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

[Out]

1/5*(-c*cos(e*x+d)+b*sin(e*x+d))/e/(b^2+c^2)^(1/2)/(b*cos(e*x+d)+c*sin(e*x+d)+(b^2+c^2)^(1/2))^3-2/15*(c*cos(e
*x+d)-b*sin(e*x+d))/(b^2+c^2)/e/(b*cos(e*x+d)+c*sin(e*x+d)+(b^2+c^2)^(1/2))^2-2/15*(c-sin(e*x+d)*(b^2+c^2)^(1/
2))/c/(b^2+c^2)/e/(c*cos(e*x+d)-b*sin(e*x+d))

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 191, normalized size of antiderivative = 1.00, 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 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]

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]

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 = 2.59, size = 420, normalized size = 2.20 \[ \frac {20 c \left (c^4-b^4\right ) \cos (2 (d+e x))-76 b^4 c-40 b^3 c^2 \sin (2 (d+e x))-152 b^2 c^3+110 b^2 c^2 \sqrt {b^2+c^2} \sin (d+e x)-6 b^2 c^2 \sqrt {b^2+c^2} \sin (5 (d+e x))+90 b c \left (b^2+c^2\right )^{3/2} \cos (d+e x)+100 c^4 \sqrt {b^2+c^2} \sin (d+e x)+5 c^4 \sqrt {b^2+c^2} \sin (3 (d+e x))+c^4 \sqrt {b^2+c^2} \sin (5 (d+e x))+10 b c^3 \sqrt {b^2+c^2} \cos (3 (d+e x))+4 b c^3 \sqrt {b^2+c^2} \cos (5 (d+e x))+10 b^4 \sqrt {b^2+c^2} \sin (d+e x)-5 b^4 \sqrt {b^2+c^2} \sin (3 (d+e x))+b^4 \sqrt {b^2+c^2} \sin (5 (d+e x))+10 b^3 c \sqrt {b^2+c^2} \cos (3 (d+e x))-4 b^3 c \sqrt {b^2+c^2} \cos (5 (d+e x))-40 b c^4 \sin (2 (d+e x))-76 c^5}{120 c e \left (b^2+c^2\right ) (c \cos (d+e x)-b \sin (d+e x))^5} \]

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)

________________________________________________________________________________________

fricas [B]  time = 0.96, size = 490, normalized size = 2.57 \[ -\frac {7 \, b^{6} + 26 \, b^{4} c^{2} + 31 \, b^{2} c^{4} + 12 \, c^{6} + 5 \, {\left (b^{6} + b^{4} c^{2} - b^{2} c^{4} - c^{6}\right )} \cos \left (e x + d\right )^{2} + 10 \, {\left (b^{5} c + 2 \, b^{3} c^{3} + b c^{5}\right )} \cos \left (e x + d\right ) \sin \left (e x + d\right ) - {\left (2 \, {\left (b^{5} - 10 \, b^{3} c^{2} + 5 \, b c^{4}\right )} \cos \left (e x + d\right )^{5} - 5 \, {\left (b^{5} - 6 \, b^{3} c^{2} + b c^{4}\right )} \cos \left (e x + d\right )^{3} + 5 \, {\left (3 \, b^{5} + 3 \, b^{3} c^{2} + 2 \, b c^{4}\right )} \cos \left (e x + d\right ) + {\left (15 \, b^{4} c + 25 \, b^{2} c^{3} + 12 \, c^{5} + 2 \, {\left (5 \, b^{4} c - 10 \, b^{2} c^{3} + c^{5}\right )} \cos \left (e x + d\right )^{4} - {\left (15 \, b^{4} c - 10 \, b^{2} c^{3} - c^{5}\right )} \cos \left (e x + d\right )^{2}\right )} \sin \left (e x + d\right )\right )} \sqrt {b^{2} + c^{2}}}{15 \, {\left ({\left (5 \, b^{8} c - 14 \, b^{4} c^{5} - 8 \, b^{2} c^{7} + c^{9}\right )} e \cos \left (e x + d\right )^{5} - 10 \, {\left (b^{8} c + b^{6} c^{3} - b^{4} c^{5} - b^{2} c^{7}\right )} e \cos \left (e x + d\right )^{3} + 5 \, {\left (b^{8} c + 2 \, b^{6} c^{3} + b^{4} c^{5}\right )} e \cos \left (e x + d\right ) - {\left ({\left (b^{9} - 8 \, b^{7} c^{2} - 14 \, b^{5} c^{4} + 5 \, b c^{8}\right )} e \cos \left (e x + d\right )^{4} - 2 \, {\left (b^{9} - 3 \, b^{7} c^{2} - 9 \, b^{5} c^{4} - 5 \, b^{3} c^{6}\right )} e \cos \left (e x + d\right )^{2} + {\left (b^{9} + 2 \, b^{7} c^{2} + b^{5} c^{4}\right )} e\right )} \sin \left (e x + d\right )\right )}} \]

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))

________________________________________________________________________________________

giac [A]  time = 0.61, size = 346, normalized size = 1.81 \[ -\frac {2 \, {\left (192 \, b^{7} + 352 \, b^{5} c^{2} + 200 \, b^{3} c^{4} + 35 \, b c^{6} + 15 \, {\left (4 \, b^{3} c^{4} + 3 \, b c^{6} + {\left (4 \, b^{2} c^{4} + c^{6}\right )} \sqrt {b^{2} + c^{2}}\right )} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{4} + 30 \, {\left (8 \, b^{4} c^{3} + 8 \, b^{2} c^{5} + c^{7} + 4 \, {\left (2 \, b^{3} c^{3} + b c^{5}\right )} \sqrt {b^{2} + c^{2}}\right )} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{3} + 20 \, {\left (24 \, b^{5} c^{2} + 32 \, b^{3} c^{4} + 9 \, b c^{6} + 2 \, {\left (12 \, b^{4} c^{2} + 10 \, b^{2} c^{4} + c^{6}\right )} \sqrt {b^{2} + c^{2}}\right )} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 10 \, {\left (48 \, b^{6} c + 76 \, b^{4} c^{3} + 31 \, b^{2} c^{5} + 2 \, c^{7} + {\left (48 \, b^{5} c + 52 \, b^{3} c^{3} + 11 \, b c^{5}\right )} \sqrt {b^{2} + c^{2}}\right )} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + {\left (192 \, b^{6} + 256 \, b^{4} c^{2} + 96 \, b^{2} c^{4} + 7 \, c^{6}\right )} \sqrt {b^{2} + c^{2}}\right )} e^{\left (-1\right )}}{15 \, {\left (c \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + b + \sqrt {b^{2} + c^{2}}\right )}^{5} c^{5}} \]

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)

________________________________________________________________________________________

maple [B]  time = 0.52, size = 496, normalized size = 2.60 \[ \frac {-\frac {2 \left (4 \sqrt {b^{2}+c^{2}}\, b^{2}+\sqrt {b^{2}+c^{2}}\, c^{2}+4 b^{3}+3 c^{2} b \right ) \left (\tan ^{4}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )}{c^{2}}-\frac {4 \left (8 b^{4}+8 b^{2} c^{2}+c^{4}+8 \sqrt {b^{2}+c^{2}}\, b^{3}+4 \sqrt {b^{2}+c^{2}}\, b \,c^{2}\right ) \left (\tan ^{3}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )}{c^{3}}-\frac {8 \left (24 \sqrt {b^{2}+c^{2}}\, b^{4}+20 \sqrt {b^{2}+c^{2}}\, b^{2} c^{2}+2 \sqrt {b^{2}+c^{2}}\, c^{4}+24 b^{5}+32 b^{3} c^{2}+9 c^{4} b \right ) \left (\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )}{3 c^{4}}-\frac {4 \left (48 b^{6}+76 b^{4} c^{2}+31 b^{2} c^{4}+2 c^{6}+48 \sqrt {b^{2}+c^{2}}\, b^{5}+52 \sqrt {b^{2}+c^{2}}\, b^{3} c^{2}+11 \sqrt {b^{2}+c^{2}}\, b \,c^{4}\right ) \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{3 c^{5}}-\frac {2 \left (192 \sqrt {b^{2}+c^{2}}\, b^{6}+256 \sqrt {b^{2}+c^{2}}\, b^{4} c^{2}+96 \sqrt {b^{2}+c^{2}}\, b^{2} c^{4}+7 \sqrt {b^{2}+c^{2}}\, c^{6}+192 b^{7}+352 b^{5} c^{2}+200 b^{3} c^{4}+35 c^{6} b \right )}{15 c^{6}}}{e \,c^{4} \left (\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )+\frac {2 \sqrt {b^{2}+c^{2}}\, \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{c}+\frac {2 b \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{c}+\frac {2 \sqrt {b^{2}+c^{2}}\, b}{c^{2}}+\frac {2 b^{2}}{c^{2}}+1\right )^{2} \left (\tan \left (\frac {d}{2}+\frac {e x}{2}\right )+\frac {\sqrt {b^{2}+c^{2}}}{c}+\frac {b}{c}\right )} \]

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*c^2*b)/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*c^4*b)/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*c^6*b))/(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)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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 >> ECL says: THROW: The catch RAT-ERR is undefined.

________________________________________________________________________________________

mupad [B]  time = 8.12, size = 592, normalized size = 3.10 \[ -\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3\,\left (\frac {32\,b^4+32\,b^2\,c^2+4\,c^4}{c^7}+\frac {\left (32\,b^3+16\,b\,c^2\right )\,\sqrt {b^2+c^2}}{c^7}\right )+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^4\,\left (\frac {8\,b^3+6\,b\,c^2}{c^6}+\frac {\left (8\,b^2+2\,c^2\right )\,\sqrt {b^2+c^2}}{c^6}\right )+\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (\frac {64\,b^6+\frac {304\,b^4\,c^2}{3}+\frac {124\,b^2\,c^4}{3}+\frac {8\,c^6}{3}}{c^9}+\frac {\sqrt {b^2+c^2}\,\left (64\,b^5+\frac {208\,b^3\,c^2}{3}+\frac {44\,b\,c^4}{3}\right )}{c^9}\right )+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (\frac {64\,b^5+\frac {256\,b^3\,c^2}{3}+24\,b\,c^4}{c^8}+\frac {\sqrt {b^2+c^2}\,\left (64\,b^4+\frac {160\,b^2\,c^2}{3}+\frac {16\,c^4}{3}\right )}{c^8}\right )+\frac {\frac {128\,b^7}{5}+\frac {704\,b^5\,c^2}{15}+\frac {80\,b^3\,c^4}{3}+\frac {14\,b\,c^6}{3}}{c^{10}}+\frac {\sqrt {b^2+c^2}\,\left (\frac {128\,b^6}{5}+\frac {512\,b^4\,c^2}{15}+\frac {64\,b^2\,c^4}{5}+\frac {14\,c^6}{15}\right )}{c^{10}}}{e\,\left (\frac {16\,b^5+20\,b^3\,c^2+5\,b\,c^4}{c^5}+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (\frac {40\,b^3+30\,b\,c^2}{c^3}+\frac {\left (40\,b^2+10\,c^2\right )\,\sqrt {b^2+c^2}}{c^3}\right )+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3\,\left (\frac {20\,b^2+10\,c^2}{c^2}+\frac {20\,b\,\sqrt {b^2+c^2}}{c^2}\right )+\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (\frac {40\,b^4+40\,b^2\,c^2+5\,c^4}{c^4}+\frac {\left (40\,b^3+20\,b\,c^2\right )\,\sqrt {b^2+c^2}}{c^4}\right )+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^4\,\left (\frac {5\,\sqrt {b^2+c^2}}{c}+\frac {5\,b}{c}\right )+\frac {\sqrt {b^2+c^2}\,\left (16\,b^4+12\,b^2\,c^2+c^4\right )}{c^5}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(tan(d/2 + (e*x)/2)^3*((32*b^4 + 4*c^4 + 32*b^2*c^2)/c^7 + ((16*b*c^2 + 32*b^3)*(b^2 + c^2)^(1/2))/c^7) + tan
(d/2 + (e*x)/2)^4*((6*b*c^2 + 8*b^3)/c^6 + ((8*b^2 + 2*c^2)*(b^2 + c^2)^(1/2))/c^6) + tan(d/2 + (e*x)/2)*((64*
b^6 + (8*c^6)/3 + (124*b^2*c^4)/3 + (304*b^4*c^2)/3)/c^9 + ((b^2 + c^2)^(1/2)*((44*b*c^4)/3 + 64*b^5 + (208*b^
3*c^2)/3))/c^9) + tan(d/2 + (e*x)/2)^2*((24*b*c^4 + 64*b^5 + (256*b^3*c^2)/3)/c^8 + ((b^2 + c^2)^(1/2)*(64*b^4
+ (16*c^4)/3 + (160*b^2*c^2)/3))/c^8) + ((14*b*c^6)/3 + (128*b^7)/5 + (80*b^3*c^4)/3 + (704*b^5*c^2)/15)/c^10
+ ((b^2 + c^2)^(1/2)*((128*b^6)/5 + (14*c^6)/15 + (64*b^2*c^4)/5 + (512*b^4*c^2)/15))/c^10)/(e*((5*b*c^4 + 16
*b^5 + 20*b^3*c^2)/c^5 + tan(d/2 + (e*x)/2)^2*((30*b*c^2 + 40*b^3)/c^3 + ((40*b^2 + 10*c^2)*(b^2 + c^2)^(1/2))
/c^3) + tan(d/2 + (e*x)/2)^5 + tan(d/2 + (e*x)/2)^3*((20*b^2 + 10*c^2)/c^2 + (20*b*(b^2 + c^2)^(1/2))/c^2) + t
an(d/2 + (e*x)/2)*((40*b^4 + 5*c^4 + 40*b^2*c^2)/c^4 + ((20*b*c^2 + 40*b^3)*(b^2 + c^2)^(1/2))/c^4) + tan(d/2
+ (e*x)/2)^4*((5*(b^2 + c^2)^(1/2))/c + (5*b)/c) + ((b^2 + c^2)^(1/2)*(16*b^4 + c^4 + 12*b^2*c^2))/c^5))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

________________________________________________________________________________________