3.2.4 \(\int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 \sqrt {c-c \sec (e+f x)}} \, dx\) [104]

Optimal. Leaf size=181 \[ -\frac {\text {ArcTan}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{4 \sqrt {2} a^3 \sqrt {c} f}+\frac {\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x)}{6 a f (a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x)}{4 f \left (a^3+a^3 \sec (e+f x)\right ) \sqrt {c-c \sec (e+f x)}} \]

[Out]

-1/8*arctan(1/2*c^(1/2)*tan(f*x+e)*2^(1/2)/(c-c*sec(f*x+e))^(1/2))/a^3/f*2^(1/2)/c^(1/2)+1/5*tan(f*x+e)/f/(a+a
*sec(f*x+e))^3/(c-c*sec(f*x+e))^(1/2)+1/6*tan(f*x+e)/a/f/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^(1/2)+1/4*tan(f*x
+e)/f/(a^3+a^3*sec(f*x+e))/(c-c*sec(f*x+e))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.28, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {4045, 3880, 209} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{4 \sqrt {2} a^3 \sqrt {c} f}+\frac {\tan (e+f x)}{4 f \left (a^3 \sec (e+f x)+a^3\right ) \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x)}{6 a f (a \sec (e+f x)+a)^2 \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x)}{5 f (a \sec (e+f x)+a)^3 \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*Sqrt[c - c*Sec[e + f*x]]),x]

[Out]

-1/4*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])]/(Sqrt[2]*a^3*Sqrt[c]*f) + Tan[e + f*x]/
(5*f*(a + a*Sec[e + f*x])^3*Sqrt[c - c*Sec[e + f*x]]) + Tan[e + f*x]/(6*a*f*(a + a*Sec[e + f*x])^2*Sqrt[c - c*
Sec[e + f*x]]) + Tan[e + f*x]/(4*f*(a^3 + a^3*Sec[e + f*x])*Sqrt[c - c*Sec[e + f*x]])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3880

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 4045

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]))

Rubi steps

\begin {align*} \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 \sqrt {c-c \sec (e+f x)}} \, dx &=\frac {\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 \sqrt {c-c \sec (e+f x)}}+\frac {\int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)}} \, dx}{2 a}\\ &=\frac {\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x)}{6 a f (a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)}}+\frac {\int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) \sqrt {c-c \sec (e+f x)}} \, dx}{4 a^2}\\ &=\frac {\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x)}{6 a f (a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x)}{4 f \left (a^3+a^3 \sec (e+f x)\right ) \sqrt {c-c \sec (e+f x)}}+\frac {\int \frac {\sec (e+f x)}{\sqrt {c-c \sec (e+f x)}} \, dx}{8 a^3}\\ &=\frac {\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x)}{6 a f (a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x)}{4 f \left (a^3+a^3 \sec (e+f x)\right ) \sqrt {c-c \sec (e+f x)}}-\frac {\text {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 )}{4 a^3 f}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{4 \sqrt {2} a^3 \sqrt {c} f}+\frac {\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x)}{6 a f (a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x)}{4 f \left (a^3+a^3 \sec (e+f x)\right ) \sqrt {c-c \sec (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains complex when optimal does not.
time = 1.61, size = 225, normalized size = 1.24 \begin {gather*} \frac {2 e^{-\frac {1}{2} i (e+f x)} \cos \left (\frac {1}{2} (e+f x)\right ) \left (-15 \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\frac {1+e^{i (e+f x)}}{\sqrt {2} \sqrt {1+e^{2 i (e+f x)}}}\right ) \cos ^5\left (\frac {1}{2} (e+f x)\right )+\frac {e^{\frac {1}{2} i (e+f x)} (67+80 \cos (e+f x)+37 \cos (2 (e+f x)))}{8 \sqrt {\sec (e+f x)}}\right ) \sec ^{\frac {7}{2}}(e+f x) \sin \left (\frac {1}{2} (e+f x)\right )}{15 a^3 f (1+\sec (e+f x))^3 \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*Sqrt[c - c*Sec[e + f*x]]),x]

[Out]

(2*Cos[(e + f*x)/2]*(-15*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh
[(1 + E^(I*(e + f*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(e + f*x))])]*Cos[(e + f*x)/2]^5 + (E^((I/2)*(e + f*x))*(67
+ 80*Cos[e + f*x] + 37*Cos[2*(e + f*x)]))/(8*Sqrt[Sec[e + f*x]]))*Sec[e + f*x]^(7/2)*Sin[(e + f*x)/2])/(15*a^3
*E^((I/2)*(e + f*x))*f*(1 + Sec[e + f*x])^3*Sqrt[c - c*Sec[e + f*x]])

________________________________________________________________________________________

Maple [A]
time = 2.48, size = 155, normalized size = 0.86

method	result	size
default	\(-\frac {\left (-1+\cos \left (f x +e \right )\right ) \left (3 \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {5}{2}}-5 \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {3}{2}}+15 \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right )+15 \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\right )}{60 a^{3} f \sqrt {\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\)	\(155\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/60/a^3/f*(-1+cos(f*x+e))*(3*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(5/2)-5*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(3/2)+15*
arctan(1/(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2))+15*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2))/(c*(-1+cos(f*x+e))/cos
(f*x+e))^(1/2)/sin(f*x+e)/(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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/(c-c*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sec(f*x + e)/((a*sec(f*x + e) + a)^3*sqrt(-c*sec(f*x + e) + c)), x)

________________________________________________________________________________________

Fricas [A]
time = 3.15, size = 435, normalized size = 2.40 \begin {gather*} \left [-\frac {15 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} + 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 (37 \, \cos \left (f x + e\right )^{3} + 40 \, \cos \left (f x + e\right )^{2} + 15 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{240 \, {\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 {15 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} + 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 (37 \, \cos \left (f x + e\right )^{3} + 40 \, \cos \left (f x + e\right )^{2} + 15 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{120 \, {\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[-1/240*(15*sqrt(2)*(cos(f*x + e)^2 + 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*(37*cos(f*x + e)^3 + 40*cos(f*x + e)^2 + 15*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)), 1/120*(15*s
qrt(2)*(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)*sqrt(c)*arctan(sqrt(2)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*co
s(f*x + e)/(sqrt(c)*sin(f*x + e)))*sin(f*x + e) - 2*(37*cos(f*x + e)^3 + 40*cos(f*x + e)^2 + 15*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))]

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 0.67, size = 119, normalized size = 0.66 \begin {gather*} \frac {\sqrt {2} {\left (\frac {15 \, \arctan \left (\frac {\sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c}}{\sqrt {c}}\right )}{\sqrt {c}} - \frac {3 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {5}{2}} c^{12} - 5 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {3}{2}} c^{13} + 15 \, \sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c} c^{14}}{c^{15}}\right )}}{120 \, a^{3} 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/(c-c*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

1/120*sqrt(2)*(15*arctan(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/sqrt(c))/sqrt(c) - (3*(c*tan(1/2*f*x + 1/2*e)^2 -
c)^(5/2)*c^12 - 5*(c*tan(1/2*f*x + 1/2*e)^2 - c)^(3/2)*c^13 + 15*sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^14)/c^15
)/(a^3*f)

________________________________________________________________________________________

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\,\sqrt {c-\frac {c}{\cos \left (e+f\,x\right )}}} \,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*(c - c/cos(e + f*x))^(1/2)),x)

[Out]

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

________________________________________________________________________________________