3.157 \(\int \frac{(a+a \sec (e+f x))^{3/2}}{c+d \sec (e+f x)} \, dx\)
Optimal. Leaf size=110 \[ \frac{2 a^{3/2} (c-d) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \tan (e+f x)}{\sqrt{c+d} \sqrt{a \sec (e+f x)+a}}\right )}{c \sqrt{d} f \sqrt{c+d}}+\frac{2 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{c f} \]
[Out]
(2*a^(3/2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(c*f) + (2*a^(3/2)*(c - d)*ArcTan[(Sqrt[a]
*Sqrt[d]*Tan[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sec[e + f*x]])])/(c*Sqrt[d]*Sqrt[c + d]*f)
________________________________________________________________________________________
Rubi [A] time = 0.240583, antiderivative size = 110, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used =
{3927, 3774, 203, 3967, 205} \[ \frac{2 a^{3/2} (c-d) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \tan (e+f x)}{\sqrt{c+d} \sqrt{a \sec (e+f x)+a}}\right )}{c \sqrt{d} f \sqrt{c+d}}+\frac{2 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{c f} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sec[e + f*x])^(3/2)/(c + d*Sec[e + f*x]),x]
[Out]
(2*a^(3/2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(c*f) + (2*a^(3/2)*(c - d)*ArcTan[(Sqrt[a]
*Sqrt[d]*Tan[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sec[e + f*x]])])/(c*Sqrt[d]*Sqrt[c + d]*f)
Rule 3927
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Dist[a/c
, Int[Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[(b*c - a*d)/c, 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] && (EqQ[a^2 - b^2, 0] || EqQ[c^2
- d^2, 0])
Rule 3774
Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(a + x^2), x], x, (b*C
ot[c + d*x])/Sqrt[a + b*Csc[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Rule 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 3967
Int[(csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)])/(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)
), x_Symbol] :> Dist[(-2*b)/f, Subst[Int[1/(b*c + a*d + d*x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x
]]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]
Rule 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rubi steps
\begin{align*} \int \frac{(a+a \sec (e+f x))^{3/2}}{c+d \sec (e+f x)} \, dx &=\frac{a \int \sqrt{a+a \sec (e+f x)} \, dx}{c}+\frac{(a c-a d) \int \frac{\sec (e+f x) \sqrt{a+a \sec (e+f x)}}{c+d \sec (e+f x)} \, dx}{c}\\ &=-\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+x^2} \, dx,x,-\frac{a \tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{c f}-\frac{\left (2 a^2 (c-d)\right ) \operatorname{Subst}\left (\int \frac{1}{a c+a d+d x^2} \, dx,x,-\frac{a \tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{c f}\\ &=\frac{2 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{c f}+\frac{2 a^{3/2} (c-d) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \tan (e+f x)}{\sqrt{c+d} \sqrt{a+a \sec (e+f x)}}\right )}{c \sqrt{d} \sqrt{c+d} f}\\ \end{align*}
Mathematica [A] time = 0.478421, size = 135, normalized size = 1.23 \[ \frac{\sqrt{2} a \sqrt{\cos (e+f x)} \sec \left (\frac{1}{2} (e+f x)\right ) \sqrt{a (\sec (e+f x)+1)} \left (\sqrt{d} \sqrt{c+d} \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (e+f x)\right )\right )+(c-d) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d} \sin \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d} \sqrt{\cos (e+f x)}}\right )\right )}{c \sqrt{d} f \sqrt{c+d}} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Sec[e + f*x])^(3/2)/(c + d*Sec[e + f*x]),x]
[Out]
(Sqrt[2]*a*(Sqrt[d]*Sqrt[c + d]*ArcSin[Sqrt[2]*Sin[(e + f*x)/2]] + (c - d)*ArcTan[(Sqrt[2]*Sqrt[d]*Sin[(e + f*
x)/2])/(Sqrt[c + d]*Sqrt[Cos[e + f*x]])])*Sqrt[Cos[e + f*x]]*Sec[(e + f*x)/2]*Sqrt[a*(1 + Sec[e + f*x])])/(c*S
qrt[d]*Sqrt[c + d]*f)
________________________________________________________________________________________
Maple [B] time = 0.217, size = 864, normalized size = 7.9 \begin{align*} \text{result too large to display} \end{align*}
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)),x)
[Out]
-1/2/f*2^(1/2)*a/((c+d)*(c-d))^(1/2)/c/(d/(c-d))^(1/2)*(2*(d/(c-d))^(1/2)*((c+d)*(c-d))^(1/2)*arctanh(1/2*2^(1
/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))-ln(-2*((-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(d
/(c-d))^(1/2)*2^(1/2)*c*sin(f*x+e)-2^(1/2)*(d/(c-d))^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(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+d*ln(-2*((-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(d/(c-d))^(1/2)*2^(1/2)*c*sin(f
*x+e)-2^(1/2)*(d/(c-d))^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(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)
)+ln(2*((-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(d/(c-d))^(1/2)*2^(1/2)*c*sin(f*x+e)-2^(1/2)*(d/(c-d))^(1/2)*(-2*
cos(f*x+e)/(1+cos(f*x+e)))^(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-d*ln(2*((-2*cos(f*x+e)/(1+cos(
f*x+e)))^(1/2)*(d/(c-d))^(1/2)*2^(1/2)*c*sin(f*x+e)-2^(1/2)*(d/(c-d))^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(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)))*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/cos(f*x+e)*a*(1+cos(f
*x+e)))^(1/2)
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}{d \sec \left (f x + e\right ) + c}\,{d x} \end{align*}
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)),x, algorithm="maxima")
[Out]
integrate((a*sec(f*x + e) + a)^(3/2)/(d*sec(f*x + e) + c), x)
________________________________________________________________________________________
Fricas [A] time = 8.84398, size = 1804, normalized size = 16.4 \begin{align*} \text{result too large to display} \end{align*}
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)),x, algorithm="fricas")
[Out]
[-((a*c - a*d)*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)) - sqrt(-a)*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*f), -(2*a^(3/2)*arctan(s
qrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) + (a*c - a*d)*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)))/
(c*f), -(2*(a*c - a*d)*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))) - sqrt(-a)*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*f), -2*((a*c - a*d)
*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^(3/2)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e)))
)/(c*f)]
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a \left (\sec{\left (e + f x \right )} + 1\right )\right )^{\frac{3}{2}}}{c + d \sec{\left (e + f x \right )}}\, dx \end{align*}
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)),x)
[Out]
Integral((a*(sec(e + f*x) + 1))**(3/2)/(c + d*sec(e + f*x)), x)
________________________________________________________________________________________
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((a+a*sec(f*x+e))^(3/2)/(c+d*sec(f*x+e)),x, algorithm="giac")
[Out]
Timed out