrt {c+d x} (d e-c f) \cos \left (a+b \sqrt {c+d x}\right )}{b d^2}-\frac {2 f (c+d x)^{3/2} \cos \left (a+b \sqrt {c+d x}\right )}{b d^2} \]

[Out]

-2*f*(d*x+c)^(3/2)*cos(a+b*(d*x+c)^(1/2))/b/d^2-12*f*sin(a+b*(d*x+c)^(1/2))/b^4/d^2+2*(-c*f+d*e)*sin(a+b*(d*x+
c)^(1/2))/b^2/d^2+6*f*(d*x+c)*sin(a+b*(d*x+c)^(1/2))/b^2/d^2+12*f*cos(a+b*(d*x+c)^(1/2))*(d*x+c)^(1/2)/b^3/d^2
-2*(-c*f+d*e)*cos(a+b*(d*x+c)^(1/2))*(d*x+c)^(1/2)/b/d^2

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Rubi [A]  time = 0.16, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3431, 3296, 2637} \[ \frac {2 (d e-c f) \sin \left (a+b \sqrt {c+d x}\right )}{b^2 d^2}+\frac {6 f (c+d x) \sin \left (a+b \sqrt {c+d x}\right )}{b^2 d^2}-\frac {12 f \sin \left (a+b \sqrt {c+d x}\right )}{b^4 d^2}+\frac {12 f \sqrt {c+d x} \cos \left (a+b \sqrt {c+d x}\right )}{b^3 d^2}-\frac {2 \sqrt {c+d x} (d e-c f) \cos \left (a+b \sqrt {c+d x}\right )}{b d^2}-\frac {2 f (c+d x)^{3/2} \cos \left (a+b \sqrt {c+d x}\right )}{b d^2} \]

Antiderivative was successfully verified.

[In]

Int[(e + f*x)*Sin[a + b*Sqrt[c + d*x]],x]

[Out]

(12*f*Sqrt[c + d*x]*Cos[a + b*Sqrt[c + d*x]])/(b^3*d^2) - (2*(d*e - c*f)*Sqrt[c + d*x]*Cos[a + b*Sqrt[c + d*x]
])/(b*d^2) - (2*f*(c + d*x)^(3/2)*Cos[a + b*Sqrt[c + d*x]])/(b*d^2) - (12*f*Sin[a + b*Sqrt[c + d*x]])/(b^4*d^2
) + (2*(d*e - c*f)*Sin[a + b*Sqrt[c + d*x]])/(b^2*d^2) + (6*f*(c + d*x)*Sin[a + b*Sqrt[c + d*x]])/(b^2*d^2)

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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 3431

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

Rubi steps

\begin {align*} \int (e+f x) \sin \left (a+b \sqrt {c+d x}\right ) \, dx &=\frac {2 \operatorname {Subst}\left (\int \left (\frac {(d e-c f) x \sin (a+b x)}{d}+\frac {f x^3 \sin (a+b x)}{d}\right ) \, dx,x,\sqrt {c+d x}\right )}{d}\\ &=\frac {(2 f) \operatorname {Subst}\left (\int x^3 \sin (a+b x) \, dx,x,\sqrt {c+d x}\right )}{d^2}+\frac {(2 (d e-c f)) \operatorname {Subst}\left (\int x \sin (a+b x) \, dx,x,\sqrt {c+d x}\right )}{d^2}\\ &=-\frac {2 (d e-c f) \sqrt {c+d x} \cos \left (a+b \sqrt {c+d x}\right )}{b d^2}-\frac {2 f (c+d x)^{3/2} \cos \left (a+b \sqrt {c+d x}\right )}{b d^2}+\frac {(6 f) \operatorname {Subst}\left (\int x^2 \cos (a+b x) \, dx,x,\sqrt {c+d x}\right )}{b d^2}+\frac {(2 (d e-c f)) \operatorname {Subst}\left (\int \cos (a+b x) \, dx,x,\sqrt {c+d x}\right )}{b d^2}\\ &=-\frac {2 (d e-c f) \sqrt {c+d x} \cos \left (a+b \sqrt {c+d x}\right )}{b d^2}-\frac {2 f (c+d x)^{3/2} \cos \left (a+b \sqrt {c+d x}\right )}{b d^2}+\frac {2 (d e-c f) \sin \left (a+b \sqrt {c+d x}\right )}{b^2 d^2}+\frac {6 f (c+d x) \sin \left (a+b \sqrt {c+d x}\right )}{b^2 d^2}-\frac {(12 f) \operatorname {Subst}\left (\int x \sin (a+b x) \, dx,x,\sqrt {c+d x}\right )}{b^2 d^2}\\ &=\frac {12 f \sqrt {c+d x} \cos \left (a+b \sqrt {c+d x}\right )}{b^3 d^2}-\frac {2 (d e-c f) \sqrt {c+d x} \cos \left (a+b \sqrt {c+d x}\right )}{b d^2}-\frac {2 f (c+d x)^{3/2} \cos \left (a+b \sqrt {c+d x}\right )}{b d^2}+\frac {2 (d e-c f) \sin \left (a+b \sqrt {c+d x}\right )}{b^2 d^2}+\frac {6 f (c+d x) \sin \left (a+b \sqrt {c+d x}\right )}{b^2 d^2}-\frac {(12 f) \operatorname {Subst}\left (\int \cos (a+b x) \, dx,x,\sqrt {c+d x}\right )}{b^3 d^2}\\ &=\frac {12 f \sqrt {c+d x} \cos \left (a+b \sqrt {c+d x}\right )}{b^3 d^2}-\frac {2 (d e-c f) \sqrt {c+d x} \cos \left (a+b \sqrt {c+d x}\right )}{b d^2}-\frac {2 f (c+d x)^{3/2} \cos \left (a+b \sqrt {c+d x}\right )}{b d^2}-\frac {12 f \sin \left (a+b \sqrt {c+d x}\right )}{b^4 d^2}+\frac {2 (d e-c f) \sin \left (a+b \sqrt {c+d x}\right )}{b^2 d^2}+\frac {6 f (c+d x) \sin \left (a+b \sqrt {c+d x}\right )}{b^2 d^2}\\ \end {align*}

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Mathematica [A]  time = 0.45, size = 85, normalized size = 0.46 \[ \frac {2 \sin \left (a+b \sqrt {c+d x}\right ) \left (b^2 (2 c f+d (e+3 f x))-6 f\right )-2 b \sqrt {c+d x} \left (b^2 d (e+f x)-6 f\right ) \cos \left (a+b \sqrt {c+d x}\right )}{b^4 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(e + f*x)*Sin[a + b*Sqrt[c + d*x]],x]

[Out]

(-2*b*Sqrt[c + d*x]*(-6*f + b^2*d*(e + f*x))*Cos[a + b*Sqrt[c + d*x]] + 2*(-6*f + b^2*(2*c*f + d*(e + 3*f*x)))
*Sin[a + b*Sqrt[c + d*x]])/(b^4*d^2)

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fricas [A]  time = 0.54, size = 86, normalized size = 0.46 \[ -\frac {2 \, {\left ({\left (b^{3} d f x + b^{3} d e - 6 \, b f\right )} \sqrt {d x + c} \cos \left (\sqrt {d x + c} b + a\right ) - {\left (3 \, b^{2} d f x + b^{2} d e + 2 \, {\left (b^{2} c - 3\right )} f\right )} \sin \left (\sqrt {d x + c} b + a\right )\right )}}{b^{4} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*((b^3*d*f*x + b^3*d*e - 6*b*f)*sqrt(d*x + c)*cos(sqrt(d*x + c)*b + a) - (3*b^2*d*f*x + b^2*d*e + 2*(b^2*c -
3)*f)*sin(sqrt(d*x + c)*b + a))/(b^4*d^2)

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giac [A]  time = 0.69, size = 219, normalized size = 1.18 \[ -\frac {2 \, {\left (\frac {{\left (\sqrt {d x + c} b \cos \left (\sqrt {d x + c} b + a\right ) - \sin \left (\sqrt {d x + c} b + a\right )\right )} e}{b} - \frac {f {\left (\frac {{\left ({\left (\sqrt {d x + c} b + a\right )} b^{2} c - a b^{2} c - {\left (\sqrt {d x + c} b + a\right )}^{3} + 3 \, {\left (\sqrt {d x + c} b + a\right )}^{2} a - 3 \, {\left (\sqrt {d x + c} b + a\right )} a^{2} + a^{3} + 6 \, \sqrt {d x + c} b\right )} \cos \left (\sqrt {d x + c} b + a\right )}{b^{2}} - \frac {{\left (b^{2} c - 3 \, {\left (\sqrt {d x + c} b + a\right )}^{2} + 6 \, {\left (\sqrt {d x + c} b + a\right )} a - 3 \, a^{2} + 6\right )} \sin \left (\sqrt {d x + c} b + a\right )}{b^{2}}\right )}}{b d}\right )}}{b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*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))*e/b - f*(((sqrt(d*x + c)*b + a)*b^2*
c - a*b^2*c - (sqrt(d*x + c)*b + a)^3 + 3*(sqrt(d*x + c)*b + a)^2*a - 3*(sqrt(d*x + c)*b + a)*a^2 + a^3 + 6*sq
rt(d*x + c)*b)*cos(sqrt(d*x + c)*b + a)/b^2 - (b^2*c - 3*(sqrt(d*x + c)*b + a)^2 + 6*(sqrt(d*x + c)*b + a)*a -
3*a^2 + 6)*sin(sqrt(d*x + c)*b + a)/b^2)/(b*d))/(b*d)

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maple [B]  time = 0.03, size = 366, normalized size = 1.98 \[ \frac {-2 c f \left (\sin \left (a +b \sqrt {d x +c}\right )-\left (a +b \sqrt {d x +c}\right ) \cos \left (a +b \sqrt {d x +c}\right )\right )+2 d e \left (\sin \left (a +b \sqrt {d x +c}\right )-\left (a +b \sqrt {d x +c}\right ) \cos \left (a +b \sqrt {d x +c}\right )\right )-2 a c f \cos \left (a +b \sqrt {d x +c}\right )+2 a d e \cos \left (a +b \sqrt {d x +c}\right )+\frac {2 f \left (-\left (a +b \sqrt {d x +c}\right )^{3} \cos \left (a +b \sqrt {d x +c}\right )+3 \left (a +b \sqrt {d x +c}\right )^{2} \sin \left (a +b \sqrt {d x +c}\right )-6 \sin \left (a +b \sqrt {d x +c}\right )+6 \left (a +b \sqrt {d x +c}\right ) \cos \left (a +b \sqrt {d x +c}\right )\right )}{b^{2}}-\frac {6 a f \left (-\left (a +b \sqrt {d x +c}\right )^{2} \cos \left (a +b \sqrt {d x +c}\right )+2 \cos \left (a +b \sqrt {d x +c}\right )+2 \left (a +b \sqrt {d x +c}\right ) \sin \left (a +b \sqrt {d x +c}\right )\right )}{b^{2}}+\frac {6 a^{2} f \left (\sin \left (a +b \sqrt {d x +c}\right )-\left (a +b \sqrt {d x +c}\right ) \cos \left (a +b \sqrt {d x +c}\right )\right )}{b^{2}}+\frac {2 a^{3} f \cos \left (a +b \sqrt {d x +c}\right )}{b^{2}}}{d^{2} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)*sin(a+b*(d*x+c)^(1/2)),x)

[Out]

2/d^2/b^2*(-c*f*(sin(a+b*(d*x+c)^(1/2))-(a+b*(d*x+c)^(1/2))*cos(a+b*(d*x+c)^(1/2)))+d*e*(sin(a+b*(d*x+c)^(1/2)
)-(a+b*(d*x+c)^(1/2))*cos(a+b*(d*x+c)^(1/2)))-a*c*f*cos(a+b*(d*x+c)^(1/2))+a*d*e*cos(a+b*(d*x+c)^(1/2))+1/b^2*
f*(-(a+b*(d*x+c)^(1/2))^3*cos(a+b*(d*x+c)^(1/2))+3*(a+b*(d*x+c)^(1/2))^2*sin(a+b*(d*x+c)^(1/2))-6*sin(a+b*(d*x
+c)^(1/2))+6*(a+b*(d*x+c)^(1/2))*cos(a+b*(d*x+c)^(1/2)))-3/b^2*a*f*(-(a+b*(d*x+c)^(1/2))^2*cos(a+b*(d*x+c)^(1/
2))+2*cos(a+b*(d*x+c)^(1/2))+2*(a+b*(d*x+c)^(1/2))*sin(a+b*(d*x+c)^(1/2)))+3/b^2*a^2*f*(sin(a+b*(d*x+c)^(1/2))
-(a+b*(d*x+c)^(1/2))*cos(a+b*(d*x+c)^(1/2)))+1/b^2*a^3*f*cos(a+b*(d*x+c)^(1/2)))

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maxima [B]  time = 0.33, size = 348, normalized size = 1.88 \[ \frac {2 \, {\left (a e \cos \left (\sqrt {d x + c} b + a\right ) - \frac {a c f \cos \left (\sqrt {d x + c} b + a\right )}{d} - {\left ({\left (\sqrt {d x + c} b + a\right )} \cos \left (\sqrt {d x + c} b + a\right ) - \sin \left (\sqrt {d x + c} b + a\right )\right )} e + \frac {{\left ({\left (\sqrt {d x + c} b + a\right )} \cos \left (\sqrt {d x + c} b + a\right ) - \sin \left (\sqrt {d x + c} b + a\right )\right )} c f}{d} + \frac {a^{3} f \cos \left (\sqrt {d x + c} b + a\right )}{b^{2} d} - \frac {3 \, {\left ({\left (\sqrt {d x + c} b + a\right )} \cos \left (\sqrt {d x + c} b + a\right ) - \sin \left (\sqrt {d x + c} b + a\right )\right )} a^{2} f}{b^{2} d} + \frac {3 \, {\left ({\left ({\left (\sqrt {d x + c} b + a\right )}^{2} - 2\right )} \cos \left (\sqrt {d x + c} b + a\right ) - 2 \, {\left (\sqrt {d x + c} b + a\right )} \sin \left (\sqrt {d x + c} b + a\right )\right )} a f}{b^{2} d} - \frac {{\left ({\left ({\left (\sqrt {d x + c} b + a\right )}^{3} - 6 \, \sqrt {d x + c} b - 6 \, a\right )} \cos \left (\sqrt {d x + c} b + a\right ) - 3 \, {\left ({\left (\sqrt {d x + c} b + a\right )}^{2} - 2\right )} \sin \left (\sqrt {d x + c} b + a\right )\right )} f}{b^{2} d}\right )}}{b^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(a*e*cos(sqrt(d*x + c)*b + a) - a*c*f*cos(sqrt(d*x + c)*b + a)/d - ((sqrt(d*x + c)*b + a)*cos(sqrt(d*x + c)*
b + a) - sin(sqrt(d*x + c)*b + a))*e + ((sqrt(d*x + c)*b + a)*cos(sqrt(d*x + c)*b + a) - sin(sqrt(d*x + c)*b +
a))*c*f/d + a^3*f*cos(sqrt(d*x + c)*b + a)/(b^2*d) - 3*((sqrt(d*x + c)*b + a)*cos(sqrt(d*x + c)*b + a) - sin(
sqrt(d*x + c)*b + a))*a^2*f/(b^2*d) + 3*(((sqrt(d*x + c)*b + a)^2 - 2)*cos(sqrt(d*x + c)*b + a) - 2*(sqrt(d*x
+ c)*b + a)*sin(sqrt(d*x + c)*b + a))*a*f/(b^2*d) - (((sqrt(d*x + c)*b + a)^3 - 6*sqrt(d*x + c)*b - 6*a)*cos(s
qrt(d*x + c)*b + a) - 3*((sqrt(d*x + c)*b + a)^2 - 2)*sin(sqrt(d*x + c)*b + a))*f/(b^2*d))/(b^2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*(c + d*x)^(1/2))*(e + f*x),x)

[Out]

int(sin(a + b*(c + d*x)^(1/2))*(e + f*x), x)

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sympy [A]  time = 0.76, size = 221, normalized size = 1.19 \[ \begin {cases} \left (e x + \frac {f x^{2}}{2}\right ) \sin {\relax (a )} & \text {for}\: b = 0 \wedge \left (b = 0 \vee d = 0\right ) \\\left (e x + \frac {f x^{2}}{2}\right ) \sin {\left (a + b \sqrt {c} \right )} & \text {for}\: d = 0 \\- \frac {2 e \sqrt {c + d x} \cos {\left (a + b \sqrt {c + d x} \right )}}{b d} - \frac {2 f x \sqrt {c + d x} \cos {\left (a + b \sqrt {c + d x} \right )}}{b d} + \frac {4 c f \sin {\left (a + b \sqrt {c + d x} \right )}}{b^{2} d^{2}} + \frac {2 e \sin {\left (a + b \sqrt {c + d x} \right )}}{b^{2} d} + \frac {6 f x \sin {\left (a + b \sqrt {c + d x} \right )}}{b^{2} d} + \frac {12 f \sqrt {c + d x} \cos {\left (a + b \sqrt {c + d x} \right )}}{b^{3} d^{2}} - \frac {12 f \sin {\left (a + b \sqrt {c + d x} \right )}}{b^{4} d^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((e*x + f*x**2/2)*sin(a), Eq(b, 0) & (Eq(b, 0) | Eq(d, 0))), ((e*x + f*x**2/2)*sin(a + b*sqrt(c)), E
q(d, 0)), (-2*e*sqrt(c + d*x)*cos(a + b*sqrt(c + d*x))/(b*d) - 2*f*x*sqrt(c + d*x)*cos(a + b*sqrt(c + d*x))/(b
*d) + 4*c*f*sin(a + b*sqrt(c + d*x))/(b**2*d**2) + 2*e*sin(a + b*sqrt(c + d*x))/(b**2*d) + 6*f*x*sin(a + b*sqr
t(c + d*x))/(b**2*d) + 12*f*sqrt(c + d*x)*cos(a + b*sqrt(c + d*x))/(b**3*d**2) - 12*f*sin(a + b*sqrt(c + d*x))
/(b**4*d**2), True))

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