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

Optimal. Leaf size=139 \[ -\frac{a^2 c \tan (e+f x) \sqrt{a \sec (e+f x)+a}}{f \sqrt{c-c \sec (e+f x)}}+\frac{a^3 c \tan (e+f x) \log (\cos (e+f x))}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{a c \tan (e+f x) (a \sec (e+f x)+a)^{3/2}}{2 f \sqrt{c-c \sec (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.262803, antiderivative size = 139, 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 = {3906, 3905, 3475} \[ -\frac{a^2 c \tan (e+f x) \sqrt{a \sec (e+f x)+a}}{f \sqrt{c-c \sec (e+f x)}}+\frac{a^3 c \tan (e+f x) \log (\cos (e+f x))}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{a c \tan (e+f x) (a \sec (e+f x)+a)^{3/2}}{2 f \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [C]  time = 1.25926, size = 164, normalized size = 1.18 \[ \frac{a^2 e^{-i (e+f x)} \left (1+e^{2 i (e+f x)}\right ) \left (\cot \left (\frac{1}{2} (e+f x)\right )+i\right ) \sec ^2(e+f x) \sqrt{a (\sec (e+f x)+1)} \sqrt{c-c \sec (e+f x)} \left (-\log \left (1+e^{2 i (e+f x)}\right )+4 \cos (e+f x)+\left (i f x-\log \left (1+e^{2 i (e+f x)}\right )\right ) \cos (2 (e+f x))+i f x+1\right )}{4 f \left (1+e^{i (e+f x)}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.307, size = 179, normalized size = 1.3 \begin{align*} -{\frac{{a}^{2}}{2\,f\cos \left ( fx+e \right ) \sin \left ( fx+e \right ) } \left ( 2\,\ln \left ({\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\,\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-3\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-4\,\cos \left ( fx+e \right ) -1 \right ) \sqrt{{\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}}\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((a+a*sec(f*x+e))^(5/2)*(c-c*sec(f*x+e))^(1/2),x)

[Out]

-1/2/f*a^2*(2*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2+2*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e
))*cos(f*x+e)^2-2*ln(2/(1+cos(f*x+e)))*cos(f*x+e)^2-3*cos(f*x+e)^2-4*cos(f*x+e)-1)*(c*(-1+cos(f*x+e))/cos(f*x+
e))^(1/2)*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)/cos(f*x+e)/sin(f*x+e)

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Maxima [B]  time = 1.84273, size = 959, normalized size = 6.9 \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))^(5/2)*(c-c*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

-((f*x + e)*a^2*cos(4*f*x + 4*e)^2 + 4*(f*x + e)*a^2*cos(2*f*x + 2*e)^2 + (f*x + e)*a^2*sin(4*f*x + 4*e)^2 + 4
*(f*x + e)*a^2*sin(2*f*x + 2*e)^2 + 4*(f*x + e)*a^2*cos(2*f*x + 2*e) + (f*x + e)*a^2 + 2*a^2*sin(2*f*x + 2*e)
- (a^2*cos(4*f*x + 4*e)^2 + 4*a^2*cos(2*f*x + 2*e)^2 + a^2*sin(4*f*x + 4*e)^2 + 4*a^2*sin(4*f*x + 4*e)*sin(2*f
*x + 2*e) + 4*a^2*sin(2*f*x + 2*e)^2 + 4*a^2*cos(2*f*x + 2*e) + a^2 + 2*(2*a^2*cos(2*f*x + 2*e) + a^2)*cos(4*f
*x + 4*e))*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) + 2*(2*(f*x + e)*a^2*cos(2*f*x + 2*e) + (f*x + e)*a
^2 + a^2*sin(2*f*x + 2*e))*cos(4*f*x + 4*e) - 4*(a^2*sin(4*f*x + 4*e) + 2*a^2*sin(2*f*x + 2*e))*cos(3/2*arctan
2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 4*(a^2*sin(4*f*x + 4*e) + 2*a^2*sin(2*f*x + 2*e))*cos(1/2*arctan2(sin
(2*f*x + 2*e), cos(2*f*x + 2*e))) + 2*(2*(f*x + e)*a^2*sin(2*f*x + 2*e) - a^2*cos(2*f*x + 2*e))*sin(4*f*x + 4*
e) + 4*(a^2*cos(4*f*x + 4*e) + 2*a^2*cos(2*f*x + 2*e) + a^2)*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e
))) + 4*(a^2*cos(4*f*x + 4*e) + 2*a^2*cos(2*f*x + 2*e) + a^2)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*
e))))*sqrt(a)*sqrt(c)/((2*(2*cos(2*f*x + 2*e) + 1)*cos(4*f*x + 4*e) + cos(4*f*x + 4*e)^2 + 4*cos(2*f*x + 2*e)^
2 + sin(4*f*x + 4*e)^2 + 4*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 4*sin(2*f*x + 2*e)^2 + 4*cos(2*f*x + 2*e) + 1)*
f)

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Fricas [A]  time = 1.64082, size = 1049, normalized size = 7.55 \begin{align*} \left [\frac{{\left (5 \, a^{2} \cos \left (f x + e\right ) + a^{2}\right )} \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 ) +{\left (a^{2} \cos \left (f x + e\right )^{2} + a^{2} \cos \left (f x + e\right )\right )} \sqrt{-a c} \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 )^{2} + f \cos \left (f x + e\right )\right )}}, \frac{{\left (5 \, a^{2} \cos \left (f x + e\right ) + a^{2}\right )} \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 ) + 2 \,{\left (a^{2} \cos \left (f x + e\right )^{2} + a^{2} \cos \left (f x + e\right )\right )} \sqrt{a c} \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 )}{2 \,{\left (f \cos \left (f x + e\right )^{2} + f \cos \left (f x + e\right )\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(5/2)*(c-c*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[1/2*((5*a^2*cos(f*x + e) + a^2)*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) + (a^2*cos(f*x + e)^2 + a^2*cos(f*x + e))*sqrt(-a*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*cos(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)^2 + f*cos(f*x + e)), 1/2*((5*a^2*cos(f*x + e) + a^2)*sq
rt((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) + 2*(a^2*cos(f*x +
e)^2 + a^2*cos(f*x + e))*sqrt(a*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)^2 + f*cos(f*x + e
))]

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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))**(5/2)*(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))^(5/2)*(c-c*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out