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

Optimal. Leaf size=140 \[ -\frac{\tan (e+f x) \tanh ^{-1}(\cos (e+f x))}{4 c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{\tan (e+f x)}{4 c f \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}}-\frac{\tan (e+f x)}{4 f \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.429864, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {3960, 3959, 3770} \[ -\frac{\tan (e+f x) \tanh ^{-1}(\cos (e+f x))}{4 c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{\tan (e+f x)}{4 c f \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}}-\frac{\tan (e+f x)}{4 f \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3959

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\sec (e+f x)}{\sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}} \, dx &=-\frac{\tan (e+f x)}{4 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}+\frac{\int \frac{\sec (e+f x)}{\sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}} \, dx}{2 c}\\ &=-\frac{\tan (e+f x)}{4 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}-\frac{\tan (e+f x)}{4 c f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac{\int \frac{\sec (e+f x)}{\sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}} \, dx}{4 c^2}\\ &=-\frac{\tan (e+f x)}{4 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}-\frac{\tan (e+f x)}{4 c f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac{\tan (e+f x) \int \csc (e+f x) \, dx}{4 c^2 \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=-\frac{\tan (e+f x)}{4 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}-\frac{\tan (e+f x)}{4 c f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}-\frac{\tanh ^{-1}(\cos (e+f x)) \tan (e+f x)}{4 c^2 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.839713, size = 91, normalized size = 0.65 \[ -\frac{\tan (e+f x) \left (3 \cos (e+f x)+8 \sin ^4\left (\frac{1}{2} (e+f x)\right ) \tanh ^{-1}\left (e^{i (e+f x)}\right )-2\right )}{4 c^2 f (\cos (e+f x)-1)^2 \sqrt{a (\sec (e+f x)+1)} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.301, size = 170, normalized size = 1.2 \begin{align*} -{\frac{-1+\cos \left ( fx+e \right ) }{16\,af \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) } \left ( 4\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-8\,\cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -5\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+4\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -2\,\cos \left ( fx+e \right ) +3 \right ) \sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}} \left ({\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\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)/(c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(1/2),x)

[Out]

-1/16/f/a*(-1+cos(f*x+e))*(4*ln(-(-1+cos(f*x+e))/sin(f*x+e))*cos(f*x+e)^2-8*cos(f*x+e)*ln(-(-1+cos(f*x+e))/sin
(f*x+e))-5*cos(f*x+e)^2+4*ln(-(-1+cos(f*x+e))/sin(f*x+e))-2*cos(f*x+e)+3)*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2
)/cos(f*x+e)^2/(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)/sin(f*x+e)

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Maxima [B]  time = 2.16308, size = 1621, normalized size = 11.58 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

1/4*((2*(4*cos(3*f*x + 3*e) - 6*cos(2*f*x + 2*e) + 4*cos(f*x + e) - 1)*cos(4*f*x + 4*e) - cos(4*f*x + 4*e)^2 +
8*(6*cos(2*f*x + 2*e) - 4*cos(f*x + e) + 1)*cos(3*f*x + 3*e) - 16*cos(3*f*x + 3*e)^2 + 12*(4*cos(f*x + e) - 1
)*cos(2*f*x + 2*e) - 36*cos(2*f*x + 2*e)^2 - 16*cos(f*x + e)^2 + 4*(2*sin(3*f*x + 3*e) - 3*sin(2*f*x + 2*e) +
2*sin(f*x + e))*sin(4*f*x + 4*e) - sin(4*f*x + 4*e)^2 + 16*(3*sin(2*f*x + 2*e) - 2*sin(f*x + e))*sin(3*f*x + 3
*e) - 16*sin(3*f*x + 3*e)^2 - 36*sin(2*f*x + 2*e)^2 + 48*sin(2*f*x + 2*e)*sin(f*x + e) - 16*sin(f*x + e)^2 + 8
*cos(f*x + e) - 1)*arctan2(sin(f*x + e), cos(f*x + e) + 1) - (2*(4*cos(3*f*x + 3*e) - 6*cos(2*f*x + 2*e) + 4*c
os(f*x + e) - 1)*cos(4*f*x + 4*e) - cos(4*f*x + 4*e)^2 + 8*(6*cos(2*f*x + 2*e) - 4*cos(f*x + e) + 1)*cos(3*f*x
+ 3*e) - 16*cos(3*f*x + 3*e)^2 + 12*(4*cos(f*x + e) - 1)*cos(2*f*x + 2*e) - 36*cos(2*f*x + 2*e)^2 - 16*cos(f*
x + e)^2 + 4*(2*sin(3*f*x + 3*e) - 3*sin(2*f*x + 2*e) + 2*sin(f*x + e))*sin(4*f*x + 4*e) - sin(4*f*x + 4*e)^2
+ 16*(3*sin(2*f*x + 2*e) - 2*sin(f*x + e))*sin(3*f*x + 3*e) - 16*sin(3*f*x + 3*e)^2 - 36*sin(2*f*x + 2*e)^2 +
48*sin(2*f*x + 2*e)*sin(f*x + e) - 16*sin(f*x + e)^2 + 8*cos(f*x + e) - 1)*arctan2(sin(f*x + e), cos(f*x + e)
- 1) - 2*(3*sin(3*f*x + 3*e) - 4*sin(2*f*x + 2*e) + 3*sin(f*x + e))*cos(4*f*x + 4*e) + 2*(3*cos(3*f*x + 3*e) -
4*cos(2*f*x + 2*e) + 3*cos(f*x + e))*sin(4*f*x + 4*e) - 2*(2*cos(2*f*x + 2*e) + 3)*sin(3*f*x + 3*e) + 4*(cos(
f*x + e) + 2)*sin(2*f*x + 2*e) + 4*cos(3*f*x + 3*e)*sin(2*f*x + 2*e) - 4*cos(2*f*x + 2*e)*sin(f*x + e) - 6*sin
(f*x + e))*sqrt(a)*sqrt(c)/((a*c^3*cos(4*f*x + 4*e)^2 + 16*a*c^3*cos(3*f*x + 3*e)^2 + 36*a*c^3*cos(2*f*x + 2*e
)^2 + 16*a*c^3*cos(f*x + e)^2 + a*c^3*sin(4*f*x + 4*e)^2 + 16*a*c^3*sin(3*f*x + 3*e)^2 + 36*a*c^3*sin(2*f*x +
2*e)^2 - 48*a*c^3*sin(2*f*x + 2*e)*sin(f*x + e) + 16*a*c^3*sin(f*x + e)^2 - 8*a*c^3*cos(f*x + e) + a*c^3 - 2*(
4*a*c^3*cos(3*f*x + 3*e) - 6*a*c^3*cos(2*f*x + 2*e) + 4*a*c^3*cos(f*x + e) - a*c^3)*cos(4*f*x + 4*e) - 8*(6*a*
c^3*cos(2*f*x + 2*e) - 4*a*c^3*cos(f*x + e) + a*c^3)*cos(3*f*x + 3*e) - 12*(4*a*c^3*cos(f*x + e) - a*c^3)*cos(
2*f*x + 2*e) - 4*(2*a*c^3*sin(3*f*x + 3*e) - 3*a*c^3*sin(2*f*x + 2*e) + 2*a*c^3*sin(f*x + e))*sin(4*f*x + 4*e)
- 16*(3*a*c^3*sin(2*f*x + 2*e) - 2*a*c^3*sin(f*x + e))*sin(3*f*x + 3*e))*f)

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Fricas [A]  time = 0.69242, size = 1138, normalized size = 8.13 \begin{align*} \left [-\frac{\sqrt{-a c}{\left (\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1\right )} \log \left (-\frac{4 \,{\left (2 \, \sqrt{-a c} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} +{\left (a c \cos \left (f x + e\right )^{2} + a c\right )} \sin \left (f x + e\right )\right )}}{{\left (\cos \left (f x + e\right )^{2} - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \,{\left (3 \, \cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{8 \,{\left (a c^{3} f \cos \left (f x + e\right )^{2} - 2 \, a c^{3} f \cos \left (f x + e\right ) + a c^{3} f\right )} \sin \left (f x + e\right )}, \frac{\sqrt{a c}{\left (\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1\right )} \arctan \left (\frac{\sqrt{a c} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{a c \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) +{\left (3 \, \cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{4 \,{\left (a c^{3} f \cos \left (f x + e\right )^{2} - 2 \, a c^{3} f \cos \left (f x + e\right ) + a 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)/(c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[-1/8*(sqrt(-a*c)*(cos(f*x + e)^2 - 2*cos(f*x + e) + 1)*log(-4*(2*sqrt(-a*c)*sqrt((a*cos(f*x + e) + a)/cos(f*x
+ e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*cos(f*x + e)^2 + (a*c*cos(f*x + e)^2 + a*c)*sin(f*x + e))/((cos
(f*x + e)^2 - 1)*sin(f*x + e)))*sin(f*x + e) - 2*(3*cos(f*x + e)^2 - 2*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)
/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((a*c^3*f*cos(f*x + e)^2 - 2*a*c^3*f*cos(f*x + e) + a*
c^3*f)*sin(f*x + e)), 1/4*(sqrt(a*c)*(cos(f*x + e)^2 - 2*cos(f*x + e) + 1)*arctan(sqrt(a*c)*sqrt((a*cos(f*x +
e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))/(a*c*sin(f*x + e)))*sin(f*x + e) + (3*cos(f*x +
e)^2 - 2*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((a*c^
3*f*cos(f*x + e)^2 - 2*a*c^3*f*cos(f*x + e) + a*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)/(c-c*sec(f*x+e))**(5/2)/(a+a*sec(f*x+e))**(1/2),x)

[Out]

Timed out

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Giac [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)/(c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out