3.358 \(\int (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)) \, dx\)

Optimal. Leaf size=37 \[ x \sqrt {b^2+c^2}+\frac {b \sin (d+e x)}{e}-\frac {c \cos (d+e x)}{e} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2637, 2638} \[ x \sqrt {b^2+c^2}+\frac {b \sin (d+e x)}{e}-\frac {c \cos (d+e x)}{e} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*x],x]

[Out]

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

Rule 2637

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

Rule 2638

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 36, normalized size = 0.97 \[ \frac {e x \sqrt {b^2+c^2}+b \sin (d+e x)-c \cos (d+e x)}{e} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*x],x]

[Out]

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

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fricas [A]  time = 0.92, size = 34, normalized size = 0.92 \[ \frac {\sqrt {b^{2} + c^{2}} e x - c \cos \left (e x + d\right ) + b \sin \left (e x + d\right )}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.12, size = 35, normalized size = 0.95 \[ -c \cos \left (x e + d\right ) e^{\left (-1\right )} + b e^{\left (-1\right )} \sin \left (x e + d\right ) + \sqrt {b^{2} + c^{2}} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*cos(x*e + d)*e^(-1) + b*e^(-1)*sin(x*e + d) + sqrt(b^2 + c^2)*x

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maple [A]  time = 0.04, size = 36, normalized size = 0.97 \[ -\frac {c \cos \left (e x +d \right )}{e}+\frac {b \sin \left (e x +d \right )}{e}+x \sqrt {b^{2}+c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 35, normalized size = 0.95 \[ \sqrt {b^{2} + c^{2}} x - \frac {c \cos \left (e x + d\right )}{e} + \frac {b \sin \left (e x + d\right )}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.67, size = 48, normalized size = 1.30 \[ x\,\sqrt {b^2+c^2}-\frac {2\,c-2\,b\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}{e\,\left ({\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*(b^2 + c^2)^(1/2) - (2*c - 2*b*tan(d/2 + (e*x)/2))/(e*(tan(d/2 + (e*x)/2)^2 + 1))

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sympy [A]  time = 0.14, size = 42, normalized size = 1.14 \[ b \left (\begin {cases} \frac {\sin {\left (d + e x \right )}}{e} & \text {for}\: e \neq 0 \\x \cos {\relax (d )} & \text {otherwise} \end {cases}\right ) + c \left (\begin {cases} - \frac {\cos {\left (d + e x \right )}}{e} & \text {for}\: e \neq 0 \\x \sin {\relax (d )} & \text {otherwise} \end {cases}\right ) + x \sqrt {b^{2} + c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*Piecewise((sin(d + e*x)/e, Ne(e, 0)), (x*cos(d), True)) + c*Piecewise((-cos(d + e*x)/e, Ne(e, 0)), (x*sin(d)
, True)) + x*sqrt(b**2 + c**2)

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