3.3.42 \(\int \frac {(g \sec (e+f x))^{3/2}}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx\) [242]

Optimal. Leaf size=167 \[ -\frac {\sqrt {2} g^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {g} \tan (e+f x)}{\sqrt {2} \sqrt {g \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\right )}{\sqrt {a} (c-d) f}+\frac {2 \sqrt {c} g^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c} \sqrt {g} \tan (e+f x)}{\sqrt {c+d} \sqrt {g \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\right )}{\sqrt {a} (c-d) \sqrt {c+d} f} \]

[Out]

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

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Rubi [A]
time = 0.38, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {4059, 3893, 214, 4050} \begin {gather*} \frac {2 \sqrt {c} g^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c} \sqrt {g} \tan (e+f x)}{\sqrt {c+d} \sqrt {a \sec (e+f x)+a} \sqrt {g \sec (e+f x)}}\right )}{\sqrt {a} f (c-d) \sqrt {c+d}}-\frac {\sqrt {2} g^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {g} \tan (e+f x)}{\sqrt {2} \sqrt {a \sec (e+f x)+a} \sqrt {g \sec (e+f x)}}\right )}{\sqrt {a} f (c-d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(g*Sec[e + f*x])^(3/2)/(Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x])),x]

[Out]

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

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 3893

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

Rule 4050

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

Rule 4059

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(3/2)/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]
*(d_.) + (c_))), x_Symbol] :> Dist[(-a)*(g/(b*c - a*d)), Int[Sqrt[g*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x]
, x] + Dist[c*(g/(b*c - a*d)), Int[Sqrt[g*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, g}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [A]
time = 0.41, size = 198, normalized size = 1.19 \begin {gather*} \frac {g \cos \left (\frac {1}{2} (e+f x)\right ) \left (2 \sqrt {c+d} \log \left (\cos \left (\frac {1}{4} (e+f x)\right )-\sin \left (\frac {1}{4} (e+f x)\right )\right )-2 \sqrt {c+d} \log \left (\cos \left (\frac {1}{4} (e+f x)\right )+\sin \left (\frac {1}{4} (e+f x)\right )\right )+\sqrt {2} \sqrt {c} \left (-\log \left (\sqrt {2} \sqrt {c+d}-2 \sqrt {c} \sin \left (\frac {1}{2} (e+f x)\right )\right )+\log \left (\sqrt {2} \sqrt {c+d}+2 \sqrt {c} \sin \left (\frac {1}{2} (e+f x)\right )\right )\right )\right ) \sqrt {g \sec (e+f x)}}{(c-d) \sqrt {c+d} f \sqrt {a (1+\sec (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(g*Sec[e + f*x])^(3/2)/(Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x])),x]

[Out]

(g*Cos[(e + f*x)/2]*(2*Sqrt[c + d]*Log[Cos[(e + f*x)/4] - Sin[(e + f*x)/4]] - 2*Sqrt[c + d]*Log[Cos[(e + f*x)/
4] + Sin[(e + f*x)/4]] + Sqrt[2]*Sqrt[c]*(-Log[Sqrt[2]*Sqrt[c + d] - 2*Sqrt[c]*Sin[(e + f*x)/2]] + Log[Sqrt[2]
*Sqrt[c + d] + 2*Sqrt[c]*Sin[(e + f*x)/2]]))*Sqrt[g*Sec[e + f*x]])/((c - d)*Sqrt[c + d]*f*Sqrt[a*(1 + Sec[e +
f*x])])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(472\) vs. \(2(132)=264\).
time = 13.86, size = 473, normalized size = 2.83

method	result	size
default	\(\frac {\left (\frac {g}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, \left (\cos ^{2}\left (f x +e \right )\right ) \left (-1+\cos \left (f x +e \right )\right )^{2} \left (\arcsinh \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \sqrt {2}\, \sqrt {\frac {c}{c -d}}\, \sqrt {\left (c +d \right ) \left (c -d \right )}+c \ln \left (-\frac {2 \left (2 \sqrt {\frac {c}{c -d}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, c \sin \left (f x +e \right )-2 \sqrt {\frac {c}{c -d}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, d \sin \left (f x +e \right )-\sqrt {\left (c +d \right ) \left (c -d \right )}\, \cos \left (f x +e \right )+c \sin \left (f x +e \right )-d \sin \left (f x +e \right )+\sqrt {\left (c +d \right ) \left (c -d \right )}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}\, \sin \left (f x +e \right )+c \cos \left (f x +e \right )-d \cos \left (f x +e \right )-c +d}\right )-c \ln \left (-\frac {2 \left (2 \sqrt {\frac {c}{c -d}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, c \sin \left (f x +e \right )-2 \sqrt {\frac {c}{c -d}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, d \sin \left (f x +e \right )+\sqrt {\left (c +d \right ) \left (c -d \right )}\, \cos \left (f x +e \right )+c \sin \left (f x +e \right )-d \sin \left (f x +e \right )-\sqrt {\left (c +d \right ) \left (c -d \right )}\right )}{c \cos \left (f x +e \right )-d \cos \left (f x +e \right )-\sqrt {\left (c +d \right ) \left (c -d \right )}\, \sin \left (f x +e \right )-c +d}\right )\right )}{f \sin \left (f x +e \right )^{4} \left (\frac {1}{\cos \left (f x +e \right )+1}\right )^{\frac {3}{2}} a \sqrt {\frac {c}{c -d}}\, \left (c -d \right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}\)	\(473\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*sec(f*x+e))^(3/2)/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/f*(g/cos(f*x+e))^(3/2)*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)*cos(f*x+e)^2*(-1+cos(f*x+e))^2*(arcsinh((-1+cos(f
*x+e))/sin(f*x+e))*2^(1/2)*(1/(c-d)*c)^(1/2)*((c+d)*(c-d))^(1/2)+c*ln(-2*(2*(1/(c-d)*c)^(1/2)*(1/(cos(f*x+e)+1
))^(1/2)*c*sin(f*x+e)-2*(1/(c-d)*c)^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*d*sin(f*x+e)-((c+d)*(c-d))^(1/2)*cos(f*x+e)
+c*sin(f*x+e)-d*sin(f*x+e)+((c+d)*(c-d))^(1/2))/(((c+d)*(c-d))^(1/2)*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d)
)-c*ln(-2*(2*(1/(c-d)*c)^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*c*sin(f*x+e)-2*(1/(c-d)*c)^(1/2)*(1/(cos(f*x+e)+1))^(1
/2)*d*sin(f*x+e)+((c+d)*(c-d))^(1/2)*cos(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-((c+d)*(c-d))^(1/2))/(c*cos(f*x+e)-d
*cos(f*x+e)-((c+d)*(c-d))^(1/2)*sin(f*x+e)-c+d)))/sin(f*x+e)^4/(1/(cos(f*x+e)+1))^(3/2)/a/(1/(c-d)*c)^(1/2)/(c
-d)/((c+d)*(c-d))^(1/2)

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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((g*sec(f*x+e))^(3/2)/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

1/2*(sqrt(2)*c*f*g*integrate(((c^2*cos(2*f*x + 2*e)^2 + c^2*sin(2*f*x + 2*e)^2 - 2*(c*d - 2*d^2)*cos(f*x + e)^
2 - (c^2 - 4*c*d)*sin(2*f*x + 2*e)*sin(f*x + e) - 2*(c*d - 2*d^2)*sin(f*x + e)^2 + (c^2 - (c^2 - 4*c*d)*cos(f*
x + e))*cos(2*f*x + 2*e) - (c^2 - 2*c*d)*cos(f*x + e))*cos(1/2*arctan2(sin(f*x + e), cos(f*x + e))) - (c^2*cos
(2*f*x + 2*e)*sin(f*x + e) - (c^2*cos(f*x + e) + c^2)*sin(2*f*x + 2*e) + (c^2 - 2*c*d)*sin(f*x + e))*sin(1/2*a
rctan2(sin(f*x + e), cos(f*x + e))))/((c^2*cos(2*f*x + 2*e)^2 + 4*d^2*cos(f*x + e)^2 + c^2*sin(2*f*x + 2*e)^2
+ 4*c*d*sin(2*f*x + 2*e)*sin(f*x + e) + 4*d^2*sin(f*x + e)^2 + 4*c*d*cos(f*x + e) + c^2 + 2*(2*c*d*cos(f*x + e
) + c^2)*cos(2*f*x + 2*e))*cos(1/2*arctan2(sin(f*x + e), cos(f*x + e)))^2 + (c^2*cos(2*f*x + 2*e)^2 + 4*d^2*co
s(f*x + e)^2 + c^2*sin(2*f*x + 2*e)^2 + 4*c*d*sin(2*f*x + 2*e)*sin(f*x + e) + 4*d^2*sin(f*x + e)^2 + 4*c*d*cos
(f*x + e) + c^2 + 2*(2*c*d*cos(f*x + e) + c^2)*cos(2*f*x + 2*e))*sin(1/2*arctan2(sin(f*x + e), cos(f*x + e)))^
2), x) + sqrt(2)*c*f*g*integrate(((2*c*d*cos(f*x + e)^2 + 2*c*d*sin(f*x + e)^2 - (c^2 - 2*c*d)*cos(2*f*x + 2*e
)^2 + c^2*cos(f*x + e) - (c^2 - 2*c*d)*sin(2*f*x + 2*e)^2 + (c^2 - 2*c*d + 4*d^2)*sin(2*f*x + 2*e)*sin(f*x + e
) - (c^2 - 2*c*d - (c^2 - 2*c*d + 4*d^2)*cos(f*x + e))*cos(2*f*x + 2*e))*cos(1/2*arctan2(sin(f*x + e), cos(f*x
+ e))) + (c^2*sin(f*x + e) + (c^2 + 2*c*d - 4*d^2)*cos(2*f*x + 2*e)*sin(f*x + e) - (c^2 - 2*c*d + (c^2 + 2*c*
d - 4*d^2)*cos(f*x + e))*sin(2*f*x + 2*e))*sin(1/2*arctan2(sin(f*x + e), cos(f*x + e))))/((c^2*cos(2*f*x + 2*e
)^2 + 4*d^2*cos(f*x + e)^2 + c^2*sin(2*f*x + 2*e)^2 + 4*c*d*sin(2*f*x + 2*e)*sin(f*x + e) + 4*d^2*sin(f*x + e)
^2 + 4*c*d*cos(f*x + e) + c^2 + 2*(2*c*d*cos(f*x + e) + c^2)*cos(2*f*x + 2*e))*cos(1/2*arctan2(sin(f*x + e), c
os(f*x + e)))^2 + (c^2*cos(2*f*x + 2*e)^2 + 4*d^2*cos(f*x + e)^2 + c^2*sin(2*f*x + 2*e)^2 + 4*c*d*sin(2*f*x +
2*e)*sin(f*x + e) + 4*d^2*sin(f*x + e)^2 + 4*c*d*cos(f*x + e) + c^2 + 2*(2*c*d*cos(f*x + e) + c^2)*cos(2*f*x +
2*e))*sin(1/2*arctan2(sin(f*x + e), cos(f*x + e)))^2), x) - sqrt(2)*d*g*log(cos(1/2*arctan2(sin(f*x + e), cos
(f*x + e)))^2 + sin(1/2*arctan2(sin(f*x + e), cos(f*x + e)))^2 + 2*sin(1/2*arctan2(sin(f*x + e), cos(f*x + e))
) + 1) + sqrt(2)*d*g*log(cos(1/2*arctan2(sin(f*x + e), cos(f*x + e)))^2 + sin(1/2*arctan2(sin(f*x + e), cos(f*
x + e)))^2 - 2*sin(1/2*arctan2(sin(f*x + e), cos(f*x + e))) + 1))*sqrt(g)/((c*d - d^2)*sqrt(a)*f)

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Fricas [A]
time = 4.42, size = 1167, normalized size = 6.99 \begin {gather*} \left [-\frac {\sqrt {2} g \sqrt {\frac {g}{a}} \log \left (\frac {2 \, \sqrt {2} \sqrt {\frac {g}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {g}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - g \cos \left (f x + e\right )^{2} + 2 \, g \cos \left (f x + e\right ) + 3 \, g}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) + \sqrt {\frac {c g}{a c + a d}} g \log \left (\frac {c^{2} g \cos \left (f x + e\right )^{3} - {\left (7 \, c^{2} + 6 \, c d\right )} g \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (c^{2} + c d\right )} \cos \left (f x + e\right )^{2} - {\left (2 \, c^{2} + 3 \, c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {c g}{a c + a d}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {g}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) + {\left (2 \, c d + d^{2}\right )} g \cos \left (f x + e\right ) + {\left (8 \, c^{2} + 8 \, c d + d^{2}\right )} g}{c^{2} \cos \left (f x + e\right )^{3} + {\left (c^{2} + 2 \, c d\right )} \cos \left (f x + e\right )^{2} + d^{2} + {\left (2 \, c d + d^{2}\right )} \cos \left (f x + e\right )}\right )}{2 \, {\left (c - d\right )} f}, \frac {2 \, \sqrt {2} g \sqrt {-\frac {g}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {-\frac {g}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {g}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{g \sin \left (f x + e\right )}\right ) - \sqrt {\frac {c g}{a c + a d}} g \log \left (\frac {c^{2} g \cos \left (f x + e\right )^{3} - {\left (7 \, c^{2} + 6 \, c d\right )} g \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (c^{2} + c d\right )} \cos \left (f x + e\right )^{2} - {\left (2 \, c^{2} + 3 \, c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {c g}{a c + a d}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {g}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) + {\left (2 \, c d + d^{2}\right )} g \cos \left (f x + e\right ) + {\left (8 \, c^{2} + 8 \, c d + d^{2}\right )} g}{c^{2} \cos \left (f x + e\right )^{3} + {\left (c^{2} + 2 \, c d\right )} \cos \left (f x + e\right )^{2} + d^{2} + {\left (2 \, c d + d^{2}\right )} \cos \left (f x + e\right )}\right )}{2 \, {\left (c - d\right )} f}, -\frac {\sqrt {2} g \sqrt {\frac {g}{a}} \log \left (\frac {2 \, \sqrt {2} \sqrt {\frac {g}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {g}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - g \cos \left (f x + e\right )^{2} + 2 \, g \cos \left (f x + e\right ) + 3 \, g}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) - 2 \, \sqrt {-\frac {c g}{a c + a d}} g \arctan \left (\frac {{\left (c \cos \left (f x + e\right )^{2} - {\left (2 \, c + d\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {c g}{a c + a d}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {g}{\cos \left (f x + e\right )}}}{2 \, c g \sin \left (f x + e\right )}\right )}{2 \, {\left (c - d\right )} f}, \frac {\sqrt {2} g \sqrt {-\frac {g}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {-\frac {g}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {g}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{g \sin \left (f x + e\right )}\right ) + \sqrt {-\frac {c g}{a c + a d}} g \arctan \left (\frac {{\left (c \cos \left (f x + e\right )^{2} - {\left (2 \, c + d\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {c g}{a c + a d}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {g}{\cos \left (f x + e\right )}}}{2 \, c g \sin \left (f x + e\right )}\right )}{{\left (c - d\right )} f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*sec(f*x+e))^(3/2)/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[-1/2*(sqrt(2)*g*sqrt(g/a)*log((2*sqrt(2)*sqrt(g/a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(g/cos(f*x + e
))*cos(f*x + e)*sin(f*x + e) - g*cos(f*x + e)^2 + 2*g*cos(f*x + e) + 3*g)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1
)) + sqrt(c*g/(a*c + a*d))*g*log((c^2*g*cos(f*x + e)^3 - (7*c^2 + 6*c*d)*g*cos(f*x + e)^2 + 4*((c^2 + c*d)*cos
(f*x + e)^2 - (2*c^2 + 3*c*d + d^2)*cos(f*x + e))*sqrt(c*g/(a*c + a*d))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e)
)*sqrt(g/cos(f*x + e))*sin(f*x + e) + (2*c*d + d^2)*g*cos(f*x + e) + (8*c^2 + 8*c*d + d^2)*g)/(c^2*cos(f*x + e
)^3 + (c^2 + 2*c*d)*cos(f*x + e)^2 + d^2 + (2*c*d + d^2)*cos(f*x + e))))/((c - d)*f), 1/2*(2*sqrt(2)*g*sqrt(-g
/a)*arctan(sqrt(2)*sqrt(-g/a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(g/cos(f*x + e))*cos(f*x + e)/(g*sin
(f*x + e))) - sqrt(c*g/(a*c + a*d))*g*log((c^2*g*cos(f*x + e)^3 - (7*c^2 + 6*c*d)*g*cos(f*x + e)^2 + 4*((c^2 +
c*d)*cos(f*x + e)^2 - (2*c^2 + 3*c*d + d^2)*cos(f*x + e))*sqrt(c*g/(a*c + a*d))*sqrt((a*cos(f*x + e) + a)/cos
(f*x + e))*sqrt(g/cos(f*x + e))*sin(f*x + e) + (2*c*d + d^2)*g*cos(f*x + e) + (8*c^2 + 8*c*d + d^2)*g)/(c^2*co
s(f*x + e)^3 + (c^2 + 2*c*d)*cos(f*x + e)^2 + d^2 + (2*c*d + d^2)*cos(f*x + e))))/((c - d)*f), -1/2*(sqrt(2)*g
*sqrt(g/a)*log((2*sqrt(2)*sqrt(g/a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(g/cos(f*x + e))*cos(f*x + e)*
sin(f*x + e) - g*cos(f*x + e)^2 + 2*g*cos(f*x + e) + 3*g)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) - 2*sqrt(-c*g
/(a*c + a*d))*g*arctan(1/2*(c*cos(f*x + e)^2 - (2*c + d)*cos(f*x + e))*sqrt(-c*g/(a*c + a*d))*sqrt((a*cos(f*x
+ e) + a)/cos(f*x + e))*sqrt(g/cos(f*x + e))/(c*g*sin(f*x + e))))/((c - d)*f), (sqrt(2)*g*sqrt(-g/a)*arctan(sq
rt(2)*sqrt(-g/a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(g/cos(f*x + e))*cos(f*x + e)/(g*sin(f*x + e))) +
sqrt(-c*g/(a*c + a*d))*g*arctan(1/2*(c*cos(f*x + e)^2 - (2*c + d)*cos(f*x + e))*sqrt(-c*g/(a*c + a*d))*sqrt((
a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(g/cos(f*x + e))/(c*g*sin(f*x + e))))/((c - d)*f)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (g \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )} \left (c + d \sec {\left (e + f x \right )}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*sec(f*x+e))**(3/2)/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))**(1/2),x)

[Out]

Integral((g*sec(e + f*x))**(3/2)/(sqrt(a*(sec(e + f*x) + 1))*(c + d*sec(e + f*x))), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*sec(f*x+e))^(3/2)/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(co

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (\frac {g}{\cos \left (e+f\,x\right )}\right )}^{3/2}}{\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}\,\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g/cos(e + f*x))^(3/2)/((a + a/cos(e + f*x))^(1/2)*(c + d/cos(e + f*x))),x)

[Out]

int((g/cos(e + f*x))^(3/2)/((a + a/cos(e + f*x))^(1/2)*(c + d/cos(e + f*x))), x)

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