3.70 \(\int \frac{\sec (e+f x) (a+a \sec (e+f x))}{(c-c \sec (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=113 \[ \frac{a \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{8 \sqrt{2} c^{5/2} f}+\frac{a \tan (e+f x)}{8 c f (c-c \sec (e+f x))^{3/2}}-\frac{a \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}} \]

[Out]

(a*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(8*Sqrt[2]*c^(5/2)*f) - (a*Tan[e + f*x])
/(2*f*(c - c*Sec[e + f*x])^(5/2)) + (a*Tan[e + f*x])/(8*c*f*(c - c*Sec[e + f*x])^(3/2))

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Rubi [A]  time = 0.154883, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3957, 3796, 3795, 203} \[ \frac{a \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{8 \sqrt{2} c^{5/2} f}+\frac{a \tan (e+f x)}{8 c f (c-c \sec (e+f x))^{3/2}}-\frac{a \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x]))/(c - c*Sec[e + f*x])^(5/2),x]

[Out]

(a*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(8*Sqrt[2]*c^(5/2)*f) - (a*Tan[e + f*x])
/(2*f*(c - c*Sec[e + f*x])^(5/2)) + (a*Tan[e + f*x])/(8*c*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 3796

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b*Cot[e + f*x]*(a
+ b*Csc[e + f*x])^m)/(a*f*(2*m + 1)), x] + Dist[(m + 1)/(a*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(
m + 1), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[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))^{5/2}} \, dx &=-\frac{a \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}-\frac{a \int \frac{\sec (e+f x)}{(c-c \sec (e+f x))^{3/2}} \, dx}{4 c}\\ &=-\frac{a \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac{a \tan (e+f x)}{8 c 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}{16 c^2}\\ &=-\frac{a \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac{a \tan (e+f x)}{8 c 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 )}{8 c^2 f}\\ &=\frac{a \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{8 \sqrt{2} c^{5/2} f}-\frac{a \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac{a \tan (e+f x)}{8 c f (c-c \sec (e+f x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 1.20918, size = 309, normalized size = 2.73 \[ \frac{a \left (-\frac{i \sqrt{2} \left (-1+e^{i (e+f x)}\right )^5 \tanh ^{-1}\left (\frac{1+e^{i (e+f x)}}{\sqrt{2} \sqrt{1+e^{2 i (e+f x)}}}\right )}{\left (1+e^{2 i (e+f x)}\right )^{5/2}}+48 \sin \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \sin ^5\left (\frac{1}{2} (e+f x)\right ) \sec ^3(e+f x)-48 \cos \left (\frac{e}{2}\right ) \cos \left (\frac{f x}{2}\right ) \sin ^5\left (\frac{1}{2} (e+f x)\right ) \sec ^3(e+f x)+56 \cot \left (\frac{e}{2}\right ) \sin ^4\left (\frac{1}{2} (e+f x)\right ) \sec ^3(e+f x)-16 \cot \left (\frac{e}{2}\right ) \sin ^2\left (\frac{1}{2} (e+f x)\right ) \sec ^3(e+f x)-56 \csc \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \sin ^3\left (\frac{1}{2} (e+f x)\right ) \sec ^3(e+f x)+16 \csc \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \sin \left (\frac{1}{2} (e+f x)\right ) \sec ^3(e+f x)\right )}{16 c^2 f (\sec (e+f x)-1)^2 \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x]))/(c - c*Sec[e + f*x])^(5/2),x]

[Out]

(a*(((-I)*Sqrt[2]*(-1 + E^(I*(e + f*x)))^5*ArcTanh[(1 + E^(I*(e + f*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(e + f*x))
])])/(1 + E^((2*I)*(e + f*x)))^(5/2) + 16*Csc[e/2]*Sec[e + f*x]^3*Sin[(f*x)/2]*Sin[(e + f*x)/2] - 16*Cot[e/2]*
Sec[e + f*x]^3*Sin[(e + f*x)/2]^2 - 56*Csc[e/2]*Sec[e + f*x]^3*Sin[(f*x)/2]*Sin[(e + f*x)/2]^3 + 56*Cot[e/2]*S
ec[e + f*x]^3*Sin[(e + f*x)/2]^4 - 48*Cos[e/2]*Cos[(f*x)/2]*Sec[e + f*x]^3*Sin[(e + f*x)/2]^5 + 48*Sec[e + f*x
]^3*Sin[e/2]*Sin[(f*x)/2]*Sin[(e + f*x)/2]^5))/(16*c^2*f*(-1 + Sec[e + f*x])^2*Sqrt[c - c*Sec[e + f*x]])

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Maple [B]  time = 0.214, size = 308, normalized size = 2.7 \begin{align*} -{\frac{a \left ( -1+\cos \left ( fx+e \right ) \right ) ^{3}}{2\,f \left ( \sin \left ( fx+e \right ) \right ) ^{5}} \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{{\frac{3}{2}}}+4\,\cos \left ( fx+e \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}\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}}} \right ) +3\, \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{3/2}+2\,\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\cos \left ( fx+e \right ) +2\,\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 ) ^{-{\frac{5}{2}}} \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{5}{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))^(5/2),x)

[Out]

-1/2*a/f*(-1+cos(f*x+e))^3*(cos(f*x+e)^2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)+4*cos(f*x+e)*(-2*cos(f*x+e)/(1+c
os(f*x+e)))^(3/2)-cos(f*x+e)^2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)-cos(f*x+e)^2*arctan(1/(-2*cos(f*x+e)/(1+co
s(f*x+e)))^(1/2))+3*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)+2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*cos(f*x+e)+2*c
os(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))^(5/2)/sin(f*x+e)^5/(-2*cos(f*x+e)/(1+cos(f*x+e)
))^(5/2)

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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))/(c-c*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.619489, size = 1042, normalized size = 9.22 \begin{align*} \left [-\frac{\sqrt{2}{\left (a \cos \left (f x + e\right )^{2} - 2 \, a \cos \left (f x + e\right ) + a\right )} \sqrt{-c} \log \left (-\frac{2 \, \sqrt{2}{\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt{-c} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} -{\left (3 \, c \cos \left (f x + e\right ) + c\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 (3 \, a \cos \left (f x + e\right )^{3} + 4 \, 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 )}}}{32 \,{\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )}, -\frac{\sqrt{2}{\left (a \cos \left (f x + e\right )^{2} - 2 \, a \cos \left (f x + e\right ) + a\right )} \sqrt{c} \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 ) - 2 \,{\left (3 \, a \cos \left (f x + e\right )^{3} + 4 \, 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 )}}}{16 \,{\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} 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))^(5/2),x, algorithm="fricas")

[Out]

[-1/32*(sqrt(2)*(a*cos(f*x + e)^2 - 2*a*cos(f*x + e) + a)*sqrt(-c)*log(-(2*sqrt(2)*(cos(f*x + e)^2 + cos(f*x +
e))*sqrt(-c)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)) - (3*c*cos(f*x + e) + c)*sin(f*x + e))/((cos(f*x + e) -
1)*sin(f*x + e)))*sin(f*x + e) - 4*(3*a*cos(f*x + e)^3 + 4*a*cos(f*x + e)^2 + a*cos(f*x + e))*sqrt((c*cos(f*x
+ e) - c)/cos(f*x + e)))/((c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) + c^3*f)*sin(f*x + e)), -1/16*(sqrt(2)*
(a*cos(f*x + e)^2 - 2*a*cos(f*x + e) + a)*sqrt(c)*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) - 2*(3*a*cos(f*x + e)^3 + 4*a*cos(f*x + e)^2 + a*cos(f*x + e))*sq
rt((c*cos(f*x + e) - c)/cos(f*x + e)))/((c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) + c^3*f)*sin(f*x + e))]

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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(sec(f*x+e)*(a+a*sec(f*x+e))/(c-c*sec(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.88553, size = 178, normalized size = 1.58 \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{{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{3}{2}} - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c} c}{c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4}}\right )}}{16 \, c 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))^(5/2),x, algorithm="giac")

[Out]

1/16*sqrt(2)*a*(arctan(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/sqrt(c))/c^(3/2) + ((c*tan(1/2*f*x + 1/2*e)^2 - c)^(
3/2) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)*c)/(c^3*tan(1/2*f*x + 1/2*e)^4))/(c*f*sgn(tan(1/2*f*x + 1/2*e)^2 - 1
)*sgn(tan(1/2*f*x + 1/2*e)))