3.6.94 \(\int \frac {1}{\sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2}} \, dx\) [594]

Optimal. Leaf size=79 \[ \frac {F\left (e+f x+\tan ^{-1}(b,c)|-\frac {b^2+c^2}{a}\right ) \sqrt {1+\frac {(c \cos (e+f x)+b \sin (e+f x))^2}{a}}}{f \sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2}} \]

[Out]

(cos(e+f*x+arctan(b,c))^2)^(1/2)/cos(e+f*x+arctan(b,c))*EllipticF(sin(e+f*x+arctan(b,c)),((-b^2-c^2)/a)^(1/2))
*(1+(c*cos(f*x+e)+b*sin(f*x+e))^2/a)^(1/2)/f/(a+(c*cos(f*x+e)+b*sin(f*x+e))^2)^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {3320, 3319, 3261} \begin {gather*} \frac {\sqrt {\frac {(b \sin (e+f x)+c \cos (e+f x))^2}{a}+1} F\left (e+f x+\tan ^{-1}(b,c)|-\frac {b^2+c^2}{a}\right )}{f \sqrt {a+(b \sin (e+f x)+c \cos (e+f x))^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a + (c*Cos[e + f*x] + b*Sin[e + f*x])^2],x]

[Out]

(EllipticF[e + f*x + ArcTan[b, c], -((b^2 + c^2)/a)]*Sqrt[1 + (c*Cos[e + f*x] + b*Sin[e + f*x])^2/a])/(f*Sqrt[
a + (c*Cos[e + f*x] + b*Sin[e + f*x])^2])

Rule 3261

Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(1/(Sqrt[a]*f))*EllipticF[e + f*x, -b/a]
, x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]

Rule 3319

Int[((a_) + (b_.)*(cos[(e_.) + (f_.)*(x_)]*(d_.) + (c_.)*sin[(e_.) + (f_.)*(x_)])^2)^(p_), x_Symbol] :> Int[(a
+ b*(Sqrt[c^2 + d^2]*Sin[ArcTan[c, d] + e + f*x])^2)^p, x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[p^2, 1/4] &
& GtQ[a, 0]

Rule 3320

Int[((a_) + (b_.)*(cos[(e_.) + (f_.)*(x_)]*(d_.) + (c_.)*sin[(e_.) + (f_.)*(x_)])^2)^(p_), x_Symbol] :> Dist[(
a + b*(c*Sin[e + f*x] + d*Cos[e + f*x])^2)^p/(1 + (b*(c*Sin[e + f*x] + d*Cos[e + f*x])^2)/a)^p, Int[(1 + (b*(c
*Sin[e + f*x] + d*Cos[e + f*x])^2)/a)^p, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[p^2, 1/4] &&  !GtQ[a, 0
]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2}} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+\frac {(c+b x)^2}{1+x^2}}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\text {Subst}\left (\int \left (\frac {i}{2 (i-x) \sqrt {a+\frac {(c+b x)^2}{1+x^2}}}+\frac {i}{2 (i+x) \sqrt {a+\frac {(c+b x)^2}{1+x^2}}}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i \text {Subst}\left (\int \frac {1}{(i-x) \sqrt {a+\frac {(c+b x)^2}{1+x^2}}} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac {i \text {Subst}\left (\int \frac {1}{(i+x) \sqrt {a+\frac {(c+b x)^2}{1+x^2}}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
time = 1.13, size = 529, normalized size = 6.70 \begin {gather*} \frac {\sqrt {2} F_1\left (\frac {1}{2};\frac {1}{2},\frac {1}{2};\frac {3}{2};\frac {2 a+b^2+c^2+b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} \sin \left (2 (e+f x)+\tan ^{-1}\left (\frac {-b^2+c^2}{2 b c}\right )\right )}{2 a+b^2+c^2-b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}}},\frac {2 a+b^2+c^2+b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} \sin \left (2 (e+f x)+\tan ^{-1}\left (\frac {-b^2+c^2}{2 b c}\right )\right )}{2 a+b^2+c^2+b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}}}\right ) \sec \left (2 (e+f x)+\tan ^{-1}\left (\frac {-b^2+c^2}{2 b c}\right )\right ) \sqrt {-\frac {b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} \left (-1+\sin \left (2 (e+f x)+\tan ^{-1}\left (\frac {-b^2+c^2}{2 b c}\right )\right )\right )}{2 a+b^2+c^2+b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}}}} \sqrt {-\frac {b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} \left (1+\sin \left (2 (e+f x)+\tan ^{-1}\left (\frac {-b^2+c^2}{2 b c}\right )\right )\right )}{2 a+b^2+c^2-b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}}}} \sqrt {2 a+b^2+c^2+b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} \sin \left (2 (e+f x)+\tan ^{-1}\left (\frac {-b^2+c^2}{2 b c}\right )\right )}}{b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} f} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/Sqrt[a + (c*Cos[e + f*x] + b*Sin[e + f*x])^2],x]

[Out]

(Sqrt[2]*AppellF1[1/2, 1/2, 1/2, 3/2, (2*a + b^2 + c^2 + b*c*Sqrt[(b^2 + c^2)^2/(b^2*c^2)]*Sin[2*(e + f*x) + A
rcTan[(-b^2 + c^2)/(2*b*c)]])/(2*a + b^2 + c^2 - b*c*Sqrt[(b^2 + c^2)^2/(b^2*c^2)]), (2*a + b^2 + c^2 + b*c*Sq
rt[(b^2 + c^2)^2/(b^2*c^2)]*Sin[2*(e + f*x) + ArcTan[(-b^2 + c^2)/(2*b*c)]])/(2*a + b^2 + c^2 + b*c*Sqrt[(b^2
+ c^2)^2/(b^2*c^2)])]*Sec[2*(e + f*x) + ArcTan[(-b^2 + c^2)/(2*b*c)]]*Sqrt[-((b*c*Sqrt[(b^2 + c^2)^2/(b^2*c^2)
]*(-1 + Sin[2*(e + f*x) + ArcTan[(-b^2 + c^2)/(2*b*c)]]))/(2*a + b^2 + c^2 + b*c*Sqrt[(b^2 + c^2)^2/(b^2*c^2)]
))]*Sqrt[-((b*c*Sqrt[(b^2 + c^2)^2/(b^2*c^2)]*(1 + Sin[2*(e + f*x) + ArcTan[(-b^2 + c^2)/(2*b*c)]]))/(2*a + b^
2 + c^2 - b*c*Sqrt[(b^2 + c^2)^2/(b^2*c^2)]))]*Sqrt[2*a + b^2 + c^2 + b*c*Sqrt[(b^2 + c^2)^2/(b^2*c^2)]*Sin[2*
(e + f*x) + ArcTan[(-b^2 + c^2)/(2*b*c)]]])/(b*c*Sqrt[(b^2 + c^2)^2/(b^2*c^2)]*f)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 5.21, size = 258013, normalized size = 3265.99

method	result	size
default	\(\text {Expression too large to display}\)	\(258013\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+(cos(f*x+e)*c+b*sin(f*x+e))^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt((c*cos(f*x + e) + b*sin(f*x + e))^2 + a), x)

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt((c*cos(f*x + e) + b*sin(f*x + e))^2 + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\sqrt {a+{\left (c\,\cos \left (e+f\,x\right )+b\,\sin \left (e+f\,x\right )\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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