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*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))

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Rubi [A]  time = 0.170474, antiderivative size = 232, normalized size of antiderivative = 1., 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 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 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 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 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])

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*}

Mathematica [F]  time = 180.014, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

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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Maple [A]  time = 2.082, size = 363, normalized size = 1.6 \begin{align*} -{\frac{1}{4\,\cos \left ( ex+d-\arctan \left ( -b,c \right ) \right ) e} \left ( -\sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) \arctan \left ({\frac{\sqrt{2}}{2}\sqrt{-\sqrt{{b}^{2}+{c}^{2}}\sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) -\sqrt{{b}^{2}+{c}^{2}}}{\frac{1}{\sqrt [4]{{b}^{2}+{c}^{2}}}}} \right ) \sqrt{2} \left ({b}^{2}+{c}^{2} \right ) +\sqrt{2}\arctan \left ({\frac{\sqrt{2}}{2}\sqrt{-\sqrt{{b}^{2}+{c}^{2}}\sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) -\sqrt{{b}^{2}+{c}^{2}}}{\frac{1}{\sqrt [4]{{b}^{2}+{c}^{2}}}}} \right ){b}^{2}+\sqrt{2}\arctan \left ({\frac{\sqrt{2}}{2}\sqrt{-\sqrt{{b}^{2}+{c}^{2}}\sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) -\sqrt{{b}^{2}+{c}^{2}}}{\frac{1}{\sqrt [4]{{b}^{2}+{c}^{2}}}}} \right ){c}^{2}+2\,\sqrt{-\sqrt{{b}^{2}+{c}^{2}}\sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) -\sqrt{{b}^{2}+{c}^{2}}} \left ({b}^{2}+{c}^{2} \right ) ^{3/4} \right ) \sqrt{-\sqrt{{b}^{2}+{c}^{2}} \left ( 1+\sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) \right ) } \left ({b}^{2}+{c}^{2} \right ) ^{-{\frac{5}{4}}}{\frac{1}{\sqrt{{({b}^{2}\sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) +{c}^{2}\sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) -{b}^{2}-{c}^{2}){\frac{1}{\sqrt{{b}^{2}+{c}^{2}}}}}}}}} \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))^(5/2),x)

[Out]

-1/4*(-sin(e*x+d-arctan(-b,c))*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))*2^(1/2)*(b^2+c^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))*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+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))*(-(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., 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))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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))^(5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError