3.2.43 \(\int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{3/2}} \, dx\) [143]

Optimal. Leaf size=104 \[ \frac {\csc (e+f x)}{2 a c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {\tanh ^{-1}(\cos (e+f x)) \tan (e+f x)}{2 a c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

1/2*csc(f*x+e)/a/c/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)-1/2*arctanh(cos(f*x+e))*tan(f*x+e)/a/c/f/(a
+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {4044, 2691, 3855} \begin {gather*} \frac {\csc (e+f x)}{2 a c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x) \tanh ^{-1}(\cos (e+f x))}{2 a c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2691

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 3855

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

Rule 4044

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]

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.80, size = 69, normalized size = 0.66 \begin {gather*} \frac {\csc (e+f x)-2 \tanh ^{-1}\left (e^{i (e+f x)}\right ) \tan (e+f x)}{2 a c f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[e + f*x] - 2*ArcTanh[E^(I*(e + f*x))]*Tan[e + f*x])/(2*a*c*f*Sqrt[a*(1 + Sec[e + f*x])]*Sqrt[c - c*Sec[e
+ f*x]])

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Maple [A]
time = 2.97, size = 133, normalized size = 1.28

method	result	size
default	\(\frac {\left (-1+\cos \left (f x +e \right )\right )^{2} \left (\left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-\ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-\cos \left (f x +e \right )\right ) \sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}}{2 f \sin \left (f x +e \right )^{3} \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \cos \left (f x +e \right ) a^{2}}\)	\(133\)
risch	\(\frac {i {\mathrm e}^{i \left (f x +e \right )}}{a c \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}-\frac {i \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{2 a c \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}+\frac {i \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{2 a c \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(347\)

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))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 613 vs. \(2 (99) = 198\).
time = 0.57, size = 613, normalized size = 5.89 \begin {gather*} \frac {{\left ({\left (2 \, {\left (2 \, \cos \left (2 \, f x + 2 \, e\right ) - 1\right )} \cos \left (4 \, f x + 4 \, e\right ) - \cos \left (4 \, f x + 4 \, e\right )^{2} - 4 \, \cos \left (2 \, f x + 2 \, e\right )^{2} - \sin \left (4 \, f x + 4 \, e\right )^{2} + 4 \, \sin \left (4 \, f x + 4 \, e\right ) \sin \left (2 \, f x + 2 \, e\right ) - 4 \, \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, \cos \left (2 \, f x + 2 \, e\right ) - 1\right )} \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) + 1\right ) - {\left (2 \, {\left (2 \, \cos \left (2 \, f x + 2 \, e\right ) - 1\right )} \cos \left (4 \, f x + 4 \, e\right ) - \cos \left (4 \, f x + 4 \, e\right )^{2} - 4 \, \cos \left (2 \, f x + 2 \, e\right )^{2} - \sin \left (4 \, f x + 4 \, e\right )^{2} + 4 \, \sin \left (4 \, f x + 4 \, e\right ) \sin \left (2 \, f x + 2 \, e\right ) - 4 \, \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, \cos \left (2 \, f x + 2 \, e\right ) - 1\right )} \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) - 1\right ) - 2 \, {\left (\sin \left (3 \, f x + 3 \, e\right ) + \sin \left (f x + e\right )\right )} \cos \left (4 \, f x + 4 \, e\right ) + 2 \, {\left (\cos \left (3 \, f x + 3 \, e\right ) + \cos \left (f x + e\right )\right )} \sin \left (4 \, f x + 4 \, e\right ) + 2 \, {\left (2 \, \cos \left (2 \, f x + 2 \, e\right ) - 1\right )} \sin \left (3 \, f x + 3 \, e\right ) - 4 \, \cos \left (3 \, f x + 3 \, e\right ) \sin \left (2 \, f x + 2 \, e\right ) - 4 \, \cos \left (f x + e\right ) \sin \left (2 \, f x + 2 \, e\right ) + 4 \, \cos \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) - 2 \, \sin \left (f x + e\right )\right )} \sqrt {a} \sqrt {c}}{2 \, {\left (a^{2} c^{2} \cos \left (4 \, f x + 4 \, e\right )^{2} + 4 \, a^{2} c^{2} \cos \left (2 \, f x + 2 \, e\right )^{2} + a^{2} c^{2} \sin \left (4 \, f x + 4 \, e\right )^{2} - 4 \, a^{2} c^{2} \sin \left (4 \, f x + 4 \, e\right ) \sin \left (2 \, f x + 2 \, e\right ) + 4 \, a^{2} c^{2} \sin \left (2 \, f x + 2 \, e\right )^{2} - 4 \, a^{2} c^{2} \cos \left (2 \, f x + 2 \, e\right ) + a^{2} c^{2} - 2 \, {\left (2 \, a^{2} c^{2} \cos \left (2 \, f x + 2 \, e\right ) - a^{2} c^{2}\right )} \cos \left (4 \, f x + 4 \, e\right )\right )} f} \end {gather*}

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))^(3/2),x, algorithm="maxima")

[Out]

1/2*((2*(2*cos(2*f*x + 2*e) - 1)*cos(4*f*x + 4*e) - cos(4*f*x + 4*e)^2 - 4*cos(2*f*x + 2*e)^2 - sin(4*f*x + 4*
e)^2 + 4*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) - 4*sin(2*f*x + 2*e)^2 + 4*cos(2*f*x + 2*e) - 1)*arctan2(sin(f*x +
e), cos(f*x + e) + 1) - (2*(2*cos(2*f*x + 2*e) - 1)*cos(4*f*x + 4*e) - cos(4*f*x + 4*e)^2 - 4*cos(2*f*x + 2*e)
^2 - sin(4*f*x + 4*e)^2 + 4*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) - 4*sin(2*f*x + 2*e)^2 + 4*cos(2*f*x + 2*e) - 1)
*arctan2(sin(f*x + e), cos(f*x + e) - 1) - 2*(sin(3*f*x + 3*e) + sin(f*x + e))*cos(4*f*x + 4*e) + 2*(cos(3*f*x
+ 3*e) + cos(f*x + e))*sin(4*f*x + 4*e) + 2*(2*cos(2*f*x + 2*e) - 1)*sin(3*f*x + 3*e) - 4*cos(3*f*x + 3*e)*si
n(2*f*x + 2*e) - 4*cos(f*x + e)*sin(2*f*x + 2*e) + 4*cos(2*f*x + 2*e)*sin(f*x + e) - 2*sin(f*x + e))*sqrt(a)*s
qrt(c)/((a^2*c^2*cos(4*f*x + 4*e)^2 + 4*a^2*c^2*cos(2*f*x + 2*e)^2 + a^2*c^2*sin(4*f*x + 4*e)^2 - 4*a^2*c^2*si
n(4*f*x + 4*e)*sin(2*f*x + 2*e) + 4*a^2*c^2*sin(2*f*x + 2*e)^2 - 4*a^2*c^2*cos(2*f*x + 2*e) + a^2*c^2 - 2*(2*a
^2*c^2*cos(2*f*x + 2*e) - a^2*c^2)*cos(4*f*x + 4*e))*f)

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Fricas [A]
time = 2.40, size = 434, normalized size = 4.17 \begin {gather*} \left [-\frac {\sqrt {-a c} {\left (\cos \left (f x + e\right )^{2} - 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 \, \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}}{4 \, {\left (a^{2} c^{2} f \cos \left (f x + e\right )^{2} - a^{2} c^{2} f\right )} \sin \left (f x + e\right )}, \frac {\sqrt {a c} {\left (\cos \left (f x + e\right )^{2} - 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 ) + \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}}{2 \, {\left (a^{2} c^{2} f \cos \left (f x + e\right )^{2} - a^{2} c^{2} f\right )} \sin \left (f x + e\right )}\right ] \end {gather*}

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))^(3/2),x, algorithm="fricas")

[Out]

[-1/4*(sqrt(-a*c)*(cos(f*x + e)^2 - 1)*log(-4*(2*sqrt(-a*c)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*co
s(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*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^2*c^2*f*cos(f*x + e)^2 - a^2*c^2*f)*sin(f*x + e)), 1/2*(sqrt(a*c)*(cos(f*x + e)^2 - 1)*a
rctan(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) + 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^2*c^2*f*cos(f*x + e)^2 - a^2*c^2*f)*sin(f*x + e))]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec {\left (e + f x \right )}}{\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \left (- c \left (\sec {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

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))**(3/2),x)

[Out]

Integral(sec(e + f*x)/((a*(sec(e + f*x) + 1))**(3/2)*(-c*(sec(e + f*x) - 1))**(3/2)), x)

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Giac [A]
time = 3.07, size = 120, normalized size = 1.15 \begin {gather*} \frac {\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c}{c} + \frac {2 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c}{c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}} - 2 \, \log \left ({\left | c \right |} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}\right ) + 2 \, \log \left ({\left | c \right |}\right ) - 1}{8 \, \sqrt {-a c} a f {\left | c \right |} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \end {gather*}

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))^(3/2),x, algorithm="giac")

[Out]

1/8*((c*tan(1/2*f*x + 1/2*e)^2 - c)/c + (2*c*tan(1/2*f*x + 1/2*e)^2 - c)/(c*tan(1/2*f*x + 1/2*e)^2) - 2*log(ab
s(c)*tan(1/2*f*x + 1/2*e)^2) + 2*log(abs(c)) - 1)/(sqrt(-a*c)*a*f*abs(c)*sgn(tan(1/2*f*x + 1/2*e)^3 + tan(1/2*
f*x + 1/2*e)))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\cos \left (e+f\,x\right )\,{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(e + f*x)*(a + a/cos(e + f*x))^(3/2)*(c - c/cos(e + f*x))^(3/2)),x)

[Out]

int(1/(cos(e + f*x)*(a + a/cos(e + f*x))^(3/2)*(c - c/cos(e + f*x))^(3/2)), x)

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