3.127 \(\int \frac {\sqrt {c-c \sec (e+f x)}}{(a+a \sec (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=140 \[ \frac {c \tan (e+f x) \log (\cos (e+f x)+1)}{a^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {c \tan (e+f x)}{a f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}}-\frac {c \tan (e+f x)}{2 f (a \sec (e+f x)+a)^{5/2} \sqrt {c-c \sec (e+f x)}} \]
[Out]
-1/2*c*tan(f*x+e)/f/(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(1/2)-c*tan(f*x+e)/a/f/(a+a*sec(f*x+e))^(3/2)/(c-c
*sec(f*x+e))^(1/2)+c*ln(1+cos(f*x+e))*tan(f*x+e)/a^2/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)
________________________________________________________________________________________
Rubi [A] time = 0.28, antiderivative size = 140, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used =
{3907, 3911, 31} \[ \frac {c \tan (e+f x) \log (\cos (e+f x)+1)}{a^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {c \tan (e+f x)}{a f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}}-\frac {c \tan (e+f x)}{2 f (a \sec (e+f x)+a)^{5/2} \sqrt {c-c \sec (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[c - c*Sec[e + f*x]]/(a + a*Sec[e + f*x])^(5/2),x]
[Out]
-(c*Tan[e + f*x])/(2*f*(a + a*Sec[e + f*x])^(5/2)*Sqrt[c - c*Sec[e + f*x]]) - (c*Tan[e + f*x])/(a*f*(a + a*Sec
[e + f*x])^(3/2)*Sqrt[c - c*Sec[e + f*x]]) + (c*Log[1 + Cos[e + f*x]]*Tan[e + f*x])/(a^2*f*Sqrt[a + a*Sec[e +
f*x]]*Sqrt[c - c*Sec[e + f*x]])
Rule 31
Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]
Rule 3907
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Simp[
(-2*a*Cot[e + f*x]*(c + d*Csc[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[1/c, Int[Sqrt[a +
b*Csc[e + f*x]]*(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] && LtQ[n, -2^(-1)]
Rule 3911
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> -Dis
t[(a*c*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*x]]), Subst[Int[((b + a*x)^(m - 1/2)*(d
+ c*x)^(n - 1/2))/x^(m + n), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] &&
EqQ[a^2 - b^2, 0] && IntegerQ[m - 1/2] && EqQ[m + n, 0]
Rubi steps
\begin {align*} \int \frac {\sqrt {c-c \sec (e+f x)}}{(a+a \sec (e+f x))^{5/2}} \, dx &=-\frac {c \tan (e+f x)}{2 f (a+a \sec (e+f x))^{5/2} \sqrt {c-c \sec (e+f x)}}+\frac {\int \frac {\sqrt {c-c \sec (e+f x)}}{(a+a \sec (e+f x))^{3/2}} \, dx}{a}\\ &=-\frac {c \tan (e+f x)}{2 f (a+a \sec (e+f x))^{5/2} \sqrt {c-c \sec (e+f x)}}-\frac {c \tan (e+f x)}{a f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}+\frac {\int \frac {\sqrt {c-c \sec (e+f x)}}{\sqrt {a+a \sec (e+f x)}} \, dx}{a^2}\\ &=-\frac {c \tan (e+f x)}{2 f (a+a \sec (e+f x))^{5/2} \sqrt {c-c \sec (e+f x)}}-\frac {c \tan (e+f x)}{a f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}+\frac {(c \tan (e+f x)) \operatorname {Subst}\left (\int \frac {1}{a+a x} \, dx,x,\cos (e+f x)\right )}{a f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ &=-\frac {c \tan (e+f x)}{2 f (a+a \sec (e+f x))^{5/2} \sqrt {c-c \sec (e+f x)}}-\frac {c \tan (e+f x)}{a f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}+\frac {c \log (1+\cos (e+f x)) \tan (e+f x)}{a^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ \end {align*}
________________________________________________________________________________________
Mathematica [C] time = 0.67, size = 151, normalized size = 1.08 \[ \frac {i \cot \left (\frac {1}{2} (e+f x)\right ) \sqrt {c-c \sec (e+f x)} \left (6 i \log \left (1+e^{i (e+f x)}\right )+\left (f x+2 i \log \left (1+e^{i (e+f x)}\right )\right ) \cos (2 (e+f x))+4 \left (2 i \log \left (1+e^{i (e+f x)}\right )+f x+i\right ) \cos (e+f x)+3 f x+3 i\right )}{2 a^2 f (\cos (e+f x)+1)^2 \sqrt {a (\sec (e+f x)+1)}} \]
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[c - c*Sec[e + f*x]]/(a + a*Sec[e + f*x])^(5/2),x]
[Out]
((I/2)*Cot[(e + f*x)/2]*(3*I + 3*f*x + Cos[2*(e + f*x)]*(f*x + (2*I)*Log[1 + E^(I*(e + f*x))]) + 4*Cos[e + f*x
]*(I + f*x + (2*I)*Log[1 + E^(I*(e + f*x))]) + (6*I)*Log[1 + E^(I*(e + f*x))])*Sqrt[c - c*Sec[e + f*x]])/(a^2*
f*(1 + Cos[e + f*x])^2*Sqrt[a*(1 + Sec[e + f*x])])
________________________________________________________________________________________
fricas [F] time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a \sec \left (f x + e\right ) + a} \sqrt {-c \sec \left (f x + e\right ) + c}}{a^{3} \sec \left (f x + e\right )^{3} + 3 \, a^{3} \sec \left (f x + e\right )^{2} + 3 \, a^{3} \sec \left (f x + e\right ) + a^{3}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c-c*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(5/2),x, algorithm="fricas")
[Out]
integral(sqrt(a*sec(f*x + e) + a)*sqrt(-c*sec(f*x + e) + c)/(a^3*sec(f*x + e)^3 + 3*a^3*sec(f*x + e)^2 + 3*a^3
*sec(f*x + e) + a^3), x)
________________________________________________________________________________________
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((c-c*sec(f*x+e))^(1/2)/(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)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (2*pi/x/2)>(-2*pi/x/2)2*sqrt(2)*(1/128*(4*sqrt(2)*a^3*c*sqrt(-a*c)*(c*tan(1/2*(f*x+exp(1)))^2-c)^2*abs(c)*
sign(tan(1/2*(f*x+exp(1)))^3+tan(1/2*(f*x+exp(1))))-16*sqrt(2)*a^3*c^2*sqrt(-a*c)*(c*tan(1/2*(f*x+exp(1)))^2-c
)*abs(c)*sign(tan(1/2*(f*x+exp(1)))^3+tan(1/2*(f*x+exp(1)))))/a^6/c^4+1/2*c*sqrt(-a*c)*sign(tan(1/2*(f*x+exp(1
)))^3+tan(1/2*(f*x+exp(1))))*ln(abs(c*tan(1/2*(f*x+exp(1)))^2+c))/sqrt(2)/a^3/abs(c))*sign(cos(f*x+exp(1)))/f
________________________________________________________________________________________
maple [A] time = 2.79, size = 152, normalized size = 1.09 \[ \frac {\left (-1+\cos \left (f x +e \right )\right )^{2} \left (8 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+7 \left (\cos ^{2}\left (f x +e \right )\right )+16 \cos \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-2 \cos \left (f x +e \right )+8 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-5\right ) \cos \left (f x +e \right ) \sqrt {\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}}{8 f \sin \left (f x +e \right )^{5} a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c-c*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(5/2),x)
[Out]
1/8/f*(-1+cos(f*x+e))^2*(8*ln(2/(1+cos(f*x+e)))*cos(f*x+e)^2+7*cos(f*x+e)^2+16*cos(f*x+e)*ln(2/(1+cos(f*x+e)))
-2*cos(f*x+e)+8*ln(2/(1+cos(f*x+e)))-5)*cos(f*x+e)*(c*(-1+cos(f*x+e))/cos(f*x+e))^(1/2)*(a*(1+cos(f*x+e))/cos(
f*x+e))^(1/2)/sin(f*x+e)^5/a^3
________________________________________________________________________________________
maxima [B] time = 1.40, size = 1165, normalized size = 8.32 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c-c*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(5/2),x, algorithm="maxima")
[Out]
-((f*x + e)*cos(4*f*x + 4*e)^2 + 16*(f*x + e)*cos(3*f*x + 3*e)^2 + 36*(f*x + e)*cos(2*f*x + 2*e)^2 + 16*(f*x +
e)*cos(f*x + e)^2 + (f*x + e)*sin(4*f*x + 4*e)^2 + 16*(f*x + e)*sin(3*f*x + 3*e)^2 + 36*(f*x + e)*sin(2*f*x +
2*e)^2 + 16*(f*x + e)*sin(f*x + e)^2 + f*x - 2*(2*(4*cos(3*f*x + 3*e) + 6*cos(2*f*x + 2*e) + 4*cos(f*x + e) +
1)*cos(4*f*x + 4*e) + cos(4*f*x + 4*e)^2 + 8*(6*cos(2*f*x + 2*e) + 4*cos(f*x + e) + 1)*cos(3*f*x + 3*e) + 16*
cos(3*f*x + 3*e)^2 + 12*(4*cos(f*x + e) + 1)*cos(2*f*x + 2*e) + 36*cos(2*f*x + 2*e)^2 + 16*cos(f*x + e)^2 + 4*
(2*sin(3*f*x + 3*e) + 3*sin(2*f*x + 2*e) + 2*sin(f*x + e))*sin(4*f*x + 4*e) + sin(4*f*x + 4*e)^2 + 16*(3*sin(2
*f*x + 2*e) + 2*sin(f*x + e))*sin(3*f*x + 3*e) + 16*sin(3*f*x + 3*e)^2 + 36*sin(2*f*x + 2*e)^2 + 48*sin(2*f*x
+ 2*e)*sin(f*x + e) + 16*sin(f*x + e)^2 + 8*cos(f*x + e) + 1)*arctan2(sin(f*x + e), cos(f*x + e) + 1) + 2*(f*x
+ 4*(f*x + e)*cos(3*f*x + 3*e) + 6*(f*x + e)*cos(2*f*x + 2*e) + 4*(f*x + e)*cos(f*x + e) + e - 2*sin(3*f*x +
3*e) - 3*sin(2*f*x + 2*e) - 2*sin(f*x + e))*cos(4*f*x + 4*e) + 8*(f*x + 6*(f*x + e)*cos(2*f*x + 2*e) + 4*(f*x
+ e)*cos(f*x + e) + e)*cos(3*f*x + 3*e) + 12*(f*x + 4*(f*x + e)*cos(f*x + e) + e)*cos(2*f*x + 2*e) + 8*(f*x +
e)*cos(f*x + e) + 2*(4*(f*x + e)*sin(3*f*x + 3*e) + 6*(f*x + e)*sin(2*f*x + 2*e) + 4*(f*x + e)*sin(f*x + e) +
2*cos(3*f*x + 3*e) + 3*cos(2*f*x + 2*e) + 2*cos(f*x + e))*sin(4*f*x + 4*e) + 4*(12*(f*x + e)*sin(2*f*x + 2*e)
+ 8*(f*x + e)*sin(f*x + e) - 1)*sin(3*f*x + 3*e) + 6*(8*(f*x + e)*sin(f*x + e) - 1)*sin(2*f*x + 2*e) + e - 4*s
in(f*x + e))*sqrt(a)*sqrt(c)/((a^3*cos(4*f*x + 4*e)^2 + 16*a^3*cos(3*f*x + 3*e)^2 + 36*a^3*cos(2*f*x + 2*e)^2
+ 16*a^3*cos(f*x + e)^2 + a^3*sin(4*f*x + 4*e)^2 + 16*a^3*sin(3*f*x + 3*e)^2 + 36*a^3*sin(2*f*x + 2*e)^2 + 48*
a^3*sin(2*f*x + 2*e)*sin(f*x + e) + 16*a^3*sin(f*x + e)^2 + 8*a^3*cos(f*x + e) + a^3 + 2*(4*a^3*cos(3*f*x + 3*
e) + 6*a^3*cos(2*f*x + 2*e) + 4*a^3*cos(f*x + e) + a^3)*cos(4*f*x + 4*e) + 8*(6*a^3*cos(2*f*x + 2*e) + 4*a^3*c
os(f*x + e) + a^3)*cos(3*f*x + 3*e) + 12*(4*a^3*cos(f*x + e) + a^3)*cos(2*f*x + 2*e) + 4*(2*a^3*sin(3*f*x + 3*
e) + 3*a^3*sin(2*f*x + 2*e) + 2*a^3*sin(f*x + e))*sin(4*f*x + 4*e) + 16*(3*a^3*sin(2*f*x + 2*e) + 2*a^3*sin(f*
x + e))*sin(3*f*x + 3*e))*f)
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {c-\frac {c}{\cos \left (e+f\,x\right )}}}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c - c/cos(e + f*x))^(1/2)/(a + a/cos(e + f*x))^(5/2),x)
[Out]
int((c - c/cos(e + f*x))^(1/2)/(a + a/cos(e + f*x))^(5/2), x)
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {- c \left (\sec {\left (e + f x \right )} - 1\right )}}{\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c-c*sec(f*x+e))**(1/2)/(a+a*sec(f*x+e))**(5/2),x)
[Out]
Integral(sqrt(-c*(sec(e + f*x) - 1))/(a*(sec(e + f*x) + 1))**(5/2), x)
________________________________________________________________________________________