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

Optimal. Leaf size=142 \[ \frac{a \tan (e+f x) \log (1-\cos (e+f x))}{c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{a \tan (e+f x)}{c f \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}}-\frac{a \tan (e+f x)}{2 f \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.271143, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3907, 3911, 31} \[ \frac{a \tan (e+f x) \log (1-\cos (e+f x))}{c^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{a \tan (e+f x)}{c f \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}}-\frac{a \tan (e+f x)}{2 f \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [C]  time = 1.23054, size = 152, normalized size = 1.07 \[ \frac{\tan \left (\frac{1}{2} (e+f x)\right ) \sqrt{a (\sec (e+f x)+1)} \left (6 \log \left (1-e^{i (e+f x)}\right )+\left (-8 \log \left (1-e^{i (e+f x)}\right )+4 i f x-4\right ) \cos (e+f x)+\left (2 \log \left (1-e^{i (e+f x)}\right )-i f x\right ) \cos (2 (e+f x))-3 i f x+3\right )}{2 c^2 f (\cos (e+f x)-1)^2 \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3 - (3*I)*f*x + Cos[e + f*x]*(-4 + (4*I)*f*x - 8*Log[1 - E^(I*(e + f*x))]) + 6*Log[1 - E^(I*(e + f*x))] + Co
s[2*(e + f*x)]*((-I)*f*x + 2*Log[1 - E^(I*(e + f*x))]))*Sqrt[a*(1 + Sec[e + f*x])]*Tan[(e + f*x)/2])/(2*c^2*f*
(-1 + Cos[e + f*x])^2*Sqrt[c - c*Sec[e + f*x]])

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Maple [A]  time = 0.303, size = 226, normalized size = 1.6 \begin{align*} -{\frac{-1+\cos \left ( fx+e \right ) }{8\,f\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( 16\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-8\,\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-7\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-32\,\cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +16\,\cos \left ( fx+e \right ) \ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) -2\,\cos \left ( fx+e \right ) +16\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -8\,\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) +5 \right ) \sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}} \left ({\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{5}{2}}}} \end{align*}

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))^(5/2),x)

[Out]

-1/8/f*(-1+cos(f*x+e))*(16*ln(-(-1+cos(f*x+e))/sin(f*x+e))*cos(f*x+e)^2-8*ln(2/(1+cos(f*x+e)))*cos(f*x+e)^2-7*
cos(f*x+e)^2-32*cos(f*x+e)*ln(-(-1+cos(f*x+e))/sin(f*x+e))+16*cos(f*x+e)*ln(2/(1+cos(f*x+e)))-2*cos(f*x+e)+16*
ln(-(-1+cos(f*x+e))/sin(f*x+e))-8*ln(2/(1+cos(f*x+e)))+5)*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)/(c*(-1+cos(f*x
+e))/cos(f*x+e))^(5/2)/sin(f*x+e)/cos(f*x+e)^2

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Maxima [B]  time = 2.41535, size = 1584, normalized size = 11.15 \begin{align*} \text{result too large to display} \end{align*}

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))^(5/2),x, algorithm="maxima")

[Out]

-((f*x + e)*cos(4*f*x + 4*e)^2 + 16*(f*x + e)*cos(3*f*x + 3*e)^2 + 36*(f*x + e)*cos(2*f*x + 2*e)^2 + 16*(f*x +
e)*cos(f*x + e)^2 + (f*x + e)*sin(4*f*x + 4*e)^2 + 16*(f*x + e)*sin(3*f*x + 3*e)^2 + 36*(f*x + e)*sin(2*f*x +
2*e)^2 + 16*(f*x + e)*sin(f*x + e)^2 + f*x + 2*(2*(4*cos(3*f*x + 3*e) - 6*cos(2*f*x + 2*e) + 4*cos(f*x + e) -
1)*cos(4*f*x + 4*e) - cos(4*f*x + 4*e)^2 + 8*(6*cos(2*f*x + 2*e) - 4*cos(f*x + e) + 1)*cos(3*f*x + 3*e) - 16*
cos(3*f*x + 3*e)^2 + 12*(4*cos(f*x + e) - 1)*cos(2*f*x + 2*e) - 36*cos(2*f*x + 2*e)^2 - 16*cos(f*x + e)^2 + 4*
(2*sin(3*f*x + 3*e) - 3*sin(2*f*x + 2*e) + 2*sin(f*x + e))*sin(4*f*x + 4*e) - sin(4*f*x + 4*e)^2 + 16*(3*sin(2
*f*x + 2*e) - 2*sin(f*x + e))*sin(3*f*x + 3*e) - 16*sin(3*f*x + 3*e)^2 - 36*sin(2*f*x + 2*e)^2 + 48*sin(2*f*x
+ 2*e)*sin(f*x + e) - 16*sin(f*x + e)^2 + 8*cos(f*x + e) - 1)*arctan2(sin(f*x + e), cos(f*x + e) - 1) + 2*(f*x
- 4*(f*x + e)*cos(3*f*x + 3*e) + 6*(f*x + e)*cos(2*f*x + 2*e) - 4*(f*x + e)*cos(f*x + e) + e + 2*sin(3*f*x +
3*e) - 3*sin(2*f*x + 2*e) + 2*sin(f*x + e))*cos(4*f*x + 4*e) - 8*(f*x + 6*(f*x + e)*cos(2*f*x + 2*e) - 4*(f*x
+ e)*cos(f*x + e) + e)*cos(3*f*x + 3*e) + 12*(f*x - 4*(f*x + e)*cos(f*x + e) + e)*cos(2*f*x + 2*e) - 8*(f*x +
e)*cos(f*x + e) - 2*(4*(f*x + e)*sin(3*f*x + 3*e) - 6*(f*x + e)*sin(2*f*x + 2*e) + 4*(f*x + e)*sin(f*x + e) +
2*cos(3*f*x + 3*e) - 3*cos(2*f*x + 2*e) + 2*cos(f*x + e))*sin(4*f*x + 4*e) - 4*(12*(f*x + e)*sin(2*f*x + 2*e)
- 8*(f*x + e)*sin(f*x + e) - 1)*sin(3*f*x + 3*e) - 6*(8*(f*x + e)*sin(f*x + e) + 1)*sin(2*f*x + 2*e) + e + 4*s
in(f*x + e))*sqrt(a)*sqrt(c)/((c^3*cos(4*f*x + 4*e)^2 + 16*c^3*cos(3*f*x + 3*e)^2 + 36*c^3*cos(2*f*x + 2*e)^2
+ 16*c^3*cos(f*x + e)^2 + c^3*sin(4*f*x + 4*e)^2 + 16*c^3*sin(3*f*x + 3*e)^2 + 36*c^3*sin(2*f*x + 2*e)^2 - 48*
c^3*sin(2*f*x + 2*e)*sin(f*x + e) + 16*c^3*sin(f*x + e)^2 - 8*c^3*cos(f*x + e) + c^3 - 2*(4*c^3*cos(3*f*x + 3*
e) - 6*c^3*cos(2*f*x + 2*e) + 4*c^3*cos(f*x + e) - c^3)*cos(4*f*x + 4*e) - 8*(6*c^3*cos(2*f*x + 2*e) - 4*c^3*c
os(f*x + e) + c^3)*cos(3*f*x + 3*e) - 12*(4*c^3*cos(f*x + e) - c^3)*cos(2*f*x + 2*e) - 4*(2*c^3*sin(3*f*x + 3*
e) - 3*c^3*sin(2*f*x + 2*e) + 2*c^3*sin(f*x + e))*sin(4*f*x + 4*e) - 16*(3*c^3*sin(2*f*x + 2*e) - 2*c^3*sin(f*
x + e))*sin(3*f*x + 3*e))*f)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{a \sec \left (f x + e\right ) + a} \sqrt{-c \sec \left (f x + e\right ) + c}}{c^{3} \sec \left (f x + e\right )^{3} - 3 \, c^{3} \sec \left (f x + e\right )^{2} + 3 \, c^{3} \sec \left (f x + e\right ) - c^{3}}, x\right ) \end{align*}

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))^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))**(5/2),x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))^(5/2),x, algorithm="giac")

[Out]

Timed out