3.78 \(\int \frac{\sec (e+f x) (a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=164 \[ -\frac{a^2 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{16 \sqrt{2} c^{7/2} f}-\frac{a^2 \tan (e+f x)}{16 c^2 f (c-c \sec (e+f x))^{3/2}}+\frac{a^2 \tan (e+f x)}{4 c f (c-c \sec (e+f x))^{5/2}}-\frac{\tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{3 f (c-c \sec (e+f x))^{7/2}} \]

[Out]

-(a^2*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(16*Sqrt[2]*c^(7/2)*f) - ((a^2 + a^2*
Sec[e + f*x])*Tan[e + f*x])/(3*f*(c - c*Sec[e + f*x])^(7/2)) + (a^2*Tan[e + f*x])/(4*c*f*(c - c*Sec[e + f*x])^
(5/2)) - (a^2*Tan[e + f*x])/(16*c^2*f*(c - c*Sec[e + f*x])^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.281871, antiderivative size = 164, 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, 3796, 3795, 203} \[ -\frac{a^2 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{16 \sqrt{2} c^{7/2} f}-\frac{a^2 \tan (e+f x)}{16 c^2 f (c-c \sec (e+f x))^{3/2}}+\frac{a^2 \tan (e+f x)}{4 c f (c-c \sec (e+f x))^{5/2}}-\frac{\tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{3 f (c-c \sec (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x])^2)/(c - c*Sec[e + f*x])^(7/2),x]

[Out]

-(a^2*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(16*Sqrt[2]*c^(7/2)*f) - ((a^2 + a^2*
Sec[e + f*x])*Tan[e + f*x])/(3*f*(c - c*Sec[e + f*x])^(7/2)) + (a^2*Tan[e + f*x])/(4*c*f*(c - c*Sec[e + f*x])^
(5/2)) - (a^2*Tan[e + f*x])/(16*c^2*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))^2}{(c-c \sec (e+f x))^{7/2}} \, dx &=-\frac{\left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{3 f (c-c \sec (e+f x))^{7/2}}-\frac{a \int \frac{\sec (e+f x) (a+a \sec (e+f x))}{(c-c \sec (e+f x))^{5/2}} \, dx}{2 c}\\ &=-\frac{\left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{3 f (c-c \sec (e+f x))^{7/2}}+\frac{a^2 \tan (e+f x)}{4 c f (c-c \sec (e+f x))^{5/2}}+\frac{a^2 \int \frac{\sec (e+f x)}{(c-c \sec (e+f x))^{3/2}} \, dx}{8 c^2}\\ &=-\frac{\left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{3 f (c-c \sec (e+f x))^{7/2}}+\frac{a^2 \tan (e+f x)}{4 c f (c-c \sec (e+f x))^{5/2}}-\frac{a^2 \tan (e+f x)}{16 c^2 f (c-c \sec (e+f x))^{3/2}}+\frac{a^2 \int \frac{\sec (e+f x)}{\sqrt{c-c \sec (e+f x)}} \, dx}{32 c^3}\\ &=-\frac{\left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{3 f (c-c \sec (e+f x))^{7/2}}+\frac{a^2 \tan (e+f x)}{4 c f (c-c \sec (e+f x))^{5/2}}-\frac{a^2 \tan (e+f x)}{16 c^2 f (c-c \sec (e+f x))^{3/2}}-\frac{a^2 \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 )}{16 c^3 f}\\ &=-\frac{a^2 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{16 \sqrt{2} c^{7/2} f}-\frac{\left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{3 f (c-c \sec (e+f x))^{7/2}}+\frac{a^2 \tan (e+f x)}{4 c f (c-c \sec (e+f x))^{5/2}}-\frac{a^2 \tan (e+f x)}{16 c^2 f (c-c \sec (e+f x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 5.69167, size = 398, normalized size = 2.43 \[ \frac{a^2 \csc \left (\frac{e}{2}\right ) e^{-\frac{1}{2} i (e+f x)} \tan \left (\frac{1}{2} (e+f x)\right ) \sec ^3\left (\frac{1}{2} (e+f x)\right ) \sec ^{\frac{3}{2}}(e+f x) (\sec (e+f x)+1)^2 \left (e^{\frac{i e}{2}} \sqrt{\sec (e+f x)} \left (e^{\frac{i f x}{2}} \sin \left (\frac{f x}{2}\right ) \left (14 \sin ^2\left (\frac{e}{2}\right ) \sin ^4\left (\frac{1}{2} (e+f x)\right )-43 \sin ^2\left (\frac{1}{2} (e+f x)\right )+34\right ) \sin ^2\left (\frac{1}{2} (e+f x)\right )-\frac{1}{8} \cos \left (\frac{e}{2}\right ) e^{\frac{i f x}{2}} \sin \left (\frac{1}{2} (e+f x)\right ) (36 \cos (e+f x)-43 \cos (2 (e+f x))-57)-\frac{7}{2} \sin (e) e^{\frac{i f x}{2}} \sin (f x) \csc \left (\frac{f x}{2}\right ) \sin ^6\left (\frac{1}{2} (e+f x)\right )+4 i e^{i f x}-4 i\right )-3 \sin \left (\frac{e}{2}\right ) \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt{1+e^{2 i (e+f x)}} \sin ^6\left (\frac{1}{2} (e+f x)\right ) \tanh ^{-1}\left (\frac{1+e^{i (e+f x)}}{\sqrt{2} \sqrt{1+e^{2 i (e+f x)}}}\right )\right )}{24 c^3 f (\sec (e+f x)-1)^3 \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x])^2)/(c - c*Sec[e + f*x])^(7/2),x]

[Out]

(a^2*Csc[e/2]*Sec[(e + f*x)/2]^3*Sec[e + f*x]^(3/2)*(1 + Sec[e + f*x])^2*(-3*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
))])]*Sin[e/2]*Sin[(e + f*x)/2]^6 + E^((I/2)*e)*Sqrt[Sec[e + f*x]]*(-4*I + (4*I)*E^(I*f*x) - (E^((I/2)*f*x)*Co
s[e/2]*(-57 + 36*Cos[e + f*x] - 43*Cos[2*(e + f*x)])*Sin[(e + f*x)/2])/8 - (7*E^((I/2)*f*x)*Csc[(f*x)/2]*Sin[e
]*Sin[f*x]*Sin[(e + f*x)/2]^6)/2 + E^((I/2)*f*x)*Sin[(f*x)/2]*Sin[(e + f*x)/2]^2*(34 - 43*Sin[(e + f*x)/2]^2 +
14*Sin[e/2]^2*Sin[(e + f*x)/2]^4)))*Tan[(e + f*x)/2])/(24*c^3*E^((I/2)*(e + f*x))*f*(-1 + Sec[e + f*x])^3*Sqr
t[c - c*Sec[e + f*x]])

________________________________________________________________________________________

Maple [B]  time = 0.236, size = 402, normalized size = 2.5 \begin{align*} -{\frac{{a}^{2} \left ( -1+\cos \left ( fx+e \right ) \right ) ^{4}}{6\,f \left ( \sin \left ( fx+e \right ) \right ) ^{7}} \left ( 5\, \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{3/2} \left ( \cos \left ( fx+e \right ) \right ) ^{3}+15\, \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{3/2}+3\,\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}} \left ( \cos \left ( fx+e \right ) \right ) ^{3}+3\,\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}}} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}+27\,\cos \left ( fx+e \right ) \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{3/2}-9\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}-9\, \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 ) +17\, \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{3/2}+9\,\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\cos \left ( fx+e \right ) +9\,\cos \left ( fx+e \right ) \arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}}} \right ) -3\,\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}-3\,\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{7}{2}}} \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{7}{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))^2/(c-c*sec(f*x+e))^(7/2),x)

[Out]

-1/6*a^2/f*(-1+cos(f*x+e))^4*(5*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)*cos(f*x+e)^3+15*cos(f*x+e)^2*(-2*cos(f*x+
e)/(1+cos(f*x+e)))^(3/2)+3*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*cos(f*x+e)^3+3*arctan(1/(-2*cos(f*x+e)/(1+cos(
f*x+e)))^(1/2))*cos(f*x+e)^3+27*cos(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)-9*cos(f*x+e)^2*(-2*cos(f*x+e)/
(1+cos(f*x+e)))^(1/2)-9*cos(f*x+e)^2*arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2))+17*(-2*cos(f*x+e)/(1+cos(f
*x+e)))^(3/2)+9*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*cos(f*x+e)+9*cos(f*x+e)*arctan(1/(-2*cos(f*x+e)/(1+cos(f*
x+e)))^(1/2))-3*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)-3*arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)))/(c*(-1+
cos(f*x+e))/cos(f*x+e))^(7/2)/sin(f*x+e)^7/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(7/2)

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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))^2/(c-c*sec(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.664644, size = 1281, normalized size = 7.81 \begin{align*} \left [-\frac{3 \, \sqrt{2}{\left (a^{2} \cos \left (f x + e\right )^{3} - 3 \, a^{2} \cos \left (f x + e\right )^{2} + 3 \, a^{2} \cos \left (f x + e\right ) - a^{2}\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 (7 \, a^{2} \cos \left (f x + e\right )^{4} + 29 \, a^{2} \cos \left (f x + e\right )^{3} + 25 \, a^{2} \cos \left (f x + e\right )^{2} + 3 \, a^{2} \cos \left (f x + e\right )\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{192 \,{\left (c^{4} f \cos \left (f x + e\right )^{3} - 3 \, c^{4} f \cos \left (f x + e\right )^{2} + 3 \, c^{4} f \cos \left (f x + e\right ) - c^{4} f\right )} \sin \left (f x + e\right )}, \frac{3 \, \sqrt{2}{\left (a^{2} \cos \left (f x + e\right )^{3} - 3 \, a^{2} \cos \left (f x + e\right )^{2} + 3 \, a^{2} \cos \left (f x + e\right ) - a^{2}\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 (7 \, a^{2} \cos \left (f x + e\right )^{4} + 29 \, a^{2} \cos \left (f x + e\right )^{3} + 25 \, a^{2} \cos \left (f x + e\right )^{2} + 3 \, a^{2} \cos \left (f x + e\right )\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{96 \,{\left (c^{4} f \cos \left (f x + e\right )^{3} - 3 \, c^{4} f \cos \left (f x + e\right )^{2} + 3 \, c^{4} f \cos \left (f x + e\right ) - c^{4} 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))^2/(c-c*sec(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

[-1/192*(3*sqrt(2)*(a^2*cos(f*x + e)^3 - 3*a^2*cos(f*x + e)^2 + 3*a^2*cos(f*x + e) - a^2)*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*(7*a^2*cos(f*x + e)^4 + 29*a^2*cos(f*x + e)^
3 + 25*a^2*cos(f*x + e)^2 + 3*a^2*cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((c^4*f*cos(f*x + e)^
3 - 3*c^4*f*cos(f*x + e)^2 + 3*c^4*f*cos(f*x + e) - c^4*f)*sin(f*x + e)), 1/96*(3*sqrt(2)*(a^2*cos(f*x + e)^3
- 3*a^2*cos(f*x + e)^2 + 3*a^2*cos(f*x + e) - a^2)*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*(7*a^2*cos(f*x + e)^4 + 29*a^2*cos(f*x + e)^3 + 25*a
^2*cos(f*x + e)^2 + 3*a^2*cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((c^4*f*cos(f*x + e)^3 - 3*c^
4*f*cos(f*x + e)^2 + 3*c^4*f*cos(f*x + e) - c^4*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))**2/(c-c*sec(f*x+e))**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 2.7277, size = 219, normalized size = 1.34 \begin{align*} -\frac{\sqrt{2} a^{2}{\left (\frac{3 \, \arctan \left (\frac{\sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c}}{\sqrt{c}}\right )}{c^{\frac{5}{2}}} + \frac{3 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{5}{2}} + 8 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{3}{2}} c - 3 \, \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c} c^{2}}{c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6}}\right )}}{96 \, 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))^2/(c-c*sec(f*x+e))^(7/2),x, algorithm="giac")

[Out]

-1/96*sqrt(2)*a^2*(3*arctan(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/sqrt(c))/c^(5/2) + (3*(c*tan(1/2*f*x + 1/2*e)^2
- c)^(5/2) + 8*(c*tan(1/2*f*x + 1/2*e)^2 - c)^(3/2)*c - 3*sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^2)/(c^5*tan(1/
2*f*x + 1/2*e)^6))/(c*f*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)*sgn(tan(1/2*f*x + 1/2*e)))