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

Optimal. Leaf size=169 \[ -\frac {5 \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{8 \sqrt {2} a^2 c^{3/2} f}-\frac {5 \tan (e+f x)}{8 a^2 f (c-c \sec (e+f x))^{3/2}}+\frac {5 \tan (e+f x)}{6 f \left (a^2 \sec (e+f x)+a^2\right ) (c-c \sec (e+f x))^{3/2}}+\frac {\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{3/2}} \]

[Out]

-5/16*arctan(1/2*c^(1/2)*tan(f*x+e)*2^(1/2)/(c-c*sec(f*x+e))^(1/2))/a^2/c^(3/2)/f*2^(1/2)-5/8*tan(f*x+e)/a^2/f
/(c-c*sec(f*x+e))^(3/2)+1/3*tan(f*x+e)/f/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^(3/2)+5/6*tan(f*x+e)/f/(a^2+a^2*s
ec(f*x+e))/(c-c*sec(f*x+e))^(3/2)

________________________________________________________________________________________

Rubi [A]  time = 0.34, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3960, 3796, 3795, 203} \[ -\frac {5 \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{8 \sqrt {2} a^2 c^{3/2} f}-\frac {5 \tan (e+f x)}{8 a^2 f (c-c \sec (e+f x))^{3/2}}+\frac {5 \tan (e+f x)}{6 f \left (a^2 \sec (e+f x)+a^2\right ) (c-c \sec (e+f x))^{3/2}}+\frac {\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(8*Sqrt[2]*a^2*c^(3/2)*f) - (5*Tan[e +
f*x])/(8*a^2*f*(c - c*Sec[e + f*x])^(3/2)) + Tan[e + f*x]/(3*f*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(3/
2)) + (5*Tan[e + f*x])/(6*f*(a^2 + a^2*Sec[e + f*x])*(c - c*Sec[e + f*x])^(3/2))

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 3796

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b*Cot[e + f*x]*(a
+ b*Csc[e + f*x])^m)/(a*f*(2*m + 1)), x] + Dist[(m + 1)/(a*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(
m + 1), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && IntegerQ[2*m]

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 (c-c \sec (e+f x))^{3/2}} \, dx &=\frac {\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}}+\frac {5 \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2}} \, dx}{6 a}\\ &=\frac {\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}}+\frac {5 \tan (e+f x)}{6 f \left (a^2+a^2 \sec (e+f x)\right ) (c-c \sec (e+f x))^{3/2}}+\frac {5 \int \frac {\sec (e+f x)}{(c-c \sec (e+f x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {5 \tan (e+f x)}{8 a^2 f (c-c \sec (e+f x))^{3/2}}+\frac {\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}}+\frac {5 \tan (e+f x)}{6 f \left (a^2+a^2 \sec (e+f x)\right ) (c-c \sec (e+f x))^{3/2}}+\frac {5 \int \frac {\sec (e+f x)}{\sqrt {c-c \sec (e+f x)}} \, dx}{16 a^2 c}\\ &=-\frac {5 \tan (e+f x)}{8 a^2 f (c-c \sec (e+f x))^{3/2}}+\frac {\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}}+\frac {5 \tan (e+f x)}{6 f \left (a^2+a^2 \sec (e+f x)\right ) (c-c \sec (e+f x))^{3/2}}-\frac {5 \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 )}{8 a^2 c f}\\ &=-\frac {5 \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{8 \sqrt {2} a^2 c^{3/2} f}-\frac {5 \tan (e+f x)}{8 a^2 f (c-c \sec (e+f x))^{3/2}}+\frac {\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}}+\frac {5 \tan (e+f x)}{6 f \left (a^2+a^2 \sec (e+f x)\right ) (c-c \sec (e+f x))^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 1.39, size = 365, normalized size = 2.16 \[ \frac {-\frac {15 i \sqrt {2} \left (-1+e^{i (e+f x)}\right )^3 \left (1+e^{i (e+f x)}\right )^4 \tanh ^{-1}\left (\frac {1+e^{i (e+f x)}}{\sqrt {2} \sqrt {1+e^{2 i (e+f x)}}}\right )}{\left (1+e^{2 i (e+f x)}\right )^{7/2}}-416 \sin \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right ) \sin ^8(e+f x) \csc \left (\frac {1}{2} (e+f x)\right ) \csc ^4(2 (e+f x))-40 \tan ^3(e+f x) \sec (e+f x)+416 \cos \left (\frac {e}{2}\right ) \cos \left (\frac {f x}{2}\right ) \sin ^3\left (\frac {1}{2} (e+f x)\right ) \cos ^4\left (\frac {1}{2} (e+f x)\right ) \sec ^4(e+f x)+32 \sin ^3\left (\frac {1}{2} (e+f x)\right ) \cos \left (\frac {1}{2} (e+f x)\right ) \sec ^4(e+f x)+48 \cot \left (\frac {e}{2}\right ) \sin ^2\left (\frac {1}{2} (e+f x)\right ) \cos ^4\left (\frac {1}{2} (e+f x)\right ) \sec ^4(e+f x)-48 \csc \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right ) \sin \left (\frac {1}{2} (e+f x)\right ) \cos ^4\left (\frac {1}{2} (e+f x)\right ) \sec ^4(e+f x)}{48 a^2 c f (\sec (e+f x)-1) (\sec (e+f x)+1)^2 \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-15*I)*Sqrt[2]*(-1 + E^(I*(e + f*x)))^3*(1 + E^(I*(e + f*x)))^4*ArcTanh[(1 + E^(I*(e + f*x)))/(Sqrt[2]*Sqrt
[1 + E^((2*I)*(e + f*x))])])/(1 + E^((2*I)*(e + f*x)))^(7/2) - 48*Cos[(e + f*x)/2]^4*Csc[e/2]*Sec[e + f*x]^4*S
in[(f*x)/2]*Sin[(e + f*x)/2] + 48*Cos[(e + f*x)/2]^4*Cot[e/2]*Sec[e + f*x]^4*Sin[(e + f*x)/2]^2 + 32*Cos[(e +
f*x)/2]*Sec[e + f*x]^4*Sin[(e + f*x)/2]^3 + 416*Cos[e/2]*Cos[(f*x)/2]*Cos[(e + f*x)/2]^4*Sec[e + f*x]^4*Sin[(e
+ f*x)/2]^3 - 416*Csc[(e + f*x)/2]*Csc[2*(e + f*x)]^4*Sin[e/2]*Sin[(f*x)/2]*Sin[e + f*x]^8 - 40*Sec[e + f*x]*
Tan[e + f*x]^3)/(48*a^2*c*f*(-1 + Sec[e + f*x])*(1 + Sec[e + f*x])^2*Sqrt[c - c*Sec[e + f*x]])

________________________________________________________________________________________

fricas [A]  time = 0.53, size = 369, normalized size = 2.18 \[ \left [-\frac {15 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {-c} \log \left (\frac {2 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {-c} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} + {\left (3 \, c \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 4 \, {\left (13 \, \cos \left (f x + e\right )^{3} - 10 \, \cos \left (f x + e\right )^{2} - 15 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{96 \, {\left (a^{2} c^{2} f \cos \left (f x + e\right )^{2} - a^{2} c^{2} f\right )} \sin \left (f x + e\right )}, \frac {15 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {c} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, {\left (13 \, \cos \left (f x + e\right )^{3} - 10 \, \cos \left (f x + e\right )^{2} - 15 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{48 \, {\left (a^{2} c^{2} f \cos \left (f x + e\right )^{2} - a^{2} c^{2} f\right )} \sin \left (f x + e\right )}\right ] \]

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))^(3/2),x, algorithm="fricas")

[Out]

[-1/96*(15*sqrt(2)*(cos(f*x + e)^2 - 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)))*s
in(f*x + e) + 4*(13*cos(f*x + e)^3 - 10*cos(f*x + e)^2 - 15*cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x +
e)))/((a^2*c^2*f*cos(f*x + e)^2 - a^2*c^2*f)*sin(f*x + e)), 1/48*(15*sqrt(2)*(cos(f*x + e)^2 - 1)*sqrt(c)*arct
an(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*(13*c
os(f*x + e)^3 - 10*cos(f*x + e)^2 - 15*cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((a^2*c^2*f*cos(
f*x + e)^2 - a^2*c^2*f)*sin(f*x + e))]

________________________________________________________________________________________

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))^(3/2),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)2/f/8*(-5/2*sqrt(c)*atan(sqrt(c*tan((f*x+exp(1))/2)^2-c)/sqrt(
c))+(-1/3*c^2*sqrt(c*tan((f*x+exp(1))/2)^2-c)*(c*tan((f*x+exp(1))/2)^2-c)+2*c^3*sqrt(c*tan((f*x+exp(1))/2)^2-c
))/c^3+1/2*c*sqrt(c*tan((f*x+exp(1))/2)^2-c)/c/tan((f*x+exp(1))/2)^2)/sqrt(2)/a^2/c^2/sign(tan((f*x+exp(1))/2)
)/sign(tan((f*x+exp(1))/2)^2-1)

________________________________________________________________________________________

maple [B]  time = 1.99, size = 320, normalized size = 1.89 \[ -\frac {\left (-1+\cos \left (f x +e \right )\right )^{2} \left (3 \left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {7}{2}} \cos \left (f x +e \right )+3 \left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {7}{2}}+3 \left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \cos \left (f x +e \right )-3 \left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {5}{2}}-5 \cos \left (f x +e \right ) \left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {3}{2}}+5 \left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {3}{2}}+15 \cos \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}+15 \cos \left (f x +e \right ) \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}}\right )-15 \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}-15 \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}}\right )\right )}{12 a^{2} f \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sin \left (f x +e \right )^{3} \left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {3}{2}}} \]

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))^(3/2),x)

[Out]

-1/12/a^2/f*(-1+cos(f*x+e))^2*(3*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(7/2)*cos(f*x+e)+3*(-2*cos(f*x+e)/(1+cos(f*x+e
)))^(7/2)+3*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2)*cos(f*x+e)-3*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2)-5*cos(f*x+e
)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)+5*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)+15*cos(f*x+e)*(-2*cos(f*x+e)/(1+
cos(f*x+e)))^(1/2)+15*cos(f*x+e)*arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2))-15*(-2*cos(f*x+e)/(1+cos(f*x+e
)))^(1/2)-15*arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)))/(c*(-1+cos(f*x+e))/cos(f*x+e))^(3/2)/sin(f*x+e)^3
/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (f x + e\right )}{{\left (a \sec \left (f x + e\right ) + a\right )}^{2} {\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]

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))^(3/2),x, algorithm="maxima")

[Out]

integrate(sec(f*x + e)/((a*sec(f*x + e) + a)^2*(-c*sec(f*x + e) + c)^(3/2)), x)

________________________________________________________________________________________

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

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))^(3/2)),x)

[Out]

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

________________________________________________________________________________________

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

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

[Out]

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

________________________________________________________________________________________