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

Optimal. Leaf size=104 \[ \frac{2 a^2 \tan (e+f x) \log (1-\sec (e+f x))}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{a^2 \tan (e+f x) \log (\cos (e+f x))}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.10332, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {3912, 72} \[ \frac{2 a^2 \tan (e+f x) \log (1-\sec (e+f x))}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{a^2 \tan (e+f x) \log (\cos (e+f x))}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3912

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Di
st[(a*c*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*x]]), Subst[Int[((a + b*x)^(m - 1/2)*(c
+ d*x)^(n - 1/2))/x, x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && E
qQ[a^2 - b^2, 0]

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

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

Mathematica [C]  time = 1.18517, size = 105, normalized size = 1.01 \[ -\frac{a \left (-1+e^{i (e+f x)}\right ) \left (4 i \log \left (1-e^{i (e+f x)}\right )-i \log \left (1+e^{2 i (e+f x)}\right )+f x\right ) \sqrt{a (\sec (e+f x)+1)}}{f \left (1+e^{i (e+f x)}\right ) \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*(-1 + E^(I*(e + f*x)))*(f*x + (4*I)*Log[1 - E^(I*(e + f*x))] - I*Log[1 + E^((2*I)*(e + f*x))])*Sqrt[a*(1
+ Sec[e + f*x])])/((1 + E^(I*(e + f*x)))*f*Sqrt[c - c*Sec[e + f*x]]))

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Maple [A]  time = 0.262, size = 149, normalized size = 1.4 \begin{align*}{\frac{a\cos \left ( fx+e \right ) }{f\sin \left ( fx+e \right ) c}\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}} \left ( \ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) +\ln \left ({\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -4\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \right ) \sqrt{{\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.84575, size = 81, normalized size = 0.78 \begin{align*} -\frac{{\left ({\left (f x + e\right )} a + a \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) - 4 \, a \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) - 1\right )\right )} \sqrt{a}}{\sqrt{c} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out