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

Optimal. Leaf size=138 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{2 \sqrt {2} a^2 \sqrt {c} f}+\frac {\tan (e+f x)}{2 f \left (a^2 \sec (e+f x)+a^2\right ) \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2 \sqrt {c-c \sec (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {3960, 3795, 203} \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{2 \sqrt {2} a^2 \sqrt {c} f}+\frac {\tan (e+f x)}{2 f \left (a^2 \sec (e+f x)+a^2\right ) \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2 \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 203

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

Rule 3795

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 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]))

Rubi steps

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

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Mathematica [C]  time = 2.08, size = 259, normalized size = 1.88 \[ \frac {2 e^{-\frac {1}{2} i (e+f x)} \sin \left (\frac {1}{2} (e+f x)\right ) \cos \left (\frac {1}{2} (e+f x)\right ) \sec ^{\frac {5}{2}}(e+f x) \left (\frac {1}{8} e^{-\frac {3}{2} i (e+f x)} \left (6 e^{i (e+f x)}+10 e^{2 i (e+f x)}+6 e^{3 i (e+f x)}+5 e^{4 i (e+f x)}+5\right ) \sqrt {\sec (e+f x)}-3 \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt {1+e^{2 i (e+f x)}} \cos ^3\left (\frac {1}{2} (e+f x)\right ) \tanh ^{-1}\left (\frac {1+e^{i (e+f x)}}{\sqrt {2} \sqrt {1+e^{2 i (e+f x)}}}\right )\right )}{3 a^2 f (\sec (e+f x)+1)^2 \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(e + f*x)/2]*(-3*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]^3 + ((5 + 6*E^(I*(e + f*x)) +
10*E^((2*I)*(e + f*x)) + 6*E^((3*I)*(e + f*x)) + 5*E^((4*I)*(e + f*x)))*Sqrt[Sec[e + f*x]])/(8*E^(((3*I)/2)*(e
+ f*x))))*Sec[e + f*x]^(5/2)*Sin[(e + f*x)/2])/(3*a^2*E^((I/2)*(e + f*x))*f*(1 + Sec[e + f*x])^2*Sqrt[c - c*S
ec[e + f*x]])

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fricas [A]  time = 0.53, size = 331, normalized size = 2.40 \[ \left [-\frac {3 \, \sqrt {2} \sqrt {-c} {\left (\cos \left (f x + e\right ) + 1\right )} \log \left (\frac {2 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {-c} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} + {\left (3 \, c \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 4 \, {\left (5 \, \cos \left (f x + e\right )^{2} + 3 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{24 \, {\left (a^{2} c f \cos \left (f x + e\right ) + a^{2} c f\right )} \sin \left (f x + e\right )}, \frac {3 \, \sqrt {2} \sqrt {c} {\left (\cos \left (f x + e\right ) + 1\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {c} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, {\left (5 \, \cos \left (f x + e\right )^{2} + 3 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{12 \, {\left (a^{2} c f \cos \left (f x + e\right ) + a^{2} c f\right )} \sin \left (f x + e\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/24*(3*sqrt(2)*sqrt(-c)*(cos(f*x + e) + 1)*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*(5*cos(f*x + e)^2 + 3*cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((a^2*c*f*cos(f*x +
e) + a^2*c*f)*sin(f*x + e)), 1/12*(3*sqrt(2)*sqrt(c)*(cos(f*x + e) + 1)*arctan(sqrt(2)*sqrt((c*cos(f*x + e) -
c)/cos(f*x + e))*cos(f*x + e)/(sqrt(c)*sin(f*x + e)))*sin(f*x + e) - 2*(5*cos(f*x + e)^2 + 3*cos(f*x + e))*sqr
t((c*cos(f*x + e) - c)/cos(f*x + e)))/((a^2*c*f*cos(f*x + e) + a^2*c*f)*sin(f*x + e))]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)2/f*((3*i*atan(-i)
-4)/12/a^2/sqrt(-c)/sqrt(2)*sign(tan((f*x+exp(1))/2))+1/4*((-1/3*c^4*sqrt(c*tan((f*x+exp(1))/2)^2-c)*(c*tan((f
*x+exp(1))/2)^2-c)+c^5*sqrt(c*tan((f*x+exp(1))/2)^2-c))/c^6-atan(sqrt(c*tan((f*x+exp(1))/2)^2-c)/sqrt(c))/sqrt
(c))/sqrt(2)/a^2/sign(tan((f*x+exp(1))/2))/sign(tan((f*x+exp(1))/2)^2-1))

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maple [A]  time = 1.82, size = 131, normalized size = 0.95 \[ \frac {\left (-1+\cos \left (f x +e \right )\right ) \left (\left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {3}{2}}-3 \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}-3 \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}}\right )\right )}{6 a^{2} f \sqrt {\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/a^2/f*(-1+cos(f*x+e))*((-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)-3*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)-3*arcta
n(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)))/(c*(-1+cos(f*x+e))/cos(f*x+e))^(1/2)/sin(f*x+e)/(-2*cos(f*x+e)/(1+c
os(f*x+e)))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (f x + e\right )}{{\left (a \sec \left (f x + e\right ) + a\right )}^{2} \sqrt {-c \sec \left (f x + e\right ) + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sec(f*x + e)/((a*sec(f*x + e) + a)^2*sqrt(-c*sec(f*x + e) + c)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\cos \left (e+f\,x\right )\,{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^2\,\sqrt {c-\frac {c}{\cos \left (e+f\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec {\left (e + f x \right )}}{\sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )} + 2 \sqrt {- c \sec {\left (e + f x \right )} + c} \sec {\left (e + f x \right )} + \sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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