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

Optimal. Leaf size=186 \[ \frac{2 \sqrt{a+b \cos (d+e x)+c \sin (d+e x)} E\left (\frac{1}{2} \left (d+e x-\tan ^{-1}(b,c)\right )|\frac{2 \sqrt{b^2+c^2}}{a+\sqrt{b^2+c^2}}\right )}{e \left (a^2-b^2-c^2\right ) \sqrt{\frac{a+b \cos (d+e x)+c \sin (d+e x)}{a+\sqrt{b^2+c^2}}}}+\frac{2 (c \cos (d+e x)-b \sin (d+e x))}{e \left (a^2-b^2-c^2\right ) \sqrt{a+b \cos (d+e x)+c \sin (d+e x)}} \]

[Out]

(2*(c*Cos[d + e*x] - b*Sin[d + e*x]))/((a^2 - b^2 - c^2)*e*Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]]) + (2*Ell
ipticE[(d + e*x - ArcTan[b, c])/2, (2*Sqrt[b^2 + c^2])/(a + Sqrt[b^2 + c^2])]*Sqrt[a + b*Cos[d + e*x] + c*Sin[
d + e*x]])/((a^2 - b^2 - c^2)*e*Sqrt[(a + b*Cos[d + e*x] + c*Sin[d + e*x])/(a + Sqrt[b^2 + c^2])])

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Rubi [A]  time = 0.103961, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {3128, 3119, 2653} \[ \frac{2 \sqrt{a+b \cos (d+e x)+c \sin (d+e x)} E\left (\frac{1}{2} \left (d+e x-\tan ^{-1}(b,c)\right )|\frac{2 \sqrt{b^2+c^2}}{a+\sqrt{b^2+c^2}}\right )}{e \left (a^2-b^2-c^2\right ) \sqrt{\frac{a+b \cos (d+e x)+c \sin (d+e x)}{a+\sqrt{b^2+c^2}}}}+\frac{2 (c \cos (d+e x)-b \sin (d+e x))}{e \left (a^2-b^2-c^2\right ) \sqrt{a+b \cos (d+e x)+c \sin (d+e x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(c*Cos[d + e*x] - b*Sin[d + e*x]))/((a^2 - b^2 - c^2)*e*Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]]) + (2*Ell
ipticE[(d + e*x - ArcTan[b, c])/2, (2*Sqrt[b^2 + c^2])/(a + Sqrt[b^2 + c^2])]*Sqrt[a + b*Cos[d + e*x] + c*Sin[
d + e*x]])/((a^2 - b^2 - c^2)*e*Sqrt[(a + b*Cos[d + e*x] + c*Sin[d + e*x])/(a + Sqrt[b^2 + c^2])])

Rule 3128

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(-3/2), x_Symbol] :> Simp[(2*(c*Cos
[d + e*x] - b*Sin[d + e*x]))/(e*(a^2 - b^2 - c^2)*Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]]), x] + Dist[1/(a^2
- b^2 - c^2), Int[Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 -
b^2 - c^2, 0]

Rule 3119

Int[Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*C
os[d + e*x] + c*Sin[d + e*x]]/Sqrt[(a + b*Cos[d + e*x] + c*Sin[d + e*x])/(a + Sqrt[b^2 + c^2])], Int[Sqrt[a/(a
+ Sqrt[b^2 + c^2]) + (Sqrt[b^2 + c^2]*Cos[d + e*x - ArcTan[b, c]])/(a + Sqrt[b^2 + c^2])], x], x] /; FreeQ[{a
, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[b^2 + c^2, 0] &&  !GtQ[a + Sqrt[b^2 + c^2], 0]

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

\begin{align*} \int \frac{1}{(a+b \cos (d+e x)+c \sin (d+e x))^{3/2}} \, dx &=\frac{2 (c \cos (d+e x)-b \sin (d+e x))}{\left (a^2-b^2-c^2\right ) e \sqrt{a+b \cos (d+e x)+c \sin (d+e x)}}+\frac{\int \sqrt{a+b \cos (d+e x)+c \sin (d+e x)} \, dx}{a^2-b^2-c^2}\\ &=\frac{2 (c \cos (d+e x)-b \sin (d+e x))}{\left (a^2-b^2-c^2\right ) e \sqrt{a+b \cos (d+e x)+c \sin (d+e x)}}+\frac{\sqrt{a+b \cos (d+e x)+c \sin (d+e x)} \int \sqrt{\frac{a}{a+\sqrt{b^2+c^2}}+\frac{\sqrt{b^2+c^2} \cos \left (d+e x-\tan ^{-1}(b,c)\right )}{a+\sqrt{b^2+c^2}}} \, dx}{\left (a^2-b^2-c^2\right ) \sqrt{\frac{a+b \cos (d+e x)+c \sin (d+e x)}{a+\sqrt{b^2+c^2}}}}\\ &=\frac{2 (c \cos (d+e x)-b \sin (d+e x))}{\left (a^2-b^2-c^2\right ) e \sqrt{a+b \cos (d+e x)+c \sin (d+e x)}}+\frac{2 E\left (\frac{1}{2} \left (d+e x-\tan ^{-1}(b,c)\right )|\frac{2 \sqrt{b^2+c^2}}{a+\sqrt{b^2+c^2}}\right ) \sqrt{a+b \cos (d+e x)+c \sin (d+e x)}}{\left (a^2-b^2-c^2\right ) e \sqrt{\frac{a+b \cos (d+e x)+c \sin (d+e x)}{a+\sqrt{b^2+c^2}}}}\\ \end{align*}

Mathematica [C]  time = 6.37446, size = 1540, normalized size = 8.28 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]]*((-2*(b^2 + c^2))/(b*c*(-a^2 + b^2 + c^2)) + (2*(a*c + b^2*Sin[d +
e*x] + c^2*Sin[d + e*x]))/(b*(-a^2 + b^2 + c^2)*(a + b*Cos[d + e*x] + c*Sin[d + e*x]))))/e - (2*a*AppellF1[1/2
, 1/2, 1/2, 3/2, -((a + Sqrt[1 + b^2/c^2]*c*Sin[d + e*x + ArcTan[b/c]])/(Sqrt[1 + b^2/c^2]*(1 - a/(Sqrt[1 + b^
2/c^2]*c))*c)), -((a + Sqrt[1 + b^2/c^2]*c*Sin[d + e*x + ArcTan[b/c]])/(Sqrt[1 + b^2/c^2]*(-1 - a/(Sqrt[1 + b^
2/c^2]*c))*c))]*Sec[d + e*x + ArcTan[b/c]]*Sqrt[(c*Sqrt[(b^2 + c^2)/c^2] - c*Sqrt[(b^2 + c^2)/c^2]*Sin[d + e*x
+ ArcTan[b/c]])/(a + c*Sqrt[(b^2 + c^2)/c^2])]*Sqrt[a + c*Sqrt[(b^2 + c^2)/c^2]*Sin[d + e*x + ArcTan[b/c]]]*S
qrt[(c*Sqrt[(b^2 + c^2)/c^2] + c*Sqrt[(b^2 + c^2)/c^2]*Sin[d + e*x + ArcTan[b/c]])/(-a + c*Sqrt[(b^2 + c^2)/c^
2])])/(Sqrt[1 + b^2/c^2]*c*(-a^2 + b^2 + c^2)*e) - (b^2*(-((c*AppellF1[-1/2, -1/2, -1/2, 1/2, -((a + b*Sqrt[1
+ c^2/b^2]*Cos[d + e*x - ArcTan[c/b]])/(b*Sqrt[1 + c^2/b^2]*(1 - a/(b*Sqrt[1 + c^2/b^2])))), -((a + b*Sqrt[1 +
c^2/b^2]*Cos[d + e*x - ArcTan[c/b]])/(b*Sqrt[1 + c^2/b^2]*(-1 - a/(b*Sqrt[1 + c^2/b^2]))))]*Sin[d + e*x - Arc
Tan[c/b]])/(b*Sqrt[1 + c^2/b^2]*Sqrt[(b*Sqrt[(b^2 + c^2)/b^2] - b*Sqrt[(b^2 + c^2)/b^2]*Cos[d + e*x - ArcTan[c
/b]])/(a + b*Sqrt[(b^2 + c^2)/b^2])]*Sqrt[a + b*Sqrt[(b^2 + c^2)/b^2]*Cos[d + e*x - ArcTan[c/b]]]*Sqrt[(b*Sqrt
[(b^2 + c^2)/b^2] + b*Sqrt[(b^2 + c^2)/b^2]*Cos[d + e*x - ArcTan[c/b]])/(-a + b*Sqrt[(b^2 + c^2)/b^2])])) - ((
2*b*(a + b*Sqrt[1 + c^2/b^2]*Cos[d + e*x - ArcTan[c/b]]))/(b^2 + c^2) - (c*Sin[d + e*x - ArcTan[c/b]])/(b*Sqrt
[1 + c^2/b^2]))/Sqrt[a + b*Sqrt[1 + c^2/b^2]*Cos[d + e*x - ArcTan[c/b]]]))/(c*(-a^2 + b^2 + c^2)*e) - (c*(-((c
*AppellF1[-1/2, -1/2, -1/2, 1/2, -((a + b*Sqrt[1 + c^2/b^2]*Cos[d + e*x - ArcTan[c/b]])/(b*Sqrt[1 + c^2/b^2]*(
1 - a/(b*Sqrt[1 + c^2/b^2])))), -((a + b*Sqrt[1 + c^2/b^2]*Cos[d + e*x - ArcTan[c/b]])/(b*Sqrt[1 + c^2/b^2]*(-
1 - a/(b*Sqrt[1 + c^2/b^2]))))]*Sin[d + e*x - ArcTan[c/b]])/(b*Sqrt[1 + c^2/b^2]*Sqrt[(b*Sqrt[(b^2 + c^2)/b^2]
- b*Sqrt[(b^2 + c^2)/b^2]*Cos[d + e*x - ArcTan[c/b]])/(a + b*Sqrt[(b^2 + c^2)/b^2])]*Sqrt[a + b*Sqrt[(b^2 + c
^2)/b^2]*Cos[d + e*x - ArcTan[c/b]]]*Sqrt[(b*Sqrt[(b^2 + c^2)/b^2] + b*Sqrt[(b^2 + c^2)/b^2]*Cos[d + e*x - Arc
Tan[c/b]])/(-a + b*Sqrt[(b^2 + c^2)/b^2])])) - ((2*b*(a + b*Sqrt[1 + c^2/b^2]*Cos[d + e*x - ArcTan[c/b]]))/(b^
2 + c^2) - (c*Sin[d + e*x - ArcTan[c/b]])/(b*Sqrt[1 + c^2/b^2]))/Sqrt[a + b*Sqrt[1 + c^2/b^2]*Cos[d + e*x - Ar
cTan[c/b]]]))/((-a^2 + b^2 + c^2)*e)

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Maple [B]  time = 8.075, size = 2388, normalized size = 12.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-(-b^2*sin(e*x+d-arctan(-b,c))-c^2*sin(e*x+d-arctan(-b,c))-a*(b^2+c^2)^(1/2))*cos(e*x+d-arctan(-b,c))^2/(b^2+
c^2)^(1/2))^(1/2)/(b^2+c^2)^(1/2)*((b^2+c^2)*cos(e*x+d-arctan(-b,c))^2/(a^2-b^2-c^2)/(-(-(b^2+c^2)^(1/2)*sin(e
*x+d-arctan(-b,c))-a)*cos(e*x+d-arctan(-b,c))^2)^(1/2)+1/(a^2-b^2-c^2)*(b^2+c^2)^(1/2)*a*(1/(b^2+c^2)^(1/2)*a-
1)*((-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2)*((-sin(e*x+d-arctan(-b,c))+1)*(b^
2+c^2)^(1/2)/(a+(b^2+c^2)^(1/2)))^(1/2)*((1+sin(e*x+d-arctan(-b,c)))*(b^2+c^2)^(1/2)/(-a+(b^2+c^2)^(1/2)))^(1/
2)/(-(-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))-a)*cos(e*x+d-arctan(-b,c))^2)^(1/2)*EllipticF(((-(b^2+c^2)^(1/2
)*sin(e*x+d-arctan(-b,c))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2),((a-(b^2+c^2)^(1/2))/(a+(b^2+c^2)^(1/2)))^(1/2))+1/(a
^2-b^2-c^2)*(b^2+c^2)*(1/(b^2+c^2)^(1/2)*a-1)*((-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))-a)/(-a+(b^2+c^2)^(1/2
)))^(1/2)*((-sin(e*x+d-arctan(-b,c))+1)*(b^2+c^2)^(1/2)/(a+(b^2+c^2)^(1/2)))^(1/2)*((1+sin(e*x+d-arctan(-b,c))
)*(b^2+c^2)^(1/2)/(-a+(b^2+c^2)^(1/2)))^(1/2)/(-(-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))-a)*cos(e*x+d-arctan(
-b,c))^2)^(1/2)*((-1/(b^2+c^2)^(1/2)*a-1)*EllipticE(((-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))-a)/(-a+(b^2+c^2
)^(1/2)))^(1/2),((a-(b^2+c^2)^(1/2))/(a+(b^2+c^2)^(1/2)))^(1/2))+EllipticF(((-(b^2+c^2)^(1/2)*sin(e*x+d-arctan
(-b,c))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2),((a-(b^2+c^2)^(1/2))/(a+(b^2+c^2)^(1/2)))^(1/2)))-(1/2*b^2+1/2*c^2)/(b^
2+c^2)^(1/2)*(1/(b^2+c^2)^(1/2)*a-1)*((-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2)
*((-sin(e*x+d-arctan(-b,c))+1)*(b^2+c^2)^(1/2)/(a+(b^2+c^2)^(1/2)))^(1/2)*((1+sin(e*x+d-arctan(-b,c)))*(b^2+c^
2)^(1/2)/(-a+(b^2+c^2)^(1/2)))^(1/2)/(-(-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))-a)*cos(e*x+d-arctan(-b,c))^2)
^(1/2)/a*EllipticPi(((-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2),-1/2*(-1/(b^2+c^
2)^(1/2)*a+1)*(b^2+c^2)^(1/2)/a,((a-(b^2+c^2)^(1/2))/(a+(b^2+c^2)^(1/2)))^(1/2))-(b^2+c^2)^(1/2)*(-b^2-c^2)*co
s(e*x+d-arctan(-b,c))^2/(a^2-b^2-c^2)/(-(-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))-a)*cos(e*x+d-arctan(-b,c))^2
*(b^2+c^2))^(1/2)+a*(b^2+c^2)/(a^2-b^2-c^2)*(1/(b^2+c^2)^(1/2)*a-1)*((-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))
-a)/(-a+(b^2+c^2)^(1/2)))^(1/2)*((-sin(e*x+d-arctan(-b,c))+1)*(b^2+c^2)^(1/2)/(a+(b^2+c^2)^(1/2)))^(1/2)*((1+s
in(e*x+d-arctan(-b,c)))*(b^2+c^2)^(1/2)/(-a+(b^2+c^2)^(1/2)))^(1/2)/(-(-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c)
)-a)*cos(e*x+d-arctan(-b,c))^2*(b^2+c^2))^(1/2)*EllipticF(((-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))-a)/(-a+(b
^2+c^2)^(1/2)))^(1/2),((a-(b^2+c^2)^(1/2))/(a+(b^2+c^2)^(1/2)))^(1/2))+2*(-(b^2+c^2)^(3/2)+2*(b^2+c^2)^(1/2)*b
^2+2*(b^2+c^2)^(1/2)*c^2)/(2*a^2-2*b^2-2*c^2)*(1/(b^2+c^2)^(1/2)*a-1)*((-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c
))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2)*((-sin(e*x+d-arctan(-b,c))+1)*(b^2+c^2)^(1/2)/(a+(b^2+c^2)^(1/2)))^(1/2)*((1
+sin(e*x+d-arctan(-b,c)))*(b^2+c^2)^(1/2)/(-a+(b^2+c^2)^(1/2)))^(1/2)/(-(-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,
c))-a)*cos(e*x+d-arctan(-b,c))^2*(b^2+c^2))^(1/2)*((-1/(b^2+c^2)^(1/2)*a-1)*EllipticE(((-(b^2+c^2)^(1/2)*sin(e
*x+d-arctan(-b,c))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2),((a-(b^2+c^2)^(1/2))/(a+(b^2+c^2)^(1/2)))^(1/2))+EllipticF((
(-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))-a)/(-a+(b^2+c^2)^(1/2)))^(1/2),((a-(b^2+c^2)^(1/2))/(a+(b^2+c^2)^(1/
2)))^(1/2)))+1/2*(b^2+c^2)*(1/(b^2+c^2)^(1/2)*a-1)*((-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))-a)/(-a+(b^2+c^2)
^(1/2)))^(1/2)*((-sin(e*x+d-arctan(-b,c))+1)*(b^2+c^2)^(1/2)/(a+(b^2+c^2)^(1/2)))^(1/2)*((1+sin(e*x+d-arctan(-
b,c)))*(b^2+c^2)^(1/2)/(-a+(b^2+c^2)^(1/2)))^(1/2)/(-(-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))-a)*cos(e*x+d-ar
ctan(-b,c))^2*(b^2+c^2))^(1/2)/a*EllipticPi(((-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))-a)/(-a+(b^2+c^2)^(1/2))
)^(1/2),-1/2*(-1/(b^2+c^2)^(1/2)*a+1)*(b^2+c^2)^(1/2)/a,((a-(b^2+c^2)^(1/2))/(a+(b^2+c^2)^(1/2)))^(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))+a*(b^2+c^2)^(1/2))/(b^2+c^2)^(1/
2))^(1/2)/e

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \cos \left (e x + d\right ) + c \sin \left (e x + d\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(e*x+d)+c*sin(e*x+d))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \cos \left (e x + d\right ) + c \sin \left (e x + d\right ) + a}}{2 \, a b \cos \left (e x + d\right ) +{\left (b^{2} - c^{2}\right )} \cos \left (e x + d\right )^{2} + a^{2} + c^{2} + 2 \,{\left (b c \cos \left (e x + d\right ) + a c\right )} \sin \left (e x + d\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(e*x+d)+c*sin(e*x+d))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*cos(e*x + d) + c*sin(e*x + d) + a)/(2*a*b*cos(e*x + d) + (b^2 - c^2)*cos(e*x + d)^2 + a^2 + c^
2 + 2*(b*c*cos(e*x + d) + a*c)*sin(e*x + d)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \cos{\left (d + e x \right )} + c \sin{\left (d + e x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \cos \left (e x + d\right ) + c \sin \left (e x + d\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(e*x+d)+c*sin(e*x+d))^(3/2),x, algorithm="giac")

[Out]

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