3.2.20 \(\int \frac {\sqrt {c-c \sec (e+f x)}}{(a+a \sec (e+f x))^{3/2}} \, dx\) [120]

Optimal. Leaf size=94 \[ -\frac {c \tan (e+f x)}{f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}+\frac {c \log (1+\cos (e+f x)) \tan (e+f x)}{a f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 94, 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 = {3992, 3996, 31} \begin {gather*} \frac {c \tan (e+f x) \log (\cos (e+f x)+1)}{a f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {c \tan (e+f x)}{f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((c*Tan[e + f*x])/(f*(a + a*Sec[e + f*x])^(3/2)*Sqrt[c - c*Sec[e + f*x]])) + (c*Log[1 + Cos[e + f*x]]*Tan[e +
f*x])/(a*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 3992

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 3996

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Dist
[(-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 {c-c \sec (e+f x)}}{(a+a \sec (e+f x))^{3/2}} \, dx &=-\frac {c \tan (e+f x)}{f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}+\frac {\int \frac {\sqrt {c-c \sec (e+f x)}}{\sqrt {a+a \sec (e+f x)}} \, dx}{a}\\ &=-\frac {c \tan (e+f x)}{f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}+\frac {(c \tan (e+f x)) \text {Subst}\left (\int \frac {1}{a+a x} \, dx,x,\cos (e+f x)\right )}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ &=-\frac {c \tan (e+f x)}{f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}+\frac {c \log (1+\cos (e+f x)) \tan (e+f x)}{a f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.60, size = 106, normalized size = 1.13 \begin {gather*} \frac {i \cot \left (\frac {1}{2} (e+f x)\right ) \left (i+f x+\cos (e+f x) \left (f x+2 i \log \left (1+e^{i (e+f x)}\right )\right )+2 i \log \left (1+e^{i (e+f x)}\right )\right ) \sec (e+f x) \sqrt {c-c \sec (e+f x)}}{f (a (1+\sec (e+f x)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.26, size = 119, normalized size = 1.27

method	result	size
default	\(-\frac {\left (2 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+\cos ^{2}\left (f x +e \right )-2 \cos \left (f x +e \right )-2 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+1\right ) \sqrt {\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, \cos \left (f x +e \right )}{2 f \sin \left (f x +e \right )^{3} a^{2}}\)	\(119\)
risch	\(\frac {\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, x}{a \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}-\frac {2 \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (f x +e \right )}{a \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}-\frac {2 i \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, {\mathrm e}^{i \left (f x +e \right )}}{a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}-\frac {2 i \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{a \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}\)	\(397\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 436 vs. \(2 (93) = 186\).
time = 0.54, size = 436, normalized size = 4.64 \begin {gather*} -\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 (a^{2} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2} \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a^{2} \sin \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) + 4 \, a^{2} \sin \left (f x + e\right )^{2} + 4 \, a^{2} \cos \left (f x + e\right ) + a^{2} + 2 \, {\left (2 \, a^{2} \cos \left (f x + e\right ) + a^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )\right )} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^(1/2)/(a+a*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)/((
a^2*cos(2*f*x + 2*e)^2 + 4*a^2*cos(f*x + e)^2 + a^2*sin(2*f*x + 2*e)^2 + 4*a^2*sin(2*f*x + 2*e)*sin(f*x + e) +
4*a^2*sin(f*x + e)^2 + 4*a^2*cos(f*x + e) + a^2 + 2*(2*a^2*cos(f*x + e) + a^2)*cos(2*f*x + 2*e))*f)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.35, size = 83, normalized size = 0.88 \begin {gather*} -\frac {\sqrt {2} {\left (\frac {2 \, \sqrt {2} \sqrt {-a c} c \log \left ({\left | -2 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, c \right |}\right )}{a^{2} {\left | c \right |}} - \frac {\sqrt {2} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} \sqrt {-a c}}{a^{2} {\left | c \right |}}\right )}}{4 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(2)*(2*sqrt(2)*sqrt(-a*c)*c*log(abs(-2*c*tan(1/2*f*x + 1/2*e)^2 - 2*c))/(a^2*abs(c)) - sqrt(2)*(c*tan
(1/2*f*x + 1/2*e)^2 - c)*sqrt(-a*c)/(a^2*abs(c)))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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