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

Optimal. Leaf size=129 \[ \frac {b \sin (d+e x)-c \cos (d+e x)}{3 e \sqrt {b^2+c^2} \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2}-\frac {c-\sqrt {b^2+c^2} \sin (d+e x)}{3 c e \sqrt {b^2+c^2} (c \cos (d+e x)-b \sin (d+e x))} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3116, 3114} \[ -\frac {c \cos (d+e x)-b \sin (d+e x)}{3 e \sqrt {b^2+c^2} \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^2}-\frac {c-\sqrt {b^2+c^2} \sin (d+e x)}{3 c e \sqrt {b^2+c^2} (c \cos (d+e x)-b \sin (d+e x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.25, size = 98, normalized size = 0.76 \[ \frac {-2 c \sqrt {b^2+c^2}+b^2 \sin ^3(d+e x)+2 b c \cos ^3(d+e x)+2 c^2 \sin (d+e x)+c^2 \sin (d+e x) \cos ^2(d+e x)}{3 c e (c \cos (d+e x)-b \sin (d+e x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.74, size = 192, normalized size = 1.49 \[ -\frac {3 \, b^{3} \cos \left (e x + d\right ) - {\left (b^{3} - 3 \, b c^{2}\right )} \cos \left (e x + d\right )^{3} + {\left (3 \, b^{2} c + 2 \, c^{3} - {\left (3 \, b^{2} c - c^{3}\right )} \cos \left (e x + d\right )^{2}\right )} \sin \left (e x + d\right ) - 2 \, {\left (b^{2} + c^{2}\right )}^{\frac {3}{2}}}{3 \, {\left ({\left (3 \, b^{4} c + 2 \, b^{2} c^{3} - c^{5}\right )} e \cos \left (e x + d\right )^{3} - 3 \, {\left (b^{4} c + b^{2} c^{3}\right )} e \cos \left (e x + d\right ) - {\left ({\left (b^{5} - 2 \, b^{3} c^{2} - 3 \, b c^{4}\right )} e \cos \left (e x + d\right )^{2} - {\left (b^{5} + b^{3} c^{2}\right )} e\right )} \sin \left (e x + d\right )\right )}} \]

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))^2,x, algorithm="fricas")

[Out]

-1/3*(3*b^3*cos(e*x + d) - (b^3 - 3*b*c^2)*cos(e*x + d)^3 + (3*b^2*c + 2*c^3 - (3*b^2*c - c^3)*cos(e*x + d)^2)
*sin(e*x + d) - 2*(b^2 + c^2)^(3/2))/((3*b^4*c + 2*b^2*c^3 - c^5)*e*cos(e*x + d)^3 - 3*(b^4*c + b^2*c^3)*e*cos
(e*x + d) - ((b^5 - 2*b^3*c^2 - 3*b*c^4)*e*cos(e*x + d)^2 - (b^5 + b^3*c^2)*e)*sin(e*x + d))

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giac [A]  time = 0.20, size = 160, normalized size = 1.24 \[ -\frac {2 \, {\left (8 \, b^{4} + 10 \, b^{2} c^{2} + 2 \, c^{4} + 3 \, {\left (2 \, b^{2} c^{2} + c^{4} + 2 \, \sqrt {b^{2} + c^{2}} b c^{2}\right )} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 3 \, {\left (4 \, b^{3} c + 3 \, b c^{3} + {\left (4 \, b^{2} c + c^{3}\right )} \sqrt {b^{2} + c^{2}}\right )} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + 2 \, {\left (4 \, b^{3} + 3 \, b c^{2}\right )} \sqrt {b^{2} + c^{2}}\right )} e^{\left (-1\right )}}{3 \, {\left (c \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + b + \sqrt {b^{2} + c^{2}}\right )}^{3} c^{3}} \]

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))^2,x, algorithm="giac")

[Out]

-2/3*(8*b^4 + 10*b^2*c^2 + 2*c^4 + 3*(2*b^2*c^2 + c^4 + 2*sqrt(b^2 + c^2)*b*c^2)*tan(1/2*x*e + 1/2*d)^2 + 3*(4
*b^3*c + 3*b*c^3 + (4*b^2*c + c^3)*sqrt(b^2 + c^2))*tan(1/2*x*e + 1/2*d) + 2*(4*b^3 + 3*b*c^2)*sqrt(b^2 + c^2)
)*e^(-1)/((c*tan(1/2*x*e + 1/2*d) + b + sqrt(b^2 + c^2))^3*c^3)

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maple [A]  time = 0.40, size = 233, normalized size = 1.81 \[ \frac {2 \left (\sqrt {b^{2}+c^{2}}+b \right ) \left (-\frac {\left (\sqrt {b^{2}+c^{2}}+b \right ) \left (\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )}{c^{2}}-\frac {\left (2 b^{2}+c^{2}+2 \sqrt {b^{2}+c^{2}}\, b \right ) \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{c^{3}}-\frac {2 \left (2 \sqrt {b^{2}+c^{2}}\, b^{2}+\sqrt {b^{2}+c^{2}}\, c^{2}+2 b^{3}+2 c^{2} b \right )}{3 c^{4}}\right )}{e \,c^{2} \left (\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )+\frac {2 \sqrt {b^{2}+c^{2}}\, \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{c}+\frac {2 b \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{c}+\frac {2 \sqrt {b^{2}+c^{2}}\, b}{c^{2}}+\frac {2 b^{2}}{c^{2}}+1\right ) \left (\tan \left (\frac {d}{2}+\frac {e x}{2}\right )+\frac {\sqrt {b^{2}+c^{2}}}{c}+\frac {b}{c}\right )} \]

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))^2,x)

[Out]

2/e*((b^2+c^2)^(1/2)+b)/c^2*(-((b^2+c^2)^(1/2)+b)/c^2*tan(1/2*d+1/2*e*x)^2-1/c^3*(2*b^2+c^2+2*(b^2+c^2)^(1/2)*
b)*tan(1/2*d+1/2*e*x)-2/3*(2*(b^2+c^2)^(1/2)*b^2+(b^2+c^2)^(1/2)*c^2+2*b^3+2*c^2*b)/c^4)/(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)/(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 = 0.00, 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))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.67, size = 274, normalized size = 2.12 \[ -\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (\frac {4\,b^2+2\,c^2}{c^4}+\frac {4\,b\,\sqrt {b^2+c^2}}{c^4}\right )+\frac {\frac {16\,b^4}{3}+\frac {20\,b^2\,c^2}{3}+\frac {4\,c^4}{3}}{c^6}+\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (\frac {8\,b^3+6\,b\,c^2}{c^5}+\frac {\left (8\,b^2+2\,c^2\right )\,\sqrt {b^2+c^2}}{c^5}\right )+\frac {\left (\frac {16\,b^3}{3}+4\,b\,c^2\right )\,\sqrt {b^2+c^2}}{c^6}}{e\,\left (\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (\frac {6\,b^2+3\,c^2}{c^2}+\frac {6\,b\,\sqrt {b^2+c^2}}{c^2}\right )+\frac {4\,b^3+3\,b\,c^2}{c^3}+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (\frac {3\,\sqrt {b^2+c^2}}{c}+\frac {3\,b}{c}\right )+\frac {\left (4\,b^2+c^2\right )\,\sqrt {b^2+c^2}}{c^3}\right )} \]

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))^2,x)

[Out]

-(tan(d/2 + (e*x)/2)^2*((4*b^2 + 2*c^2)/c^4 + (4*b*(b^2 + c^2)^(1/2))/c^4) + ((16*b^4)/3 + (4*c^4)/3 + (20*b^2
*c^2)/3)/c^6 + tan(d/2 + (e*x)/2)*((6*b*c^2 + 8*b^3)/c^5 + ((8*b^2 + 2*c^2)*(b^2 + c^2)^(1/2))/c^5) + ((4*b*c^
2 + (16*b^3)/3)*(b^2 + c^2)^(1/2))/c^6)/(e*(tan(d/2 + (e*x)/2)*((6*b^2 + 3*c^2)/c^2 + (6*b*(b^2 + c^2)^(1/2))/
c^2) + (3*b*c^2 + 4*b^3)/c^3 + tan(d/2 + (e*x)/2)^3 + tan(d/2 + (e*x)/2)^2*((3*(b^2 + c^2)^(1/2))/c + (3*b)/c)
+ ((4*b^2 + c^2)*(b^2 + c^2)^(1/2))/c^3))

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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))**2,x)

[Out]

Timed out

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