3.239 \(\int \frac{(g \sec (e+f x))^{3/2} \sqrt{a+a \sec (e+f x)}}{c+d \sec (e+f x)} \, dx\)
Optimal. Leaf size=149 \[ \frac{2 \sqrt{a} g^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{g} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a} \sqrt{g \sec (e+f x)}}\right )}{d f}-\frac{2 \sqrt{a} \sqrt{c} g^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c} \sqrt{g} \tan (e+f x)}{\sqrt{c+d} \sqrt{a \sec (e+f x)+a} \sqrt{g \sec (e+f x)}}\right )}{d f \sqrt{c+d}} \]
[Out]
(2*Sqrt[a]*g^(3/2)*ArcTanh[(Sqrt[a]*Sqrt[g]*Tan[e + f*x])/(Sqrt[g*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]])])/(d
*f) - (2*Sqrt[a]*Sqrt[c]*g^(3/2)*ArcTanh[(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]])])/(d*Sqrt[c + d]*f)
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Rubi [A] time = 0.56998, antiderivative size = 149, normalized size of antiderivative = 1.,
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 =
{3970, 3802, 208, 3965} \[ \frac{2 \sqrt{a} g^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{g} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a} \sqrt{g \sec (e+f x)}}\right )}{d f}-\frac{2 \sqrt{a} \sqrt{c} g^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c} \sqrt{g} \tan (e+f x)}{\sqrt{c+d} \sqrt{a \sec (e+f x)+a} \sqrt{g \sec (e+f x)}}\right )}{d f \sqrt{c+d}} \]
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]
(2*Sqrt[a]*g^(3/2)*ArcTanh[(Sqrt[a]*Sqrt[g]*Tan[e + f*x])/(Sqrt[g*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]])])/(d
*f) - (2*Sqrt[a]*Sqrt[c]*g^(3/2)*ArcTanh[(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]])])/(d*Sqrt[c + d]*f)
Rule 3970
Int[((csc[(e_.) + (f_.)*(x_)]*(g_.))^(3/2)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)])/(csc[(e_.) + (f_.)*(x_)
]*(d_.) + (c_)), x_Symbol] :> Dist[g/d, Int[Sqrt[g*Csc[e + f*x]]*Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(c*g)
/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]
Rule 3802
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b*d)
/f, Subst[Int[1/(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] /; F
reeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && !GtQ[(a*d)/b, 0]
Rule 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rule 3965
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]
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 \sqrt{g \sec (e+f x)} \sqrt{a+a \sec (e+f x)} \, dx}{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}{d}\\ &=-\frac{\left (2 a g^2\right ) \operatorname{Subst}\left (\int \frac{1}{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 )}{d f}+\frac{\left (2 a c g^2\right ) \operatorname{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 )}{d f}\\ &=\frac{2 \sqrt{a} g^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{g} \tan (e+f x)}{\sqrt{g \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\right )}{d f}-\frac{2 \sqrt{a} \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 )}{d \sqrt{c+d} f}\\ \end{align*}
Mathematica [C] time = 1.32184, size = 427, normalized size = 2.87 \[ \frac{\left (\sqrt{2}-2 i\right ) g^2 \sec \left (\frac{1}{2} (e+f x)\right ) \sqrt{a (\sec (e+f x)+1)} \left (i \left (2 \sqrt{c+d} \log \left (2 \sin \left (\frac{1}{2} (e+f x)\right )+\sqrt{2}\right )+2 \sqrt{c} \log \left (\sqrt{2} \sqrt{c+d}-2 \sqrt{c} \sin \left (\frac{1}{2} (e+f x)\right )\right )-2 \sqrt{c} \log \left (\sqrt{2} \sqrt{c+d}+2 \sqrt{c} \sin \left (\frac{1}{2} (e+f x)\right )\right )-\sqrt{c+d} \log \left (-\sqrt{2} \sin \left (\frac{1}{2} (e+f x)\right )-\sqrt{2} \cos \left (\frac{1}{2} (e+f x)\right )+2\right )-\sqrt{c+d} \log \left (-\sqrt{2} \sin \left (\frac{1}{2} (e+f x)\right )+\sqrt{2} \cos \left (\frac{1}{2} (e+f x)\right )+2\right )\right )+2 \sqrt{c+d} \tan ^{-1}\left (\frac{\cos \left (\frac{1}{4} (e+f x)\right )-\left (\sqrt{2}-1\right ) \sin \left (\frac{1}{4} (e+f x)\right )}{\left (1+\sqrt{2}\right ) \cos \left (\frac{1}{4} (e+f x)\right )-\sin \left (\frac{1}{4} (e+f x)\right )}\right )+2 \sqrt{c+d} \tan ^{-1}\left (\frac{\cos \left (\frac{1}{4} (e+f x)\right )-\left (1+\sqrt{2}\right ) \sin \left (\frac{1}{4} (e+f x)\right )}{\left (\sqrt{2}-1\right ) \cos \left (\frac{1}{4} (e+f x)\right )-\sin \left (\frac{1}{4} (e+f x)\right )}\right )\right )}{4 \left (\sqrt{2}+i\right ) d f \sqrt{c+d} \sqrt{g \sec (e+f x)}} \]
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]
((-2*I + Sqrt[2])*g^2*(2*Sqrt[c + d]*ArcTan[(Cos[(e + f*x)/4] - (-1 + Sqrt[2])*Sin[(e + f*x)/4])/((1 + Sqrt[2]
)*Cos[(e + f*x)/4] - Sin[(e + f*x)/4])] + 2*Sqrt[c + d]*ArcTan[(Cos[(e + f*x)/4] - (1 + Sqrt[2])*Sin[(e + f*x)
/4])/((-1 + Sqrt[2])*Cos[(e + f*x)/4] - Sin[(e + f*x)/4])] + I*(2*Sqrt[c + d]*Log[Sqrt[2] + 2*Sin[(e + f*x)/2]
] - Sqrt[c + d]*Log[2 - Sqrt[2]*Cos[(e + f*x)/2] - Sqrt[2]*Sin[(e + f*x)/2]] - Sqrt[c + d]*Log[2 + Sqrt[2]*Cos
[(e + f*x)/2] - Sqrt[2]*Sin[(e + f*x)/2]] + 2*Sqrt[c]*Log[Sqrt[2]*Sqrt[c + d] - 2*Sqrt[c]*Sin[(e + f*x)/2]] -
2*Sqrt[c]*Log[Sqrt[2]*Sqrt[c + d] + 2*Sqrt[c]*Sin[(e + f*x)/2]]))*Sec[(e + f*x)/2]*Sqrt[a*(1 + Sec[e + f*x])])
/(4*(I + Sqrt[2])*d*Sqrt[c + d]*f*Sqrt[g*Sec[e + f*x]])
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Maple [B] time = 0.379, size = 568, normalized size = 3.8 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((g*sec(f*x+e))^(3/2)*(a+a*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e)),x)
[Out]
-2/f*(c-d)/((c+d)*(c-d))^(1/2)/(c-d+((c+d)*(c-d))^(1/2))/(-c+d+((c+d)*(c-d))^(1/2))/(c/(c-d))^(1/2)*(g/cos(f*x
+e))^(3/2)*cos(f*x+e)^2*(-1+cos(f*x+e))^2*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*(((c+d)*(c-d))^(1/2)*arctanh(1
/2*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)+1-sin(f*x+e)))*(c/(c-d))^(1/2)-((c+d)*(c-d))^(1/2)*arctanh(1/2*(1/(1+c
os(f*x+e)))^(1/2)*(cos(f*x+e)+1+sin(f*x+e)))*(c/(c-d))^(1/2)+c*ln(2*(-2*(1/(1+cos(f*x+e)))^(1/2)*(c/(c-d))^(1/
2)*c*sin(f*x+e)+2*(1/(1+cos(f*x+e)))^(1/2)*(c/(c-d))^(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/(1+cos(f*x+e)))^(1/2)*(c/(c-d))^(1/2)*c*sin(f*x+e)-2*(1/(1+cos(f*x+e)))^(1/2)*(c/(c-d))^(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)))/sin(f*x+e)^4/(1/(1+cos(f*x+e)))^(3/2)
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((g*sec(f*x+e))^(3/2)*(a+a*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e)),x, algorithm="maxima")
[Out]
Exception raised: RuntimeError
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Fricas [A] time = 21.0628, size = 2753, normalized size = 18.48 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((g*sec(f*x+e))^(3/2)*(a+a*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e)),x, algorithm="fricas")
[Out]
[1/2*(sqrt(a*c*g/(c + d))*g*log((a*c^2*g*cos(f*x + e)^3 - (7*a*c^2 + 6*a*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(a*c*g/(c + d))*sqrt((a*cos(f*x + e) + a)/cos(f*x +
e))*sqrt(g/cos(f*x + e))*sin(f*x + e) + (2*a*c*d + a*d^2)*g*cos(f*x + e) + (8*a*c^2 + 8*a*c*d + a*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))) + sqrt(a*g)*g*log((a*g*co
s(f*x + e)^3 - 7*a*g*cos(f*x + e)^2 - 4*sqrt(a*g)*(cos(f*x + e)^2 - 2*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/
cos(f*x + e))*sqrt(g/cos(f*x + e))*sin(f*x + e) + 8*a*g)/(cos(f*x + e)^3 + cos(f*x + e)^2)))/(d*f), -1/2*(2*sq
rt(-a*c*g/(c + d))*g*arctan(1/2*(c*cos(f*x + e)^2 - (2*c + d)*cos(f*x + e))*sqrt(-a*c*g/(c + d))*sqrt((a*cos(f
*x + e) + a)/cos(f*x + e))*sqrt(g/cos(f*x + e))/(a*c*g*sin(f*x + e))) - sqrt(a*g)*g*log((a*g*cos(f*x + e)^3 -
7*a*g*cos(f*x + e)^2 - 4*sqrt(a*g)*(cos(f*x + e)^2 - 2*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*s
qrt(g/cos(f*x + e))*sin(f*x + e) + 8*a*g)/(cos(f*x + e)^3 + cos(f*x + e)^2)))/(d*f), 1/2*(2*sqrt(-a*g)*g*arcta
n(2*sqrt(-a*g)*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)/(a*g*cos
(f*x + e)^2 - a*g*cos(f*x + e) - 2*a*g)) + sqrt(a*c*g/(c + d))*g*log((a*c^2*g*cos(f*x + e)^3 - (7*a*c^2 + 6*a*
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(a*c*g/(c + d)
)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(g/cos(f*x + e))*sin(f*x + e) + (2*a*c*d + a*d^2)*g*cos(f*x + e)
+ (8*a*c^2 + 8*a*c*d + a*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))))/(d*f), (sqrt(-a*g)*g*arctan(2*sqrt(-a*g)*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)/(a*g*cos(f*x + e)^2 - a*g*cos(f*x + e) - 2*a*g)) - sqrt(-a*c*g/(c + d))*g*arcta
n(1/2*(c*cos(f*x + e)^2 - (2*c + d)*cos(f*x + e))*sqrt(-a*c*g/(c + d))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))
*sqrt(g/cos(f*x + e))/(a*c*g*sin(f*x + e))))/(d*f)]
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((g*sec(f*x+e))**(3/2)*(a+a*sec(f*x+e))**(1/2)/(c+d*sec(f*x+e)),x)
[Out]
Timed out
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Giac [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((g*sec(f*x+e))^(3/2)*(a+a*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e)),x, algorithm="giac")
[Out]
Timed out