3.431 \(\int (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x))^{5/2} \, dx\)

Optimal. Leaf size=190 \[ -\frac{2 (c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}{5 e}-\frac{16 \sqrt{b^2+c^2} (c \cos (d+e x)-b \sin (d+e x)) \sqrt{\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)}}{15 e}-\frac{64 \left (b^2+c^2\right ) (c \cos (d+e x)-b \sin (d+e x))}{15 e \sqrt{\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)}} \]

[Out]

(-64*(b^2 + c^2)*(c*Cos[d + e*x] - b*Sin[d + e*x]))/(15*e*Sqrt[Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*
x]]) - (16*Sqrt[b^2 + c^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]])/(15*e) - (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])^(3/2
))/(5*e)

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Rubi [A]  time = 0.122111, antiderivative size = 190, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {3113, 3112} \[ -\frac{2 (c \cos (d+e x)-b \sin (d+e x)) \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}{5 e}-\frac{16 \sqrt{b^2+c^2} (c \cos (d+e x)-b \sin (d+e x)) \sqrt{\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)}}{15 e}-\frac{64 \left (b^2+c^2\right ) (c \cos (d+e x)-b \sin (d+e x))}{15 e \sqrt{\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*x])^(5/2),x]

[Out]

(-64*(b^2 + c^2)*(c*Cos[d + e*x] - b*Sin[d + e*x]))/(15*e*Sqrt[Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*
x]]) - (16*Sqrt[b^2 + c^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]])/(15*e) - (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])^(3/2
))/(5*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 )^{5/2} \, dx &=-\frac{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 )^{3/2}}{5 e}+\frac{1}{5} \left (8 \sqrt{b^2+c^2}\right ) \int \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2} \, dx\\ &=-\frac{16 \sqrt{b^2+c^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)}}{15 e}-\frac{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 )^{3/2}}{5 e}+\frac{1}{15} \left (32 \left (b^2+c^2\right )\right ) \int \sqrt{\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)} \, dx\\ &=-\frac{64 \left (b^2+c^2\right ) (c \cos (d+e x)-b \sin (d+e x))}{15 e \sqrt{\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)}}-\frac{16 \sqrt{b^2+c^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)}}{15 e}-\frac{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 )^{3/2}}{5 e}\\ \end{align*}

Mathematica [C]  time = 33.0913, size = 11771, normalized size = 61.95 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*x])^(5/2),x]

[Out]

Result too large to show

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Maple [A]  time = 1.45, size = 200, normalized size = 1.1 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) \right ) \left ( \sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) -1 \right ) \left ( 3\, \left ( \sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) \right ) ^{2}{b}^{2}+3\,{c}^{2} \left ( \sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) \right ) ^{2}+14\,{b}^{2}\sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) +14\,{c}^{2}\sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) +43\,{b}^{2}+43\,{c}^{2} \right ) }{15\,\cos \left ( ex+d-\arctan \left ( -b,c \right ) \right ) e}\sqrt{{b}^{2}+{c}^{2}}{\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))^(5/2),x)

[Out]

2/15*(1+sin(e*x+d-arctan(-b,c)))*(b^2+c^2)^(1/2)*(sin(e*x+d-arctan(-b,c))-1)*(3*sin(e*x+d-arctan(-b,c))^2*b^2+
3*c^2*sin(e*x+d-arctan(-b,c))^2+14*b^2*sin(e*x+d-arctan(-b,c))+14*c^2*sin(e*x+d-arctan(-b,c))+43*b^2+43*c^2)/c
os(e*x+d-arctan(-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))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.95729, size = 463, normalized size = 2.44 \begin{align*} \frac{2 \,{\left (3 \,{\left (b^{3} - 3 \, b c^{2}\right )} \cos \left (e x + d\right )^{3} +{\left (29 \, b^{3} + 38 \, b c^{2}\right )} \cos \left (e x + d\right ) +{\left (29 \, b^{2} c + 32 \, c^{3} + 3 \,{\left (3 \, b^{2} c - c^{3}\right )} \cos \left (e x + d\right )^{2}\right )} \sin \left (e x + d\right ) +{\left (22 \, b c \cos \left (e x + d\right ) \sin \left (e x + d\right ) + 11 \,{\left (b^{2} - c^{2}\right )} \cos \left (e x + d\right )^{2} - 43 \, b^{2} - 32 \, c^{2}\right )} \sqrt{b^{2} + c^{2}}\right )} \sqrt{b \cos \left (e x + d\right ) + c \sin \left (e x + d\right ) + \sqrt{b^{2} + c^{2}}}}{15 \,{\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))^(5/2),x, algorithm="fricas")

[Out]

2/15*(3*(b^3 - 3*b*c^2)*cos(e*x + d)^3 + (29*b^3 + 38*b*c^2)*cos(e*x + d) + (29*b^2*c + 32*c^3 + 3*(3*b^2*c -
c^3)*cos(e*x + d)^2)*sin(e*x + d) + (22*b*c*cos(e*x + d)*sin(e*x + d) + 11*(b^2 - c^2)*cos(e*x + d)^2 - 43*b^2
- 32*c^2)*sqrt(b^2 + c^2))*sqrt(b*cos(e*x + d) + c*sin(e*x + d) + sqrt(b^2 + c^2))/(c*e*cos(e*x + d) - b*e*si
n(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))**(5/2),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))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError