3.361 \(\int \frac{1}{(\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x))^3} \, dx\)

Optimal. Leaf size=191 \[ -\frac{2 (c \cos (d+e x)-b \sin (d+e x))}{15 e \left (b^2+c^2\right ) \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2}-\frac{c \cos (d+e x)-b \sin (d+e x)}{5 e \sqrt{b^2+c^2} \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^3}-\frac{2 \left (c-\sqrt{b^2+c^2} \sin (d+e x)\right )}{15 c e \left (b^2+c^2\right ) (c \cos (d+e x)-b \sin (d+e x))} \]

[Out]

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

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Rubi [A]  time = 0.132685, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {3116, 3114} \[ -\frac{2 (c \cos (d+e x)-b \sin (d+e x))}{15 e \left (b^2+c^2\right ) \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2}-\frac{c \cos (d+e x)-b \sin (d+e x)}{5 e \sqrt{b^2+c^2} \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^3}-\frac{2 \left (c-\sqrt{b^2+c^2} \sin (d+e x)\right )}{15 c e \left (b^2+c^2\right ) (c \cos (d+e x)-b \sin (d+e x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3116

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)/(a*e*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n +
1)), 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] && LtQ[n, -1]

Rule 3114

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

Rubi steps

\begin{align*} \int \frac{1}{\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)}{5 \sqrt{b^2+c^2} e \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^3}+\frac{2 \int \frac{1}{\left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2} \, dx}{5 \sqrt{b^2+c^2}}\\ &=-\frac{c \cos (d+e x)-b \sin (d+e x)}{5 \sqrt{b^2+c^2} e \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^3}-\frac{2 (c \cos (d+e x)-b \sin (d+e x))}{15 \left (b^2+c^2\right ) e \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2}+\frac{2 \int \frac{1}{\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)} \, dx}{15 \left (b^2+c^2\right )}\\ &=-\frac{c \cos (d+e x)-b \sin (d+e x)}{5 \sqrt{b^2+c^2} e \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^3}-\frac{2 (c \cos (d+e x)-b \sin (d+e x))}{15 \left (b^2+c^2\right ) e \left (\sqrt{b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2}-\frac{2 \left (c-\sqrt{b^2+c^2} \sin (d+e x)\right )}{15 c \left (b^2+c^2\right ) e (c \cos (d+e x)-b \sin (d+e x))}\\ \end{align*}

Mathematica [B]  time = 2.77742, size = 420, normalized size = 2.2 \[ \frac{100 c^4 \sqrt{b^2+c^2} \sin (d+e x)+5 c^4 \sqrt{b^2+c^2} \sin (3 (d+e x))+c^4 \sqrt{b^2+c^2} \sin (5 (d+e x))+110 b^2 c^2 \sqrt{b^2+c^2} \sin (d+e x)-40 b^3 c^2 \sin (2 (d+e x))-6 b^2 c^2 \sqrt{b^2+c^2} \sin (5 (d+e x))+10 b^4 \sqrt{b^2+c^2} \sin (d+e x)-5 b^4 \sqrt{b^2+c^2} \sin (3 (d+e x))+b^4 \sqrt{b^2+c^2} \sin (5 (d+e x))+10 b c^3 \sqrt{b^2+c^2} \cos (3 (d+e x))+4 b c^3 \sqrt{b^2+c^2} \cos (5 (d+e x))+90 b c \left (b^2+c^2\right )^{3/2} \cos (d+e x)+20 c \left (c^4-b^4\right ) \cos (2 (d+e x))+10 b^3 c \sqrt{b^2+c^2} \cos (3 (d+e x))-4 b^3 c \sqrt{b^2+c^2} \cos (5 (d+e x))-152 b^2 c^3-76 b^4 c-40 b c^4 \sin (2 (d+e x))-76 c^5}{120 c e \left (b^2+c^2\right ) (c \cos (d+e x)-b \sin (d+e x))^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-76*b^4*c - 152*b^2*c^3 - 76*c^5 + 90*b*c*(b^2 + c^2)^(3/2)*Cos[d + e*x] + 20*c*(-b^4 + c^4)*Cos[2*(d + e*x)]
+ 10*b^3*c*Sqrt[b^2 + c^2]*Cos[3*(d + e*x)] + 10*b*c^3*Sqrt[b^2 + c^2]*Cos[3*(d + e*x)] - 4*b^3*c*Sqrt[b^2 +
c^2]*Cos[5*(d + e*x)] + 4*b*c^3*Sqrt[b^2 + c^2]*Cos[5*(d + e*x)] + 10*b^4*Sqrt[b^2 + c^2]*Sin[d + e*x] + 110*b
^2*c^2*Sqrt[b^2 + c^2]*Sin[d + e*x] + 100*c^4*Sqrt[b^2 + c^2]*Sin[d + e*x] - 40*b^3*c^2*Sin[2*(d + e*x)] - 40*
b*c^4*Sin[2*(d + e*x)] - 5*b^4*Sqrt[b^2 + c^2]*Sin[3*(d + e*x)] + 5*c^4*Sqrt[b^2 + c^2]*Sin[3*(d + e*x)] + b^4
*Sqrt[b^2 + c^2]*Sin[5*(d + e*x)] - 6*b^2*c^2*Sqrt[b^2 + c^2]*Sin[5*(d + e*x)] + c^4*Sqrt[b^2 + c^2]*Sin[5*(d
+ e*x)])/(120*c*(b^2 + c^2)*e*(c*Cos[d + e*x] - b*Sin[d + e*x])^5)

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Maple [B]  time = 0.194, size = 496, normalized size = 2.6 \begin{align*} 2\,{\frac{1}{{c}^{4}e} \left ( -{\frac{ \left ( 4\,\sqrt{{b}^{2}+{c}^{2}}{b}^{2}+\sqrt{{b}^{2}+{c}^{2}}{c}^{2}+4\,{b}^{3}+3\,b{c}^{2} \right ) \left ( \tan \left ( d/2+1/2\,ex \right ) \right ) ^{4}}{{c}^{2}}}-2\,{\frac{ \left ( 8\,{b}^{4}+8\,{b}^{2}{c}^{2}+{c}^{4}+8\,\sqrt{{b}^{2}+{c}^{2}}{b}^{3}+4\,\sqrt{{b}^{2}+{c}^{2}}b{c}^{2} \right ) \left ( \tan \left ( d/2+1/2\,ex \right ) \right ) ^{3}}{{c}^{3}}}-4/3\,{\frac{ \left ( 24\,\sqrt{{b}^{2}+{c}^{2}}{b}^{4}+20\,\sqrt{{b}^{2}+{c}^{2}}{b}^{2}{c}^{2}+2\,\sqrt{{b}^{2}+{c}^{2}}{c}^{4}+24\,{b}^{5}+32\,{b}^{3}{c}^{2}+9\,b{c}^{4} \right ) \left ( \tan \left ( d/2+1/2\,ex \right ) \right ) ^{2}}{{c}^{4}}}-2/3\,{\frac{ \left ( 48\,{b}^{6}+76\,{b}^{4}{c}^{2}+31\,{b}^{2}{c}^{4}+2\,{c}^{6}+48\,\sqrt{{b}^{2}+{c}^{2}}{b}^{5}+52\,\sqrt{{b}^{2}+{c}^{2}}{b}^{3}{c}^{2}+11\,\sqrt{{b}^{2}+{c}^{2}}b{c}^{4} \right ) \tan \left ( d/2+1/2\,ex \right ) }{{c}^{5}}}-1/15\,{\frac{192\,\sqrt{{b}^{2}+{c}^{2}}{b}^{6}+256\,\sqrt{{b}^{2}+{c}^{2}}{b}^{4}{c}^{2}+96\,\sqrt{{b}^{2}+{c}^{2}}{b}^{2}{c}^{4}+7\,\sqrt{{b}^{2}+{c}^{2}}{c}^{6}+192\,{b}^{7}+352\,{b}^{5}{c}^{2}+200\,{b}^{3}{c}^{4}+35\,b{c}^{6}}{{c}^{6}}} \right ) \left ( \left ( \tan \left ( d/2+1/2\,ex \right ) \right ) ^{2}+2\,{\frac{\sqrt{{b}^{2}+{c}^{2}}\tan \left ( d/2+1/2\,ex \right ) }{c}}+2\,{\frac{b\tan \left ( d/2+1/2\,ex \right ) }{c}}+2\,{\frac{\sqrt{{b}^{2}+{c}^{2}}b}{{c}^{2}}}+2\,{\frac{{b}^{2}}{{c}^{2}}}+1 \right ) ^{-2} \left ( \tan \left ( d/2+1/2\,ex \right ) +{\frac{\sqrt{{b}^{2}+{c}^{2}}}{c}}+{\frac{b}{c}} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/e/c^4*(-(4*(b^2+c^2)^(1/2)*b^2+(b^2+c^2)^(1/2)*c^2+4*b^3+3*b*c^2)/c^2*tan(1/2*d+1/2*e*x)^4-2*(8*b^4+8*b^2*c^
2+c^4+8*(b^2+c^2)^(1/2)*b^3+4*(b^2+c^2)^(1/2)*b*c^2)/c^3*tan(1/2*d+1/2*e*x)^3-4/3*(24*(b^2+c^2)^(1/2)*b^4+20*(
b^2+c^2)^(1/2)*b^2*c^2+2*(b^2+c^2)^(1/2)*c^4+24*b^5+32*b^3*c^2+9*b*c^4)/c^4*tan(1/2*d+1/2*e*x)^2-2/3*(48*b^6+7
6*b^4*c^2+31*b^2*c^4+2*c^6+48*(b^2+c^2)^(1/2)*b^5+52*(b^2+c^2)^(1/2)*b^3*c^2+11*(b^2+c^2)^(1/2)*b*c^4)/c^5*tan
(1/2*d+1/2*e*x)-1/15/c^6*(192*(b^2+c^2)^(1/2)*b^6+256*(b^2+c^2)^(1/2)*b^4*c^2+96*(b^2+c^2)^(1/2)*b^2*c^4+7*(b^
2+c^2)^(1/2)*c^6+192*b^7+352*b^5*c^2+200*b^3*c^4+35*b*c^6))/(tan(1/2*d+1/2*e*x)^2+2/c*(b^2+c^2)^(1/2)*tan(1/2*
d+1/2*e*x)+2*b/c*tan(1/2*d+1/2*e*x)+2/c^2*(b^2+c^2)^(1/2)*b+2/c^2*b^2+1)^2/(tan(1/2*d+1/2*e*x)+1/c*(b^2+c^2)^(
1/2)+b/c)

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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(1/(b*cos(e*x+d)+c*sin(e*x+d)+(b^2+c^2)^(1/2))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.98511, size = 1080, normalized size = 5.65 \begin{align*} -\frac{7 \, b^{6} + 26 \, b^{4} c^{2} + 31 \, b^{2} c^{4} + 12 \, c^{6} + 5 \,{\left (b^{6} + b^{4} c^{2} - b^{2} c^{4} - c^{6}\right )} \cos \left (e x + d\right )^{2} + 10 \,{\left (b^{5} c + 2 \, b^{3} c^{3} + b c^{5}\right )} \cos \left (e x + d\right ) \sin \left (e x + d\right ) -{\left (2 \,{\left (b^{5} - 10 \, b^{3} c^{2} + 5 \, b c^{4}\right )} \cos \left (e x + d\right )^{5} - 5 \,{\left (b^{5} - 6 \, b^{3} c^{2} + b c^{4}\right )} \cos \left (e x + d\right )^{3} + 5 \,{\left (3 \, b^{5} + 3 \, b^{3} c^{2} + 2 \, b c^{4}\right )} \cos \left (e x + d\right ) +{\left (15 \, b^{4} c + 25 \, b^{2} c^{3} + 12 \, c^{5} + 2 \,{\left (5 \, b^{4} c - 10 \, b^{2} c^{3} + c^{5}\right )} \cos \left (e x + d\right )^{4} -{\left (15 \, b^{4} c - 10 \, b^{2} c^{3} - c^{5}\right )} \cos \left (e x + d\right )^{2}\right )} \sin \left (e x + d\right )\right )} \sqrt{b^{2} + c^{2}}}{15 \,{\left ({\left (5 \, b^{8} c - 14 \, b^{4} c^{5} - 8 \, b^{2} c^{7} + c^{9}\right )} e \cos \left (e x + d\right )^{5} - 10 \,{\left (b^{8} c + b^{6} c^{3} - b^{4} c^{5} - b^{2} c^{7}\right )} e \cos \left (e x + d\right )^{3} + 5 \,{\left (b^{8} c + 2 \, b^{6} c^{3} + b^{4} c^{5}\right )} e \cos \left (e x + d\right ) -{\left ({\left (b^{9} - 8 \, b^{7} c^{2} - 14 \, b^{5} c^{4} + 5 \, b c^{8}\right )} e \cos \left (e x + d\right )^{4} - 2 \,{\left (b^{9} - 3 \, b^{7} c^{2} - 9 \, b^{5} c^{4} - 5 \, b^{3} c^{6}\right )} e \cos \left (e x + d\right )^{2} +{\left (b^{9} + 2 \, b^{7} c^{2} + b^{5} c^{4}\right )} e\right )} \sin \left (e x + d\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*cos(e*x+d)+c*sin(e*x+d)+(b^2+c^2)^(1/2))^3,x, algorithm="fricas")

[Out]

-1/15*(7*b^6 + 26*b^4*c^2 + 31*b^2*c^4 + 12*c^6 + 5*(b^6 + b^4*c^2 - b^2*c^4 - c^6)*cos(e*x + d)^2 + 10*(b^5*c
+ 2*b^3*c^3 + b*c^5)*cos(e*x + d)*sin(e*x + d) - (2*(b^5 - 10*b^3*c^2 + 5*b*c^4)*cos(e*x + d)^5 - 5*(b^5 - 6*
b^3*c^2 + b*c^4)*cos(e*x + d)^3 + 5*(3*b^5 + 3*b^3*c^2 + 2*b*c^4)*cos(e*x + d) + (15*b^4*c + 25*b^2*c^3 + 12*c
^5 + 2*(5*b^4*c - 10*b^2*c^3 + c^5)*cos(e*x + d)^4 - (15*b^4*c - 10*b^2*c^3 - c^5)*cos(e*x + d)^2)*sin(e*x + d
))*sqrt(b^2 + c^2))/((5*b^8*c - 14*b^4*c^5 - 8*b^2*c^7 + c^9)*e*cos(e*x + d)^5 - 10*(b^8*c + b^6*c^3 - b^4*c^5
- b^2*c^7)*e*cos(e*x + d)^3 + 5*(b^8*c + 2*b^6*c^3 + b^4*c^5)*e*cos(e*x + d) - ((b^9 - 8*b^7*c^2 - 14*b^5*c^4
+ 5*b*c^8)*e*cos(e*x + d)^4 - 2*(b^9 - 3*b^7*c^2 - 9*b^5*c^4 - 5*b^3*c^6)*e*cos(e*x + d)^2 + (b^9 + 2*b^7*c^2
+ b^5*c^4)*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(1/(b*cos(e*x+d)+c*sin(e*x+d)+(b**2+c**2)**(1/2))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.14242, size = 467, normalized size = 2.45 \begin{align*} -\frac{2 \,{\left (192 \, b^{7} + 352 \, b^{5} c^{2} + 200 \, b^{3} c^{4} + 35 \, b c^{6} + 15 \,{\left (4 \, b^{3} c^{4} + 3 \, b c^{6} +{\left (4 \, b^{2} c^{4} + c^{6}\right )} \sqrt{b^{2} + c^{2}}\right )} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{4} + 30 \,{\left (8 \, b^{4} c^{3} + 8 \, b^{2} c^{5} + c^{7} + 4 \,{\left (2 \, b^{3} c^{3} + b c^{5}\right )} \sqrt{b^{2} + c^{2}}\right )} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{3} + 20 \,{\left (24 \, b^{5} c^{2} + 32 \, b^{3} c^{4} + 9 \, b c^{6} + 2 \,{\left (12 \, b^{4} c^{2} + 10 \, b^{2} c^{4} + c^{6}\right )} \sqrt{b^{2} + c^{2}}\right )} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + 10 \,{\left (48 \, b^{6} c + 76 \, b^{4} c^{3} + 31 \, b^{2} c^{5} + 2 \, c^{7} +{\left (48 \, b^{5} c + 52 \, b^{3} c^{3} + 11 \, b c^{5}\right )} \sqrt{b^{2} + c^{2}}\right )} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) +{\left (192 \, b^{6} + 256 \, b^{4} c^{2} + 96 \, b^{2} c^{4} + 7 \, c^{6}\right )} \sqrt{b^{2} + c^{2}}\right )} e^{\left (-1\right )}}{15 \,{\left (c \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + b + \sqrt{b^{2} + c^{2}}\right )}^{5} c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*cos(e*x+d)+c*sin(e*x+d)+(b^2+c^2)^(1/2))^3,x, algorithm="giac")

[Out]

-2/15*(192*b^7 + 352*b^5*c^2 + 200*b^3*c^4 + 35*b*c^6 + 15*(4*b^3*c^4 + 3*b*c^6 + (4*b^2*c^4 + c^6)*sqrt(b^2 +
c^2))*tan(1/2*x*e + 1/2*d)^4 + 30*(8*b^4*c^3 + 8*b^2*c^5 + c^7 + 4*(2*b^3*c^3 + b*c^5)*sqrt(b^2 + c^2))*tan(1
/2*x*e + 1/2*d)^3 + 20*(24*b^5*c^2 + 32*b^3*c^4 + 9*b*c^6 + 2*(12*b^4*c^2 + 10*b^2*c^4 + c^6)*sqrt(b^2 + c^2))
*tan(1/2*x*e + 1/2*d)^2 + 10*(48*b^6*c + 76*b^4*c^3 + 31*b^2*c^5 + 2*c^7 + (48*b^5*c + 52*b^3*c^3 + 11*b*c^5)*
sqrt(b^2 + c^2))*tan(1/2*x*e + 1/2*d) + (192*b^6 + 256*b^4*c^2 + 96*b^2*c^4 + 7*c^6)*sqrt(b^2 + c^2))*e^(-1)/(
(c*tan(1/2*x*e + 1/2*d) + b + sqrt(b^2 + c^2))^5*c^5)