3.144 \(\int \frac{\sec (e+f x)}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=146 \[ \frac{3 \csc (e+f x)}{8 a c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{3 \tan (e+f x) \tanh ^{-1}(\cos (e+f x))}{8 a c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{\tan (e+f x)}{4 f (a \sec (e+f x)+a)^{3/2} (c-c \sec (e+f x))^{5/2}} \]

[Out]

(3*Csc[e + f*x])/(8*a*c^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - Tan[e + f*x]/(4*f*(a + a*Sec[
e + f*x])^(3/2)*(c - c*Sec[e + f*x])^(5/2)) - (3*ArcTanh[Cos[e + f*x]]*Tan[e + f*x])/(8*a*c^2*f*Sqrt[a + a*Sec
[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])

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Rubi [A]  time = 0.340645, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {3960, 3959, 2611, 3770} \[ \frac{3 \csc (e+f x)}{8 a c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{3 \tan (e+f x) \tanh ^{-1}(\cos (e+f x))}{8 a c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{\tan (e+f x)}{4 f (a \sec (e+f x)+a)^{3/2} (c-c \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]/((a + a*Sec[e + f*x])^(3/2)*(c - c*Sec[e + f*x])^(5/2)),x]

[Out]

(3*Csc[e + f*x])/(8*a*c^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - Tan[e + f*x]/(4*f*(a + a*Sec[
e + f*x])^(3/2)*(c - c*Sec[e + f*x])^(5/2)) - (3*ArcTanh[Cos[e + f*x]]*Tan[e + f*x])/(8*a*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 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

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))^{3/2} (c-c \sec (e+f x))^{5/2}} \, dx &=-\frac{\tan (e+f x)}{4 f (a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{5/2}}+\frac{3 \int \frac{\sec (e+f x)}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{3/2}} \, dx}{4 c}\\ &=-\frac{\tan (e+f x)}{4 f (a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{5/2}}-\frac{(3 \tan (e+f x)) \int \cot ^2(e+f x) \csc (e+f x) \, dx}{4 a c^2 \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=\frac{3 \csc (e+f x)}{8 a c^2 f \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))^{3/2} (c-c \sec (e+f x))^{5/2}}+\frac{(3 \tan (e+f x)) \int \csc (e+f x) \, dx}{8 a c^2 \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=\frac{3 \csc (e+f x)}{8 a c^2 f \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))^{3/2} (c-c \sec (e+f x))^{5/2}}-\frac{3 \tanh ^{-1}(\cos (e+f x)) \tan (e+f x)}{8 a c^2 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 1.50492, size = 122, normalized size = 0.84 \[ -\frac{\tan (e+f x) \left (-2 \cos (e+f x)+5 \cos (2 (e+f x))+24 \sin ^2\left (\frac{1}{2} (e+f x)\right ) \sin ^2(e+f x) \tanh ^{-1}\left (e^{i (e+f x)}\right )+1\right )}{16 a c^2 f (\cos (e+f x)-1)^2 (\cos (e+f x)+1) \sqrt{a (\sec (e+f x)+1)} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])^(3/2)*(c - c*Sec[e + f*x])^(5/2)),x]

[Out]

-((1 - 2*Cos[e + f*x] + 5*Cos[2*(e + f*x)] + 24*ArcTanh[E^(I*(e + f*x))]*Sin[(e + f*x)/2]^2*Sin[e + f*x]^2)*Ta
n[e + f*x])/(16*a*c^2*f*(-1 + Cos[e + f*x])^2*(1 + Cos[e + f*x])*Sqrt[a*(1 + Sec[e + f*x])]*Sqrt[c - c*Sec[e +
f*x]])

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Maple [A]  time = 0.269, size = 211, normalized size = 1.5 \begin{align*}{\frac{ \left ( -1+\cos \left ( fx+e \right ) \right ) ^{2}}{32\,f{a}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{3} \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( 12\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -5\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}-12\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-15\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-12\,\cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +9\,\cos \left ( fx+e \right ) +12\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \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)/(a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^(5/2),x)

[Out]

1/32/f/a^2*(-1+cos(f*x+e))^2*(12*cos(f*x+e)^3*ln(-(-1+cos(f*x+e))/sin(f*x+e))-5*cos(f*x+e)^3-12*ln(-(-1+cos(f*
x+e))/sin(f*x+e))*cos(f*x+e)^2-15*cos(f*x+e)^2-12*cos(f*x+e)*ln(-(-1+cos(f*x+e))/sin(f*x+e))+9*cos(f*x+e)+12*l
n(-(-1+cos(f*x+e))/sin(f*x+e))+3)*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)/sin(f*x+e)^3/cos(f*x+e)^2/(c*(-1+cos(f
*x+e))/cos(f*x+e))^(5/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 0.744877, size = 1315, normalized size = 9.01 \begin{align*} \left [-\frac{3 \,{\left (\cos \left (f x + e\right )^{3} - \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right ) + 1\right )} \sqrt{-a c} \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 (5 \, \cos \left (f x + e\right )^{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 )}}}{16 \,{\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{3 \,{\left (\cos \left (f x + e\right )^{3} - \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right ) + 1\right )} \sqrt{a c} \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 (5 \, \cos \left (f x + e\right )^{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^{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))^(3/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

[-1/16*(3*(cos(f*x + e)^3 - cos(f*x + e)^2 - cos(f*x + e) + 1)*sqrt(-a*c)*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)*s
in(f*x + e))/((cos(f*x + e)^2 - 1)*sin(f*x + e)))*sin(f*x + e) - 2*(5*cos(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^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)), 1/8*(3*(cos(f*x + e)^3
- cos(f*x + e)^2 - cos(f*x + e) + 1)*sqrt(a*c)*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) + (5*cos(f*x + e)^3 - cos(f*x + e)^2 - 2*c
os(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^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))**(3/2)/(c-c*sec(f*x+e))**(5/2),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))^(3/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Timed out