3.69 \(\int \frac{\sec (e+f x) (a+a \sec (e+f x))}{(c-c \sec (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=76 \[ \frac{a \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{\sqrt{2} c^{3/2} f}-\frac{a \tan (e+f x)}{f (c-c \sec (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.114423, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used = {3957, 3795, 203} \[ \frac{a \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{\sqrt{2} c^{3/2} f}-\frac{a \tan (e+f x)}{f (c-c \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3957

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

Rule 3795

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

Rubi steps

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

Mathematica [C]  time = 1.10001, size = 246, normalized size = 3.24 \[ \frac{a \left (\frac{i \sqrt{2} \left (-1+e^{i (e+f x)}\right )^3 \tanh ^{-1}\left (\frac{1+e^{i (e+f x)}}{\sqrt{2} \sqrt{1+e^{2 i (e+f x)}}}\right )}{\left (1+e^{2 i (e+f x)}\right )^{3/2}}+8 \sin \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \sin ^3\left (\frac{1}{2} (e+f x)\right ) \sec ^2(e+f x)-8 \cos \left (\frac{e}{2}\right ) \cos \left (\frac{f x}{2}\right ) \sin ^3\left (\frac{1}{2} (e+f x)\right ) \sec ^2(e+f x)+4 \cot \left (\frac{e}{2}\right ) \sin ^2\left (\frac{1}{2} (e+f x)\right ) \sec ^2(e+f x)-4 \csc \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \sin \left (\frac{1}{2} (e+f x)\right ) \sec ^2(e+f x)\right )}{2 c f (\sec (e+f x)-1) \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*((I*Sqrt[2]*(-1 + E^(I*(e + f*x)))^3*ArcTanh[(1 + E^(I*(e + f*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(e + f*x))])]
)/(1 + E^((2*I)*(e + f*x)))^(3/2) - 4*Csc[e/2]*Sec[e + f*x]^2*Sin[(f*x)/2]*Sin[(e + f*x)/2] + 4*Cot[e/2]*Sec[e
+ f*x]^2*Sin[(e + f*x)/2]^2 - 8*Cos[e/2]*Cos[(f*x)/2]*Sec[e + f*x]^2*Sin[(e + f*x)/2]^3 + 8*Sec[e + f*x]^2*Si
n[e/2]*Sin[(f*x)/2]*Sin[(e + f*x)/2]^3))/(2*c*f*(-1 + Sec[e + f*x])*Sqrt[c - c*Sec[e + f*x]])

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Maple [B]  time = 0.199, size = 164, normalized size = 2.2 \begin{align*} 2\,{\frac{a \left ( -1+\cos \left ( fx+e \right ) \right ) ^{2}}{f \left ( \sin \left ( fx+e \right ) \right ) ^{3}} \left ( \sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\cos \left ( fx+e \right ) +\cos \left ( fx+e \right ) \arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}}} \right ) +\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}-\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}}} \right ) \right ) \left ({\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{-3/2} \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{-3/2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))/(c-c*sec(f*x+e))^(3/2),x)

[Out]

2*a/f*(-1+cos(f*x+e))^2*((-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*cos(f*x+e)+cos(f*x+e)*arctan(1/(-2*cos(f*x+e)/(1
+cos(f*x+e)))^(1/2))+(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)-arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)))/(c*(
-1+cos(f*x+e))/cos(f*x+e))^(3/2)/sin(f*x+e)^3/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sec \left (f x + e\right ) + a\right )} \sec \left (f x + e\right )}{{\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))/(c-c*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sec(f*x + e) + a)*sec(f*x + e)/(-c*sec(f*x + e) + c)^(3/2), x)

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Fricas [B]  time = 0.606074, size = 864, normalized size = 11.37 \begin{align*} \left [\frac{\sqrt{2}{\left (a c \cos \left (f x + e\right ) - a c\right )} \sqrt{-\frac{1}{c}} \log \left (\frac{2 \, \sqrt{2}{\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \sqrt{-\frac{1}{c}} +{\left (3 \, \cos \left (f x + e\right ) + 1\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 4 \,{\left (a \cos \left (f x + e\right )^{2} + a \cos \left (f x + e\right )\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{4 \,{\left (c^{2} f \cos \left (f x + e\right ) - c^{2} f\right )} \sin \left (f x + e\right )}, -\frac{\frac{\sqrt{2}{\left (a c \cos \left (f x + e\right ) - a c\right )} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt{c} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right )}{\sqrt{c}} - 2 \,{\left (a \cos \left (f x + e\right )^{2} + a \cos \left (f x + e\right )\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{2 \,{\left (c^{2} f \cos \left (f x + e\right ) - c^{2} f\right )} \sin \left (f x + e\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))/(c-c*sec(f*x+e))**(3/2),x)

[Out]

a*(Integral(sec(e + f*x)/(-c*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x) + c*sqrt(-c*sec(e + f*x) + c)), x) + Integ
ral(sec(e + f*x)**2/(-c*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x) + c*sqrt(-c*sec(e + f*x) + c)), x))

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Giac [A]  time = 1.59768, size = 142, normalized size = 1.87 \begin{align*} \frac{\sqrt{2} a{\left (\frac{\arctan \left (\frac{\sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c}}{\sqrt{c}}\right )}{c^{\frac{3}{2}}} + \frac{\sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c}}{c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2}}\right )}}{2 \, f \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*a*(arctan(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/sqrt(c))/c^(3/2) + sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)
/(c^2*tan(1/2*f*x + 1/2*e)^2))/(f*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)*sgn(tan(1/2*f*x + 1/2*e)))