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

Optimal. Leaf size=96 \[ \frac {a \tan (e+f x) \log (1-\cos (e+f x))}{c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {a \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}} \]

[Out]

-a*tan(f*x+e)/f/(c-c*sec(f*x+e))^(3/2)/(a+a*sec(f*x+e))^(1/2)+a*ln(1-cos(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 = 0.18, antiderivative size = 96, 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 = {3907, 3911, 31} \[ \frac {a \tan (e+f x) \log (1-\cos (e+f x))}{c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {a \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^(3/2))) + (a*Log[1 - Cos[e + f*x]]*Tan[e +
f*x])/(c*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 3907

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Simp[
(-2*a*Cot[e + f*x]*(c + d*Csc[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[1/c, Int[Sqrt[a +
b*Csc[e + f*x]]*(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] && 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 {\sqrt {a+a \sec (e+f x)}}{(c-c \sec (e+f x))^{3/2}} \, dx &=-\frac {a \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac {\int \frac {\sqrt {a+a \sec (e+f x)}}{\sqrt {c-c \sec (e+f x)}} \, dx}{c}\\ &=-\frac {a \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac {(a \tan (e+f x)) \operatorname {Subst}\left (\int \frac {1}{-c+c x} \, dx,x,\cos (e+f x)\right )}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ &=-\frac {a \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac {a \log (1-\cos (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 = 1.04, size = 107, normalized size = 1.11 \[ \frac {\tan \left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) \sqrt {a (\sec (e+f x)+1)} \left (-2 \log \left (1-e^{i (e+f x)}\right )+\left (2 \log \left (1-e^{i (e+f x)}\right )-i f x\right ) \cos (e+f x)+i f x-1\right )}{f (c-c \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a \sec \left (f x + e\right ) + a} \sqrt {-c \sec \left (f x + e\right ) + c}}{c^{2} \sec \left (f x + e\right )^{2} - 2 \, c^{2} \sec \left (f x + e\right ) + c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(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((a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 2.55, size = 164, normalized size = 1.71 \[ -\frac {\left (-1+\cos \left (f x +e \right )\right ) \left (4 \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \cos \left (f x +e \right )-2 \cos \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-\cos \left (f x +e \right )-4 \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+2 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-1\right ) \sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}}{2 f \cos \left (f x +e \right ) \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sin \left (f x +e \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/f*(-1+cos(f*x+e))*(4*ln(-(-1+cos(f*x+e))/sin(f*x+e))*cos(f*x+e)-2*cos(f*x+e)*ln(2/(1+cos(f*x+e)))-cos(f*x
+e)-4*ln(-(-1+cos(f*x+e))/sin(f*x+e))+2*ln(2/(1+cos(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)

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maxima [B]  time = 1.03, size = 399, normalized size = 4.16 \[ -\frac {{\left ({\left (f x + e\right )} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, {\left (f x + e\right )} \cos \left (f x + e\right )^{2} + {\left (f x + e\right )} \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, {\left (f x + e\right )} \sin \left (f x + e\right )^{2} + f x + 2 \, {\left (2 \, {\left (2 \, \cos \left (f x + e\right ) - 1\right )} \cos \left (2 \, f x + 2 \, e\right ) - \cos \left (2 \, f x + 2 \, e\right )^{2} - 4 \, \cos \left (f x + e\right )^{2} - \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, \sin \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) - 4 \, \sin \left (f x + e\right )^{2} + 4 \, \cos \left (f x + e\right ) - 1\right )} \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) - 1\right ) + 2 \, {\left (f x - 2 \, {\left (f x + e\right )} \cos \left (f x + e\right ) + e + \sin \left (f x + e\right )\right )} \cos \left (2 \, f x + 2 \, e\right ) - 4 \, {\left (f x + e\right )} \cos \left (f x + e\right ) - 2 \, {\left (2 \, {\left (f x + e\right )} \sin \left (f x + e\right ) + \cos \left (f x + e\right )\right )} \sin \left (2 \, f x + 2 \, e\right ) + e + 2 \, \sin \left (f x + e\right )\right )} \sqrt {a} \sqrt {c}}{{\left (c^{2} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, c^{2} \cos \left (f x + e\right )^{2} + c^{2} \sin \left (2 \, f x + 2 \, e\right )^{2} - 4 \, c^{2} \sin \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) + 4 \, c^{2} \sin \left (f x + e\right )^{2} - 4 \, c^{2} \cos \left (f x + e\right ) + c^{2} - 2 \, {\left (2 \, c^{2} \cos \left (f x + e\right ) - c^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^(1/2)/(c - c/cos(e + f*x))^(3/2),x)

[Out]

int((a + a/cos(e + f*x))^(1/2)/(c - c/cos(e + f*x))^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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