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

Optimal. Leaf size=43 \[ -\frac {a \tan (e+f x)}{2 f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 43, 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 = {3953} \[ -\frac {a \tan (e+f x)}{2 f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3953

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

Rubi steps

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

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Mathematica [A]  time = 0.39, size = 69, normalized size = 1.60 \[ -\frac {(2 \cos (e+f x)-1) \tan \left (\frac {1}{2} (e+f x)\right ) \sqrt {a (\sec (e+f x)+1)}}{2 c^2 f (\cos (e+f x)-1)^2 \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.44, size = 106, normalized size = 2.47 \[ \frac {{\left (2 \, \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{2 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/2*(2*cos(f*x + e)^2 - cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*
x + e))/((c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) + c^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)*(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*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)1/2*a^2*(1/4*(2*a*
(-a*tan(1/2*(f*x+exp(1)))^2+a)-a^2)/(-a*tan(1/2*(f*x+exp(1)))^2)^2+1/4)*sign(cos(f*x+exp(1)))/c^2/sqrt(-a*c)/f
/abs(a)/sign(tan(1/2*(f*x+exp(1)))^2-1)/sign(tan(1/2*(f*x+exp(1))))

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maple [A]  time = 2.12, size = 70, normalized size = 1.63 \[ -\frac {\left (3 \cos \left (f x +e \right )-1\right ) \sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \sin \left (f x +e \right )}{8 f \cos \left (f x +e \right )^{2} \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/f*(3*cos(f*x+e)-1)*(a*(1+cos(f*x+e))/cos(f*x+e))^(1/2)*sin(f*x+e)/cos(f*x+e)^2/(c*(-1+cos(f*x+e))/cos(f*x
+e))^(5/2)

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maxima [B]  time = 1.22, size = 758, normalized size = 17.63 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

2*((sin(4*f*x + 4*e) + 2*sin(2*f*x + 2*e))*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (sin(4*f*x +
4*e) + 2*sin(2*f*x + 2*e))*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - cos(2*f*x + 2*e)*sin(4*f*x
+ 4*e) + cos(4*f*x + 4*e)*sin(2*f*x + 2*e) - (cos(4*f*x + 4*e) + 2*cos(2*f*x + 2*e) + 1)*sin(3/2*arctan2(sin(2
*f*x + 2*e), cos(2*f*x + 2*e))) - (cos(4*f*x + 4*e) + 2*cos(2*f*x + 2*e) + 1)*sin(1/2*arctan2(sin(2*f*x + 2*e)
, cos(2*f*x + 2*e))) + sin(2*f*x + 2*e))*sqrt(a)*sqrt(c)/((c^3*cos(4*f*x + 4*e)^2 + 36*c^3*cos(2*f*x + 2*e)^2
+ 16*c^3*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 16*c^3*cos(1/2*arctan2(sin(2*f*x + 2*e), cos
(2*f*x + 2*e)))^2 + c^3*sin(4*f*x + 4*e)^2 + 12*c^3*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 36*c^3*sin(2*f*x + 2*e
)^2 + 16*c^3*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 16*c^3*sin(1/2*arctan2(sin(2*f*x + 2*e),
cos(2*f*x + 2*e)))^2 + 12*c^3*cos(2*f*x + 2*e) + c^3 + 2*(6*c^3*cos(2*f*x + 2*e) + c^3)*cos(4*f*x + 4*e) - 8*
(c^3*cos(4*f*x + 4*e) + 6*c^3*cos(2*f*x + 2*e) - 4*c^3*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) +
c^3)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 8*(c^3*cos(4*f*x + 4*e) + 6*c^3*cos(2*f*x + 2*e) +
c^3)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 8*(c^3*sin(4*f*x + 4*e) + 6*c^3*sin(2*f*x + 2*e)
- 4*c^3*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*
e))) - 8*(c^3*sin(4*f*x + 4*e) + 6*c^3*sin(2*f*x + 2*e))*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))
*f)

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mupad [B]  time = 6.61, size = 203, normalized size = 4.72 \[ \frac {\sqrt {c-\frac {c}{\cos \left (e+f\,x\right )}}\,\left (\frac {{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}\,4{}\mathrm {i}}{c^3\,f}+\frac {{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}\,4{}\mathrm {i}}{c^3\,f}-\frac {\cos \left (e+f\,x\right )\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}\,4{}\mathrm {i}}{c^3\,f}\right )}{{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,10{}\mathrm {i}-{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sin \left (2\,e+2\,f\,x\right )\,8{}\mathrm {i}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sin \left (3\,e+3\,f\,x\right )\,2{}\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^(1/2)/(cos(e + f*x)*(c - c/cos(e + f*x))^(5/2)),x)

[Out]

((c - c/cos(e + f*x))^(1/2)*((exp(e*3i + f*x*3i)*(a + a/cos(e + f*x))^(1/2)*4i)/(c^3*f) + (exp(e*3i + f*x*3i)*
cos(2*e + 2*f*x)*(a + a/cos(e + f*x))^(1/2)*4i)/(c^3*f) - (cos(e + f*x)*exp(e*3i + f*x*3i)*(a + a/cos(e + f*x)
)^(1/2)*4i)/(c^3*f)))/(exp(e*3i + f*x*3i)*sin(e + f*x)*10i - exp(e*3i + f*x*3i)*sin(2*e + 2*f*x)*8i + exp(e*3i
+ f*x*3i)*sin(3*e + 3*f*x)*2i)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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