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]

-2/3*cot(f*x+e)^3*(a+a*sec(f*x+e))^(3/2)/a/c^4/f+2/5*cot(f*x+e)^5*(a+a*sec(f*x+e))^(5/2)/a^2/c^4/f-2/7*cot(f*x
+e)^7*(a+a*sec(f*x+e))^(7/2)/a^3/c^4/f+2*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))*a^(1/2)/c^4/f+2*cot
(f*x+e)*(a+a*sec(f*x+e))^(1/2)/c^4/f

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 174, normalized size of antiderivative = 1.00, 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 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])

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 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 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])

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 = 0.24, size = 78, normalized size = 0.45 \[ -\frac {2 \sqrt {\cos (e+f x)} \tan \left (\frac {1}{2} (e+f x)\right ) \sqrt {a (\sec (e+f x)+1)} \, _2F_1\left (-\frac {7}{2},-\frac {7}{2};-\frac {5}{2};2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )}{7 c^4 f (\cos (e+f x)-1)^4} \]

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)

________________________________________________________________________________________

fricas [A]  time = 0.64, size = 475, normalized size = 2.73 \[ \left [\frac {105 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )^{2} + 3 \, \cos \left (f x + e\right ) - 1\right )} \sqrt {-a} \log \left (-\frac {8 \, a \cos \left (f x + e\right )^{3} - 4 \, {\left (2 \, \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - 7 \, a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right ) + 1}\right ) \sin \left (f x + e\right ) + 4 \, {\left (176 \, \cos \left (f x + e\right )^{4} - 406 \, \cos \left (f x + e\right )^{3} + 350 \, \cos \left (f x + e\right )^{2} - 105 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{210 \, {\left (c^{4} f \cos \left (f x + e\right )^{3} - 3 \, c^{4} f \cos \left (f x + e\right )^{2} + 3 \, c^{4} f \cos \left (f x + e\right ) - c^{4} f\right )} \sin \left (f x + e\right )}, \frac {105 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )^{2} + 3 \, \cos \left (f x + e\right ) - 1\right )} \sqrt {a} \arctan \left (\frac {2 \, \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{2 \, a \cos \left (f x + e\right )^{2} + a \cos \left (f x + e\right ) - a}\right ) \sin \left (f x + e\right ) + 2 \, {\left (176 \, \cos \left (f x + e\right )^{4} - 406 \, \cos \left (f x + e\right )^{3} + 350 \, \cos \left (f x + e\right )^{2} - 105 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{105 \, {\left (c^{4} f \cos \left (f x + e\right )^{3} - 3 \, c^{4} f \cos \left (f x + e\right )^{2} + 3 \, c^{4} f \cos \left (f x + e\right ) - c^{4} f\right )} \sin \left (f x + e\right )}\right ] \]

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))]

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

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]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)4*sqrt(2)*(1/1680*(-1575*a*sqrt(-a)*(sqr
t(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^12+7140*a^2*sqrt(-a)*(sqrt(-a*tan(1/2*(f*x+exp
(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^10-16415*a^3*sqrt(-a)*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*
tan(1/2*(f*x+exp(1))))^8+19880*a^4*sqrt(-a)*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1)))
)^6-14637*a^5*sqrt(-a)*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^4+5684*a^6*sqrt(-a)
*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^2-1037*a^7*sqrt(-a))/c^4/((sqrt(-a*tan(1/
2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^2-a)^7-1/4*a*sqrt(-a)*ln(abs(2*(sqrt(-a*tan(1/2*(f*x+exp(
1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^2-4*sqrt(2)*abs(a)-6*a)/abs(2*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sq
rt(-a)*tan(1/2*(f*x+exp(1))))^2+4*sqrt(2)*abs(a)-6*a))/sqrt(2)/c^4/abs(a))*sign(cos(f*x+exp(1)))/f

________________________________________________________________________________________

maple [B]  time = 2.20, size = 402, normalized size = 2.31 \[ \frac {\sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \left (1+\cos \left (f x +e \right )\right )^{2} \left (-105 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right )+315 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right )-315 \cos \left (f x +e \right ) \sin \left (f x +e \right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right )+352 \left (\cos ^{4}\left (f x +e \right )\right )+105 \sqrt {2}\, \sin \left (f x +e \right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}-812 \left (\cos ^{3}\left (f x +e \right )\right )+700 \left (\cos ^{2}\left (f x +e \right )\right )-210 \cos \left (f x +e \right )\right )}{105 c^{4} f \sin \left (f x +e \right )^{5} \left (-1+\cos \left (f x +e \right )\right )} \]

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*(a*(1+cos(f*x+e))/cos(f*x+e))^(1/2)*(1+cos(f*x+e))^2*(-105*cos(f*x+e)^3*sin(f*x+e)*2^(1/2)*(-2*cos
(f*x+e)/(1+cos(f*x+e)))^(1/2)*arctanh(1/2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))+
315*sin(f*x+e)*cos(f*x+e)^2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*arctanh(1/2*(-2*cos(f*x+e)/(1+cos(f*x
+e)))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))-315*cos(f*x+e)*sin(f*x+e)*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1
/2)*arctanh(1/2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))+352*cos(f*x+e)^4+105*2^(1/
2)*sin(f*x+e)*arctanh(1/2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))*(-2*cos(f*x+e)/(
1+cos(f*x+e)))^(1/2)-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))

________________________________________________________________________________________

maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}}{{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sqrt {a \sec {\left (e + f x \right )} + a}}{\sec ^{4}{\left (e + f x \right )} - 4 \sec ^{3}{\left (e + f x \right )} + 6 \sec ^{2}{\left (e + f x \right )} - 4 \sec {\left (e + f x \right )} + 1}\, dx}{c^{4}} \]

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

________________________________________________________________________________________