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

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

[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])

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Rubi [F]  time = 0.697165, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{1}{\sqrt{a+(c \cos (e+f x)+b \sin (e+f x))^2}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

((I/2)*Defer[Subst][Defer[Int][1/((I - x)*Sqrt[a + (c + b*x)^2/(1 + x^2)]), x], x, Tan[e + f*x]])/f + ((I/2)*D
efer[Subst][Defer[Int][1/((I + x)*Sqrt[a + (c + b*x)^2/(1 + x^2)]), x], x, Tan[e + f*x]])/f

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{a+(c \cos (e+f x)+b \sin (e+f x))^2}} \, dx &=\frac{\operatorname{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{\operatorname{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 \operatorname{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 \operatorname{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*}

Mathematica [C]  time = 1.6031, size = 529, normalized size = 6.7 \[ \frac{\sqrt{2} \sec \left (\tan ^{-1}\left (\frac{c^2-b^2}{2 b c}\right )+2 (e+f x)\right ) \sqrt{-\frac{b c \sqrt{\frac{\left (b^2+c^2\right )^2}{b^2 c^2}} \left (\sin \left (\tan ^{-1}\left (\frac{c^2-b^2}{2 b c}\right )+2 (e+f x)\right )-1\right )}{2 a+b c \sqrt{\frac{\left (b^2+c^2\right )^2}{b^2 c^2}}+b^2+c^2}} \sqrt{-\frac{b c \sqrt{\frac{\left (b^2+c^2\right )^2}{b^2 c^2}} \left (\sin \left (\tan ^{-1}\left (\frac{c^2-b^2}{2 b c}\right )+2 (e+f x)\right )+1\right )}{2 a-b c \sqrt{\frac{\left (b^2+c^2\right )^2}{b^2 c^2}}+b^2+c^2}} \sqrt{2 a+b c \sqrt{\frac{\left (b^2+c^2\right )^2}{b^2 c^2}} \sin \left (\tan ^{-1}\left (\frac{c^2-b^2}{2 b c}\right )+2 (e+f x)\right )+b^2+c^2} F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{2};\frac{3}{2};\frac{b^2+c \sqrt{\frac{\left (b^2+c^2\right )^2}{b^2 c^2}} \sin \left (2 (e+f x)+\tan ^{-1}\left (\frac{c^2-b^2}{2 b c}\right )\right ) b+c^2+2 a}{b^2-c \sqrt{\frac{\left (b^2+c^2\right )^2}{b^2 c^2}} b+c^2+2 a},\frac{b^2+c \sqrt{\frac{\left (b^2+c^2\right )^2}{b^2 c^2}} \sin \left (2 (e+f x)+\tan ^{-1}\left (\frac{c^2-b^2}{2 b c}\right )\right ) b+c^2+2 a}{b^2+c \sqrt{\frac{\left (b^2+c^2\right )^2}{b^2 c^2}} b+c^2+2 a}\right )}{b c f \sqrt{\frac{\left (b^2+c^2\right )^2}{b^2 c^2}}} \]

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]  time = 0.871, size = 257865, normalized size = 3264.1 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

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]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{\sqrt{2 \, b c \cos \left (f x + e\right ) \sin \left (f x + e\right ) -{\left (b^{2} - c^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2} + a}}, x\right ) \end{align*}

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]

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

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

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]

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

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

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)