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

Optimal. Leaf size=76 \[ \frac {a \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{\sqrt {2} c^{3/2} f}-\frac {a \tan (e+f x)}{f (c-c \sec (e+f x))^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3957, 3795, 203} \[ \frac {a \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{\sqrt {2} c^{3/2} f}-\frac {a \tan (e+f x)}{f (c-c \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(Sqrt[2]*c^(3/2)*f) - (a*Tan[e + f*x])/(
f*(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 3957

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 1.15, size = 246, normalized size = 3.24 \[ \frac {a \left (\frac {i \sqrt {2} \left (-1+e^{i (e+f x)}\right )^3 \tanh ^{-1}\left (\frac {1+e^{i (e+f x)}}{\sqrt {2} \sqrt {1+e^{2 i (e+f x)}}}\right )}{\left (1+e^{2 i (e+f x)}\right )^{3/2}}+8 \sin \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right ) \sin ^3\left (\frac {1}{2} (e+f x)\right ) \sec ^2(e+f x)-8 \cos \left (\frac {e}{2}\right ) \cos \left (\frac {f x}{2}\right ) \sin ^3\left (\frac {1}{2} (e+f x)\right ) \sec ^2(e+f x)+4 \cot \left (\frac {e}{2}\right ) \sin ^2\left (\frac {1}{2} (e+f x)\right ) \sec ^2(e+f x)-4 \csc \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right ) \sin \left (\frac {1}{2} (e+f x)\right ) \sec ^2(e+f x)\right )}{2 c f (\sec (e+f x)-1) \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*((I*Sqrt[2]*(-1 + E^(I*(e + f*x)))^3*ArcTanh[(1 + E^(I*(e + f*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(e + f*x))])]
)/(1 + E^((2*I)*(e + f*x)))^(3/2) - 4*Csc[e/2]*Sec[e + f*x]^2*Sin[(f*x)/2]*Sin[(e + f*x)/2] + 4*Cot[e/2]*Sec[e
+ f*x]^2*Sin[(e + f*x)/2]^2 - 8*Cos[e/2]*Cos[(f*x)/2]*Sec[e + f*x]^2*Sin[(e + f*x)/2]^3 + 8*Sec[e + f*x]^2*Si
n[e/2]*Sin[(f*x)/2]*Sin[(e + f*x)/2]^3))/(2*c*f*(-1 + Sec[e + f*x])*Sqrt[c - c*Sec[e + f*x]])

________________________________________________________________________________________

fricas [B]  time = 0.53, size = 342, normalized size = 4.50 \[ \left [\frac {\sqrt {2} {\left (a c \cos \left (f x + e\right ) - a c\right )} \sqrt {-\frac {1}{c}} \log \left (\frac {2 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \sqrt {-\frac {1}{c}} + {\left (3 \, \cos \left (f x + e\right ) + 1\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 4 \, {\left (a \cos \left (f x + e\right )^{2} + a \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{4 \, {\left (c^{2} f \cos \left (f x + e\right ) - c^{2} f\right )} \sin \left (f x + e\right )}, -\frac {\frac {\sqrt {2} {\left (a c \cos \left (f x + e\right ) - a c\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {c} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right )}{\sqrt {c}} - 2 \, {\left (a \cos \left (f x + e\right )^{2} + a \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{2 \, {\left (c^{2} f \cos \left (f x + e\right ) - c^{2} f\right )} \sin \left (f x + e\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(sqrt(2)*(a*c*cos(f*x + e) - a*c)*sqrt(-1/c)*log((2*sqrt(2)*(cos(f*x + e)^2 + cos(f*x + e))*sqrt((c*cos(f
*x + e) - c)/cos(f*x + e))*sqrt(-1/c) + (3*cos(f*x + e) + 1)*sin(f*x + e))/((cos(f*x + e) - 1)*sin(f*x + e)))*
sin(f*x + e) + 4*(a*cos(f*x + e)^2 + a*cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((c^2*f*cos(f*x
+ e) - c^2*f)*sin(f*x + e)), -1/2*(sqrt(2)*(a*c*cos(f*x + e) - a*c)*arctan(sqrt(2)*sqrt((c*cos(f*x + e) - c)/c
os(f*x + e))*cos(f*x + e)/(sqrt(c)*sin(f*x + e)))*sin(f*x + e)/sqrt(c) - 2*(a*cos(f*x + e)^2 + a*cos(f*x + e))
*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((c^2*f*cos(f*x + e) - 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))/(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*a*(1/2*sqrt(c)*atan(sqrt(c*tan((f*x+exp(1))/2)^2-c)/sqrt(c
))+1/2*c*sqrt(c*tan((f*x+exp(1))/2)^2-c)/c/tan((f*x+exp(1))/2)^2)/sqrt(2)/c^2/sign(tan((f*x+exp(1))/2))/sign(t
an((f*x+exp(1))/2)^2-1)

________________________________________________________________________________________

maple [B]  time = 1.38, size = 164, normalized size = 2.16 \[ \frac {2 a \left (-1+\cos \left (f x +e \right )\right )^{2} \left (\cos \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}+\cos \left (f x +e \right ) \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}}\right )+\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}-\arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}}\right )\right )}{f \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sin \left (f x +e \right )^{3} \left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a/f*(-1+cos(f*x+e))^2*(cos(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)+cos(f*x+e)*arctan(1/(-2*cos(f*x+e)/(1
+cos(f*x+e)))^(1/2))+(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)-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 \[ \int \frac {{\left (a \sec \left (f x + e\right ) + a\right )} \sec \left (f x + e\right )}{{\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(Integral(sec(e + f*x)/(-c*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x) + c*sqrt(-c*sec(e + f*x) + c)), x) + Integ
ral(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))

________________________________________________________________________________________