3.128 \(\int \frac {\sec (e+f x) (a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=145 \[ -\frac {4 a^3 \tan (e+f x) \log (1-\sec (e+f x))}{c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {2 a^2 \tan (e+f x) \sqrt {a \sec (e+f x)+a}}{c f \sqrt {c-c \sec (e+f x)}}-\frac {a \tan (e+f x) (a \sec (e+f x)+a)^{3/2}}{f (c-c \sec (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.42, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3954, 3955, 3952} \[ -\frac {2 a^2 \tan (e+f x) \sqrt {a \sec (e+f x)+a}}{c f \sqrt {c-c \sec (e+f x)}}-\frac {4 a^3 \tan (e+f x) \log (1-\sec (e+f x))}{c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {a \tan (e+f x) (a \sec (e+f x)+a)^{3/2}}{f (c-c \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x])^(5/2))/(c - c*Sec[e + f*x])^(3/2),x]

[Out]

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

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

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Mathematica [C]  time = 1.50, size = 188, normalized size = 1.30 \[ \frac {a^2 \tan \left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) \sqrt {a (\sec (e+f x)+1)} \left (-4 \log \left (1-e^{i (e+f x)}\right )+2 \log \left (1+e^{2 i (e+f x)}\right )+\left (8 \log \left (1-e^{i (e+f x)}\right )-4 \log \left (1+e^{2 i (e+f x)}\right )-5\right ) \cos (e+f x)+\left (2 \log \left (1+e^{2 i (e+f x)}\right )-4 \log \left (1-e^{i (e+f x)}\right )\right ) \cos (2 (e+f x))+1\right )}{c f (\cos (e+f x)-1) \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x])^(5/2))/(c - c*Sec[e + f*x])^(3/2),x]

[Out]

(a^2*(1 - 4*Log[1 - E^(I*(e + f*x))] + Cos[e + f*x]*(-5 + 8*Log[1 - E^(I*(e + f*x))] - 4*Log[1 + E^((2*I)*(e +
f*x))]) + 2*Log[1 + E^((2*I)*(e + f*x))] + Cos[2*(e + f*x)]*(-4*Log[1 - E^(I*(e + f*x))] + 2*Log[1 + E^((2*I)
*(e + f*x))]))*Sec[e + f*x]*Sqrt[a*(1 + Sec[e + f*x])]*Tan[(e + f*x)/2])/(c*f*(-1 + Cos[e + f*x])*Sqrt[c - c*S
ec[e + f*x]])

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fricas [F]  time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{2} \sec \left (f x + e\right )^{3} + 2 \, a^{2} \sec \left (f x + e\right )^{2} + a^{2} \sec \left (f x + e\right )\right )} \sqrt {a \sec \left (f x + e\right ) + a} \sqrt {-c \sec \left (f x + e\right ) + c}}{c^{2} \sec \left (f x + e\right )^{2} - 2 \, c^{2} \sec \left (f x + e\right ) + c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral((a^2*sec(f*x + e)^3 + 2*a^2*sec(f*x + e)^2 + a^2*sec(f*x + e))*sqrt(a*sec(f*x + e) + a)*sqrt(-c*sec(f
*x + e) + c)/(c^2*sec(f*x + e)^2 - 2*c^2*sec(f*x + e) + c^2), x)

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

[Out]

Timed out

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maple [B]  time = 2.20, size = 289, normalized size = 1.99 \[ -\frac {\left (-1+\cos \left (f x +e \right )\right ) \left (4 \ln \left (-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+4 \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right )-8 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-3 \left (\cos ^{2}\left (f x +e \right )\right )-4 \cos \left (f x +e \right ) \ln \left (-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-4 \cos \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+8 \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \cos \left (f x +e \right )-2 \cos \left (f x +e \right )+1\right ) \sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, a^{2}}{f \cos \left (f x +e \right )^{2} \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sin \left (f x +e \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(3/2),x)

[Out]

-1/f*(-1+cos(f*x+e))*(4*ln(-(-sin(f*x+e)-1+cos(f*x+e))/sin(f*x+e))*cos(f*x+e)^2+4*ln(-(-1+cos(f*x+e)+sin(f*x+e
))/sin(f*x+e))*cos(f*x+e)^2-8*cos(f*x+e)^2*ln(-(-1+cos(f*x+e))/sin(f*x+e))-3*cos(f*x+e)^2-4*cos(f*x+e)*ln(-(-s
in(f*x+e)-1+cos(f*x+e))/sin(f*x+e))-4*cos(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+8*ln(-(-1+cos(f*x+
e))/sin(f*x+e))*cos(f*x+e)-2*cos(f*x+e)+1)*(a*(1+cos(f*x+e))/cos(f*x+e))^(1/2)/cos(f*x+e)^2/(c*(-1+cos(f*x+e))
/cos(f*x+e))^(3/2)/sin(f*x+e)*a^2

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maxima [B]  time = 0.74, size = 2035, normalized size = 14.03 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

2*(8*a^2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^
2 + 8*a^2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))
^2 + 8*a^2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))
)^2 + 8*a^2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)
))^2 - 2*a^2*cos(2*f*x + 2*e)*sin(4*f*x + 4*e) + 2*a^2*cos(4*f*x + 4*e)*sin(2*f*x + 2*e) + 2*a^2*sin(2*f*x + 2
*e) + 2*(a^2*cos(4*f*x + 4*e)^2 + 4*a^2*cos(2*f*x + 2*e)^2 + a^2*sin(4*f*x + 4*e)^2 + 4*a^2*sin(4*f*x + 4*e)*s
in(2*f*x + 2*e) + 4*a^2*sin(2*f*x + 2*e)^2 + 4*a^2*cos(2*f*x + 2*e) + a^2 + 2*(2*a^2*cos(2*f*x + 2*e) + a^2)*c
os(4*f*x + 4*e))*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) - 4*(a^2*cos(4*f*x + 4*e)^2 + 4*a^2*cos(2*f*x
+ 2*e)^2 + 4*a^2*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 4*a^2*cos(1/2*arctan2(sin(2*f*x + 2
*e), cos(2*f*x + 2*e)))^2 + a^2*sin(4*f*x + 4*e)^2 + 4*a^2*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 4*a^2*sin(2*f*x
+ 2*e)^2 + 4*a^2*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 4*a^2*sin(1/2*arctan2(sin(2*f*x + 2
*e), cos(2*f*x + 2*e)))^2 + 4*a^2*cos(2*f*x + 2*e) + a^2 + 2*(2*a^2*cos(2*f*x + 2*e) + a^2)*cos(4*f*x + 4*e) -
4*(a^2*cos(4*f*x + 4*e) + 2*a^2*cos(2*f*x + 2*e) - 2*a^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))
+ a^2)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 4*(a^2*cos(4*f*x + 4*e) + 2*a^2*cos(2*f*x + 2*e
) + a^2)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 4*(a^2*sin(4*f*x + 4*e) + 2*a^2*sin(2*f*x + 2*
e) - 2*a^2*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))) - 4*(a^2*sin(4*f*x + 4*e) + 2*a^2*sin(2*f*x + 2*e))*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*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) + (16*a^2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(
2*f*x + 2*e))) + 5*a^2*sin(4*f*x + 4*e) + 6*a^2*sin(2*f*x + 2*e) - 8*(a^2*cos(4*f*x + 4*e) + 2*a^2*cos(2*f*x +
2*e) + a^2)*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*
e))) + (5*a^2*sin(4*f*x + 4*e) + 6*a^2*sin(2*f*x + 2*e) - 8*(a^2*cos(4*f*x + 4*e) + 2*a^2*cos(2*f*x + 2*e) + a
^2)*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (1
6*a^2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 5
*a^2*cos(4*f*x + 4*e) - 6*a^2*cos(2*f*x + 2*e) - 5*a^2 - 8*(a^2*sin(4*f*x + 4*e) + 2*a^2*sin(2*f*x + 2*e))*arc
tan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (5*a^2*co
s(4*f*x + 4*e) + 6*a^2*cos(2*f*x + 2*e) + 5*a^2 + 8*(a^2*sin(4*f*x + 4*e) + 2*a^2*sin(2*f*x + 2*e))*arctan2(si
n(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)*sqrt(c)/(
(c^2*cos(4*f*x + 4*e)^2 + 4*c^2*cos(2*f*x + 2*e)^2 + 4*c^2*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))
)^2 + 4*c^2*cos(1/2*arctan2(sin(2*f*x + 2*e), 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*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))
)^2 + 4*c^2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 4*c^2*cos(2*f*x + 2*e) + c^2 + 2*(2*c^2*c
os(2*f*x + 2*e) + c^2)*cos(4*f*x + 4*e) - 4*(c^2*cos(4*f*x + 4*e) + 2*c^2*cos(2*f*x + 2*e) - 2*c^2*cos(1/2*arc
tan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + c^2)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 4*(c^2
*cos(4*f*x + 4*e) + 2*c^2*cos(2*f*x + 2*e) + c^2)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 4*(c^
2*sin(4*f*x + 4*e) + 2*c^2*sin(2*f*x + 2*e) - 2*c^2*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))) - 4*(c^2*sin(4*f*x + 4*e) + 2*c^2*sin(2*f*x + 2*e))*sin(1/2*a
rctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*f)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{\cos \left (e+f\,x\right )\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^(5/2)/(cos(e + f*x)*(c - c/cos(e + f*x))^(3/2)),x)

[Out]

int((a + a/cos(e + f*x))^(5/2)/(cos(e + f*x)*(c - c/cos(e + f*x))^(3/2)), 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)*(a+a*sec(f*x+e))**(5/2)/(c-c*sec(f*x+e))**(3/2),x)

[Out]

Timed out

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