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

Optimal. Leaf size=203 \[ -\frac{35 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{64 \sqrt{2} a^2 c^{5/2} f}-\frac{35 \tan (e+f x)}{64 a^2 c f (c-c \sec (e+f x))^{3/2}}-\frac{35 \tan (e+f x)}{48 a^2 f (c-c \sec (e+f x))^{5/2}}+\frac{7 \tan (e+f x)}{6 f \left (a^2 \sec (e+f x)+a^2\right ) (c-c \sec (e+f x))^{5/2}}+\frac{\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{5/2}} \]

[Out]

(-35*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(64*Sqrt[2]*a^2*c^(5/2)*f) - (35*Tan[e
+ f*x])/(48*a^2*f*(c - c*Sec[e + f*x])^(5/2)) + Tan[e + f*x]/(3*f*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])
^(5/2)) + (7*Tan[e + f*x])/(6*f*(a^2 + a^2*Sec[e + f*x])*(c - c*Sec[e + f*x])^(5/2)) - (35*Tan[e + f*x])/(64*a
^2*c*f*(c - c*Sec[e + f*x])^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.388975, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {3960, 3796, 3795, 203} \[ -\frac{35 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{64 \sqrt{2} a^2 c^{5/2} f}-\frac{35 \tan (e+f x)}{64 a^2 c f (c-c \sec (e+f x))^{3/2}}-\frac{35 \tan (e+f x)}{48 a^2 f (c-c \sec (e+f x))^{5/2}}+\frac{7 \tan (e+f x)}{6 f \left (a^2 \sec (e+f x)+a^2\right ) (c-c \sec (e+f x))^{5/2}}+\frac{\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-35*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(64*Sqrt[2]*a^2*c^(5/2)*f) - (35*Tan[e
+ f*x])/(48*a^2*f*(c - c*Sec[e + f*x])^(5/2)) + Tan[e + f*x]/(3*f*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])
^(5/2)) + (7*Tan[e + f*x])/(6*f*(a^2 + a^2*Sec[e + f*x])*(c - c*Sec[e + f*x])^(5/2)) - (35*Tan[e + f*x])/(64*a
^2*c*f*(c - c*Sec[e + f*x])^(3/2))

Rule 3960

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

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))^{5/2}} \, dx &=\frac{\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2}}+\frac{7 \int \frac{\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^{5/2}} \, dx}{6 a}\\ &=\frac{\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2}}+\frac{7 \tan (e+f x)}{6 f \left (a^2+a^2 \sec (e+f x)\right ) (c-c \sec (e+f x))^{5/2}}+\frac{35 \int \frac{\sec (e+f x)}{(c-c \sec (e+f x))^{5/2}} \, dx}{12 a^2}\\ &=-\frac{35 \tan (e+f x)}{48 a^2 f (c-c \sec (e+f x))^{5/2}}+\frac{\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2}}+\frac{7 \tan (e+f x)}{6 f \left (a^2+a^2 \sec (e+f x)\right ) (c-c \sec (e+f x))^{5/2}}+\frac{35 \int \frac{\sec (e+f x)}{(c-c \sec (e+f x))^{3/2}} \, dx}{32 a^2 c}\\ &=-\frac{35 \tan (e+f x)}{48 a^2 f (c-c \sec (e+f x))^{5/2}}+\frac{\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2}}+\frac{7 \tan (e+f x)}{6 f \left (a^2+a^2 \sec (e+f x)\right ) (c-c \sec (e+f x))^{5/2}}-\frac{35 \tan (e+f x)}{64 a^2 c f (c-c \sec (e+f x))^{3/2}}+\frac{35 \int \frac{\sec (e+f x)}{\sqrt{c-c \sec (e+f x)}} \, dx}{128 a^2 c^2}\\ &=-\frac{35 \tan (e+f x)}{48 a^2 f (c-c \sec (e+f x))^{5/2}}+\frac{\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2}}+\frac{7 \tan (e+f x)}{6 f \left (a^2+a^2 \sec (e+f x)\right ) (c-c \sec (e+f x))^{5/2}}-\frac{35 \tan (e+f x)}{64 a^2 c f (c-c \sec (e+f x))^{3/2}}-\frac{35 \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 )}{64 a^2 c^2 f}\\ &=-\frac{35 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{64 \sqrt{2} a^2 c^{5/2} f}-\frac{35 \tan (e+f x)}{48 a^2 f (c-c \sec (e+f x))^{5/2}}+\frac{\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2}}+\frac{7 \tan (e+f x)}{6 f \left (a^2+a^2 \sec (e+f x)\right ) (c-c \sec (e+f x))^{5/2}}-\frac{35 \tan (e+f x)}{64 a^2 c f (c-c \sec (e+f x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 2.38116, size = 434, normalized size = 2.14 \[ \frac{\cot ^4(e+f x) \left (\frac{105 i \sqrt{2} \left (-1+e^{i (e+f x)}\right )^5 \left (1+e^{i (e+f x)}\right )^4 \tanh ^{-1}\left (\frac{1+e^{i (e+f x)}}{\sqrt{2} \sqrt{1+e^{2 i (e+f x)}}}\right )}{\left (1+e^{2 i (e+f x)}\right )^{9/2}}-3648 \csc \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \sin ^9(e+f x) \csc \left (\frac{1}{2} (e+f x)\right ) \csc ^5(2 (e+f x))-5504 \sin \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \sin \left (\frac{1}{2} (e+f x)\right ) \sin ^9(e+f x) \csc ^5(2 (e+f x))-13312 \sin ^2\left (\frac{1}{2} (e+f x)\right ) \sin ^8(e+f x) \csc ^5(2 (e+f x))+2752 \cos \left (\frac{e}{2}\right ) \cos \left (\frac{f x}{2}\right ) \sin ^5\left (\frac{1}{2} (e+f x)\right ) \cos ^4\left (\frac{1}{2} (e+f x)\right ) \sec ^5(e+f x)+256 \sin ^5\left (\frac{1}{2} (e+f x)\right ) \cos \left (\frac{1}{2} (e+f x)\right ) \sec ^5(e+f x)+114 \cot \left (\frac{e}{2}\right ) \tan ^4(e+f x) \sec (e+f x)-192 \cot \left (\frac{e}{2}\right ) \sin ^2\left (\frac{1}{2} (e+f x)\right ) \cos ^4\left (\frac{1}{2} (e+f x)\right ) \sec ^5(e+f x)+192 \csc \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \sin \left (\frac{1}{2} (e+f x)\right ) \cos ^4\left (\frac{1}{2} (e+f x)\right ) \sec ^5(e+f x)\right )}{384 a^2 c^2 f \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cot[e + f*x]^4*(((105*I)*Sqrt[2]*(-1 + E^(I*(e + f*x)))^5*(1 + E^(I*(e + f*x)))^4*ArcTanh[(1 + E^(I*(e + f*x)
))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(e + f*x))])])/(1 + E^((2*I)*(e + f*x)))^(9/2) + 192*Cos[(e + f*x)/2]^4*Csc[e/2]
*Sec[e + f*x]^5*Sin[(f*x)/2]*Sin[(e + f*x)/2] - 192*Cos[(e + f*x)/2]^4*Cot[e/2]*Sec[e + f*x]^5*Sin[(e + f*x)/2
]^2 + 256*Cos[(e + f*x)/2]*Sec[e + f*x]^5*Sin[(e + f*x)/2]^5 + 2752*Cos[e/2]*Cos[(f*x)/2]*Cos[(e + f*x)/2]^4*S
ec[e + f*x]^5*Sin[(e + f*x)/2]^5 - 13312*Csc[2*(e + f*x)]^5*Sin[(e + f*x)/2]^2*Sin[e + f*x]^8 - 3648*Csc[e/2]*
Csc[(e + f*x)/2]*Csc[2*(e + f*x)]^5*Sin[(f*x)/2]*Sin[e + f*x]^9 - 5504*Csc[2*(e + f*x)]^5*Sin[e/2]*Sin[(f*x)/2
]*Sin[(e + f*x)/2]*Sin[e + f*x]^9 + 114*Cot[e/2]*Sec[e + f*x]*Tan[e + f*x]^4))/(384*a^2*c^2*f*Sqrt[c - c*Sec[e
+ f*x]])

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Maple [B]  time = 0.242, size = 551, normalized size = 2.7 \begin{align*} \text{result too large to display} \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))^(5/2),x)

[Out]

1/48/a^2/f*(-1+cos(f*x+e))^3*(21*cos(f*x+e)^2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(9/2)+12*cos(f*x+e)*(-2*cos(f*x+e
)/(1+cos(f*x+e)))^(9/2)+15*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(7/2)*cos(f*x+e)^2-9*(-2*cos(f*x+e)/(1+cos(f*x+e)))^
(9/2)-30*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(7/2)*cos(f*x+e)-21*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2)*cos(f*x+e)^2+
15*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(7/2)+42*cos(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2)+35*cos(f*x+e)^2*(-2
*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)-21*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2)-70*cos(f*x+e)*(-2*cos(f*x+e)/(1+cos(
f*x+e)))^(3/2)-105*cos(f*x+e)^2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)-105*cos(f*x+e)^2*arctan(1/(-2*cos(f*x+e)/
(1+cos(f*x+e)))^(1/2))+35*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)+210*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*cos(f*
x+e)+210*cos(f*x+e)*arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2))-105*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)-10
5*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))^2/(c-c*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.759568, size = 1239, normalized size = 6.1 \begin{align*} \left [-\frac{105 \, \sqrt{2}{\left (\cos \left (f x + e\right )^{3} - \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right ) + 1\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 (43 \, \cos \left (f x + e\right )^{4} - 161 \, \cos \left (f x + e\right )^{3} - 35 \, \cos \left (f x + e\right )^{2} + 105 \, \cos \left (f x + e\right )\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{768 \,{\left (a^{2} c^{3} f \cos \left (f x + e\right )^{3} - a^{2} c^{3} f \cos \left (f x + e\right )^{2} - a^{2} c^{3} f \cos \left (f x + e\right ) + a^{2} c^{3} f\right )} \sin \left (f x + e\right )}, \frac{105 \, \sqrt{2}{\left (\cos \left (f x + e\right )^{3} - \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right ) + 1\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 (43 \, \cos \left (f x + e\right )^{4} - 161 \, \cos \left (f x + e\right )^{3} - 35 \, \cos \left (f x + e\right )^{2} + 105 \, \cos \left (f x + e\right )\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{384 \,{\left (a^{2} c^{3} f \cos \left (f x + e\right )^{3} - a^{2} c^{3} f \cos \left (f x + e\right )^{2} - a^{2} c^{3} f \cos \left (f x + e\right ) + a^{2} 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))^2/(c-c*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

[-1/768*(105*sqrt(2)*(cos(f*x + e)^3 - cos(f*x + e)^2 - cos(f*x + e) + 1)*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*(43*cos(f*x + e)^4 - 161*cos(f*x + e)^3 - 35*cos(f*x + e)^2
+ 105*cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((a^2*c^3*f*cos(f*x + e)^3 - a^2*c^3*f*cos(f*x +
e)^2 - a^2*c^3*f*cos(f*x + e) + a^2*c^3*f)*sin(f*x + e)), 1/384*(105*sqrt(2)*(cos(f*x + e)^3 - cos(f*x + e)^2
- cos(f*x + e) + 1)*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*(43*cos(f*x + e)^4 - 161*cos(f*x + e)^3 - 35*cos(f*x + e)^2 + 105*cos(f*x + e))*sqr
t((c*cos(f*x + e) - c)/cos(f*x + e)))/((a^2*c^3*f*cos(f*x + e)^3 - a^2*c^3*f*cos(f*x + e)^2 - a^2*c^3*f*cos(f*
x + e) + a^2*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))**2/(c-c*sec(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.64403, size = 265, normalized size = 1.31 \begin{align*} -\frac{\sqrt{2}{\left (105 \, \sqrt{c} \arctan \left (\frac{\sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c}}{\sqrt{c}}\right ) + \frac{8 \,{\left ({\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{3}{2}} c^{2} - 9 \, \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c} c^{3}\right )}}{c^{3}} - \frac{3 \,{\left (13 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{3}{2}} c + 11 \, \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c} c^{2}\right )}}{c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4}}\right )}}{384 \, a^{2} c^{3} 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))^(5/2),x, algorithm="giac")

[Out]

-1/384*sqrt(2)*(105*sqrt(c)*arctan(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/sqrt(c)) + 8*((c*tan(1/2*f*x + 1/2*e)^2
- c)^(3/2)*c^2 - 9*sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^3)/c^3 - 3*(13*(c*tan(1/2*f*x + 1/2*e)^2 - c)^(3/2)*c
+ 11*sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^2)/(c^2*tan(1/2*f*x + 1/2*e)^4))/(a^2*c^3*f*sgn(tan(1/2*f*x + 1/2*e)
^2 - 1)*sgn(tan(1/2*f*x + 1/2*e)))