3.1.16 \(\int \frac {\sqrt {c+d \sec (e+f x)}}{a+b \cos (e+f x)} \, dx\) [16]

Optimal. Leaf size=213 \[ \frac {2 \sqrt {c+d} \cot (e+f x) F\left (\text {ArcSin}\left (\frac {\sqrt {c+d \sec (e+f x)}}{\sqrt {c+d}}\right )|\frac {c+d}{c-d}\right ) \sqrt {\frac {d (1-\sec (e+f x))}{c+d}} \sqrt {-\frac {d (1+\sec (e+f x))}{c-d}}}{a f}+\frac {2 (a c-b d) \Pi \left (\frac {2 a}{a+b};\text {ArcSin}\left (\frac {\sqrt {1-\sec (e+f x)}}{\sqrt {2}}\right )|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sec (e+f x)}{c+d}} \tan (e+f x)}{a (a+b) f \sqrt {c+d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}} \]

[Out]

2*cot(f*x+e)*EllipticF((c+d*sec(f*x+e))^(1/2)/(c+d)^(1/2),((c+d)/(c-d))^(1/2))*(c+d)^(1/2)*(d*(1-sec(f*x+e))/(
c+d))^(1/2)*(-d*(1+sec(f*x+e))/(c-d))^(1/2)/a/f+2*(a*c-b*d)*EllipticPi(1/2*(1-sec(f*x+e))^(1/2)*2^(1/2),2*a/(a
+b),2^(1/2)*(d/(c+d))^(1/2))*((c+d*sec(f*x+e))/(c+d))^(1/2)*tan(f*x+e)/a/(a+b)/f/(c+d*sec(f*x+e))^(1/2)/(-tan(
f*x+e)^2)^(1/2)

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Rubi [A]
time = 0.25, antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2908, 4054, 3917, 4058} \begin {gather*} \frac {2 (a c-b d) \tan (e+f x) \sqrt {\frac {c+d \sec (e+f x)}{c+d}} \Pi \left (\frac {2 a}{a+b};\text {ArcSin}\left (\frac {\sqrt {1-\sec (e+f x)}}{\sqrt {2}}\right )|\frac {2 d}{c+d}\right )}{a f (a+b) \sqrt {-\tan ^2(e+f x)} \sqrt {c+d \sec (e+f x)}}+\frac {2 \sqrt {c+d} \cot (e+f x) \sqrt {\frac {d (1-\sec (e+f x))}{c+d}} \sqrt {-\frac {d (\sec (e+f x)+1)}{c-d}} F\left (\text {ArcSin}\left (\frac {\sqrt {c+d \sec (e+f x)}}{\sqrt {c+d}}\right )|\frac {c+d}{c-d}\right )}{a f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*Sec[e + f*x]]/(a + b*Cos[e + f*x]),x]

[Out]

(2*Sqrt[c + d]*Cot[e + f*x]*EllipticF[ArcSin[Sqrt[c + d*Sec[e + f*x]]/Sqrt[c + d]], (c + d)/(c - d)]*Sqrt[(d*(
1 - Sec[e + f*x]))/(c + d)]*Sqrt[-((d*(1 + Sec[e + f*x]))/(c - d))])/(a*f) + (2*(a*c - b*d)*EllipticPi[(2*a)/(
a + b), ArcSin[Sqrt[1 - Sec[e + f*x]]/Sqrt[2]], (2*d)/(c + d)]*Sqrt[(c + d*Sec[e + f*x])/(c + d)]*Tan[e + f*x]
)/(a*(a + b)*f*Sqrt[c + d*Sec[e + f*x]]*Sqrt[-Tan[e + f*x]^2])

Rule 2908

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int
[(b + a*Csc[e + f*x])^m*((c + d*Csc[e + f*x])^n/Csc[e + f*x]^m), x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !In
tegerQ[n] && IntegerQ[m]

Rule 3917

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

Rule 4054

Int[(csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)])/(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)
), x_Symbol] :> Dist[b/d, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(b*c - a*d)/d, Int[Csc[e +
f*x]/(Sqrt[a + b*Csc[e + f*x]]*(c + d*Csc[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 4058

Int[csc[(e_.) + (f_.)*(x_)]/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))
), x_Symbol] :> Simp[-2*(Cot[e + f*x]/(f*(c + d)*Sqrt[a + b*Csc[e + f*x]]*Sqrt[-Cot[e + f*x]^2]))*Sqrt[(a + b*
Csc[e + f*x])/(a + b)]*EllipticPi[2*(d/(c + d)), ArcSin[Sqrt[1 - Csc[e + f*x]]/Sqrt[2]], 2*(b/(a + b))], x] /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps

\begin {align*} \int \frac {\sqrt {c+d \sec (e+f x)}}{a+b \cos (e+f x)} \, dx &=\int \frac {\sec (e+f x) \sqrt {c+d \sec (e+f x)}}{b+a \sec (e+f x)} \, dx\\ &=\frac {d \int \frac {\sec (e+f x)}{\sqrt {c+d \sec (e+f x)}} \, dx}{a}-\frac {(-a c+b d) \int \frac {\sec (e+f x)}{(b+a \sec (e+f x)) \sqrt {c+d \sec (e+f x)}} \, dx}{a}\\ &=\frac {2 \sqrt {c+d} \cot (e+f x) F\left (\sin ^{-1}\left (\frac {\sqrt {c+d \sec (e+f x)}}{\sqrt {c+d}}\right )|\frac {c+d}{c-d}\right ) \sqrt {\frac {d (1-\sec (e+f x))}{c+d}} \sqrt {-\frac {d (1+\sec (e+f x))}{c-d}}}{a f}+\frac {2 (a c-b d) \Pi \left (\frac {2 a}{a+b};\sin ^{-1}\left (\frac {\sqrt {1-\sec (e+f x)}}{\sqrt {2}}\right )|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sec (e+f x)}{c+d}} \tan (e+f x)}{a (a+b) f \sqrt {c+d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 6.90, size = 184, normalized size = 0.86 \begin {gather*} \frac {4 \cos ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}} \sqrt {\frac {d+c \cos (e+f x)}{(c+d) (1+\cos (e+f x))}} \left (-\left ((a+b) (c-d) F\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {c-d}{c+d}\right )\right )+2 (a c-b d) \Pi \left (\frac {-a+b}{a+b};\text {ArcSin}\left (\tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {c-d}{c+d}\right )\right ) \sqrt {c+d \sec (e+f x)}}{(a-b) (a+b) f (d+c \cos (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + d*Sec[e + f*x]]/(a + b*Cos[e + f*x]),x]

[Out]

(4*Cos[(e + f*x)/2]^2*Sqrt[Cos[e + f*x]/(1 + Cos[e + f*x])]*Sqrt[(d + c*Cos[e + f*x])/((c + d)*(1 + Cos[e + f*
x]))]*(-((a + b)*(c - d)*EllipticF[ArcSin[Tan[(e + f*x)/2]], (c - d)/(c + d)]) + 2*(a*c - b*d)*EllipticPi[(-a
+ b)/(a + b), ArcSin[Tan[(e + f*x)/2]], (c - d)/(c + d)])*Sqrt[c + d*Sec[e + f*x]])/((a - b)*(a + b)*f*(d + c*
Cos[e + f*x]))

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Maple [A]
time = 3.73, size = 357, normalized size = 1.68

method	result	size
default	\(-\frac {2 \sqrt {\frac {d +c \cos \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {d +c \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \left (c +d \right )}}\, \left (\cos \left (f x +e \right )+1\right )^{2} \left (\EllipticF \left (\frac {\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}, \sqrt {\frac {c -d}{c +d}}\right ) a c -\EllipticF \left (\frac {\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}, \sqrt {\frac {c -d}{c +d}}\right ) a d +\EllipticF \left (\frac {\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}, \sqrt {\frac {c -d}{c +d}}\right ) b c -\EllipticF \left (\frac {\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}, \sqrt {\frac {c -d}{c +d}}\right ) b d -2 \EllipticPi \left (\frac {\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}, -\frac {a -b}{a +b}, \sqrt {\frac {c -d}{c +d}}\right ) a c +2 \EllipticPi \left (\frac {\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}, -\frac {a -b}{a +b}, \sqrt {\frac {c -d}{c +d}}\right ) b d \right ) \left (\cos \left (f x +e \right )-1\right )}{f \left (d +c \cos \left (f x +e \right )\right ) \sin \left (f x +e \right )^{2} \left (a -b \right ) \left (a +b \right )}\)	\(357\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/f*((d+c*cos(f*x+e))/cos(f*x+e))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((d+c*cos(f*x+e))/(cos(f*x+e)+1)/(c
+d))^(1/2)*(cos(f*x+e)+1)^2*(EllipticF((cos(f*x+e)-1)/sin(f*x+e),((c-d)/(c+d))^(1/2))*a*c-EllipticF((cos(f*x+e
)-1)/sin(f*x+e),((c-d)/(c+d))^(1/2))*a*d+EllipticF((cos(f*x+e)-1)/sin(f*x+e),((c-d)/(c+d))^(1/2))*b*c-Elliptic
F((cos(f*x+e)-1)/sin(f*x+e),((c-d)/(c+d))^(1/2))*b*d-2*EllipticPi((cos(f*x+e)-1)/sin(f*x+e),-(a-b)/(a+b),((c-d
)/(c+d))^(1/2))*a*c+2*EllipticPi((cos(f*x+e)-1)/sin(f*x+e),-(a-b)/(a+b),((c-d)/(c+d))^(1/2))*b*d)*(cos(f*x+e)-
1)/(d+c*cos(f*x+e))/sin(f*x+e)^2/(a-b)/(a+b)

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

[Out]

integrate(sqrt(d*sec(f*x + e) + c)/(b*cos(f*x + e) + a), x)

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Fricas [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((c+d*sec(f*x+e))^(1/2)/(a+b*cos(f*x+e)),x, algorithm="fricas")

[Out]

integral(sqrt(d*sec(f*x + e) + c)/(b*cos(f*x + e) + a), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {c + d \sec {\left (e + f x \right )}}}{a + b \cos {\left (e + f x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(c + d*sec(e + f*x))/(a + b*cos(e + f*x)), x)

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

[Out]

integrate(sqrt(d*sec(f*x + e) + c)/(b*cos(f*x + e) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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