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

Optimal. Leaf size=103 \[ \frac{8 a^2 \cot (e+f x) \sqrt{a \sec (e+f x)+a}}{c f}+\frac{2 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{c f}-\frac{2 a^3 \tan (e+f x)}{c f \sqrt{a \sec (e+f x)+a}} \]

[Out]

(2*a^(5/2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(c*f) + (8*a^2*Cot[e + f*x]*Sqrt[a + a*Sec
[e + f*x]])/(c*f) - (2*a^3*Tan[e + f*x])/(c*f*Sqrt[a + a*Sec[e + f*x]])

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Rubi [A]  time = 0.171966, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3904, 3887, 461, 203} \[ \frac{8 a^2 \cot (e+f x) \sqrt{a \sec (e+f x)+a}}{c f}+\frac{2 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{c f}-\frac{2 a^3 \tan (e+f x)}{c f \sqrt{a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^(5/2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(c*f) + (8*a^2*Cot[e + f*x]*Sqrt[a + a*Sec
[e + f*x]])/(c*f) - (2*a^3*Tan[e + f*x])/(c*f*Sqrt[a + a*Sec[e + f*x]])

Rule 3904

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

Rule 3887

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

Rule 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(a+a \sec (e+f x))^{5/2}}{c-c \sec (e+f x)} \, dx &=-\frac{\int \cot ^2(e+f x) (a+a \sec (e+f x))^{7/2} \, dx}{a c}\\ &=\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{\left (2+a x^2\right )^2}{x^2 \left (1+a x^2\right )} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{c f}\\ &=\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \left (a+\frac{4}{x^2}-\frac{a}{1+a x^2}\right ) \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{c f}\\ &=\frac{8 a^2 \cot (e+f x) \sqrt{a+a \sec (e+f x)}}{c f}-\frac{2 a^3 \tan (e+f x)}{c f \sqrt{a+a \sec (e+f x)}}-\frac{\left (2 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{1+a x^2} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{c f}\\ &=\frac{2 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{c f}+\frac{8 a^2 \cot (e+f x) \sqrt{a+a \sec (e+f x)}}{c f}-\frac{2 a^3 \tan (e+f x)}{c f \sqrt{a+a \sec (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.680392, size = 96, normalized size = 0.93 \[ \frac{2 a^3 \tan (e+f x) \sec (e+f x) \left ((5 \cos (e+f x)-1) \sqrt{\sec (e+f x)-1}-(\cos (e+f x)-1) \tan ^{-1}\left (\sqrt{\sec (e+f x)-1}\right )\right )}{c f (\sec (e+f x)-1)^{3/2} \sqrt{a (\sec (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*(-(ArcTan[Sqrt[-1 + Sec[e + f*x]]]*(-1 + Cos[e + f*x])) + (-1 + 5*Cos[e + f*x])*Sqrt[-1 + Sec[e + f*x]]
)*Sec[e + f*x]*Tan[e + f*x])/(c*f*(-1 + Sec[e + f*x])^(3/2)*Sqrt[a*(1 + Sec[e + f*x])])

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Maple [A]  time = 0.227, size = 120, normalized size = 1.2 \begin{align*} -{\frac{{a}^{2}}{fc\sin \left ( fx+e \right ) } \left ( \sqrt{2}\sin \left ( fx+e \right ){\it Artanh} \left ({\frac{\sqrt{2}\sin \left ( fx+e \right ) }{2\,\cos \left ( fx+e \right ) }\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}} \right ) \sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}-10\,\cos \left ( fx+e \right ) +2 \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)),x)

[Out]

-1/c/f*a^2*(2^(1/2)*sin(f*x+e)*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))
*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)-10*cos(f*x+e)+2)*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)/sin(f*x+e)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}{c \sec \left (f x + e\right ) - c}\,{d x} \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)),x, algorithm="maxima")

[Out]

-integrate((a*sec(f*x + e) + a)^(5/2)/(c*sec(f*x + e) - c), x)

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Fricas [A]  time = 1.47104, size = 724, normalized size = 7.03 \begin{align*} \left [\frac{\sqrt{-a} a^{2} \log \left (-\frac{8 \, a \cos \left (f x + e\right )^{3} - 4 \,{\left (2 \, \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt{-a} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - 7 \, a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right ) + 1}\right ) \sin \left (f x + e\right ) + 4 \,{\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 )}}}{2 \, c f \sin \left (f x + e\right )}, \frac{a^{\frac{5}{2}} \arctan \left (\frac{2 \, \sqrt{a} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{2 \, a \cos \left (f x + e\right )^{2} + a \cos \left (f x + e\right ) - a}\right ) \sin \left (f x + e\right ) + 2 \,{\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 )}}}{c f \sin \left (f x + e\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)),x, algorithm="fricas")

[Out]

[1/2*(sqrt(-a)*a^2*log(-(8*a*cos(f*x + e)^3 - 4*(2*cos(f*x + e)^2 - cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e
) + a)/cos(f*x + e))*sin(f*x + e) - 7*a*cos(f*x + e) + a)/(cos(f*x + e) + 1))*sin(f*x + e) + 4*(5*a^2*cos(f*x
+ e) - a^2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e)))/(c*f*sin(f*x + e)), (a^(5/2)*arctan(2*sqrt(a)*sqrt((a*cos
(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e)/(2*a*cos(f*x + e)^2 + a*cos(f*x + e) - a))*sin(f*x + e)
+ 2*(5*a^2*cos(f*x + e) - a^2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e)))/(c*f*sin(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)),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)),x, algorithm="giac")

[Out]

Timed out