3.356 \(\int (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x))^3 \, dx\)

Optimal. Leaf size=178 \[ \frac{5 b \left (b^2+c^2\right ) \sin (d+e x)}{2 e}-\frac{5 c \left (b^2+c^2\right ) \cos (d+e x)}{2 e}-\frac{(c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2}{3 e}-\frac{5 \sqrt{b^2+c^2} (c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )}{6 e}+\frac{5}{2} x \left (b^2+c^2\right )^{3/2} \]

[Out]

(5*(b^2 + c^2)^(3/2)*x)/2 - (5*c*(b^2 + c^2)*Cos[d + e*x])/(2*e) + (5*b*(b^2 + c^2)*Sin[d + e*x])/(2*e) - (5*S
qrt[b^2 + c^2]*(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]))/(6*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)/(3*e)

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Rubi [A]  time = 0.101832, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3113, 2637, 2638} \[ \frac{5 b \left (b^2+c^2\right ) \sin (d+e x)}{2 e}-\frac{5 c \left (b^2+c^2\right ) \cos (d+e x)}{2 e}-\frac{(c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2}{3 e}-\frac{5 \sqrt{b^2+c^2} (c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )}{6 e}+\frac{5}{2} x \left (b^2+c^2\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(b^2 + c^2)^(3/2)*x)/2 - (5*c*(b^2 + c^2)*Cos[d + e*x])/(2*e) + (5*b*(b^2 + c^2)*Sin[d + e*x])/(2*e) - (5*S
qrt[b^2 + c^2]*(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]))/(6*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)/(3*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 )^3 \, 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}{3 e}+\frac{1}{3} \left (5 \sqrt{b^2+c^2}\right ) \int \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2 \, dx\\ &=-\frac{5 \sqrt{b^2+c^2} (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 )}{6 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}{3 e}+\frac{1}{2} \left (5 \left (b^2+c^2\right )\right ) \int \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right ) \, dx\\ &=\frac{5}{2} \left (b^2+c^2\right )^{3/2} x-\frac{5 \sqrt{b^2+c^2} (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 )}{6 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}{3 e}+\frac{1}{2} \left (5 b \left (b^2+c^2\right )\right ) \int \cos (d+e x) \, dx+\frac{1}{2} \left (5 c \left (b^2+c^2\right )\right ) \int \sin (d+e x) \, dx\\ &=\frac{5}{2} \left (b^2+c^2\right )^{3/2} x-\frac{5 c \left (b^2+c^2\right ) \cos (d+e x)}{2 e}+\frac{5 b \left (b^2+c^2\right ) \sin (d+e x)}{2 e}-\frac{5 \sqrt{b^2+c^2} (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 )}{6 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}{3 e}\\ \end{align*}

Mathematica [C]  time = 0.661051, size = 163, normalized size = 0.92 \[ \frac{30 (b-i c) (b+i c) \sqrt{b^2+c^2} (d+e x)+45 b \left (b^2+c^2\right ) \sin (d+e x)+9 \left (b^2-c^2\right ) \sqrt{b^2+c^2} \sin (2 (d+e x))+b \left (b^2-3 c^2\right ) \sin (3 (d+e x))-45 c \left (b^2+c^2\right ) \cos (d+e x)-18 b c \sqrt{b^2+c^2} \cos (2 (d+e x))+c \left (c^2-3 b^2\right ) \cos (3 (d+e x))}{12 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(30*(b - I*c)*(b + I*c)*Sqrt[b^2 + c^2]*(d + e*x) - 45*c*(b^2 + c^2)*Cos[d + e*x] - 18*b*c*Sqrt[b^2 + c^2]*Cos
[2*(d + e*x)] + c*(-3*b^2 + c^2)*Cos[3*(d + e*x)] + 45*b*(b^2 + c^2)*Sin[d + e*x] + 9*(b^2 - c^2)*Sqrt[b^2 + c
^2]*Sin[2*(d + e*x)] + b*(b^2 - 3*c^2)*Sin[3*(d + e*x)])/(12*e)

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Maple [A]  time = 0.085, size = 250, normalized size = 1.4 \begin{align*}{\frac{1}{e} \left ({\frac{{b}^{3} \left ( 2+ \left ( \cos \left ( ex+d \right ) \right ) ^{2} \right ) \sin \left ( ex+d \right ) }{3}}- \left ( \cos \left ( ex+d \right ) \right ) ^{3}{b}^{2}c+3\,\sqrt{{b}^{2}+{c}^{2}}{b}^{2} \left ( 1/2\,\sin \left ( ex+d \right ) \cos \left ( ex+d \right ) +1/2\,ex+d/2 \right ) +b{c}^{2} \left ( \sin \left ( ex+d \right ) \right ) ^{3}-3\,\sqrt{{b}^{2}+{c}^{2}}bc \left ( \cos \left ( ex+d \right ) \right ) ^{2}+3\,\sin \left ( ex+d \right ){b}^{3}+3\,b{c}^{2}\sin \left ( ex+d \right ) -{\frac{{c}^{3} \left ( 2+ \left ( \sin \left ( ex+d \right ) \right ) ^{2} \right ) \cos \left ( ex+d \right ) }{3}}+3\,\sqrt{{b}^{2}+{c}^{2}}{c}^{2} \left ( -1/2\,\sin \left ( ex+d \right ) \cos \left ( ex+d \right ) +1/2\,ex+d/2 \right ) -3\,\cos \left ( ex+d \right ){b}^{2}c-3\,\cos \left ( ex+d \right ){c}^{3}+\sqrt{{b}^{2}+{c}^{2}}{b}^{2} \left ( ex+d \right ) +\sqrt{{b}^{2}+{c}^{2}}{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))^3,x)

[Out]

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

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Maxima [A]  time = 1.00097, size = 279, normalized size = 1.57 \begin{align*} -\frac{b^{2} c \cos \left (e x + d\right )^{3}}{e} + \frac{b c^{2} \sin \left (e x + d\right )^{3}}{e} - \frac{{\left (\sin \left (e x + d\right )^{3} - 3 \, \sin \left (e x + d\right )\right )} b^{3}}{3 \, e} + \frac{{\left (\cos \left (e x + d\right )^{3} - 3 \, \cos \left (e x + d\right )\right )} c^{3}}{3 \, e} +{\left (b^{2} + c^{2}\right )}^{\frac{3}{2}} x - 3 \,{\left (b^{2} + c^{2}\right )}{\left (\frac{c \cos \left (e x + d\right )}{e} - \frac{b \sin \left (e x + d\right )}{e}\right )} - \frac{3}{4} \,{\left (\frac{4 \, 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}}{e} - \frac{{\left (2 \, e x + 2 \, d - \sin \left (2 \, e x + 2 \, d\right )\right )} c^{2}}{e}\right )} \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))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.25459, size = 342, normalized size = 1.92 \begin{align*} -\frac{2 \,{\left (3 \, b^{2} c - c^{3}\right )} \cos \left (e x + d\right )^{3} + 6 \,{\left (3 \, b^{2} c + 4 \, c^{3}\right )} \cos \left (e x + d\right ) - 2 \,{\left (11 \, b^{3} + 12 \, b c^{2} +{\left (b^{3} - 3 \, b c^{2}\right )} \cos \left (e x + d\right )^{2}\right )} \sin \left (e x + d\right ) + 3 \,{\left (6 \, b c \cos \left (e x + d\right )^{2} - 5 \,{\left (b^{2} + c^{2}\right )} e x - 3 \,{\left (b^{2} - c^{2}\right )} \cos \left (e x + d\right ) \sin \left (e x + d\right )\right )} \sqrt{b^{2} + c^{2}}}{6 \, 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))^3,x, algorithm="fricas")

[Out]

-1/6*(2*(3*b^2*c - c^3)*cos(e*x + d)^3 + 6*(3*b^2*c + 4*c^3)*cos(e*x + d) - 2*(11*b^3 + 12*b*c^2 + (b^3 - 3*b*
c^2)*cos(e*x + d)^2)*sin(e*x + d) + 3*(6*b*c*cos(e*x + d)^2 - 5*(b^2 + c^2)*e*x - 3*(b^2 - c^2)*cos(e*x + d)*s
in(e*x + d))*sqrt(b^2 + c^2))/e

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Sympy [A]  time = 1.881, size = 415, normalized size = 2.33 \begin{align*} \begin{cases} \frac{2 b^{3} \sin ^{3}{\left (d + e x \right )}}{3 e} + \frac{b^{3} \sin{\left (d + e x \right )} \cos ^{2}{\left (d + e x \right )}}{e} + \frac{3 b^{3} \sin{\left (d + e x \right )}}{e} - \frac{b^{2} c \cos ^{3}{\left (d + e x \right )}}{e} - \frac{3 b^{2} c \cos{\left (d + e x \right )}}{e} + \frac{3 b^{2} x \sqrt{b^{2} + c^{2}} \sin ^{2}{\left (d + e x \right )}}{2} + \frac{3 b^{2} x \sqrt{b^{2} + c^{2}} \cos ^{2}{\left (d + e x \right )}}{2} + b^{2} x \sqrt{b^{2} + c^{2}} + \frac{3 b^{2} \sqrt{b^{2} + c^{2}} \sin{\left (d + e x \right )} \cos{\left (d + e x \right )}}{2 e} + \frac{b c^{2} \sin ^{3}{\left (d + e x \right )}}{e} + \frac{3 b c^{2} \sin{\left (d + e x \right )}}{e} + \frac{3 b c \sqrt{b^{2} + c^{2}} \sin ^{2}{\left (d + e x \right )}}{e} - \frac{c^{3} \sin ^{2}{\left (d + e x \right )} \cos{\left (d + e x \right )}}{e} - \frac{2 c^{3} \cos ^{3}{\left (d + e x \right )}}{3 e} - \frac{3 c^{3} \cos{\left (d + e x \right )}}{e} + \frac{3 c^{2} x \sqrt{b^{2} + c^{2}} \sin ^{2}{\left (d + e x \right )}}{2} + \frac{3 c^{2} x \sqrt{b^{2} + c^{2}} \cos ^{2}{\left (d + e x \right )}}{2} + c^{2} x \sqrt{b^{2} + c^{2}} - \frac{3 c^{2} \sqrt{b^{2} + c^{2}} \sin{\left (d + e x \right )} \cos{\left (d + e x \right )}}{2 e} & \text{for}\: e \neq 0 \\x \left (b \cos{\left (d \right )} + c \sin{\left (d \right )} + \sqrt{b^{2} + c^{2}}\right )^{3} & \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))**3,x)

[Out]

Piecewise((2*b**3*sin(d + e*x)**3/(3*e) + b**3*sin(d + e*x)*cos(d + e*x)**2/e + 3*b**3*sin(d + e*x)/e - b**2*c
*cos(d + e*x)**3/e - 3*b**2*c*cos(d + e*x)/e + 3*b**2*x*sqrt(b**2 + c**2)*sin(d + e*x)**2/2 + 3*b**2*x*sqrt(b*
*2 + c**2)*cos(d + e*x)**2/2 + b**2*x*sqrt(b**2 + c**2) + 3*b**2*sqrt(b**2 + c**2)*sin(d + e*x)*cos(d + e*x)/(
2*e) + b*c**2*sin(d + e*x)**3/e + 3*b*c**2*sin(d + e*x)/e + 3*b*c*sqrt(b**2 + c**2)*sin(d + e*x)**2/e - c**3*s
in(d + e*x)**2*cos(d + e*x)/e - 2*c**3*cos(d + e*x)**3/(3*e) - 3*c**3*cos(d + e*x)/e + 3*c**2*x*sqrt(b**2 + c*
*2)*sin(d + e*x)**2/2 + 3*c**2*x*sqrt(b**2 + c**2)*cos(d + e*x)**2/2 + c**2*x*sqrt(b**2 + c**2) - 3*c**2*sqrt(
b**2 + c**2)*sin(d + e*x)*cos(d + e*x)/(2*e), Ne(e, 0)), (x*(b*cos(d) + c*sin(d) + sqrt(b**2 + c**2))**3, True
))

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Giac [A]  time = 1.15752, size = 269, normalized size = 1.51 \begin{align*} -\frac{3}{2} \, \sqrt{b^{2} + c^{2}} b c \cos \left (2 \, x e + 2 \, d\right ) e^{\left (-1\right )} - \frac{1}{12} \,{\left (3 \, b^{2} c - c^{3}\right )} \cos \left (3 \, x e + 3 \, d\right ) e^{\left (-1\right )} - \frac{15}{4} \,{\left (b^{2} c + c^{3}\right )} \cos \left (x e + d\right ) e^{\left (-1\right )} + \frac{1}{12} \,{\left (b^{3} - 3 \, b c^{2}\right )} e^{\left (-1\right )} \sin \left (3 \, x e + 3 \, d\right ) + \frac{3}{4} \,{\left (\sqrt{b^{2} + c^{2}} b^{2} - \sqrt{b^{2} + c^{2}} c^{2}\right )} e^{\left (-1\right )} \sin \left (2 \, x e + 2 \, d\right ) + \frac{15}{4} \,{\left (b^{3} + b c^{2}\right )} e^{\left (-1\right )} \sin \left (x e + d\right ) +{\left (b^{2} + c^{2}\right )}^{\frac{3}{2}} x + \frac{3}{2} \,{\left (\sqrt{b^{2} + c^{2}} b^{2} + \sqrt{b^{2} + c^{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))^3,x, algorithm="giac")

[Out]

-3/2*sqrt(b^2 + c^2)*b*c*cos(2*x*e + 2*d)*e^(-1) - 1/12*(3*b^2*c - c^3)*cos(3*x*e + 3*d)*e^(-1) - 15/4*(b^2*c
+ c^3)*cos(x*e + d)*e^(-1) + 1/12*(b^3 - 3*b*c^2)*e^(-1)*sin(3*x*e + 3*d) + 3/4*(sqrt(b^2 + c^2)*b^2 - sqrt(b^
2 + c^2)*c^2)*e^(-1)*sin(2*x*e + 2*d) + 15/4*(b^3 + b*c^2)*e^(-1)*sin(x*e + d) + (b^2 + c^2)^(3/2)*x + 3/2*(sq
rt(b^2 + c^2)*b^2 + sqrt(b^2 + c^2)*c^2)*x