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

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

[Out]

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

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Rubi [A]  time = 0.424006, 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} \[ -\frac{\tan (e+f x) \tanh ^{-1}(\cos (e+f x))}{4 a^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{4 a f (a \sec (e+f x)+a)^{3/2} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{4 f (a \sec (e+f x)+a)^{5/2} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Tan[e + f*x]/(4*f*(a + a*Sec[e + f*x])^(5/2)*Sqrt[c - c*Sec[e + f*x]]) + Tan[e + f*x]/(4*a*f*(a + a*Sec[e + f*
x])^(3/2)*Sqrt[c - c*Sec[e + f*x]]) - (ArcTanh[Cos[e + f*x]]*Tan[e + f*x])/(4*a^2*f*Sqrt[a + a*Sec[e + f*x]]*S
qrt[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)}{(a+a \sec (e+f x))^{5/2} \sqrt{c-c \sec (e+f x)}} \, dx &=\frac{\tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2} \sqrt{c-c \sec (e+f x)}}+\frac{\int \frac{\sec (e+f x)}{(a+a \sec (e+f x))^{3/2} \sqrt{c-c \sec (e+f x)}} \, dx}{2 a}\\ &=\frac{\tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{4 a f (a+a \sec (e+f x))^{3/2} \sqrt{c-c \sec (e+f x)}}+\frac{\int \frac{\sec (e+f x)}{\sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}} \, dx}{4 a^2}\\ &=\frac{\tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{4 a f (a+a \sec (e+f x))^{3/2} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x) \int \csc (e+f x) \, dx}{4 a^2 \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=\frac{\tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{4 a f (a+a \sec (e+f x))^{3/2} \sqrt{c-c \sec (e+f x)}}-\frac{\tanh ^{-1}(\cos (e+f x)) \tan (e+f x)}{4 a^2 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 1.05336, size = 91, normalized size = 0.65 \[ -\frac{\tan (e+f x) \left (3 \cos (e+f x)+8 \cos ^4\left (\frac{1}{2} (e+f x)\right ) \tanh ^{-1}\left (e^{i (e+f x)}\right )+2\right )}{4 a^2 f (\cos (e+f x)+1)^2 \sqrt{a (\sec (e+f x)+1)} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2 + 8*ArcTanh[E^(I*(e + f*x))]*Cos[(e + f*x)/2]^4 + 3*Cos[e + f*x])*Tan[e + f*x])/(4*a^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 = 0.3, size = 164, normalized size = 1.2 \begin{align*} -{\frac{ \left ( -1+\cos \left ( fx+e \right ) \right ) ^{3}}{16\,f{a}^{3} \left ( \sin \left ( fx+e \right ) \right ) ^{5}}\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \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}+5\, \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 ) -2\,\cos \left ( fx+e \right ) +4\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -3 \right ){\frac{1}{\sqrt{{\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)/(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(1/2),x)

[Out]

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

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Maxima [B]  time = 2.14867, size = 1608, normalized size = 11.49 \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)/(a+a*sec(f*x+e))^(5/2)/(c-c*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*
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*(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*si
n(f*x + e))*sqrt(a)*sqrt(c)/((a^3*c*cos(4*f*x + 4*e)^2 + 16*a^3*c*cos(3*f*x + 3*e)^2 + 36*a^3*c*cos(2*f*x + 2*
e)^2 + 16*a^3*c*cos(f*x + e)^2 + a^3*c*sin(4*f*x + 4*e)^2 + 16*a^3*c*sin(3*f*x + 3*e)^2 + 36*a^3*c*sin(2*f*x +
2*e)^2 + 48*a^3*c*sin(2*f*x + 2*e)*sin(f*x + e) + 16*a^3*c*sin(f*x + e)^2 + 8*a^3*c*cos(f*x + e) + a^3*c + 2*
(4*a^3*c*cos(3*f*x + 3*e) + 6*a^3*c*cos(2*f*x + 2*e) + 4*a^3*c*cos(f*x + e) + a^3*c)*cos(4*f*x + 4*e) + 8*(6*a
^3*c*cos(2*f*x + 2*e) + 4*a^3*c*cos(f*x + e) + a^3*c)*cos(3*f*x + 3*e) + 12*(4*a^3*c*cos(f*x + e) + a^3*c)*cos
(2*f*x + 2*e) + 4*(2*a^3*c*sin(3*f*x + 3*e) + 3*a^3*c*sin(2*f*x + 2*e) + 2*a^3*c*sin(f*x + e))*sin(4*f*x + 4*e
) + 16*(3*a^3*c*sin(2*f*x + 2*e) + 2*a^3*c*sin(f*x + e))*sin(3*f*x + 3*e))*f)

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Fricas [A]  time = 0.693415, 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^{3} c f \cos \left (f x + e\right )^{2} + 2 \, a^{3} c f \cos \left (f x + e\right ) + a^{3} c 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^{3} c f \cos \left (f x + e\right )^{2} + 2 \, a^{3} c f \cos \left (f x + e\right ) + a^{3} c 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))^(5/2)/(c-c*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^3*c*f*cos(f*x + e)^2 + 2*a^3*c*f*cos(f*x + e) + a^
3*c*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^3*
c*f*cos(f*x + e)^2 + 2*a^3*c*f*cos(f*x + e) + a^3*c*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))**(5/2)/(c-c*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)/(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out