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

Optimal. Leaf size=145 \[ -\frac {c^3 \tan (e+f x) \log (\sec (e+f x)+1)}{a^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {c^2 \tan (e+f x) \sqrt {c-c \sec (e+f x)}}{a f (a \sec (e+f x)+a)^{3/2}}+\frac {c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{2 f (a \sec (e+f x)+a)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.44, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {3954, 3952} \[ -\frac {c^3 \tan (e+f x) \log (\sec (e+f x)+1)}{a^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {c^2 \tan (e+f x) \sqrt {c-c \sec (e+f x)}}{a f (a \sec (e+f x)+a)^{3/2}}+\frac {c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{2 f (a \sec (e+f x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c^3*Log[1 + Sec[e + f*x]]*Tan[e + f*x])/(a^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])) - (c^2*S
qrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(a*f*(a + a*Sec[e + f*x])^(3/2)) + (c*(c - c*Sec[e + f*x])^(3/2)*Tan[e +
f*x])/(2*f*(a + a*Sec[e + f*x])^(5/2))

Rule 3952

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

Rule 3954

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 - 1/2, 0
] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{5/2}} \, dx &=\frac {c (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{2 f (a+a \sec (e+f x))^{5/2}}-\frac {c \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{3/2}}{(a+a \sec (e+f x))^{3/2}} \, dx}{a}\\ &=-\frac {c^2 \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{a f (a+a \sec (e+f x))^{3/2}}+\frac {c (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{2 f (a+a \sec (e+f x))^{5/2}}+\frac {c^2 \int \frac {\sec (e+f x) \sqrt {c-c \sec (e+f x)}}{\sqrt {a+a \sec (e+f x)}} \, dx}{a^2}\\ &=-\frac {c^3 \log (1+\sec (e+f x)) \tan (e+f x)}{a^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {c^2 \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{a f (a+a \sec (e+f x))^{3/2}}+\frac {c (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{2 f (a+a \sec (e+f x))^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 1.39, size = 178, normalized size = 1.23 \[ \frac {c^2 \cot \left (\frac {1}{2} (e+f x)\right ) \sqrt {c-c \sec (e+f x)} \left (6 \log \left (1+e^{i (e+f x)}\right )-3 \log \left (1+e^{2 i (e+f x)}\right )+\left (8 \log \left (1+e^{i (e+f x)}\right )-4 \log \left (1+e^{2 i (e+f x)}\right )\right ) \cos (e+f x)+\left (2 \log \left (1+e^{i (e+f x)}\right )-\log \left (1+e^{2 i (e+f x)}\right )\right ) \cos (2 (e+f x))-4\right )}{2 a^2 f (\cos (e+f x)+1)^2 \sqrt {a (\sec (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*Cot[(e + f*x)/2]*(-4 + 6*Log[1 + E^(I*(e + f*x))] + Cos[e + f*x]*(8*Log[1 + E^(I*(e + f*x))] - 4*Log[1 +
E^((2*I)*(e + f*x))]) + Cos[2*(e + f*x)]*(2*Log[1 + E^(I*(e + f*x))] - Log[1 + E^((2*I)*(e + f*x))]) - 3*Log[1
+ E^((2*I)*(e + f*x))])*Sqrt[c - c*Sec[e + f*x]])/(2*a^2*f*(1 + Cos[e + f*x])^2*Sqrt[a*(1 + Sec[e + f*x])])

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fricas [F]  time = 1.10, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c^{2} \sec \left (f x + e\right )^{3} - 2 \, c^{2} \sec \left (f x + e\right )^{2} + c^{2} \sec \left (f x + e\right )\right )} \sqrt {a \sec \left (f x + e\right ) + a} \sqrt {-c \sec \left (f x + e\right ) + c}}{a^{3} \sec \left (f x + e\right )^{3} + 3 \, a^{3} \sec \left (f x + e\right )^{2} + 3 \, a^{3} \sec \left (f x + e\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 2.08, size = 281, normalized size = 1.94 \[ -\frac {\left (2 \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \ln \left (-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right )+4 \cos \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+4 \cos \left (f x +e \right ) \ln \left (-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-2 \cos \left (f x +e \right )+2 \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+2 \ln \left (-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+3\right ) \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}}{2 f \sin \left (f x +e \right )^{5} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/f*(2*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2+2*ln(-(-sin(f*x+e)-1+cos(f*x+e))/sin(f*x+e))
*cos(f*x+e)^2-cos(f*x+e)^2+4*cos(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+4*cos(f*x+e)*ln(-(-sin(f*x+
e)-1+cos(f*x+e))/sin(f*x+e))-2*cos(f*x+e)+2*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+2*ln(-(-sin(f*x+e)-1+co
s(f*x+e))/sin(f*x+e))+3)*(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 = 0.47, size = 133, normalized size = 0.92 \[ -\frac {\frac {2 \, c^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{\sqrt {-a} a^{2}} + \frac {2 \, c^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{\sqrt {-a} a^{2}} - \frac {\frac {2 \, \sqrt {-a} c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {\sqrt {-a} c^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}}{a^{3}}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(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)/(cos(f*x + e) + 1) + 1)/(sqrt(-a)*a^2) + 2*c^(5/2)*log(sin(f*x + e)/(cos(f*x
+ e) + 1) - 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 \[ \int \frac {{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{\cos \left (e+f\,x\right )\,{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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