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

Optimal. Leaf size=116 \[ \frac{3 b \sqrt{b^2+c^2} \sin (d+e x)}{2 e}-\frac{3 c \sqrt{b^2+c^2} \cos (d+e x)}{2 e}-\frac{(c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )}{2 e}+\frac{3}{2} x \left (b^2+c^2\right ) \]

[Out]

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

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Rubi [A]  time = 0.0584102, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3113, 2637, 2638} \[ \frac{3 b \sqrt{b^2+c^2} \sin (d+e x)}{2 e}-\frac{3 c \sqrt{b^2+c^2} \cos (d+e x)}{2 e}-\frac{(c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )}{2 e}+\frac{3}{2} x \left (b^2+c^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3113

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(n_), x_Symbol] :> -Simp[((c*Cos[d
+ e*x] - b*Sin[d + e*x])*(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n - 1))/(e*n), x] + Dist[(a*(2*n - 1))/n, Int[
(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[a^2 - b^2 - c^2, 0]
&& GtQ[n, 0]

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 )^2 \, dx &=-\frac{(c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )}{2 e}+\frac{1}{2} \left (3 \sqrt{b^2+c^2}\right ) \int \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right ) \, dx\\ &=\frac{3}{2} \left (b^2+c^2\right ) x-\frac{(c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )}{2 e}+\frac{1}{2} \left (3 b \sqrt{b^2+c^2}\right ) \int \cos (d+e x) \, dx+\frac{1}{2} \left (3 c \sqrt{b^2+c^2}\right ) \int \sin (d+e x) \, dx\\ &=\frac{3}{2} \left (b^2+c^2\right ) x-\frac{3 c \sqrt{b^2+c^2} \cos (d+e x)}{2 e}+\frac{3 b \sqrt{b^2+c^2} \sin (d+e x)}{2 e}-\frac{(c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )}{2 e}\\ \end{align*}

Mathematica [A]  time = 0.222119, size = 111, normalized size = 0.96 \[ \frac{8 b \sqrt{b^2+c^2} \sin (d+e x)-8 c \sqrt{b^2+c^2} \cos (d+e x)+b^2 \sin (2 (d+e x))+6 b^2 d+6 b^2 e x-2 b c \cos (2 (d+e x))-c^2 \sin (2 (d+e x))+6 c^2 d+6 c^2 e x}{4 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*b^2*d + 6*c^2*d + 6*b^2*e*x + 6*c^2*e*x - 8*c*Sqrt[b^2 + c^2]*Cos[d + e*x] - 2*b*c*Cos[2*(d + e*x)] + 8*b*S
qrt[b^2 + c^2]*Sin[d + e*x] + b^2*Sin[2*(d + e*x)] - c^2*Sin[2*(d + e*x)])/(4*e)

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Maple [A]  time = 0.061, size = 124, normalized size = 1.1 \begin{align*}{\frac{1}{e} \left ({b}^{2} \left ({\frac{\sin \left ( ex+d \right ) \cos \left ( ex+d \right ) }{2}}+{\frac{ex}{2}}+{\frac{d}{2}} \right ) - \left ( \cos \left ( ex+d \right ) \right ) ^{2}bc+{c}^{2} \left ( -{\frac{\sin \left ( ex+d \right ) \cos \left ( ex+d \right ) }{2}}+{\frac{ex}{2}}+{\frac{d}{2}} \right ) +2\,\sqrt{{b}^{2}+{c}^{2}}b\sin \left ( ex+d \right ) -2\,\sqrt{{b}^{2}+{c}^{2}}c\cos \left ( ex+d \right ) +{b}^{2} \left ( ex+d \right ) +{c}^{2} \left ( ex+d \right ) \right ) } \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))^2,x)

[Out]

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

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Maxima [A]  time = 0.986031, size = 153, normalized size = 1.32 \begin{align*} b^{2} x + c^{2} x - \frac{b c \cos \left (e x + d\right )^{2}}{e} + \frac{{\left (2 \, e x + 2 \, d + \sin \left (2 \, e x + 2 \, d\right )\right )} b^{2}}{4 \, e} + \frac{{\left (2 \, e x + 2 \, d - \sin \left (2 \, e x + 2 \, d\right )\right )} c^{2}}{4 \, e} - 2 \, \sqrt{b^{2} + c^{2}}{\left (\frac{c \cos \left (e x + d\right )}{e} - \frac{b \sin \left (e x + d\right )}{e}\right )} \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))^2,x, algorithm="maxima")

[Out]

b^2*x + c^2*x - b*c*cos(e*x + d)^2/e + 1/4*(2*e*x + 2*d + sin(2*e*x + 2*d))*b^2/e + 1/4*(2*e*x + 2*d - sin(2*e
*x + 2*d))*c^2/e - 2*sqrt(b^2 + c^2)*(c*cos(e*x + d)/e - b*sin(e*x + d)/e)

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Fricas [A]  time = 2.19603, size = 196, normalized size = 1.69 \begin{align*} -\frac{2 \, b c \cos \left (e x + d\right )^{2} - 3 \,{\left (b^{2} + c^{2}\right )} e x -{\left (b^{2} - c^{2}\right )} \cos \left (e x + d\right ) \sin \left (e x + d\right ) + 4 \, \sqrt{b^{2} + c^{2}}{\left (c \cos \left (e x + d\right ) - b \sin \left (e x + d\right )\right )}}{2 \, 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))^2,x, algorithm="fricas")

[Out]

-1/2*(2*b*c*cos(e*x + d)^2 - 3*(b^2 + c^2)*e*x - (b^2 - c^2)*cos(e*x + d)*sin(e*x + d) + 4*sqrt(b^2 + c^2)*(c*
cos(e*x + d) - b*sin(e*x + d)))/e

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Sympy [A]  time = 0.617558, size = 192, normalized size = 1.66 \begin{align*} \begin{cases} \frac{b^{2} x \sin ^{2}{\left (d + e x \right )}}{2} + \frac{b^{2} x \cos ^{2}{\left (d + e x \right )}}{2} + b^{2} x + \frac{b^{2} \sin{\left (d + e x \right )} \cos{\left (d + e x \right )}}{2 e} + \frac{b c \sin ^{2}{\left (d + e x \right )}}{e} + \frac{2 b \sqrt{b^{2} + c^{2}} \sin{\left (d + e x \right )}}{e} + \frac{c^{2} x \sin ^{2}{\left (d + e x \right )}}{2} + \frac{c^{2} x \cos ^{2}{\left (d + e x \right )}}{2} + c^{2} x - \frac{c^{2} \sin{\left (d + e x \right )} \cos{\left (d + e x \right )}}{2 e} - \frac{2 c \sqrt{b^{2} + c^{2}} \cos{\left (d + e x \right )}}{e} & \text{for}\: e \neq 0 \\x \left (b \cos{\left (d \right )} + c \sin{\left (d \right )} + \sqrt{b^{2} + c^{2}}\right )^{2} & \text{otherwise} \end{cases} \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))**2,x)

[Out]

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

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Giac [A]  time = 1.14036, size = 124, normalized size = 1.07 \begin{align*} -\frac{1}{2} \, b c \cos \left (2 \, x e + 2 \, d\right ) e^{\left (-1\right )} - 2 \, \sqrt{b^{2} + c^{2}} c \cos \left (x e + d\right ) e^{\left (-1\right )} + 2 \, \sqrt{b^{2} + c^{2}} b e^{\left (-1\right )} \sin \left (x e + d\right ) + \frac{1}{4} \,{\left (b^{2} - c^{2}\right )} e^{\left (-1\right )} \sin \left (2 \, x e + 2 \, d\right ) + \frac{3}{2} \,{\left (b^{2} + c^{2}\right )} 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))^2,x, algorithm="giac")

[Out]

-1/2*b*c*cos(2*x*e + 2*d)*e^(-1) - 2*sqrt(b^2 + c^2)*c*cos(x*e + d)*e^(-1) + 2*sqrt(b^2 + c^2)*b*e^(-1)*sin(x*
e + d) + 1/4*(b^2 - c^2)*e^(-1)*sin(2*x*e + 2*d) + 3/2*(b^2 + c^2)*x