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

Optimal. Leaf size=99 \[ -\frac {a^2 \tan (e+f x) \log (1-\sec (e+f x))}{c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {a \tan (e+f x) \sqrt {a \sec (e+f x)+a}}{f (c-c \sec (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {3954, 3952} \[ -\frac {a^2 \tan (e+f x) \log (1-\sec (e+f x))}{c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {a \tan (e+f x) \sqrt {a \sec (e+f x)+a}}{f (c-c \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [C]  time = 0.70, size = 134, normalized size = 1.35 \[ -\frac {a \tan \left (\frac {1}{2} (e+f x)\right ) \sqrt {a (\sec (e+f x)+1)} \left (-2 \log \left (1-e^{i (e+f x)}\right )+\log \left (1+e^{2 i (e+f x)}\right )+\left (2 \log \left (1-e^{i (e+f x)}\right )-\log \left (1+e^{2 i (e+f x)}\right )\right ) \cos (e+f x)+2\right )}{c f (\cos (e+f x)-1) \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*(2 - 2*Log[1 - E^(I*(e + f*x))] + Cos[e + f*x]*(2*Log[1 - E^(I*(e + f*x))] - Log[1 + E^((2*I)*(e + f*x))]
) + Log[1 + E^((2*I)*(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*Sec[e + f*x]]))

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fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a \sec \left (f x + e\right )^{2} + a \sec \left (f x + e\right )\right )} \sqrt {a \sec \left (f x + e\right ) + a} \sqrt {-c \sec \left (f x + e\right ) + c}}{c^{2} \sec \left (f x + e\right )^{2} - 2 \, c^{2} \sec \left (f x + e\right ) + c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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maple [B]  time = 1.89, size = 246, normalized size = 2.48 \[ \frac {\left (-1+\cos \left (f x +e \right )\right ) \left (2 \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \cos \left (f x +e \right )-\cos \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-\cos \left (f x +e \right ) \ln \left (-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+\cos \left (f x +e \right )-2 \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+\ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+\ln \left (-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+1\right ) \sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, a}{f \cos \left (f x +e \right ) \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sin \left (f x +e \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(-1+cos(f*x+e))*(2*ln(-(-1+cos(f*x+e))/sin(f*x+e))*cos(f*x+e)-cos(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/si
n(f*x+e))-cos(f*x+e)*ln(-(-sin(f*x+e)-1+cos(f*x+e))/sin(f*x+e))+cos(f*x+e)-2*ln(-(-1+cos(f*x+e))/sin(f*x+e))+l
n(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+ln(-(-sin(f*x+e)-1+cos(f*x+e))/sin(f*x+e))+1)*(a*(1+cos(f*x+e))/cos(
f*x+e))^(1/2)/cos(f*x+e)/(c*(-1+cos(f*x+e))/cos(f*x+e))^(3/2)/sin(f*x+e)*a

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maxima [A]  time = 0.74, size = 122, normalized size = 1.23 \[ \frac {\frac {\sqrt {-a} a \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c^{\frac {3}{2}}} + \frac {\sqrt {-a} a \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{\frac {3}{2}}} - \frac {2 \, \sqrt {-a} a \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{\frac {3}{2}}} + \frac {\sqrt {-a} a {\left (\cos \left (f x + e\right ) + 1\right )}^{2}}{c^{\frac {3}{2}} \sin \left (f x + e\right )^{2}}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \sec {\left (e + f x \right )}}{\left (- c \left (\sec {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**(3/2)/(c-c*sec(f*x+e))**(3/2),x)

[Out]

Integral((a*(sec(e + f*x) + 1))**(3/2)*sec(e + f*x)/(-c*(sec(e + f*x) - 1))**(3/2), x)

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