3.442 \(\int \frac {1}{(-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x))^{5/2}} \, dx\)

Optimal. Leaf size=232 \[ -\frac {3 \tan ^{-1}\left (\frac {\sqrt [4]{b^2+c^2} \sin \left (-\tan ^{-1}(b,c)+d+e x\right )}{\sqrt {2} \sqrt {\sqrt {b^2+c^2} \cos \left (-\tan ^{-1}(b,c)+d+e x\right )-\sqrt {b^2+c^2}}}\right )}{16 \sqrt {2} e \left (b^2+c^2\right )^{5/4}}-\frac {3 (c \cos (d+e x)-b \sin (d+e x))}{16 e \left (b^2+c^2\right ) \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}+\frac {c \cos (d+e x)-b \sin (d+e x)}{4 e \sqrt {b^2+c^2} \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}} \]

[Out]

-3/32*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)^(5/4)/e*2^(1/2)-3/16*(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))^(3/2)+1/4*(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))^(5
/2)/(b^2+c^2)^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3116, 3115, 2649, 204} \[ -\frac {3 \tan ^{-1}\left (\frac {\sqrt [4]{b^2+c^2} \sin \left (-\tan ^{-1}(b,c)+d+e x\right )}{\sqrt {2} \sqrt {\sqrt {b^2+c^2} \cos \left (-\tan ^{-1}(b,c)+d+e x\right )-\sqrt {b^2+c^2}}}\right )}{16 \sqrt {2} e \left (b^2+c^2\right )^{5/4}}-\frac {3 (c \cos (d+e x)-b \sin (d+e x))}{16 e \left (b^2+c^2\right ) \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}+\frac {c \cos (d+e x)-b \sin (d+e x)}{4 e \sqrt {b^2+c^2} \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}} \]

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]

(-3*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]*Co
s[d + e*x - ArcTan[b, c]]])])/(16*Sqrt[2]*(b^2 + c^2)^(5/4)*e) + (c*Cos[d + e*x] - b*Sin[d + e*x])/(4*Sqrt[b^2
+ c^2]*e*(-Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*x])^(5/2)) - (3*(c*Cos[d + e*x] - b*Sin[d + e*x]))/
(16*(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 )^{5/2}} \, dx &=\frac {c \cos (d+e x)-b \sin (d+e x)}{4 \sqrt {b^2+c^2} e \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}}-\frac {3 \int \frac {1}{\left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}} \, dx}{8 \sqrt {b^2+c^2}}\\ &=\frac {c \cos (d+e x)-b \sin (d+e x)}{4 \sqrt {b^2+c^2} e \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}}-\frac {3 (c \cos (d+e x)-b \sin (d+e x))}{16 \left (b^2+c^2\right ) e \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}+\frac {3 \int \frac {1}{\sqrt {-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)}} \, dx}{32 \left (b^2+c^2\right )}\\ &=\frac {c \cos (d+e x)-b \sin (d+e x)}{4 \sqrt {b^2+c^2} e \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}}-\frac {3 (c \cos (d+e x)-b \sin (d+e x))}{16 \left (b^2+c^2\right ) e \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}+\frac {3 \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}{32 \left (b^2+c^2\right )}\\ &=\frac {c \cos (d+e x)-b \sin (d+e x)}{4 \sqrt {b^2+c^2} e \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}}-\frac {3 (c \cos (d+e x)-b \sin (d+e x))}{16 \left (b^2+c^2\right ) e \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}-\frac {3 \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 )}{16 \left (b^2+c^2\right ) e}\\ &=-\frac {3 \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 )}{16 \sqrt {2} \left (b^2+c^2\right )^{5/4} e}+\frac {c \cos (d+e x)-b \sin (d+e x)}{4 \sqrt {b^2+c^2} e \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}}-\frac {3 (c \cos (d+e x)-b \sin (d+e x))}{16 \left (b^2+c^2\right ) 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 = 180.06, 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])^(-5/2),x]

[Out]

$Aborted

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fricas [B]  time = 1.20, size = 655, normalized size = 2.82 \[ \frac {\frac {3 \, \sqrt {\frac {1}{2}} {\left (5 \, b^{4} c \cos \left (e x + d\right ) + {\left (5 \, b^{4} c - 10 \, b^{2} c^{3} + c^{5}\right )} \cos \left (e x + d\right )^{5} - 10 \, {\left (b^{4} c - b^{2} c^{3}\right )} \cos \left (e x + d\right )^{3} - {\left (b^{5} + {\left (b^{5} - 10 \, b^{3} c^{2} + 5 \, b c^{4}\right )} \cos \left (e x + d\right )^{4} - 2 \, {\left (b^{5} - 5 \, b^{3} c^{2}\right )} \cos \left (e x + d\right )^{2}\right )} \sin \left (e x + d\right )\right )} \arctan \left (-\frac {\sqrt {\frac {1}{2}} {\left (b \cos \left (e x + d\right ) + c \sin \left (e x + d\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}}}}{{\left (b^{2} + c^{2}\right )}^{\frac {1}{4}} {\left (c \cos \left (e x + d\right ) - b \sin \left (e x + d\right )\right )}}\right )}{{\left (b^{2} + c^{2}\right )}^{\frac {1}{4}}} + {\left (3 \, {\left (b^{4} - 6 \, b^{2} c^{2} + c^{4}\right )} \cos \left (e x + d\right )^{4} - 7 \, b^{4} - 26 \, b^{2} c^{2} - 16 \, c^{4} - 6 \, {\left (2 \, b^{4} - 3 \, b^{2} c^{2} - c^{4}\right )} \cos \left (e x + d\right )^{2} + 12 \, {\left ({\left (b^{3} c - b c^{3}\right )} \cos \left (e x + d\right )^{3} - {\left (2 \, b^{3} c + b c^{3}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right ) + 2 \, {\left ({\left (b^{3} - 3 \, b c^{2}\right )} \cos \left (e x + d\right )^{3} - 3 \, {\left (3 \, b^{3} + 2 \, b c^{2}\right )} \cos \left (e x + d\right ) - {\left (9 \, b^{2} c + 8 \, c^{3} - {\left (3 \, b^{2} c - c^{3}\right )} \cos \left (e x + d\right )^{2}\right )} \sin \left (e x + d\right )\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}}}}{16 \, {\left ({\left (5 \, b^{6} c - 5 \, b^{4} c^{3} - 9 \, b^{2} c^{5} + c^{7}\right )} e \cos \left (e x + d\right )^{5} - 10 \, {\left (b^{6} c - b^{2} c^{5}\right )} e \cos \left (e x + d\right )^{3} + 5 \, {\left (b^{6} c + b^{4} c^{3}\right )} e \cos \left (e x + d\right ) - {\left ({\left (b^{7} - 9 \, b^{5} c^{2} - 5 \, b^{3} c^{4} + 5 \, b c^{6}\right )} e \cos \left (e x + d\right )^{4} - 2 \, {\left (b^{7} - 4 \, b^{5} c^{2} - 5 \, b^{3} c^{4}\right )} e \cos \left (e x + d\right )^{2} + {\left (b^{7} + b^{5} 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))^(5/2),x, algorithm="fricas")

[Out]

1/16*(3*sqrt(1/2)*(5*b^4*c*cos(e*x + d) + (5*b^4*c - 10*b^2*c^3 + c^5)*cos(e*x + d)^5 - 10*(b^4*c - b^2*c^3)*c
os(e*x + d)^3 - (b^5 + (b^5 - 10*b^3*c^2 + 5*b*c^4)*cos(e*x + d)^4 - 2*(b^5 - 5*b^3*c^2)*cos(e*x + d)^2)*sin(e
*x + d))*arctan(-sqrt(1/2)*(b*cos(e*x + d) + c*sin(e*x + d) + sqrt(b^2 + c^2))*sqrt(b*cos(e*x + d) + c*sin(e*x
+ d) - sqrt(b^2 + c^2))/((b^2 + c^2)^(1/4)*(c*cos(e*x + d) - b*sin(e*x + d))))/(b^2 + c^2)^(1/4) + (3*(b^4 -
6*b^2*c^2 + c^4)*cos(e*x + d)^4 - 7*b^4 - 26*b^2*c^2 - 16*c^4 - 6*(2*b^4 - 3*b^2*c^2 - c^4)*cos(e*x + d)^2 + 1
2*((b^3*c - b*c^3)*cos(e*x + d)^3 - (2*b^3*c + b*c^3)*cos(e*x + d))*sin(e*x + d) + 2*((b^3 - 3*b*c^2)*cos(e*x
+ d)^3 - 3*(3*b^3 + 2*b*c^2)*cos(e*x + d) - (9*b^2*c + 8*c^3 - (3*b^2*c - c^3)*cos(e*x + d)^2)*sin(e*x + d))*s
qrt(b^2 + c^2))*sqrt(b*cos(e*x + d) + c*sin(e*x + d) - sqrt(b^2 + c^2)))/((5*b^6*c - 5*b^4*c^3 - 9*b^2*c^5 + c
^7)*e*cos(e*x + d)^5 - 10*(b^6*c - b^2*c^5)*e*cos(e*x + d)^3 + 5*(b^6*c + b^4*c^3)*e*cos(e*x + d) - ((b^7 - 9*
b^5*c^2 - 5*b^3*c^4 + 5*b*c^6)*e*cos(e*x + d)^4 - 2*(b^7 - 4*b^5*c^2 - 5*b^3*c^4)*e*cos(e*x + d)^2 + (b^7 + b^
5*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))^(5/2),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 0Evaluation time: 0.87sym2poly/r2sym(const gen & e,const index_m & i,const vecteur
& l) Error: Bad Argument Value

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maple [A]  time = 0.45, size = 363, normalized size = 1.56 \[ -\frac {\left (-\sin \left (e x +d -\arctan \left (-b , c\right )\right ) \sqrt {2}\, \arctan \left (\frac {\sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )-\sqrt {b^{2}+c^{2}}}\, \sqrt {2}}{2 \left (b^{2}+c^{2}\right )^{\frac {1}{4}}}\right ) \left (b^{2}+c^{2}\right )+2 \sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )-\sqrt {b^{2}+c^{2}}}\, \left (b^{2}+c^{2}\right )^{\frac {3}{4}}+\sqrt {2}\, \arctan \left (\frac {\sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )-\sqrt {b^{2}+c^{2}}}\, \sqrt {2}}{2 \left (b^{2}+c^{2}\right )^{\frac {1}{4}}}\right ) b^{2}+\sqrt {2}\, \arctan \left (\frac {\sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )-\sqrt {b^{2}+c^{2}}}\, \sqrt {2}}{2 \left (b^{2}+c^{2}\right )^{\frac {1}{4}}}\right ) c^{2}\right ) \sqrt {-\sqrt {b^{2}+c^{2}}\, \left (1+\sin \left (e x +d -\arctan \left (-b , c\right )\right )\right )}}{4 \left (b^{2}+c^{2}\right )^{\frac {5}{4}} \cos \left (e x +d -\arctan \left (-b , c\right )\right ) \sqrt {\frac {b^{2} \sin \left (e x +d -\arctan \left (-b , c\right )\right )+c^{2} \sin \left (e x +d -\arctan \left (-b , c\right )\right )-b^{2}-c^{2}}{\sqrt {b^{2}+c^{2}}}}\, e} \]

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))^(5/2),x)

[Out]

-1/4*(-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)/(b^2+c^2)^(5/4)/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))^(5/2),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 = -0.00 \[ \int \frac {1}{{\left (b\,\cos \left (d+e\,x\right )+c\,\sin \left (d+e\,x\right )-\sqrt {b^2+c^2}\right )}^{5/2}} \,d x \]

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))^(5/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \cos {\left (d + e x \right )} + c \sin {\left (d + e x \right )} - \sqrt {b^{2} + c^{2}}\right )^{\frac {5}{2}}}\, dx \]

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))**(5/2),x)

[Out]

Integral((b*cos(d + e*x) + c*sin(d + e*x) - sqrt(b**2 + c**2))**(-5/2), x)

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