3.2.58 \(\int \frac {(a+a \sec (e+f x))^{3/2}}{(c+d \sec (e+f x))^2} \, dx\) [158]
Optimal. Leaf size=229 \[ \frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{c^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {a^{5/2} \left (c^2-3 c d-2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{c^2 \sqrt {d} (c+d)^{3/2} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {a^2 (c-d) \tan (e+f x)}{c (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \]
[Out]
a^2*(c-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^(5/2)*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^(5/2)*(c^2-3*c*d-2*d^2)*arctanh(d^
(1/2)*(a-a*sec(f*x+e))^(1/2)/a^(1/2)/(c+d)^(1/2))*tan(f*x+e)/c^2/(c+d)^(3/2)/f/d^(1/2)/(a-a*sec(f*x+e))^(1/2)/
(a+a*sec(f*x+e))^(1/2)
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Rubi [A]
time = 0.17, antiderivative size = 229, 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, 156, 162,
65, 212, 214} \begin {gather*} \frac {a^{5/2} \left (c^2-3 c d-2 d^2\right ) \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 \sqrt {d} f (c+d)^{3/2} \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {2 a^{5/2} \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^2 (c-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[(a + a*Sec[e + f*x])^(3/2)/(c + d*Sec[e + f*x])^2,x]
[Out]
(2*a^(5/2)*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^(5/2)*(c^2 - 3*c*d - 2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d
])]*Tan[e + f*x])/(c^2*Sqrt[d]*(c + d)^(3/2)*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) + (a^2*(c -
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 156
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(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*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
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 {(a+a \sec (e+f x))^{3/2}}{(c+d \sec (e+f x))^2} \, dx &=-\frac {\left (a^2 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {a+a x}{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^2 (c-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^2 (c+d)+\frac {1}{2} a^2 (c-d) x}{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^2 (c-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^3 \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^3 \left (c^2-3 c d-2 d^2\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 )}{2 c^2 (c+d) f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {a^2 (c-d) \tan (e+f x)}{c (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))}+\frac {\left (2 a^2 \tan (e+f x)\right ) \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 (a^2 \left (c^2-3 c d-2 d^2\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^{5/2} \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^{5/2} \left (c^2-3 c d-2 d^2\right ) \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 \sqrt {d} (c+d)^{3/2} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {a^2 (c-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 = 24.92, size = 2862, normalized size = 12.50 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + a*Sec[e + f*x])^(3/2)/(c + d*Sec[e + f*x])^2,x]
[Out]
((d + c*Cos[e + f*x])^2*Sec[(e + f*x)/2]^3*Sec[e + f*x]*(a*(1 + Sec[e + f*x]))^(3/2)*(((c - d)*Sin[(e + f*x)/2
])/(2*c^2*(c + d)) + (-(c*d*Sin[(e + f*x)/2]) + d^2*Sin[(e + f*x)/2])/(2*c^2*(c + d)*(d + c*Cos[e + f*x]))))/(
f*(c + d*Sec[e + f*x])^2) - (Sqrt[3 - 2*Sqrt[2]]*Cos[(e + f*x)/4]^2*(d + c*Cos[e + f*x])^2*(c*(3*c + d)*Ellipt
icF[ArcSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] - 4*(c + d)^2*EllipticPi[-3 + 2*Sqrt[2], Ar
cSin[Tan[(e + f*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] - (c^2 - 3*c*d - 2*d^2)*(EllipticPi[-(((-3 + 2*Sq
rt[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]^3*((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]])/(4*(c + d)*(d + c*Cos[e + f*x])) + (d*Cos
[(3*(e + f*x))/2]*Sqrt[Sec[e + f*x]])/(4*c*(c + d)*(d + c*Cos[e + f*x])))*Sec[e + f*x]*(a*(1 + Sec[e + f*x]))^
(3/2)*Sqrt[1 + (-3 + 2*Sqrt[2])*Tan[(e + f*x)/4]^2]*Sqrt[1 - (3 + 2*Sqrt[2])*Tan[(e + f*x)/4]^2])/(c^2*(c + d)
^2*f*(c + d*Sec[e + f*x])^2*((Sqrt[3 - 2*Sqrt[2]]*(3 + 2*Sqrt[2])*(c*(3*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]/Sqr
t[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] - (c^2 - 3*c*d - 2*d^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]] + 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]]*Tan[(e + f*x)/4]*Sqrt[1 + (-3 + 2*Sqrt[2])*Tan[(e + f*x)/4]^2])/(4*c^2*(
c + d)^2*Sqrt[1 - (3 + 2*Sqrt[2])*Tan[(e + f*x)/4]^2]) - (Sqrt[3 - 2*Sqrt[2]]*(-3 + 2*Sqrt[2])*(c*(3*c + d)*El
lipticF[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]] - (c^2 - 3*c*d - 2*d^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]] + 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]]*Tan[(e + f*x)/4]*Sqrt[1 - (3 + 2*Sqrt[2])*T
an[(e + f*x)/4]^2])/(4*c^2*(c + d)^2*Sqrt[1 + (-3 + 2*Sqrt[2])*Tan[(e + f*x)/4]^2]) + (Sqrt[3 - 2*Sqrt[2]]*Cos
[(e + f*x)/4]*(c*(3*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]] - (c^2 - 3*c*d -
2*d^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]] + 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[1 + (-3 + 2*Sqrt[2])*Tan[(e + f*x)/4]^2]*Sqrt[1 - (3 + 2*Sqrt[2])*Tan[(e + f*x)/4]^2])/(2*c^2*(c + d)^
2) - (Sqrt[3 - 2*Sqrt[2]]*Cos[(e + f*x)/4]^2*(c*(3*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]] - (c^2 - 3*c*d - 2*d^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]] + 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]]))*Se
c[e + f*x]^(3/2)*Sin[e + f*x]*Sqrt[1 + (-3 + 2*Sqrt[2])*Tan[(e + f*x)/4]^2]*Sqrt[1 - (3 + 2*Sqrt[2])*Tan[(e +
f*x)/4]^2])/(2*c^2*(c + d)^2) - (Sqrt[3 - 2*Sqrt[2]]*Cos[(e + f*x)/4]^2*Sqrt[Sec[e + f*x]]*Sqrt[1 + (-3 + 2*Sq
rt[2])*Tan[(e + f*x)/4]^2]*Sqrt[1 - (3 + 2*Sqrt[2])*Tan[(e + f*x)/4]^2]*((c*(3*c + d)*Sec[(e + f*x)/4]^2)/(4*S
qrt[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*S
qrt[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]))) - (c^2 - 3*c*d - 2*d^2)*(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*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])...
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(62282\) vs.
\(2(199)=398\).
time = 9.26, size = 62283, normalized size = 271.98
| | |
| method |
result |
size |
| | |
| default |
\(\text {Expression too large to display}\) |
\(62283\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sec(f*x+e))^(3/2)/(c+d*sec(f*x+e))^2,x,method=_RETURNVERBOSE)
[Out]
result too large to display
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Maxima [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((a+a*sec(f*x+e))^(3/2)/(c+d*sec(f*x+e))^2,x, algorithm="maxima")
[Out]
Timed out
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Fricas [A]
time = 16.61, size = 1726, normalized size = 7.54 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(f*x+e))^(3/2)/(c+d*sec(f*x+e))^2,x, algorithm="fricas")
[Out]
[1/2*(2*(a*c^2 - a*c*d)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) - (a*c^2*d - 3*a*c*d
^2 - 2*a*d^3 + (a*c^3 - 3*a*c^2*d - 2*a*c*d^2)*cos(f*x + e)^2 + (a*c^3 - 2*a*c^2*d - 5*a*c*d^2 - 2*a*d^3)*cos(
f*x + e))*sqrt(-a/(c*d + d^2))*log((2*(c*d + d^2)*sqrt(-a/(c*d + d^2))*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*(a*c*d + a*d^2 + (a*c^2 + a*c*d)*cos(f*x + e)^2 + (a*c^2 + 2*a*c*d + a*d^2)*c
os(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*(a*c^2 - a*c*d)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*
cos(f*x + e)*sin(f*x + e) - 4*(a*c*d + a*d^2 + (a*c^2 + a*c*d)*cos(f*x + e)^2 + (a*c^2 + 2*a*c*d + a*d^2)*cos(
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))) - (a*c^2
*d - 3*a*c*d^2 - 2*a*d^3 + (a*c^3 - 3*a*c^2*d - 2*a*c*d^2)*cos(f*x + e)^2 + (a*c^3 - 2*a*c^2*d - 5*a*c*d^2 - 2
*a*d^3)*cos(f*x + e))*sqrt(-a/(c*d + d^2))*log((2*(c*d + d^2)*sqrt(-a/(c*d + d^2))*sqrt((a*cos(f*x + e) + a)/c
os(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)))/((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), ((a*c^2 - a*c*d)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x +
e) - (a*c^2*d - 3*a*c*d^2 - 2*a*d^3 + (a*c^3 - 3*a*c^2*d - 2*a*c*d^2)*cos(f*x + e)^2 + (a*c^3 - 2*a*c^2*d - 5
*a*c*d^2 - 2*a*d^3)*cos(f*x + e))*sqrt(a/(c*d + d^2))*arctan((c + d)*sqrt(a/(c*d + d^2))*sqrt((a*cos(f*x + e)
+ a)/cos(f*x + e))*cos(f*x + e)/(a*sin(f*x + e))) + (a*c*d + a*d^2 + (a*c^2 + a*c*d)*cos(f*x + e)^2 + (a*c^2 +
2*a*c*d + a*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), ((a*c^2 - a*c*d)*sqrt((a*cos(f*x + e) + a)/c
os(f*x + e))*cos(f*x + e)*sin(f*x + e) - (a*c^2*d - 3*a*c*d^2 - 2*a*d^3 + (a*c^3 - 3*a*c^2*d - 2*a*c*d^2)*cos(
f*x + e)^2 + (a*c^3 - 2*a*c^2*d - 5*a*c*d^2 - 2*a*d^3)*cos(f*x + e))*sqrt(a/(c*d + d^2))*arctan((c + d)*sqrt(a
/(c*d + d^2))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(a*sin(f*x + e))) - 2*(a*c*d + a*d^2 + (a*c
^2 + a*c*d)*cos(f*x + e)^2 + (a*c^2 + 2*a*c*d + a*d^2)*cos(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))))/((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 {\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}{\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))**(3/2)/(c+d*sec(f*x+e))**2,x)
[Out]
Integral((a*(sec(e + f*x) + 1))**(3/2)/(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. 688 vs.
\(2 (199) = 398\).
time = 1.66, size = 688, normalized size = 3.00 \begin {gather*} -\frac {\frac {\sqrt {-a} a^{2} \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 ) \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{c^{2} {\left | a \right |}} - \frac {\sqrt {2} {\left (\sqrt {-a} a^{2} c^{2} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) - 3 \, \sqrt {-a} a^{2} c d \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) - 2 \, \sqrt {-a} a^{2} d^{2} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )\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 (\sqrt {2} c^{3} + \sqrt {2} c^{2} d\right )} \sqrt {-c d - d^{2}} a} + \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^{2} c \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + 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^{2} d \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + \sqrt {-a} a^{3} c \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) - \sqrt {-a} a^{3} d \mathrm {sgn}\left (\cos \left (f x + e\right )\right )\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 (\sqrt {2} c^{2} + \sqrt {2} c d\right )}}}{f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(f*x+e))^(3/2)/(c+d*sec(f*x+e))^2,x, algorithm="giac")
[Out]
-(sqrt(-a)*a^2*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)*a
bs(a) - 6*a)/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))*sgn(cos(f*x + e))/(c^2*abs(a)) - sqrt(2)*(sqrt(-a)*a^2*c^2*sgn(cos(f*x + e)) - 3*sqrt(-a)*a^2*c*d*sgn(
cos(f*x + e)) - 2*sqrt(-a)*a^2*d^2*sgn(cos(f*x + e)))*arctan(1/4*sqrt(2)*((sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqr
t(-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))/((sqrt(2)*c^3 + sqrt(2)*c^2*d)*sqrt(-c*d - d^2)*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^2*c*sgn(cos(f*x + e)) + 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^2*d*sgn(cos(f*x + e)) + sqrt(-a)*a^3*c*sg
n(cos(f*x + e)) - sqrt(-a)*a^3*d*sgn(cos(f*x + e)))/(((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)*ta
n(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*t
an(1/2*f*x + 1/2*e)^2 + a))^2*a*d + a^2*c - a^2*d)*(sqrt(2)*c^2 + sqrt(2)*c*d)))/f
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}}{{\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))^(3/2)/(c + d/cos(e + f*x))^2,x)
[Out]
int((a + a/cos(e + f*x))^(3/2)/(c + d/cos(e + f*x))^2, x)
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