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

Optimal. Leaf size=139 \[ -\frac {4 c^3 \tan (e+f x) \log (\sec (e+f x)+1)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {2 c^2 \tan (e+f x) \sqrt {c-c \sec (e+f x)}}{f \sqrt {a \sec (e+f x)+a}}-\frac {c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{2 f \sqrt {a \sec (e+f x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.41, antiderivative size = 139, 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 = {3955, 3952} \[ -\frac {2 c^2 \tan (e+f x) \sqrt {c-c \sec (e+f x)}}{f \sqrt {a \sec (e+f x)+a}}-\frac {4 c^3 \tan (e+f x) \log (\sec (e+f x)+1)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{2 f \sqrt {a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*c^3*Log[1 + Sec[e + f*x]]*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - (2*c^2*Sqr
t[c - c*Sec[e + f*x]]*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]) - (c*(c - c*Sec[e + f*x])^(3/2)*Tan[e + f*x])
/(2*f*Sqrt[a + a*Sec[e + f*x]])

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 3955

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

Rubi steps

\begin {align*} \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{5/2}}{\sqrt {a+a \sec (e+f x)}} \, dx &=-\frac {c (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{2 f \sqrt {a+a \sec (e+f x)}}+(2 c) \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{3/2}}{\sqrt {a+a \sec (e+f x)}} \, dx\\ &=-\frac {2 c^2 \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)}}-\frac {c (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{2 f \sqrt {a+a \sec (e+f x)}}+\left (4 c^2\right ) \int \frac {\sec (e+f x) \sqrt {c-c \sec (e+f x)}}{\sqrt {a+a \sec (e+f x)}} \, dx\\ &=-\frac {4 c^3 \log (1+\sec (e+f x)) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {2 c^2 \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)}}-\frac {c (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{2 f \sqrt {a+a \sec (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 1.48, size = 141, normalized size = 1.01 \[ \frac {c^2 \cot \left (\frac {1}{2} (e+f x)\right ) \sec ^2(e+f x) \sqrt {c-c \sec (e+f x)} \left (8 \log \left (1+e^{i (e+f x)}\right )-4 \log \left (1+e^{2 i (e+f x)}\right )-6 \cos (e+f x)+\left (8 \log \left (1+e^{i (e+f x)}\right )-4 \log \left (1+e^{2 i (e+f x)}\right )\right ) \cos (2 (e+f x))+1\right )}{2 f \sqrt {a (\sec (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c^{2} \sec \left (f x + e\right )^{3} - 2 \, c^{2} \sec \left (f x + e\right )^{2} + c^{2} \sec \left (f x + e\right )\right )} \sqrt {-c \sec \left (f x + e\right ) + c}}{\sqrt {a \sec \left (f x + e\right ) + a}}, x\right ) \]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)8*c^2*(1/4*(4*c^2*sqrt(-a*c)*(c*tan(1/2*(f*x+exp(1)))^2-c)+c^3*sqrt(-a*c)+3*c*sqrt(-a*c)*(c*tan(1/2*(f*x+e
xp(1)))^2-c)^2)/(c*tan(1/2*(f*x+exp(1)))^2-c)^2/a/abs(c)-1/2*c*sqrt(-a*c)*ln(c*tan(1/2*(f*x+exp(1)))^2-c)/a/ab
s(c))*sign(tan(1/2*(f*x+exp(1)))^3+tan(1/2*(f*x+exp(1))))/f

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maple [A]  time = 2.06, size = 165, normalized size = 1.19 \[ -\frac {\left (8 \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+8 \ln \left (-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+7 \left (\cos ^{2}\left (f x +e \right )\right )+6 \cos \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}}{2 f \sin \left (f x +e \right ) \left (-1+\cos \left (f x +e \right )\right )^{2} a} \]

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))^(1/2),x)

[Out]

-1/2/f*(8*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2+8*ln(-(-sin(f*x+e)-1+cos(f*x+e))/sin(f*x+e))
*cos(f*x+e)^2+7*cos(f*x+e)^2+6*cos(f*x+e)-1)*cos(f*x+e)*(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)*(a*(1+cos(f*x+e))
/cos(f*x+e))^(1/2)/sin(f*x+e)/(-1+cos(f*x+e))^2/a

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maxima [B]  time = 0.60, size = 737, normalized size = 5.30 \[ \frac {2 \, {\left (c^{2} \cos \left (2 \, f x + 2 \, e\right ) \sin \left (4 \, f x + 4 \, e\right ) - c^{2} \cos \left (4 \, f x + 4 \, e\right ) \sin \left (2 \, f x + 2 \, e\right ) - c^{2} \sin \left (2 \, f x + 2 \, e\right ) + 2 \, {\left (c^{2} \cos \left (4 \, f x + 4 \, e\right )^{2} + 4 \, c^{2} \cos \left (2 \, f x + 2 \, e\right )^{2} + c^{2} \sin \left (4 \, f x + 4 \, e\right )^{2} + 4 \, c^{2} \sin \left (4 \, f x + 4 \, e\right ) \sin \left (2 \, f x + 2 \, e\right ) + 4 \, c^{2} \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) + c^{2} + 2 \, {\left (2 \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) + c^{2}\right )} \cos \left (4 \, f x + 4 \, e\right )\right )} \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) - 4 \, {\left (c^{2} \cos \left (4 \, f x + 4 \, e\right )^{2} + 4 \, c^{2} \cos \left (2 \, f x + 2 \, e\right )^{2} + c^{2} \sin \left (4 \, f x + 4 \, e\right )^{2} + 4 \, c^{2} \sin \left (4 \, f x + 4 \, e\right ) \sin \left (2 \, f x + 2 \, e\right ) + 4 \, c^{2} \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) + c^{2} + 2 \, {\left (2 \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) + c^{2}\right )} \cos \left (4 \, f x + 4 \, e\right )\right )} \arctan \left (\sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ), \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) - 3 \, {\left (c^{2} \sin \left (4 \, f x + 4 \, e\right ) + 2 \, c^{2} \sin \left (2 \, f x + 2 \, e\right )\right )} \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 3 \, {\left (c^{2} \sin \left (4 \, f x + 4 \, e\right ) + 2 \, c^{2} \sin \left (2 \, f x + 2 \, e\right )\right )} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 3 \, {\left (c^{2} \cos \left (4 \, f x + 4 \, e\right ) + 2 \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) + c^{2}\right )} \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 3 \, {\left (c^{2} \cos \left (4 \, f x + 4 \, e\right ) + 2 \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) + c^{2}\right )} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {a} \sqrt {c}}{{\left (a \cos \left (4 \, f x + 4 \, e\right )^{2} + 4 \, a \cos \left (2 \, f x + 2 \, e\right )^{2} + a \sin \left (4 \, f x + 4 \, e\right )^{2} + 4 \, a \sin \left (4 \, f x + 4 \, e\right ) \sin \left (2 \, f x + 2 \, e\right ) + 4 \, a \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, {\left (2 \, a \cos \left (2 \, f x + 2 \, e\right ) + a\right )} \cos \left (4 \, f x + 4 \, e\right ) + 4 \, a \cos \left (2 \, f x + 2 \, e\right ) + a\right )} 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))^(1/2),x, algorithm="maxima")

[Out]

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

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

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))^(1/2)),x)

[Out]

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

[Out]

Timed out

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