3.146 \(\int \frac {\sec (e+f x) (c-c \sec (e+f x))^{3/2}}{(a+a \sec (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=42 \[ \frac {\tan (e+f x) (c-c \sec (e+f x))^{3/2}}{4 f (a \sec (e+f x)+a)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {3950} \[ \frac {\tan (e+f x) (c-c \sec (e+f x))^{3/2}}{4 f (a \sec (e+f x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(c - c*Sec[e + f*x])^(3/2))/(a + a*Sec[e + f*x])^(5/2),x]

[Out]

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

Rule 3950

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] /
; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] && NeQ[2*m
+ 1, 0]

Rubi steps

\begin {align*} \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{3/2}}{(a+a \sec (e+f x))^{5/2}} \, dx &=\frac {(c-c \sec (e+f x))^{3/2} \tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 68, normalized size = 1.62 \[ \frac {c \cos (e+f x) \csc \left (\frac {1}{2} (e+f x)\right ) \sec ^3\left (\frac {1}{2} (e+f x)\right ) \sqrt {c-c \sec (e+f x)}}{4 a^2 f \sqrt {a (\sec (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c - c*Sec[e + f*x])^(3/2))/(a + a*Sec[e + f*x])^(5/2),x]

[Out]

(c*Cos[e + f*x]*Csc[(e + f*x)/2]*Sec[(e + f*x)/2]^3*Sqrt[c - c*Sec[e + f*x]])/(4*a^2*f*Sqrt[a*(1 + Sec[e + f*x
])])

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fricas [B]  time = 0.44, size = 95, normalized size = 2.26 \[ \frac {c \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2}}{{\left (a^{3} f \cos \left (f x + e\right )^{2} + 2 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^(3/2)/(a+a*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

c*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^3*f*cos(f
*x + e)^2 + 2*a^3*f*cos(f*x + e) + a^3*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)*(c-c*sec(f*x+e))^(3/2)/(a+a*sec(f*x+e))^(5/2),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)1/2*c^2*(1/2*(c*tan(1/2*(f*x+exp(1)))^2-c)^2*sign(tan(1/2*(f*x
+exp(1)))^3+tan(1/2*(f*x+exp(1))))+c*(c*tan(1/2*(f*x+exp(1)))^2-c)*sign(tan(1/2*(f*x+exp(1)))^3+tan(1/2*(f*x+e
xp(1)))))/a^2/sqrt(-a*c)/c/f/abs(c)

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maple [B]  time = 2.06, size = 75, normalized size = 1.79 \[ -\frac {\sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \left (\cos ^{2}\left (f x +e \right )\right ) \left (-1+\cos \left (f x +e \right )\right )^{3} \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}}}{4 f \sin \left (f x +e \right )^{5} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(c-c*sec(f*x+e))^(3/2)/(a+a*sec(f*x+e))^(5/2),x)

[Out]

-1/4/f*(a*(1+cos(f*x+e))/cos(f*x+e))^(1/2)*cos(f*x+e)^2*(-1+cos(f*x+e))^3*(c*(-1+cos(f*x+e))/cos(f*x+e))^(3/2)
/sin(f*x+e)^5/a^3

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maxima [B]  time = 0.47, size = 98, normalized size = 2.33 \[ -\frac {\sqrt {-a} c^{\frac {3}{2}} {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )} {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )} \sin \left (f x + e\right )^{4}}{4 \, {\left (a^{3} - \frac {a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} f {\left (\cos \left (f x + e\right ) + 1\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^(3/2)/(a+a*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-1/4*sqrt(-a)*c^(3/2)*(sin(f*x + e)/(cos(f*x + e) + 1) + 1)*(sin(f*x + e)/(cos(f*x + e) + 1) - 1)*sin(f*x + e)
^4/((a^3 - a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)*f*(cos(f*x + e) + 1)^4)

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mupad [B]  time = 3.51, size = 119, normalized size = 2.83 \[ -\frac {2\,c\,\sqrt {\frac {c\,\left (\cos \left (e+f\,x\right )-1\right )}{\cos \left (e+f\,x\right )}}\,\left (\sin \left (e+f\,x\right )+2\,\sin \left (2\,e+2\,f\,x\right )+\sin \left (3\,e+3\,f\,x\right )\right )}{a^2\,f\,\sqrt {\frac {a\,\left (\cos \left (e+f\,x\right )+1\right )}{\cos \left (e+f\,x\right )}}\,\left (4\,\cos \left (2\,e+2\,f\,x\right )-4\,\cos \left (e+f\,x\right )+4\,\cos \left (3\,e+3\,f\,x\right )+\cos \left (4\,e+4\,f\,x\right )-5\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c/cos(e + f*x))^(3/2)/(cos(e + f*x)*(a + a/cos(e + f*x))^(5/2)),x)

[Out]

-(2*c*((c*(cos(e + f*x) - 1))/cos(e + f*x))^(1/2)*(sin(e + f*x) + 2*sin(2*e + 2*f*x) + sin(3*e + 3*f*x)))/(a^2
*f*((a*(cos(e + f*x) + 1))/cos(e + f*x))^(1/2)*(4*cos(2*e + 2*f*x) - 4*cos(e + f*x) + 4*cos(3*e + 3*f*x) + cos
(4*e + 4*f*x) - 5))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- c \left (\sec {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}} \sec {\left (e + f x \right )}}{\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))**(3/2)/(a+a*sec(f*x+e))**(5/2),x)

[Out]

Integral((-c*(sec(e + f*x) - 1))**(3/2)*sec(e + f*x)/(a*(sec(e + f*x) + 1))**(5/2), x)

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