3.2.52 \(\int \frac {\sqrt {a+a \sec (e+f x)}}{(c+d \sec (e+f x))^2} \, dx\) [152]

Optimal. Leaf size=219 \[ \frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{c^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {a^{3/2} \sqrt {d} (3 c+2 d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{c^2 (c+d)^{3/2} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {a d \tan (e+f x)}{c (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \]

[Out]

-a*d*tan(f*x+e)/c/(c+d)/f/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2)+2*a^(3/2)*arctanh((a-a*sec(f*x+e))^(1/2)/a^(
1/2))*tan(f*x+e)/c^2/f/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)-a^(3/2)*(3*c+2*d)*arctanh(d^(1/2)*(a-a*se
c(f*x+e))^(1/2)/a^(1/2)/(c+d)^(1/2))*d^(1/2)*tan(f*x+e)/c^2/(c+d)^(3/2)/f/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+
e))^(1/2)

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Rubi [A]
time = 0.16, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4025, 105, 162, 65, 212, 214} \begin {gather*} -\frac {a^{3/2} \sqrt {d} (3 c+2 d) \tan (e+f x) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{c^2 f (c+d)^{3/2} \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {2 a^{3/2} \tan (e+f x) \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{c^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}-\frac {a d \tan (e+f x)}{c f (c+d) \sqrt {a \sec (e+f x)+a} (c+d \sec (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Sec[e + f*x]]/(c + d*Sec[e + f*x])^2,x]

[Out]

(2*a^(3/2)*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]]*Tan[e + f*x])/(c^2*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*
Sec[e + f*x]]) - (a^(3/2)*Sqrt[d]*(3*c + 2*d)*ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d])
]*Tan[e + f*x])/(c^2*(c + d)^(3/2)*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (a*d*Tan[e + f*x])/(
c*(c + d)*f*Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x]))

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 105

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])

Rule 162

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 4025

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Di
st[a^2*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]])), Subst[Int[(a + b*x)^(m - 1/2)*((c
+ d*x)^n/(x*Sqrt[a - b*x])), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && IntegerQ[m - 1/2]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+a \sec (e+f x)}}{(c+d \sec (e+f x))^2} \, dx &=-\frac {\left (a^2 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a-a x} (c+d x)^2} \, dx,x,\sec (e+f x)\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=-\frac {a d \tan (e+f x)}{c (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))}-\frac {(a \tan (e+f x)) \text {Subst}\left (\int \frac {a (c+d)-\frac {a d x}{2}}{x \sqrt {a-a x} (c+d x)} \, dx,x,\sec (e+f x)\right )}{c (c+d) f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=-\frac {a d \tan (e+f x)}{c (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))}-\frac {\left (a^2 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a-a x}} \, dx,x,\sec (e+f x)\right )}{c^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {\left (a \left (\frac {a c d}{2}+a d (c+d)\right ) \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a-a x} (c+d x)} \, dx,x,\sec (e+f x)\right )}{c^2 (c+d) f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=-\frac {a d \tan (e+f x)}{c (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))}+\frac {(2 a \tan (e+f x)) \text {Subst}\left (\int \frac {1}{1-\frac {x^2}{a}} \, dx,x,\sqrt {a-a \sec (e+f x)}\right )}{c^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {\left (2 \left (\frac {a c d}{2}+a d (c+d)\right ) \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{c+d-\frac {d x^2}{a}} \, dx,x,\sqrt {a-a \sec (e+f x)}\right )}{c^2 (c+d) f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{c^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {a^{3/2} \sqrt {d} (3 c+2 d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{c^2 (c+d)^{3/2} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {a d \tan (e+f x)}{c (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 28.73, size = 2907, normalized size = 13.27 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[a + a*Sec[e + f*x]]/(c + d*Sec[e + f*x])^2,x]

[Out]

((d + c*Cos[e + f*x])^2*Sec[(e + f*x)/2]*Sec[e + f*x]^2*Sqrt[a*(1 + Sec[e + f*x])]*(-((d*Sin[(e + f*x)/2])/(c^
2*(c + d))) + (d^2*Sin[(e + f*x)/2])/(c^2*(c + d)*(d + c*Cos[e + f*x]))))/(f*(c + d*Sec[e + f*x])^2) - (2*Sqrt
[2]*Cos[(e + f*x)/4]^2*Sqrt[(-1 + Sqrt[2] - (-2 + Sqrt[2])*Cos[(e + f*x)/2])/(1 + Cos[(e + f*x)/2])]*(d + c*Co
s[e + f*x])^2*(c*(2*c + d)*EllipticF[ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] - 4*(c + d
)^2*EllipticPi[-3 + 2*Sqrt[2], ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] + d*(3*c + 2*d)*
(EllipticPi[-(((-3 + 2*Sqrt[2])*(c + d))/(3*c + 2*Sqrt[2]*Sqrt[c*(c - d)] - d)), ArcSin[Tan[(e + f*x)/4]/Sqrt[
3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] + EllipticPi[((-3 + 2*Sqrt[2])*(c + d))/(-3*c + 2*Sqrt[2]*Sqrt[c*(c - d)] +
d), ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]]))*Sec[(e + f*x)/2]*((Cos[(e + f*x)/2]*Sqrt[
Sec[e + f*x]])/(2*(c + d)*(d + c*Cos[e + f*x])) + (Cos[(3*(e + f*x))/2]*Sqrt[Sec[e + f*x]])/(2*(c + d)*(d + c*
Cos[e + f*x])) + (d*Cos[(3*(e + f*x))/2]*Sqrt[Sec[e + f*x]])/(2*c*(c + d)*(d + c*Cos[e + f*x])))*Sec[e + f*x]^
2*Sqrt[a*(1 + Sec[e + f*x])]*Sqrt[3 - 2*Sqrt[2] - Tan[(e + f*x)/4]^2])/(c^2*(c + d)^2*f*(c + d*Sec[e + f*x])^2
*((Sqrt[(-1 + Sqrt[2] - (-2 + Sqrt[2])*Cos[(e + f*x)/2])/(1 + Cos[(e + f*x)/2])]*(c*(2*c + d)*EllipticF[ArcSin
[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] - 4*(c + d)^2*EllipticPi[-3 + 2*Sqrt[2], ArcSin[Tan[(
e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] + d*(3*c + 2*d)*(EllipticPi[-(((-3 + 2*Sqrt[2])*(c + d))/(3
*c + 2*Sqrt[2]*Sqrt[c*(c - d)] - d)), ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] + Ellipti
cPi[((-3 + 2*Sqrt[2])*(c + d))/(-3*c + 2*Sqrt[2]*Sqrt[c*(c - d)] + d), ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt
[2]]], 17 - 12*Sqrt[2]]))*Sqrt[Sec[e + f*x]]*Tan[(e + f*x)/4])/(Sqrt[2]*c^2*(c + d)^2*Sqrt[3 - 2*Sqrt[2] - Tan
[(e + f*x)/4]^2]) + (Sqrt[2]*Cos[(e + f*x)/4]*Sqrt[(-1 + Sqrt[2] - (-2 + Sqrt[2])*Cos[(e + f*x)/2])/(1 + Cos[(
e + f*x)/2])]*(c*(2*c + d)*EllipticF[ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] - 4*(c + d
)^2*EllipticPi[-3 + 2*Sqrt[2], ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] + d*(3*c + 2*d)*
(EllipticPi[-(((-3 + 2*Sqrt[2])*(c + d))/(3*c + 2*Sqrt[2]*Sqrt[c*(c - d)] - d)), ArcSin[Tan[(e + f*x)/4]/Sqrt[
3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] + EllipticPi[((-3 + 2*Sqrt[2])*(c + d))/(-3*c + 2*Sqrt[2]*Sqrt[c*(c - d)] +
d), ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]]))*Sqrt[Sec[e + f*x]]*Sin[(e + f*x)/4]*Sqrt[
3 - 2*Sqrt[2] - Tan[(e + f*x)/4]^2])/(c^2*(c + d)^2) - (Sqrt[2]*Cos[(e + f*x)/4]^2*(c*(2*c + d)*EllipticF[ArcS
in[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] - 4*(c + d)^2*EllipticPi[-3 + 2*Sqrt[2], ArcSin[Tan
[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] + d*(3*c + 2*d)*(EllipticPi[-(((-3 + 2*Sqrt[2])*(c + d))/
(3*c + 2*Sqrt[2]*Sqrt[c*(c - d)] - d)), ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] + Ellip
ticPi[((-3 + 2*Sqrt[2])*(c + d))/(-3*c + 2*Sqrt[2]*Sqrt[c*(c - d)] + d), ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sq
rt[2]]], 17 - 12*Sqrt[2]]))*Sqrt[Sec[e + f*x]]*(((-2 + Sqrt[2])*Sin[(e + f*x)/2])/(2*(1 + Cos[(e + f*x)/2])) +
((-1 + Sqrt[2] - (-2 + Sqrt[2])*Cos[(e + f*x)/2])*Sin[(e + f*x)/2])/(2*(1 + Cos[(e + f*x)/2])^2))*Sqrt[3 - 2*
Sqrt[2] - Tan[(e + f*x)/4]^2])/(c^2*(c + d)^2*Sqrt[(-1 + Sqrt[2] - (-2 + Sqrt[2])*Cos[(e + f*x)/2])/(1 + Cos[(
e + f*x)/2])]) - (Sqrt[2]*Cos[(e + f*x)/4]^2*Sqrt[(-1 + Sqrt[2] - (-2 + Sqrt[2])*Cos[(e + f*x)/2])/(1 + Cos[(e
+ f*x)/2])]*(c*(2*c + d)*EllipticF[ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] - 4*(c + d)
^2*EllipticPi[-3 + 2*Sqrt[2], ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] + d*(3*c + 2*d)*(
EllipticPi[-(((-3 + 2*Sqrt[2])*(c + d))/(3*c + 2*Sqrt[2]*Sqrt[c*(c - d)] - d)), ArcSin[Tan[(e + f*x)/4]/Sqrt[3
- 2*Sqrt[2]]], 17 - 12*Sqrt[2]] + EllipticPi[((-3 + 2*Sqrt[2])*(c + d))/(-3*c + 2*Sqrt[2]*Sqrt[c*(c - d)] + d
), ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]]))*Sec[e + f*x]^(3/2)*Sin[e + f*x]*Sqrt[3 - 2
*Sqrt[2] - Tan[(e + f*x)/4]^2])/(c^2*(c + d)^2) - (2*Sqrt[2]*Cos[(e + f*x)/4]^2*Sqrt[(-1 + Sqrt[2] - (-2 + Sqr
t[2])*Cos[(e + f*x)/2])/(1 + Cos[(e + f*x)/2])]*Sqrt[Sec[e + f*x]]*Sqrt[3 - 2*Sqrt[2] - Tan[(e + f*x)/4]^2]*((
c*(2*c + d)*Sec[(e + f*x)/4]^2)/(4*Sqrt[3 - 2*Sqrt[2]]*Sqrt[1 - Tan[(e + f*x)/4]^2/(3 - 2*Sqrt[2])]*Sqrt[1 - (
(17 - 12*Sqrt[2])*Tan[(e + f*x)/4]^2)/(3 - 2*Sqrt[2])]) - ((c + d)^2*Sec[(e + f*x)/4]^2)/(Sqrt[3 - 2*Sqrt[2]]*
Sqrt[1 - Tan[(e + f*x)/4]^2/(3 - 2*Sqrt[2])]*Sqrt[1 - ((17 - 12*Sqrt[2])*Tan[(e + f*x)/4]^2)/(3 - 2*Sqrt[2])]*
(1 - ((-3 + 2*Sqrt[2])*Tan[(e + f*x)/4]^2)/(3 - 2*Sqrt[2]))) + d*(3*c + 2*d)*(Sec[(e + f*x)/4]^2/(4*Sqrt[3 - 2
*Sqrt[2]]*Sqrt[1 - Tan[(e + f*x)/4]^2/(3 - 2*Sqrt[2])]*Sqrt[1 - ((17 - 12*Sqrt[2])*Tan[(e + f*x)/4]^2)/(3 - 2*
Sqrt[2])]*(1 + ((-3 + 2*Sqrt[2])*(c + d)*Tan[(e + f*x)/4]^2)/((3 - 2*Sqrt[2])*(3*c + 2*Sqrt[2]*Sqrt[c*(c - d)]
- d)))) + Sec[(e + f*x)/4]^2/(4*Sqrt[3 - 2*Sqr...

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(97142\) vs. \(2(189)=378\).
time = 0.92, size = 97143, normalized size = 443.58

method	result	size
default	\(\text {Expression too large to display}\)	\(97143\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a*sec(f*x + e) + a)/(d*sec(f*x + e) + c)^2, x)

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Fricas [A]
time = 7.84, size = 1499, normalized size = 6.84 \begin {gather*} \left [-\frac {2 \, c d \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left ({\left (3 \, c^{2} + 2 \, c d\right )} \cos \left (f x + e\right )^{2} + 3 \, c d + 2 \, d^{2} + {\left (3 \, c^{2} + 5 \, c d + 2 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {a d}{c + d}} \log \left (\frac {2 \, {\left (c + d\right )} \sqrt {-\frac {a d}{c + d}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (a c + 2 \, a d\right )} \cos \left (f x + e\right )^{2} - a d + {\left (a c + a d\right )} \cos \left (f x + e\right )}{c \cos \left (f x + e\right )^{2} + {\left (c + d\right )} \cos \left (f x + e\right ) + d}\right ) - 2 \, {\left ({\left (c^{2} + c d\right )} \cos \left (f x + e\right )^{2} + c d + d^{2} + {\left (c^{2} + 2 \, c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right )}{2 \, {\left ({\left (c^{4} + c^{3} d\right )} f \cos \left (f x + e\right )^{2} + {\left (c^{4} + 2 \, c^{3} d + c^{2} d^{2}\right )} f \cos \left (f x + e\right ) + {\left (c^{3} d + c^{2} d^{2}\right )} f\right )}}, -\frac {2 \, c d \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 4 \, {\left ({\left (c^{2} + c d\right )} \cos \left (f x + e\right )^{2} + c d + d^{2} + {\left (c^{2} + 2 \, c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right ) - {\left ({\left (3 \, c^{2} + 2 \, c d\right )} \cos \left (f x + e\right )^{2} + 3 \, c d + 2 \, d^{2} + {\left (3 \, c^{2} + 5 \, c d + 2 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {a d}{c + d}} \log \left (\frac {2 \, {\left (c + d\right )} \sqrt {-\frac {a d}{c + d}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (a c + 2 \, a d\right )} \cos \left (f x + e\right )^{2} - a d + {\left (a c + a d\right )} \cos \left (f x + e\right )}{c \cos \left (f x + e\right )^{2} + {\left (c + d\right )} \cos \left (f x + e\right ) + d}\right )}{2 \, {\left ({\left (c^{4} + c^{3} d\right )} f \cos \left (f x + e\right )^{2} + {\left (c^{4} + 2 \, c^{3} d + c^{2} d^{2}\right )} f \cos \left (f x + e\right ) + {\left (c^{3} d + c^{2} d^{2}\right )} f\right )}}, -\frac {c d \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left ({\left (3 \, c^{2} + 2 \, c d\right )} \cos \left (f x + e\right )^{2} + 3 \, c d + 2 \, d^{2} + {\left (3 \, c^{2} + 5 \, c d + 2 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a d}{c + d}} \arctan \left (\frac {{\left (c + d\right )} \sqrt {\frac {a d}{c + d}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{a d \sin \left (f x + e\right )}\right ) - {\left ({\left (c^{2} + c d\right )} \cos \left (f x + e\right )^{2} + c d + d^{2} + {\left (c^{2} + 2 \, c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right )}{{\left (c^{4} + c^{3} d\right )} f \cos \left (f x + e\right )^{2} + {\left (c^{4} + 2 \, c^{3} d + c^{2} d^{2}\right )} f \cos \left (f x + e\right ) + {\left (c^{3} d + c^{2} d^{2}\right )} f}, -\frac {c d \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 2 \, {\left ({\left (c^{2} + c d\right )} \cos \left (f x + e\right )^{2} + c d + d^{2} + {\left (c^{2} + 2 \, c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right ) - {\left ({\left (3 \, c^{2} + 2 \, c d\right )} \cos \left (f x + e\right )^{2} + 3 \, c d + 2 \, d^{2} + {\left (3 \, c^{2} + 5 \, c d + 2 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a d}{c + d}} \arctan \left (\frac {{\left (c + d\right )} \sqrt {\frac {a d}{c + d}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{a d \sin \left (f x + e\right )}\right )}{{\left (c^{4} + c^{3} d\right )} f \cos \left (f x + e\right )^{2} + {\left (c^{4} + 2 \, c^{3} d + c^{2} d^{2}\right )} f \cos \left (f x + e\right ) + {\left (c^{3} d + c^{2} d^{2}\right )} f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*c*d*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) - ((3*c^2 + 2*c*d)*cos(f*x + e)
^2 + 3*c*d + 2*d^2 + (3*c^2 + 5*c*d + 2*d^2)*cos(f*x + e))*sqrt(-a*d/(c + d))*log((2*(c + d)*sqrt(-a*d/(c + d)
)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + (a*c + 2*a*d)*cos(f*x + e)^2 - a*d + (a*
c + a*d)*cos(f*x + e))/(c*cos(f*x + e)^2 + (c + d)*cos(f*x + e) + d)) - 2*((c^2 + c*d)*cos(f*x + e)^2 + c*d +
d^2 + (c^2 + 2*c*d + d^2)*cos(f*x + e))*sqrt(-a)*log((2*a*cos(f*x + e)^2 - 2*sqrt(-a)*sqrt((a*cos(f*x + e) + a
)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)))/((c^4 + c^3*d)*f*cos(f*x
+ e)^2 + (c^4 + 2*c^3*d + c^2*d^2)*f*cos(f*x + e) + (c^3*d + c^2*d^2)*f), -1/2*(2*c*d*sqrt((a*cos(f*x + e) + a
)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + 4*((c^2 + c*d)*cos(f*x + e)^2 + c*d + d^2 + (c^2 + 2*c*d + d^2)*co
s(f*x + e))*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) - ((3*
c^2 + 2*c*d)*cos(f*x + e)^2 + 3*c*d + 2*d^2 + (3*c^2 + 5*c*d + 2*d^2)*cos(f*x + e))*sqrt(-a*d/(c + d))*log((2*
(c + d)*sqrt(-a*d/(c + d))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + (a*c + 2*a*d)*c
os(f*x + e)^2 - a*d + (a*c + a*d)*cos(f*x + e))/(c*cos(f*x + e)^2 + (c + d)*cos(f*x + e) + d)))/((c^4 + c^3*d)
*f*cos(f*x + e)^2 + (c^4 + 2*c^3*d + c^2*d^2)*f*cos(f*x + e) + (c^3*d + c^2*d^2)*f), -(c*d*sqrt((a*cos(f*x + e
) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) - ((3*c^2 + 2*c*d)*cos(f*x + e)^2 + 3*c*d + 2*d^2 + (3*c^2 + 5*
c*d + 2*d^2)*cos(f*x + e))*sqrt(a*d/(c + d))*arctan((c + d)*sqrt(a*d/(c + d))*sqrt((a*cos(f*x + e) + a)/cos(f*
x + e))*cos(f*x + e)/(a*d*sin(f*x + e))) - ((c^2 + c*d)*cos(f*x + e)^2 + c*d + d^2 + (c^2 + 2*c*d + d^2)*cos(f
*x + e))*sqrt(-a)*log((2*a*cos(f*x + e)^2 - 2*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*si
n(f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)))/((c^4 + c^3*d)*f*cos(f*x + e)^2 + (c^4 + 2*c^3*d + c^2*d
^2)*f*cos(f*x + e) + (c^3*d + c^2*d^2)*f), -(c*d*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x
+ e) + 2*((c^2 + c*d)*cos(f*x + e)^2 + c*d + d^2 + (c^2 + 2*c*d + d^2)*cos(f*x + e))*sqrt(a)*arctan(sqrt((a*co
s(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) - ((3*c^2 + 2*c*d)*cos(f*x + e)^2 + 3*c*d +
2*d^2 + (3*c^2 + 5*c*d + 2*d^2)*cos(f*x + e))*sqrt(a*d/(c + d))*arctan((c + d)*sqrt(a*d/(c + d))*sqrt((a*cos(
f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(a*d*sin(f*x + e))))/((c^4 + c^3*d)*f*cos(f*x + e)^2 + (c^4 + 2*c^3*d
+ c^2*d^2)*f*cos(f*x + e) + (c^3*d + c^2*d^2)*f)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )}}{\left (c + d \sec {\left (e + f x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*(sec(e + f*x) + 1))/(c + d*sec(e + f*x))**2, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 617 vs. \(2 (189) = 378\).
time = 1.39, size = 617, normalized size = 2.82 \begin {gather*} -\frac {\sqrt {2} {\left (\frac {\sqrt {2} {\left (3 \, \sqrt {-a} a c d + 2 \, \sqrt {-a} a d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} c - {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} d + a c + 3 \, a d\right )}}{4 \, \sqrt {-c d - d^{2}} a}\right )}{{\left (c^{3} + c^{2} d\right )} \sqrt {-c d - d^{2}} a} + \frac {\sqrt {2} \sqrt {-a} a \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right )}{c^{2} {\left | a \right |}} - \frac {4 \, {\left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} \sqrt {-a} a c d + 3 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} \sqrt {-a} a d^{2} + \sqrt {-a} a^{2} c d - \sqrt {-a} a^{2} d^{2}\right )}}{{\left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{4} c - {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{4} d + 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} a c + 6 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} a d + a^{2} c - a^{2} d\right )} {\left (c^{3} - c d^{2}\right )}}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(2)*(sqrt(2)*(3*sqrt(-a)*a*c*d + 2*sqrt(-a)*a*d^2)*arctan(1/4*sqrt(2)*((sqrt(-a)*tan(1/2*f*x + 1/2*e)
- sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^2*c - (sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2
+ a))^2*d + a*c + 3*a*d)/(sqrt(-c*d - d^2)*a))/((c^3 + c^2*d)*sqrt(-c*d - d^2)*a) + sqrt(2)*sqrt(-a)*a*log(abs
(2*(sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^2 - 4*sqrt(2)*abs(a) - 6*a)/abs(2*(sq
rt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^2 + 4*sqrt(2)*abs(a) - 6*a))/(c^2*abs(a)) -
4*((sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^2*sqrt(-a)*a*c*d + 3*(sqrt(-a)*tan(1
/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^2*sqrt(-a)*a*d^2 + sqrt(-a)*a^2*c*d - sqrt(-a)*a^2*d^2)
/(((sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^4*c - (sqrt(-a)*tan(1/2*f*x + 1/2*e)
- sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^4*d + 2*(sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2
+ a))^2*a*c + 6*(sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^2*a*d + a^2*c - a^2*d)*
(c^3 - c*d^2)))*sgn(cos(f*x + e))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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