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

Optimal. Leaf size=130 \[ \frac{8 \sqrt{b^2+c^2} (c \cos (d+e x)-b \sin (d+e x))}{3 e \sqrt{-\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)}}-\frac{2 (c \cos (d+e x)-b \sin (d+e x)) \sqrt{-\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)}}{3 e} \]

[Out]

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

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Rubi [A]  time = 0.0807832, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {3113, 3112} \[ \frac{8 \sqrt{b^2+c^2} (c \cos (d+e x)-b \sin (d+e x))}{3 e \sqrt{-\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)}}-\frac{2 (c \cos (d+e x)-b \sin (d+e x)) \sqrt{-\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)}}{3 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*Sqrt[b^2 + c^2]*(c*Cos[d + e*x] - b*Sin[d + e*x]))/(3*e*Sqrt[-Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d +
e*x]]) - (2*(c*Cos[d + e*x] - b*Sin[d + e*x])*Sqrt[-Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*x]])/(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 3112

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

Rubi steps

\begin{align*} \int \left (-\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2} \, dx &=-\frac{2 (c \cos (d+e x)-b \sin (d+e x)) \sqrt{-\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)}}{3 e}-\frac{1}{3} \left (4 \sqrt{b^2+c^2}\right ) \int \sqrt{-\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)} \, dx\\ &=\frac{8 \sqrt{b^2+c^2} (c \cos (d+e x)-b \sin (d+e x))}{3 e \sqrt{-\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)}}-\frac{2 (c \cos (d+e x)-b \sin (d+e x)) \sqrt{-\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)}}{3 e}\\ \end{align*}

Mathematica [C]  time = 21.2547, size = 11512, normalized size = 88.55 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [A]  time = 1.591, size = 130, normalized size = 1. \begin{align*}{\frac{ \left ( 2\,\sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) -2 \right ) \left ({b}^{2}+{c}^{2} \right ) \left ( 1+\sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) \right ) \left ( \sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) -5 \right ) }{3\,\cos \left ( ex+d-\arctan \left ( -b,c \right ) \right ) e}{\frac{1}{\sqrt{{({b}^{2}\sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) +{c}^{2}\sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) -{b}^{2}-{c}^{2}){\frac{1}{\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))^(3/2),x)

[Out]

2/3*(sin(e*x+d-arctan(-b,c))-1)*(b^2+c^2)*(1+sin(e*x+d-arctan(-b,c)))*(sin(e*x+d-arctan(-b,c))-5)/cos(e*x+d-ar
ctan(-b,c))/((b^2*sin(e*x+d-arctan(-b,c))+c^2*sin(e*x+d-arctan(-b,c))-b^2-c^2)/(b^2+c^2)^(1/2))^(1/2)/e

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \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))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.88337, size = 313, normalized size = 2.41 \begin{align*} \frac{2 \,{\left (2 \, b c \cos \left (e x + d\right ) \sin \left (e x + d\right ) +{\left (b^{2} - c^{2}\right )} \cos \left (e x + d\right )^{2} - 5 \, b^{2} - 4 \, c^{2} - 4 \, \sqrt{b^{2} + c^{2}}{\left (b \cos \left (e x + d\right ) + c \sin \left (e x + d\right )\right )}\right )} \sqrt{b \cos \left (e x + d\right ) + c \sin \left (e x + d\right ) - \sqrt{b^{2} + c^{2}}}}{3 \,{\left (c e \cos \left (e x + d\right ) - b e \sin \left (e x + d\right )\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))^(3/2),x, algorithm="fricas")

[Out]

2/3*(2*b*c*cos(e*x + d)*sin(e*x + d) + (b^2 - c^2)*cos(e*x + d)^2 - 5*b^2 - 4*c^2 - 4*sqrt(b^2 + c^2)*(b*cos(e
*x + d) + c*sin(e*x + d)))*sqrt(b*cos(e*x + d) + c*sin(e*x + d) - sqrt(b^2 + c^2))/(c*e*cos(e*x + d) - b*e*sin
(e*x + d))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**(3/2),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError