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]

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

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Rubi [A]  time = 0.0153018, antiderivative size = 37, normalized size of antiderivative = 1., 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*}

Mathematica [A]  time = 0.0373104, 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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Maple [A]  time = 0.005, size = 36, normalized size = 1. \begin{align*} -{\frac{c\cos \left ( ex+d \right ) }{e}}+{\frac{b\sin \left ( ex+d \right ) }{e}}+x\sqrt{{b}^{2}+{c}^{2}} \end{align*}

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 = 0.984325, size = 47, normalized size = 1.27 \begin{align*} \sqrt{b^{2} + c^{2}} x - \frac{c \cos \left (e x + d\right )}{e} + \frac{b \sin \left (e x + d\right )}{e} \end{align*}

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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Fricas [A]  time = 2.14202, size = 80, normalized size = 2.16 \begin{align*} \frac{\sqrt{b^{2} + c^{2}} e x - c \cos \left (e x + d\right ) + b \sin \left (e x + d\right )}{e} \end{align*}

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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Sympy [A]  time = 0.163549, size = 42, normalized size = 1.14 \begin{align*} b \left (\begin{cases} \frac{\sin{\left (d + e x \right )}}{e} & \text{for}\: e \neq 0 \\x \cos{\left (d \right )} & \text{otherwise} \end{cases}\right ) + c \left (\begin{cases} - \frac{\cos{\left (d + e x \right )}}{e} & \text{for}\: e \neq 0 \\x \sin{\left (d \right )} & \text{otherwise} \end{cases}\right ) + x \sqrt{b^{2} + c^{2}} \end{align*}

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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Giac [A]  time = 1.13188, size = 47, normalized size = 1.27 \begin{align*} -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 \end{align*}

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