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

Optimal. Leaf size=93 \[ \frac{a c^2 \tan (e+f x) \log (\cos (e+f x))}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{a c \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{f \sqrt{a \sec (e+f x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.178326, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3906, 3905, 3475} \[ \frac{a c^2 \tan (e+f x) \log (\cos (e+f x))}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{a c \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{f \sqrt{a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3906

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Simp
[(2*a*c*Cot[e + f*x]*(c + d*Csc[e + f*x])^(n - 1))/(f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[c, Int[Sq
rt[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] && GtQ[n, 1/2]

Rule 3905

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

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [C]  time = 1.22626, size = 99, normalized size = 1.06 \[ \frac{i c \left (\cot \left (\frac{1}{2} (e+f x)\right )+i\right ) \sqrt{a (\sec (e+f x)+1)} \sqrt{c-c \sec (e+f x)} \left (\left (f x+i \log \left (1+e^{2 i (e+f x)}\right )\right ) \cos (e+f x)+i\right )}{f \left (1+e^{i (e+f x)}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.297, size = 164, normalized size = 1.8 \begin{align*} -{\frac{\cos \left ( fx+e \right ) }{f\sin \left ( fx+e \right ) \left ( -1+\cos \left ( fx+e \right ) \right ) } \left ( \cos \left ( fx+e \right ) \ln \left ({\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +\cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -\cos \left ( fx+e \right ) \ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) +\cos \left ( fx+e \right ) +1 \right ) \left ({\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{3}{2}}}\sqrt{{\frac{a \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((c-c*sec(f*x+e))^(3/2)*(a+a*sec(f*x+e))^(1/2),x)

[Out]

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

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Maxima [B]  time = 1.79602, size = 328, normalized size = 3.53 \begin{align*} -\frac{{\left ({\left (f x + e\right )} c \cos \left (2 \, f x + 2 \, e\right )^{2} +{\left (f x + e\right )} c \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \,{\left (f x + e\right )} c \cos \left (2 \, f x + 2 \, e\right ) + 2 \, c \cos \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) \sin \left (2 \, f x + 2 \, e\right ) +{\left (f x + e\right )} c -{\left (c \cos \left (2 \, f x + 2 \, e\right )^{2} + c \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, c \cos \left (2 \, f x + 2 \, e\right ) + c\right )} \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) - 2 \,{\left (c \cos \left (2 \, f x + 2 \, e\right ) + c\right )} \sin \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt{a} \sqrt{c}}{{\left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.65703, size = 887, normalized size = 9.54 \begin{align*} \left [-\frac{2 \, c \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - \sqrt{-a c}{\left (c \cos \left (f x + e\right ) + c\right )} \log \left (\frac{a c \cos \left (f x + e\right )^{4} -{\left (\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )\right )} \sqrt{-a c} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) + a c}{2 \, \cos \left (f x + e\right )^{2}}\right )}{2 \,{\left (f \cos \left (f x + e\right ) + f\right )}}, -\frac{c \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - \sqrt{a c}{\left (c \cos \left (f x + e\right ) + c\right )} \arctan \left (\frac{\sqrt{a c} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{a c \cos \left (f x + e\right )^{2} + a c}\right )}{f \cos \left (f x + e\right ) + f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*c*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*sin(f*x + e) - sqrt
(-a*c)*(c*cos(f*x + e) + c)*log(1/2*(a*c*cos(f*x + e)^4 - (cos(f*x + e)^3 + cos(f*x + e))*sqrt(-a*c)*sqrt((a*c
os(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*sin(f*x + e) + a*c)/cos(f*x + e)^2))/(f
*cos(f*x + e) + f), -(c*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*sin(f*
x + e) - sqrt(a*c)*(c*cos(f*x + e) + c)*arctan(sqrt(a*c)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f
*x + e) - c)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e)/(a*c*cos(f*x + e)^2 + a*c)))/(f*cos(f*x + e) + f)]

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

[Out]

Timed out