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

Optimal. Leaf size=168 \[ \frac{10 \sqrt{2} a^3 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{c^{3/2} f}-\frac{10 a^3 \tan (e+f x)}{c f \sqrt{c-c \sec (e+f x)}}-\frac{5 \tan (e+f x) \left (a^3 \sec (e+f x)+a^3\right )}{3 c f \sqrt{c-c \sec (e+f x)}}-\frac{a \tan (e+f x) (a \sec (e+f x)+a)^2}{f (c-c \sec (e+f x))^{3/2}} \]

[Out]

(10*Sqrt[2]*a^3*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(c^(3/2)*f) - (a*(a + a*Sec
[e + f*x])^2*Tan[e + f*x])/(f*(c - c*Sec[e + f*x])^(3/2)) - (10*a^3*Tan[e + f*x])/(c*f*Sqrt[c - c*Sec[e + f*x]
]) - (5*(a^3 + a^3*Sec[e + f*x])*Tan[e + f*x])/(3*c*f*Sqrt[c - c*Sec[e + f*x]])

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Rubi [A]  time = 0.336487, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {3957, 3956, 3795, 203} \[ \frac{10 \sqrt{2} a^3 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{c^{3/2} f}-\frac{10 a^3 \tan (e+f x)}{c f \sqrt{c-c \sec (e+f x)}}-\frac{5 \tan (e+f x) \left (a^3 \sec (e+f x)+a^3\right )}{3 c f \sqrt{c-c \sec (e+f x)}}-\frac{a \tan (e+f x) (a \sec (e+f x)+a)^2}{f (c-c \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(10*Sqrt[2]*a^3*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(c^(3/2)*f) - (a*(a + a*Sec
[e + f*x])^2*Tan[e + f*x])/(f*(c - c*Sec[e + f*x])^(3/2)) - (10*a^3*Tan[e + f*x])/(c*f*Sqrt[c - c*Sec[e + f*x]
]) - (5*(a^3 + a^3*Sec[e + f*x])*Tan[e + f*x])/(3*c*f*Sqrt[c - c*Sec[e + f*x]])

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 3956

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.)
+ (a_)], x_Symbol] :> Simp[(-2*d*Cot[e + f*x]*(c + d*Csc[e + f*x])^(n - 1))/(f*(2*n - 1)*Sqrt[a + b*Csc[e + f*
x]]), x] + Dist[(2*c*(2*n - 1))/(2*n - 1), Int[(Csc[e + f*x]*(c + d*Csc[e + f*x])^(n - 1))/Sqrt[a + b*Csc[e +
f*x]], 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]

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))^3}{(c-c \sec (e+f x))^{3/2}} \, dx &=-\frac{a (a+a \sec (e+f x))^2 \tan (e+f x)}{f (c-c \sec (e+f x))^{3/2}}-\frac{(5 a) \int \frac{\sec (e+f x) (a+a \sec (e+f x))^2}{\sqrt{c-c \sec (e+f x)}} \, dx}{2 c}\\ &=-\frac{a (a+a \sec (e+f x))^2 \tan (e+f x)}{f (c-c \sec (e+f x))^{3/2}}-\frac{5 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 c f \sqrt{c-c \sec (e+f x)}}-\frac{\left (5 a^2\right ) \int \frac{\sec (e+f x) (a+a \sec (e+f x))}{\sqrt{c-c \sec (e+f x)}} \, dx}{c}\\ &=-\frac{a (a+a \sec (e+f x))^2 \tan (e+f x)}{f (c-c \sec (e+f x))^{3/2}}-\frac{10 a^3 \tan (e+f x)}{c f \sqrt{c-c \sec (e+f x)}}-\frac{5 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 c f \sqrt{c-c \sec (e+f x)}}-\frac{\left (10 a^3\right ) \int \frac{\sec (e+f x)}{\sqrt{c-c \sec (e+f x)}} \, dx}{c}\\ &=-\frac{a (a+a \sec (e+f x))^2 \tan (e+f x)}{f (c-c \sec (e+f x))^{3/2}}-\frac{10 a^3 \tan (e+f x)}{c f \sqrt{c-c \sec (e+f x)}}-\frac{5 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 c f \sqrt{c-c \sec (e+f x)}}+\frac{\left (20 a^3\right ) \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{10 \sqrt{2} a^3 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{c^{3/2} f}-\frac{a (a+a \sec (e+f x))^2 \tan (e+f x)}{f (c-c \sec (e+f x))^{3/2}}-\frac{10 a^3 \tan (e+f x)}{c f \sqrt{c-c \sec (e+f x)}}-\frac{5 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 c f \sqrt{c-c \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 2.96077, size = 324, normalized size = 1.93 \[ -\frac{a^3 \csc \left (\frac{e}{2}\right ) e^{-\frac{1}{2} i (e+f x)} \tan ^3\left (\frac{1}{2} (e+f x)\right ) \sec ^2\left (\frac{1}{2} (e+f x)\right ) (\sec (e+f x)+1)^3 \left (-\frac{i \left (-1+e^{i e}\right ) e^{\frac{i f x}{2}} \left (-24 e^{i (e+f x)}+34 e^{2 i (e+f x)}-24 e^{3 i (e+f x)}+19 e^{4 i (e+f x)}+19\right ) \sqrt{\sec (e+f x)}}{2 \left (-1+e^{i (e+f x)}\right )^2 \left (1+e^{2 i (e+f x)}\right )}-15 \sin \left (\frac{e}{2}\right ) \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt{1+e^{2 i (e+f x)}} \sec \left (\frac{1}{2} (e+f x)\right ) \tanh ^{-1}\left (\frac{1+e^{i (e+f x)}}{\sqrt{2} \sqrt{1+e^{2 i (e+f x)}}}\right )\right )}{3 c f (\sec (e+f x)-1) \sec ^{\frac{3}{2}}(e+f x) \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*Csc[e/2]*Sec[(e + f*x)/2]^2*(1 + Sec[e + f*x])^3*(((-I/2)*E^((I/2)*f*x)*(-1 + E^(I*e))*(19 - 24*E^(I*(e
+ f*x)) + 34*E^((2*I)*(e + f*x)) - 24*E^((3*I)*(e + f*x)) + 19*E^((4*I)*(e + f*x)))*Sqrt[Sec[e + f*x]])/((-1 +
E^(I*(e + f*x)))^2*(1 + E^((2*I)*(e + f*x)))) - 15*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*Sqrt[1 + E
^((2*I)*(e + f*x))]*ArcTanh[(1 + E^(I*(e + f*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(e + f*x))])]*Sec[(e + f*x)/2]*Si
n[e/2])*Tan[(e + f*x)/2]^3)/(3*c*E^((I/2)*(e + f*x))*f*(-1 + Sec[e + f*x])*Sec[e + f*x]^(3/2)*Sqrt[c - c*Sec[e
+ f*x]])

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Maple [A]  time = 0.232, size = 157, normalized size = 0.9 \begin{align*} -{\frac{{a}^{3}\sin \left ( fx+e \right ) }{3\,f \left ( \cos \left ( fx+e \right ) \right ) ^{3}} \left ( 15\,\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}}} \right ) \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{3/2} \left ( \cos \left ( fx+e \right ) \right ) ^{2}-15\,\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}}} \right ) \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{3/2}+38\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-24\,\cos \left ( fx+e \right ) -2 \right ) \left ({\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{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))^3/(c-c*sec(f*x+e))^(3/2),x)

[Out]

-1/3*a^3/f*(15*arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2))*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)*cos(f*x+e)^
2-15*arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2))*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)+38*cos(f*x+e)^2-24*co
s(f*x+e)-2)*sin(f*x+e)/cos(f*x+e)^3/(c*(-1+cos(f*x+e))/cos(f*x+e))^(3/2)

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

[Out]

Timed out

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Fricas [A]  time = 0.655823, size = 1067, normalized size = 6.35 \begin{align*} \left [\frac{15 \, \sqrt{2}{\left (a^{3} c \cos \left (f x + e\right )^{2} - a^{3} c \cos \left (f x + e\right )\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 ) + 2 \,{\left (19 \, a^{3} \cos \left (f x + e\right )^{3} + 7 \, a^{3} \cos \left (f x + e\right )^{2} - 13 \, a^{3} \cos \left (f x + e\right ) - a^{3}\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{3 \,{\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}, -\frac{2 \,{\left (\frac{15 \, \sqrt{2}{\left (a^{3} c \cos \left (f x + e\right )^{2} - a^{3} c \cos \left (f x + e\right )\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}} -{\left (19 \, a^{3} \cos \left (f x + e\right )^{3} + 7 \, a^{3} \cos \left (f x + e\right )^{2} - 13 \, a^{3} \cos \left (f x + e\right ) - a^{3}\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}\right )}}{3 \,{\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right )\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))^3/(c-c*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

[1/3*(15*sqrt(2)*(a^3*c*cos(f*x + e)^2 - a^3*c*cos(f*x + e))*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) + 2*(19*a^3*cos(f*x + e)^3 + 7*a^3*cos(f*x + e)^2 - 13*a^3*cos(f*x + e) - a
^3)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((c^2*f*cos(f*x + e)^2 - c^2*f*cos(f*x + e))*sin(f*x + e)), -2/3*
(15*sqrt(2)*(a^3*c*cos(f*x + e)^2 - a^3*c*cos(f*x + e))*arctan(sqrt(2)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))
*cos(f*x + e)/(sqrt(c)*sin(f*x + e)))*sin(f*x + e)/sqrt(c) - (19*a^3*cos(f*x + e)^3 + 7*a^3*cos(f*x + e)^2 - 1
3*a^3*cos(f*x + e) - a^3)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((c^2*f*cos(f*x + e)^2 - c^2*f*cos(f*x + e)
)*sin(f*x + e))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \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{3 \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 + \int \frac{3 \sec ^{3}{\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 ^{4}{\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))**3/(c-c*sec(f*x+e))**(3/2),x)

[Out]

a**3*(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) + In
tegral(3*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) + Integ
ral(3*sec(e + f*x)**3/(-c*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x) + c*sqrt(-c*sec(e + f*x) + c)), x) + Integral
(sec(e + f*x)**4/(-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 = 1.82238, size = 300, normalized size = 1.79 \begin{align*} \frac{2 \, a^{3} c^{2}{\left (\frac{15 \, \sqrt{2} \arctan \left (\frac{\sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c}}{\sqrt{c}}\right )}{c^{\frac{7}{2}} \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 )} + \frac{2 \, \sqrt{2}{\left (6 \, c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 7 \, c\right )}}{{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{3}{2}} c^{3} \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 )} + \frac{3 \, \sqrt{2} \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c}}{c^{4} \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 ) \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2}}\right )}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*a^3*c^2*(15*sqrt(2)*arctan(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/sqrt(c))/(c^(7/2)*sgn(tan(1/2*f*x + 1/2*e)^2
- 1)*sgn(tan(1/2*f*x + 1/2*e))) + 2*sqrt(2)*(6*c*tan(1/2*f*x + 1/2*e)^2 - 7*c)/((c*tan(1/2*f*x + 1/2*e)^2 - c
)^(3/2)*c^3*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)*sgn(tan(1/2*f*x + 1/2*e))) + 3*sqrt(2)*sqrt(c*tan(1/2*f*x + 1/2*e)
^2 - c)/(c^4*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)*sgn(tan(1/2*f*x + 1/2*e))*tan(1/2*f*x + 1/2*e)^2))/f