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

Optimal. Leaf size=98 \[ \frac {c^3 \tan (e+f x) \log (\cos (e+f x)+1)}{a^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {2 c^3 \tan (e+f x)}{f (a \sec (e+f x)+a)^{5/2} \sqrt {c-c \sec (e+f x)}} \]

[Out]

-2*c^3*tan(f*x+e)/f/(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(1/2)+c^3*ln(1+cos(f*x+e))*tan(f*x+e)/a^2/f/(a+a*s
ec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3910, 3911, 31} \[ \frac {c^3 \tan (e+f x) \log (\cos (e+f x)+1)}{a^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {2 c^3 \tan (e+f x)}{f (a \sec (e+f x)+a)^{5/2} \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c^3*Tan[e + f*x])/(f*(a + a*Sec[e + f*x])^(5/2)*Sqrt[c - c*Sec[e + f*x]]) + (c^3*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 3910

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(5/2)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Si
mp[(-8*a^3*Cot[e + f*x]*(c + d*Csc[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[a^2/c^2, Int
[Sqrt[a + b*Csc[e + f*x]]*(c + d*Csc[e + f*x])^(n + 2), 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 {(c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{5/2}} \, dx &=-\frac {2 c^3 \tan (e+f x)}{f (a+a \sec (e+f x))^{5/2} \sqrt {c-c \sec (e+f x)}}+\frac {c^2 \int \frac {\sqrt {c-c \sec (e+f x)}}{\sqrt {a+a \sec (e+f x)}} \, dx}{a^2}\\ &=-\frac {2 c^3 \tan (e+f x)}{f (a+a \sec (e+f x))^{5/2} \sqrt {c-c \sec (e+f x)}}+\frac {\left (c^3 \tan (e+f x)\right ) \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 {2 c^3 \tan (e+f x)}{f (a+a \sec (e+f x))^{5/2} \sqrt {c-c \sec (e+f x)}}+\frac {c^3 \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*}

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Mathematica [C]  time = 1.57, size = 154, normalized size = 1.57 \[ \frac {i c^2 \cot \left (\frac {1}{2} (e+f x)\right ) \sqrt {c-c \sec (e+f x)} \left (6 i \log \left (1+e^{i (e+f x)}\right )+\left (f x+2 i \log \left (1+e^{i (e+f x)}\right )\right ) \cos (2 (e+f x))+4 \left (2 i \log \left (1+e^{i (e+f x)}\right )+f x+2 i\right ) \cos (e+f x)+3 f x+4 i\right )}{2 a^2 f (\cos (e+f x)+1)^2 \sqrt {a (\sec (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*c^2*Cot[(e + f*x)/2]*(4*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]*(2*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])])

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c^{2} \sec \left (f x + e\right )^{2} - 2 \, c^{2} \sec \left (f x + e\right ) + c^{2}\right )} \sqrt {a \sec \left (f x + e\right ) + a} \sqrt {-c \sec \left (f x + e\right ) + c}}{a^{3} \sec \left (f x + e\right )^{3} + 3 \, a^{3} \sec \left (f x + e\right )^{2} + 3 \, a^{3} \sec \left (f x + e\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c^2*sec(f*x + e)^2 - 2*c^2*sec(f*x + e) + c^2)*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)

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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((c-c*sec(f*x+e))^(5/2)/(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)2*(-1/2*c^3*sqrt(-
a*c)*sign(tan(1/2*(f*x+exp(1)))^3+tan(1/2*(f*x+exp(1))))*ln(2*abs(c))/a^3/abs(c)+1/2*c^3*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))/a^3/abs(c)+1/4*sqrt(-a*c)*(c*ta
n(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))))/a^3/c)/f

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maple [A]  time = 2.75, size = 144, normalized size = 1.47 \[ \frac {\left (2 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+3 \left (\cos ^{2}\left (f x +e \right )\right )+4 \cos \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-2 \cos \left (f x +e \right )+2 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-1\right ) \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}}{2 f \sin \left (f x +e \right )^{5} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.71, size = 102, normalized size = 1.04 \[ \frac {\frac {2 \, c^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{\sqrt {-a} a^{2}} + \frac {\frac {2 \, \sqrt {-a} c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {\sqrt {-a} c^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}}{a^{3}}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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