3.49 \(\int \frac{\sqrt{a+a \sec (e+f x)}}{(c-c \sec (e+f x))^4} \, dx\)

Optimal. Leaf size=174 \[ -\frac{2 \cot ^7(e+f x) (a \sec (e+f x)+a)^{7/2}}{7 a^3 c^4 f}+\frac{2 \cot ^5(e+f x) (a \sec (e+f x)+a)^{5/2}}{5 a^2 c^4 f}-\frac{2 \cot ^3(e+f x) (a \sec (e+f x)+a)^{3/2}}{3 a c^4 f}+\frac{2 \cot (e+f x) \sqrt{a \sec (e+f x)+a}}{c^4 f}+\frac{2 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{c^4 f} \]

[Out]

(2*Sqrt[a]*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(c^4*f) + (2*Cot[e + f*x]*Sqrt[a + a*Sec[e
+ f*x]])/(c^4*f) - (2*Cot[e + f*x]^3*(a + a*Sec[e + f*x])^(3/2))/(3*a*c^4*f) + (2*Cot[e + f*x]^5*(a + a*Sec[e
+ f*x])^(5/2))/(5*a^2*c^4*f) - (2*Cot[e + f*x]^7*(a + a*Sec[e + f*x])^(7/2))/(7*a^3*c^4*f)

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Rubi [A]  time = 0.176576, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3904, 3887, 325, 203} \[ -\frac{2 \cot ^7(e+f x) (a \sec (e+f x)+a)^{7/2}}{7 a^3 c^4 f}+\frac{2 \cot ^5(e+f x) (a \sec (e+f x)+a)^{5/2}}{5 a^2 c^4 f}-\frac{2 \cot ^3(e+f x) (a \sec (e+f x)+a)^{3/2}}{3 a c^4 f}+\frac{2 \cot (e+f x) \sqrt{a \sec (e+f x)+a}}{c^4 f}+\frac{2 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{c^4 f} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Sec[e + f*x]]/(c - c*Sec[e + f*x])^4,x]

[Out]

(2*Sqrt[a]*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(c^4*f) + (2*Cot[e + f*x]*Sqrt[a + a*Sec[e
+ f*x]])/(c^4*f) - (2*Cot[e + f*x]^3*(a + a*Sec[e + f*x])^(3/2))/(3*a*c^4*f) + (2*Cot[e + f*x]^5*(a + a*Sec[e
+ f*x])^(5/2))/(5*a^2*c^4*f) - (2*Cot[e + f*x]^7*(a + a*Sec[e + f*x])^(7/2))/(7*a^3*c^4*f)

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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{\sqrt{a+a \sec (e+f x)}}{(c-c \sec (e+f x))^4} \, dx &=\frac{\int \cot ^8(e+f x) (a+a \sec (e+f x))^{9/2} \, dx}{a^4 c^4}\\ &=-\frac{2 \operatorname{Subst}\left (\int \frac{1}{x^8 \left (1+a x^2\right )} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{a^3 c^4 f}\\ &=-\frac{2 \cot ^7(e+f x) (a+a \sec (e+f x))^{7/2}}{7 a^3 c^4 f}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{x^6 \left (1+a x^2\right )} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{a^2 c^4 f}\\ &=\frac{2 \cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 a^2 c^4 f}-\frac{2 \cot ^7(e+f x) (a+a \sec (e+f x))^{7/2}}{7 a^3 c^4 f}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{x^4 \left (1+a x^2\right )} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{a c^4 f}\\ &=-\frac{2 \cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{3 a c^4 f}+\frac{2 \cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 a^2 c^4 f}-\frac{2 \cot ^7(e+f x) (a+a \sec (e+f x))^{7/2}}{7 a^3 c^4 f}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+a x^2\right )} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{c^4 f}\\ &=\frac{2 \cot (e+f x) \sqrt{a+a \sec (e+f x)}}{c^4 f}-\frac{2 \cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{3 a c^4 f}+\frac{2 \cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 a^2 c^4 f}-\frac{2 \cot ^7(e+f x) (a+a \sec (e+f x))^{7/2}}{7 a^3 c^4 f}-\frac{(2 a) \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^4 f}\\ &=\frac{2 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{c^4 f}+\frac{2 \cot (e+f x) \sqrt{a+a \sec (e+f x)}}{c^4 f}-\frac{2 \cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{3 a c^4 f}+\frac{2 \cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 a^2 c^4 f}-\frac{2 \cot ^7(e+f x) (a+a \sec (e+f x))^{7/2}}{7 a^3 c^4 f}\\ \end{align*}

Mathematica [C]  time = 0.249655, size = 78, normalized size = 0.45 \[ -\frac{2 \sqrt{\cos (e+f x)} \tan \left (\frac{1}{2} (e+f x)\right ) \sqrt{a (\sec (e+f x)+1)} \text{Hypergeometric2F1}\left (-\frac{7}{2},-\frac{7}{2},-\frac{5}{2},2 \sin ^2\left (\frac{1}{2} (e+f x)\right )\right )}{7 c^4 f (\cos (e+f x)-1)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + a*Sec[e + f*x]]/(c - c*Sec[e + f*x])^4,x]

[Out]

(-2*Sqrt[Cos[e + f*x]]*Hypergeometric2F1[-7/2, -7/2, -5/2, 2*Sin[(e + f*x)/2]^2]*Sqrt[a*(1 + Sec[e + f*x])]*Ta
n[(e + f*x)/2])/(7*c^4*f*(-1 + Cos[e + f*x])^4)

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Maple [B]  time = 0.37, size = 402, normalized size = 2.3 \begin{align*} -{\frac{ \left ( 1+\cos \left ( fx+e \right ) \right ) ^{2}}{105\,f{c}^{4} \left ( \sin \left ( fx+e \right ) \right ) ^{5} \left ( -1+\cos \left ( fx+e \right ) \right ) }\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}} \left ( 105\,\sqrt{2} \left ( \cos \left ( fx+e \right ) \right ) ^{3}\sin \left ( fx+e \right ) \sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}} \right ) -315\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) \sqrt{2}\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}} \right ) +315\,\sqrt{2}\cos \left ( fx+e \right ) \sin \left ( fx+e \right ) \sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}} \right ) -105\,\sqrt{2}\sin \left ( fx+e \right ){\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( fx+e \right ) }{\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 ) }}}-352\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}+812\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}-700\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+210\,\cos \left ( fx+e \right ) \right ) } \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))^4,x)

[Out]

-1/105/c^4/f*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*(1+cos(f*x+e))^2*(105*2^(1/2)*cos(f*x+e)^3*sin(f*x+e)*(-2*c
os(f*x+e)/(1+cos(f*x+e)))^(1/2)*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)
)-315*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*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))+315*2^(1/2)*cos(f*x+e)*sin(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^
(1/2)*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))-105*2^(1/2)*sin(f*x+e)*a
rctanh(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)-352*cos(f*x+e)^4+812*cos(f*x+e)^3-700*cos(f*x+e)^2+210*cos(f*x+e))/sin(f*x+e)^5/(-1+cos(f*x+e))

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Maxima [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))^4,x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.64564, size = 1245, normalized size = 7.16 \begin{align*} \left [\frac{105 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )^{2} + 3 \, \cos \left (f x + e\right ) - 1\right )} \sqrt{-a} \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 (176 \, \cos \left (f x + e\right )^{4} - 406 \, \cos \left (f x + e\right )^{3} + 350 \, \cos \left (f x + e\right )^{2} - 105 \, \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{210 \,{\left (c^{4} f \cos \left (f x + e\right )^{3} - 3 \, c^{4} f \cos \left (f x + e\right )^{2} + 3 \, c^{4} f \cos \left (f x + e\right ) - c^{4} f\right )} \sin \left (f x + e\right )}, \frac{105 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )^{2} + 3 \, \cos \left (f x + e\right ) - 1\right )} \sqrt{a} \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 (176 \, \cos \left (f x + e\right )^{4} - 406 \, \cos \left (f x + e\right )^{3} + 350 \, \cos \left (f x + e\right )^{2} - 105 \, \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{105 \,{\left (c^{4} f \cos \left (f x + e\right )^{3} - 3 \, c^{4} f \cos \left (f x + e\right )^{2} + 3 \, c^{4} f \cos \left (f x + e\right ) - c^{4} f\right )} \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))^(1/2)/(c-c*sec(f*x+e))^4,x, algorithm="fricas")

[Out]

[1/210*(105*(cos(f*x + e)^3 - 3*cos(f*x + e)^2 + 3*cos(f*x + e) - 1)*sqrt(-a)*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*(176*cos(f*x + e)^4 - 406*cos(f*x + e)^3 + 350*cos(f*x + e)^2 - 10
5*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e)))/((c^4*f*cos(f*x + e)^3 - 3*c^4*f*cos(f*x + e)^2 + 3*c
^4*f*cos(f*x + e) - c^4*f)*sin(f*x + e)), 1/105*(105*(cos(f*x + e)^3 - 3*cos(f*x + e)^2 + 3*cos(f*x + e) - 1)*
sqrt(a)*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*(176*cos(f*x + e)^4 - 406*cos(f*x + e)^3 + 350*cos(f*x + e)^2 - 105*c
os(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e)))/((c^4*f*cos(f*x + e)^3 - 3*c^4*f*cos(f*x + e)^2 + 3*c^4*
f*cos(f*x + e) - c^4*f)*sin(f*x + e))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sqrt{a \sec{\left (e + f x \right )} + a}}{\sec ^{4}{\left (e + f x \right )} - 4 \sec ^{3}{\left (e + f x \right )} + 6 \sec ^{2}{\left (e + f x \right )} - 4 \sec{\left (e + f x \right )} + 1}\, dx}{c^{4}} \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))**4,x)

[Out]

Integral(sqrt(a*sec(e + f*x) + a)/(sec(e + f*x)**4 - 4*sec(e + f*x)**3 + 6*sec(e + f*x)**2 - 4*sec(e + f*x) +
1), x)/c**4

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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))^4,x, algorithm="giac")

[Out]

Timed out