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

Optimal. Leaf size=164 \[ \frac {\tan ^{-1}\left (\frac {\sqrt [4]{b^2+c^2} \sin \left (-\tan ^{-1}(b,c)+d+e x\right )}{\sqrt {2} \sqrt {\sqrt {b^2+c^2} \cos \left (-\tan ^{-1}(b,c)+d+e x\right )-\sqrt {b^2+c^2}}}\right )}{2 \sqrt {2} e \left (b^2+c^2\right )^{3/4}}+\frac {c \cos (d+e x)-b \sin (d+e x)}{2 e \sqrt {b^2+c^2} \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3116, 3115, 2649, 204} \[ \frac {\tan ^{-1}\left (\frac {\sqrt [4]{b^2+c^2} \sin \left (-\tan ^{-1}(b,c)+d+e x\right )}{\sqrt {2} \sqrt {\sqrt {b^2+c^2} \cos \left (-\tan ^{-1}(b,c)+d+e x\right )-\sqrt {b^2+c^2}}}\right )}{2 \sqrt {2} e \left (b^2+c^2\right )^{3/4}}+\frac {c \cos (d+e x)-b \sin (d+e x)}{2 e \sqrt {b^2+c^2} \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}} \]

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]

ArcTan[((b^2 + c^2)^(1/4)*Sin[d + e*x - ArcTan[b, c]])/(Sqrt[2]*Sqrt[-Sqrt[b^2 + c^2] + Sqrt[b^2 + c^2]*Cos[d
+ e*x - ArcTan[b, c]]])]/(2*Sqrt[2]*(b^2 + c^2)^(3/4)*e) + (c*Cos[d + e*x] - b*Sin[d + e*x])/(2*Sqrt[b^2 + c^2
]*e*(-Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*x])^(3/2))

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 3115

Int[1/Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]], x_Symbol] :> Int[1/Sqrt[a +
Sqrt[b^2 + c^2]*Cos[d + e*x - ArcTan[b, c]]], 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 )^{3/2}} \, dx &=\frac {c \cos (d+e x)-b \sin (d+e x)}{2 \sqrt {b^2+c^2} e \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}-\frac {\int \frac {1}{\sqrt {-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)}} \, dx}{4 \sqrt {b^2+c^2}}\\ &=\frac {c \cos (d+e x)-b \sin (d+e x)}{2 \sqrt {b^2+c^2} e \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}-\frac {\int \frac {1}{\sqrt {-\sqrt {b^2+c^2}+\sqrt {b^2+c^2} \cos \left (d+e x-\tan ^{-1}(b,c)\right )}} \, dx}{4 \sqrt {b^2+c^2}}\\ &=\frac {c \cos (d+e x)-b \sin (d+e x)}{2 \sqrt {b^2+c^2} e \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-2 \sqrt {b^2+c^2}-x^2} \, dx,x,-\frac {\sqrt {b^2+c^2} \sin \left (d+e x-\tan ^{-1}(b,c)\right )}{\sqrt {-\sqrt {b^2+c^2}+\sqrt {b^2+c^2} \cos \left (d+e x-\tan ^{-1}(b,c)\right )}}\right )}{2 \sqrt {b^2+c^2} e}\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt [4]{b^2+c^2} \sin \left (d+e x-\tan ^{-1}(b,c)\right )}{\sqrt {2} \sqrt {-\sqrt {b^2+c^2}+\sqrt {b^2+c^2} \cos \left (d+e x-\tan ^{-1}(b,c)\right )}}\right )}{2 \sqrt {2} \left (b^2+c^2\right )^{3/4} e}+\frac {c \cos (d+e x)-b \sin (d+e x)}{2 \sqrt {b^2+c^2} e \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}\\ \end {align*}

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Mathematica [F]  time = 180.00, size = 0, normalized size = 0.00 \[ \text {\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

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fricas [B]  time = 0.86, size = 442, normalized size = 2.70 \[ \frac {{\left (3 \, \sqrt {2} b^{2} c \cos \left (e x + d\right ) - \sqrt {2} {\left (3 \, b^{2} c - c^{3}\right )} \cos \left (e x + d\right )^{3} - {\left (\sqrt {2} b^{3} - \sqrt {2} {\left (b^{3} - 3 \, b c^{2}\right )} \cos \left (e x + d\right )^{2}\right )} \sin \left (e x + d\right )\right )} {\left (b^{2} + c^{2}\right )}^{\frac {1}{4}} \arctan \left (-\frac {{\left (b^{2} + c^{2}\right )}^{\frac {1}{4}} \sqrt {b \cos \left (e x + d\right ) + c \sin \left (e x + d\right ) - \sqrt {b^{2} + c^{2}}} {\left ({\left (\sqrt {2} b \cos \left (e x + d\right ) + \sqrt {2} c \sin \left (e x + d\right )\right )} \sqrt {b^{2} + c^{2}} + \sqrt {2} {\left (b^{2} + c^{2}\right )}\right )}}{2 \, {\left ({\left (b^{2} c + c^{3}\right )} \cos \left (e x + d\right ) - {\left (b^{3} + b c^{2}\right )} \sin \left (e x + d\right )\right )}}\right ) - 2 \, {\left (2 \, {\left (b^{3} + b c^{2}\right )} \cos \left (e x + d\right ) + 2 \, {\left (b^{2} c + c^{3}\right )} \sin \left (e x + d\right ) + {\left (2 \, b c \cos \left (e x + d\right ) \sin \left (e x + d\right ) + {\left (b^{2} - c^{2}\right )} \cos \left (e x + d\right )^{2} + b^{2} + 2 \, c^{2}\right )} \sqrt {b^{2} + c^{2}}\right )} \sqrt {b \cos \left (e x + d\right ) + c \sin \left (e x + d\right ) - \sqrt {b^{2} + c^{2}}}}{4 \, {\left ({\left (3 \, b^{4} c + 2 \, b^{2} c^{3} - c^{5}\right )} e \cos \left (e x + d\right )^{3} - 3 \, {\left (b^{4} c + b^{2} c^{3}\right )} e \cos \left (e x + d\right ) - {\left ({\left (b^{5} - 2 \, b^{3} c^{2} - 3 \, b c^{4}\right )} e \cos \left (e x + d\right )^{2} - {\left (b^{5} + b^{3} c^{2}\right )} e\right )} \sin \left (e x + d\right )\right )}} \]

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))^(3/2),x, algorithm="fricas")

[Out]

1/4*((3*sqrt(2)*b^2*c*cos(e*x + d) - sqrt(2)*(3*b^2*c - c^3)*cos(e*x + d)^3 - (sqrt(2)*b^3 - sqrt(2)*(b^3 - 3*
b*c^2)*cos(e*x + d)^2)*sin(e*x + d))*(b^2 + c^2)^(1/4)*arctan(-1/2*(b^2 + c^2)^(1/4)*sqrt(b*cos(e*x + d) + c*s
in(e*x + d) - sqrt(b^2 + c^2))*((sqrt(2)*b*cos(e*x + d) + sqrt(2)*c*sin(e*x + d))*sqrt(b^2 + c^2) + sqrt(2)*(b
^2 + c^2))/((b^2*c + c^3)*cos(e*x + d) - (b^3 + b*c^2)*sin(e*x + d))) - 2*(2*(b^3 + b*c^2)*cos(e*x + d) + 2*(b
^2*c + c^3)*sin(e*x + d) + (2*b*c*cos(e*x + d)*sin(e*x + d) + (b^2 - c^2)*cos(e*x + d)^2 + b^2 + 2*c^2)*sqrt(b
^2 + c^2))*sqrt(b*cos(e*x + d) + c*sin(e*x + d) - sqrt(b^2 + c^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 [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Simp
lification assuming b near 0sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argum
ent Value

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maple [B]  time = 0.36, size = 363, normalized size = 2.21 \[ \frac {\left (-\sin \left (e x +d -\arctan \left (-b , c\right )\right ) \sqrt {2}\, \arctan \left (\frac {\sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )-\sqrt {b^{2}+c^{2}}}\, \sqrt {2}}{2 \left (b^{2}+c^{2}\right )^{\frac {1}{4}}}\right ) \left (b^{2}+c^{2}\right )+2 \sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )-\sqrt {b^{2}+c^{2}}}\, \left (b^{2}+c^{2}\right )^{\frac {3}{4}}+\sqrt {2}\, \arctan \left (\frac {\sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )-\sqrt {b^{2}+c^{2}}}\, \sqrt {2}}{2 \left (b^{2}+c^{2}\right )^{\frac {1}{4}}}\right ) b^{2}+\sqrt {2}\, \arctan \left (\frac {\sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )-\sqrt {b^{2}+c^{2}}}\, \sqrt {2}}{2 \left (b^{2}+c^{2}\right )^{\frac {1}{4}}}\right ) c^{2}\right ) \sqrt {-\sqrt {b^{2}+c^{2}}\, \left (1+\sin \left (e x +d -\arctan \left (-b , c\right )\right )\right )}}{4 \left (b^{2}+c^{2}\right )^{\frac {7}{4}} \cos \left (e x +d -\arctan \left (-b , c\right )\right ) \sqrt {\frac {b^{2} \sin \left (e x +d -\arctan \left (-b , c\right )\right )+c^{2} \sin \left (e x +d -\arctan \left (-b , c\right )\right )-b^{2}-c^{2}}{\sqrt {b^{2}+c^{2}}}}\, e} \]

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))^(3/2),x)

[Out]

1/4/(b^2+c^2)^(7/4)*(-sin(e*x+d-arctan(-b,c))*2^(1/2)*arctan(1/2*(-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))-(b^
2+c^2)^(1/2))^(1/2)*2^(1/2)/(b^2+c^2)^(1/4))*(b^2+c^2)+2*(-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))-(b^2+c^2)^(
1/2))^(1/2)*(b^2+c^2)^(3/4)+2^(1/2)*arctan(1/2*(-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))-(b^2+c^2)^(1/2))^(1/2
)*2^(1/2)/(b^2+c^2)^(1/4))*b^2+2^(1/2)*arctan(1/2*(-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))-(b^2+c^2)^(1/2))^(
1/2)*2^(1/2)/(b^2+c^2)^(1/4))*c^2)*(-(b^2+c^2)^(1/2)*(1+sin(e*x+d-arctan(-b,c))))^(1/2)/cos(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 = 0.00, 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))^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (b\,\cos \left (d+e\,x\right )+c\,\sin \left (d+e\,x\right )-\sqrt {b^2+c^2}\right )}^{3/2}} \,d x \]

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))^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \cos {\left (d + e x \right )} + c \sin {\left (d + e x \right )} - \sqrt {b^{2} + c^{2}}\right )^{\frac {3}{2}}}\, dx \]

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))**(3/2),x)

[Out]

Integral((b*cos(d + e*x) + c*sin(d + e*x) - sqrt(b**2 + c**2))**(-3/2), x)

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